/ Vf,, ''m' PUBLICAriONS OF SOWER, POTTS i CO., PHILADELPHIA. THE Normal Educational Series OP School and College Text-Books. " Every child that comes into the world has a ri^ht to an education." " The dearest interest of a nation is the education of its children." The art of Taaching, as well as all other arts, is making very rapid progress in this very progressive age. The remarkable growth of^NormsU Schools, organized to instruct in the best methods of teaching, and employing as professors the most able and advanced educators in the country, has given an immense impetus to the ad- vancement of this most honorable and useful of professions, and almost revolutionized the whole art of teaching. These great changes create a necessity for text-books adapted to them, and the publishers of the above series have taken great pains to meet this necessity, fiy the aid of their improved text-books, the work of the school-room, instead of being a drudgery, Vecomes pleasant to teachers and pupils, and they as well as parents are delighted with the rapid progress made with them. Raub's Normal Primary Speller. Raub's Normal Speller. BY PROF. A. H. RADB, PRINCIPAL OP PEKWSYLVANIA CENTRAL NORMAL SCHOOL, LOCK HAVEN. Ttaese clemcDtary works are admirably arranged and classified. Simple and easy, yet loflcal and compreheasive. they never fail to make- ready and correct spellers. r Fewsmith's Elementary Grammar. Fewsmith's Grammar of Eng. Language. BY WM. FEWSMITH, A.M., AND EDGAR A. SINGER. The Hniform testimony of teachers who have introduced these grammars is, that they have been most agreeably surprised at their effects upon pupils. They are easy to understand by the youngest pupil, and the lessons before dreaded become a delight to teacher and pupils. Ex- traordinary care has been taken in grading every lesson, modeling rules and definitions after B definite and uniform plan, and making every word and Mmtence an example of grammati- cal accuracy. They only need a trial to supersede all others. - PUBLICATIONS OF SOWER, POTTS i CO., PHILADELPHIA THE Normal Series of Mathematics. BY EDWARD BROOKS, A. M., Ph.D., PRINCIPAL OF PENNSYLVANIA STATE NORMAL SCHOOL AT MILLERSVILLE? This Series has had an extraordinary success, and is used in very many of the best Normal Schools, Seminaries and Public Schools in the country. Wherever known, the yirorks receive the highest commendation. BROOKS'S NORMAL STANDARD SERIES. The Standard Series is a full couise intended for Schools and Classes having ample time for a thorough study of the Science. It consists of the following four books ; Brooks's Normal Primary Arithmetic. Brooks's Normal Elementary Arithmetic. Brooks's New Normal Mental Arithmetic. Brooks's New Normal Written Arithmetic. The Primary contains Mental and Written Exercises for very young pupils. Its treatment Is very plain, easy and progressive. The Elementary will fumisli a practical business education in a shorter time and with less labor than any other, and is emphatically the work for those pupils who must be qualified for com- mon Business in one or two terms. Key, 'so cts. The New Mental is a philosophical and com^prehensive treatise upon the Analysis of Numbers. It is easily mastered by young pupils, and those who accomplish it are ready to grapple with the most difficult problems. It makes logical thinkers on all subjects. AH who use'it say they cannot be induced to dispense with it. Key, 'sS cts. The New Written is a thoroughly practical work full of business applications. (See Bills and Ac- counts, Taxes, Banking, Exchange, Custom-House Duties, Insurance, Building Associations, etc.) Its treatment is novel, very successful in the school-room and popular among the best educators. Key, •$!. BROOKS'S NORMAL UNION SERIES. The Union Series is a condensed course, complete in Two Books, as follows : Brooks's 'Normal Union Arithmetic. Part 1, Brooks's Normal Union Arithmetic. Complete, For convenience of certain graded Schools In cities the latter work may be had bound up in two books, as follows : Brooks's Normal Union Arithmetic. Part 2. Brooks's Normal Union Arithmetic. Part 3. In the Union, Mental and Written Arithmetic are so combined that the pupil may obtain a thorough course ^in arithmetical analysis while becoming familiar with the application of tlie science to practical business. This union is here made not a mere nominal one, but a scien- tific reality. Key, «$i. _^ Brooks's Normal Higher Arithmetic. Original, complete and practical. It abounds with striking novelties, presented with the utmost clearness and simplicity, all ;alculated to make the student a master of the theory ofArith' metie. It also represents the actual business as practiced in the counting-houses of mer- chants, custom-houses, banks and all kinds of incorporated companies. Cornell University Library The original of tliis book is in the Cornell University Library. There are no known copyright restrictions in the United States on the use of the text. http://www.archive.org/details/cu31 924031 21 8989 „, .Cornell University Library arV19280 The normal elementary algebra: ,. 3 1924 031 218 989 olin,anx THE NORMAL Elementary Algebra; COSTAINLfrG THE First Principles of the Science, DEVELOPKD WITH CONCISENESS AND SIMPLICITY, FOR Common Schools, Academies, Seminaries and Normal Schools. REVISED EDITION. •By EDWAED BEOOKS, A.M., Ph.D., LATE PttlNCIPAL OF PENNSTLVAHIA STATE NOBMAl SCHOOL, AND AUTHOR OP THE NOBMAL SEBIE8 or ARITHMETICS, "NOBMAL GEOMETRY AJSD TRIGONOMETRY," "philosophy OF ABTTHMETIC," ETC. '^ McUhemotical studies cultivate clearness of thought^ acuteness of analysis and accitaracy of ea^ession." PHILADELPHIA: CHRISTOPHER SOWER COMPANY, 530 Market St. and 523 Minor St. Entered according to Act of Congress, In the year 1871, by EDWARD BROOKS, In the Office of the Librarian of Congress, at Washington Becopyright, 1888, by Edwabd Bsooks. WebtCOTT & TaoK3nli, Oiatim Pnm 1^ Sureotapera, PhUada. SHEKMAIt 4 Co., Phiiamlphi*. CONTENTS. k-fTBODTJOTION . . SECTIOlSr I. DEFINITIONS AND EXPLANATIONS. PAGE Deflnitions 19 Symbols of Quantity 20 Symbols of Operation 20 Symbols of Belatlon 22 Algebraic Expressions 23 PAGB Exercises in Numeration and No- tation 24 Numerical Values 25 Positive and Negative Quanti- ties 28 SECTION 11. FUNDAMENTAL OPEEATIONS. Addition 29 Subtraction 34 Use of the Parenttiesis 38 Multiplication 41 Division 46 Principles of Division 51 SECTION III. COMPOSITION AND FACTORING. Composition 55 I Factoring 58 I Greatest Common Divisor- Least Common Muitiple SECTION IV. FRACTIONS. Definitions 73 Principles 73 Signs of -Fractions 74 Ileduction 75 Addition 82 Subtraction 84 Multiplication 86 Division 89 Complex Fractions 92 Vanisliing Fractions 94 Definitions 96 Transformations 97 Solution of Equations lOO Special Artifices 103 Problems with one Unknown Quantity 106 Equations of two Unknown Quantities 122 Elimination 122 Problems with two Unknown Quantities 129 SECTION V. SIMPLE EQUATIONS. Equations of tliree or more Un- known Quantities 135 Problems with three or more Un- known Quantities ■ 138 Zero and Infinity 141 Generalization 142 Negative Solution 146 Discussion of Problems 148 Courier Problem 149 Indeterminate and Impossible Problems 153 CONTENTS. SECTION VI. INVOLUTION, EVOLUTION AND RADICALS. PAGE Definitions 155 Powers of Monomials 156 Powers of Fractions 157 Powers of Polynomials 158 Binomial Theorem 162 Evolution, Definitions 168 Roots of Monomials 169 Bctuare Root of Polynomials 171 Square Root of Numbers..... 173 Cube Root of Polynomials. .-;...L... 176 PAGB Cube Root of Numbers 177 Second Method of Cube Boot 180 Reduction of Radicals 183 Addition of Radicals 187 Subtraction of Radicals 189 Multiplication of Radicals., 190 Division of Radicals 192 Rationalization 196 Imaginary Quantities... 198 Radical Equations 200 SECTION VII. QUADRATIC Definitions ,. 203 Pure Quadratics 203 Problems producing Pure Quad- ratics 206 Affected Quadratics 208 First Method of Completing the Square 208 Second Method of Completing the Square 212 EQUATIONS. Equations in the Quadratic Form 21i Quadratic Equations containing two Unknown Quantities 220 Problems producing Quadratics with two Unknown Quanti- ties 227 Principles of Quadratics 230 Imaginary Root 236 SECTION VIII. RATIO AND PROPORTION. Definitions of Ratio 238 Problems in Ratio 240 Definitions of Proportion 242 Theorems of Proportion 243 Problems in Ratio and Propor- tion 248 SECTION IX. PROGRESSIONS. Definitions 251 Arithmetical Progression 251 Table of Formulas 257 Problems in Arithmetical Pro- gression 259 Geometrical Progression 282 Infinite Series ,...: 265 Table of Formulas 288 Problems in Geometrical Pro- gression 269 MiSCELLANEOTJS EXAMPLES ; 272 SUPPLEMENT. Inequalities and Indetekminates. 289 Higher Equations 296 Theoey op Exponents 303 Logarithms 314 Peemhtations and Combinations... 323 Binomial Theoeem 330 PREFACE. EDtrcAnoN aims at mental culture and practical skill ; and for the attainment of both of these objects the mathematical sciences have in all ages been highly valued. They train the mind to logical methods of thought, give vigor and intensity to its operations, and lead to the import- ant habit of resting only in certainty of results; while as instruments of investigation they stand pre-eminent. Among the three fundamental branches of mathematics, Algebra occu- pies a prominent position in view of both of these objects. As a piethod of calculation it is the most powerful of them all, and for giving mental acnteness and the habit of analytic thought it is unequaled. With the advance of education this science is growing in popularity, and is being introduced into our best public schools, as well as academies and semi- naries. Many teachers are beginning to see that a knowledge of elementary algebra is worth more than a knowledge of higher arithmetic, and are omitting the arithmetic, when necessary, for tlie algebra. This has in- creased the demand for good text-books upon the subject; and to assist in meeting this demand the present work has been prepared. Some of its general and special features will be briefly stated. General Features. — This work is not a mere collection of problems and solutions, but the evolution of a carefully-matured plan, the embodi- ment of an ideal formed by long and thoughtful experience in the school- room. Attention is called to its extent, its matter and its method. JEktent.—The work embraces just about as many topics as it is thought the ordinary pupil in elementary algebra should be required to study. These topics have been presented, not superficially, but with comparative thoroughness, so that the knowledge given may be of actual use in calcu- lation, and afford a basis for the study of a higher work if desired. While not presenting quite as much as some teachers might prefer, the author has been careful not to make the work too elementary. Superficial scholarship is one of the growing evils of our country, and teachers and text-books are responsible for it. It should never be forgotten th^t it in better to know much <>f a fcvi thing's than to know a little of many things. While endeavoring to avoid superficiality, the author has been caieful to &o simplify tHe subje-M; as to render it suitable to those lieginning the b PREFACE. m study. The obj.'cV has been to hit the golden mean, and thus adapt it to the wants of the majority of pupils and teachers. McUter. — In the development of the various topics, care has been taken to combine in due proportion theory and practice. The French works on algebra are very complete in the discussion of principles, but are deficient in matter for the attainment of skill in their application. The English works abound with practical examples, but are usually less complete in their theoretical discussions. I have endeavored to combine both of these features, by giving an ample collection of problems, as well as a thorough discussion of principles. The aim has been to make the work both philosophical and practical. •The problems were prepared with especial reference to the principles and methods which they are designed to illustrate. They are often so related that one problem prepares the way for the succeeding one, thus making a problem which alone might be quite difficult, comparatively easy. The miscellaneous examples at the close of the book embrace a choice selection from American and English works, especially from the excellent collection given by Todhunter in his Elementary Algebra. Method of Treatment, — In the treatment of the various topics I have aimed to simplify the subject without impairing its logical completeness and thoroughness. In this respect the skill of an author is particularly shown, and in this consist the principal merits of a text-book. The difficulty of a subject is not so much in itself as in the manner in which it is presented ; special pains have therefore been taken to pass so gradu- ally from the simple to the complex as to make easy what otherwise might be complicated and difficult. I have also been careful to give conciseness, clearness and simplicity to its methods of explanation. There is a simple and direct method of stating a solution or demonstration that is much more readily compre- hended by the learner than a talking about it in a popular sort of style. The scientific method is usually the simplest method. Textrbooks on algebra have been especially defective in this respect. The methods of mental arithmetic have created a revolution in the forms of explanation in the science of numbers, giving beauty and simplicity to that which was before awkward and complicated. Geometry, coming to us as the prod- uct of the Greek mind, is characterized by the simplicity and elegance of its demonstrations ; while algebra, mainly the product of modern thought has been less clear, logical and finished in its methods of development. I have endeavored to make an improvement in this respect, by combining the simple and natural analytical methods of mental arithmetic with the elegance of form and logical exactness of the geometrical methods. Special Feattjbbs. — The principal merits of the work are supposed to oonsiet in its methods of treatment, its solutions, discussions and explanar PREFACE. 7 tions — in brief, in the general spirit that pervades it, giving Bimplicity and unity to the work as a whole. ' There are, however, several special features to which the attention of teachers is respectfully invited : 1. The definition of algebra seems simpler and more complete than any which the author has met. 2. The dass^eation of algebraic symbols undei three distinct hesds 18 different from anything heretofore presented/ 3. The method of explaining addition and subtraction by means of an auxiliary quantity is a new feature which merits notice. 4. The new topic, called Composition, as a synthetic process correlative to the analytic process of Factoring, will no doubt arrest attention. 5. The variety of the cases in factoring, the demonstration of the divisi- bility of a"— &" by a — b, and the explanation of the greatest common divisor and least common multiple, solicit notice. 6. The treatment of involution and evolution possesses some points of novelty, and especial attention is called to the second method of cube root 7. Particular attention is invited to the discussion of the Courier, Problen. and to the development of the properties of the incomplet* Quadratic equa tion. 8. The variety and appropriateness of the problems, the frequent gen- eralization from a particular to a general problem, and the variation of old problems, are designed to add interest to the study and give discipline to the student. 9. Unusual care has been given to the typography of the work, in order to make it attractive and interesting to the pupil. The introduction of the small symbols for addition, subtraction, multiplication and division, as used in the best English works, is regarded as a great improvement. Encouraged by the approval of many of the best teachers of the country who have used my former works, and yielding to their wishes as well as my own inclination, I have found time and strength, amid the cares and duties of a large institution, to prepare the present work ; and I now send it forth, trusting that it may afford discipline and knowledge to many youth- ful minds of the present generation, and convey a kindly remembranct of my own labors to teachers and pupils of the future. Statu Nobmal School, 1 EDWARD BROOKS. MiUtrmille, Pa., May, 1871. •• SUGGESTIONS TO TEACHERS. The following suggestions are respectfully made for the benefit of 70ung or inexperienced teachers : 1. It will be well to make frequent use of the inductive method of teaching, as suggested in the Introduction, leading pupils from the ideas and methods of arithmetic to those of algebra. 2. Drill thoroughly upon the fundamental operations, especially upon the use of the minus sign, the use of exponents in multiplication and division, and the methods of factoring. Slowness here is speed afterward. 3. Where there are two solutions, the teacher may select the one he prefers, and drill the pupils thoroughly upon it before they attempt the other. The first of two given solutions is usually regarded as the simplest, though not always the best. 4. Frequently require pupils to change a particular problem into a general one, as on page 107, and also, to make special problems out of general ones. Have pupils to make problems by changing the conditions of given problems, or by using a required condition and requiring some given oonuition, as in problems on page 111. Pupils should be encour- aged to form new problems, and to originate new methods of explanation and solution. We should always aim to make thinkers of our pupils rather than mathematical m,achines. 5. A Shorter Course. — While this work is the author's ideal of the ex- teiit of an elementary algebra, yet it may be used by teachers who desire a shorter course. For such the following omissions may be made without impairing the unity of the subject : Omit the second methods of Greatest Common Divisor and Leaist Com- mon Multiple, the Supplement to Simple Equations, Imaginary Quan- tities, Principles of Quadratics, Quadratics of Two Unknown Quantities, and the Miscellaneous Examples at the close of the work. A etill shorter ■course may be attained by the further omission of the latter half of the examples under each subject. 6. In conclusion, the author suggests to teachers of public schools to give their pupils a course in elementary algebra before completing higher arithmetic. His own experience is, that pupils cannot thoroughl r under- 6tand arithmetic until they have studied algebra. THE NORMAL EJJIMENTARY ALGEBRA. INTRODUCTION. The object of the exercises in the Introduction is to lead pupils from the ideas and operations of Arithmetic to those of Algebra. LESSON I. 4 1. Henry's number of apples, increased by three times his number, equals 24 ; how many apples has he ? SoiiUTioir. By arithmetic this problem is solved in the following manner : Henr'jfi number, plus three tiroes his number, equals 24 ; hence, 4 times Henry's number equals 24, and once Henry's number equals one-fourth of 24, or 6. Abbreviated Solution. If we represent the expression Henry's number by some character, as the letter x, the solution will be much shorter ; thus : X plus 3 times x equals 24 ; hence, 4 times x equals 24, and once x equals 6, Algebraic Solution.- If we use Zx and 4a; to denote " 3 times x" and " 4 times x," the sign = for the word " equals," and the sign + foi the word " plus," the solution ^vill be purelj algebraic ; thus : a;+3» = 24; 4a; = 24, - ■X — &. Ans. 9 10 INTBODUCTION. Symbols. — It will be seen that thia last solation is the same as the first, except that we use characters instead of words. These characters are called symbols. The method of solving problems by means of symbols is called Algebra. Addition is denoted by the symbol +, read pliis ; subtraction is denoted by the symbol — , read minus. The expressions 2x, Sx, Ax, etc., mean 2 times x, 3 tinves x 4 times x, etc. One-half of x, one-third of x, two-thirds of x, etc, are denoted by ^x, ^x, fa;, etc., or -> -> — > etc. The symbol = denotes equality, and is read equals, or it equal to. The expression a; + 3a; = 24 is called an equation. The pupil will now express the following in algebraic symbols : 2. Three times John's number of apples equals 27. 8. One-half of Mary's number of peaches equals 12. 4. A's number of books, plus three times his number, equals 16^ 5. B's jiumber of dollars, minus half his number, equals 18. 6. Two-thirds of C's fortune, minus one-half of his fortune, equals $50. Note. — The pupil may be led to see, from the solution given, which is the unknown quantity and which the hwvm quantity, and how each is represented. Also, that the symbols of Algebra are of three classes : symbols of quantity, symbols of operation, and symbols of relation. LESSON II. 1. John has a certain number of peaches, and James has three times as many, and they both have 40 ; how many has each? SoLiTTiON. Let X equal John's number ; opeeatioit. then, since James has 3 times as many as Let a; = John's nimiber. John, 5x wUl equal James' number, and 3a; = Jam es' number, since they together have 40, x + Zx will a; + 3a; = 40 equal 40, or, adding, 4a; will equal 40. If 4a; = 40 4a; equals 40, v will equal one-fourth of ^ in t i, > i .„ \- 1, • in r I, . ». 1 o a; = 10, John's number 40, which 13 10, Johns numBer; and Sx q„. ^n t > i •1 n *• in on T I u 3a;-30, James' number will equal I times JO, or 30, James number. INTRODUCTION. 11 2. Mary's age is twice Sarah's, and the ?um of their ages i« 36 years ; what is the age of each ? Am. Sarah, 12 years ; Mary, 24 years. 3. A man bought a coat and a vest for |40, and the coal cost 4 times as much as the vest ; required the cost of each. Am. Coat, $32 ; vest, |8. 4. In a mixture of 860 bushels of grain there is 5 times aa much wheat as com ; how many bushels of each ? Am. Wheat, 800 bushels ; corn, 60 bushels. 5. Divide the number 144 into two parts, such that the larger part will be 5 times the smaller part. Am. 120 ; 24. 6. The sum of two numbers is 120, and the larger number is 4 times the smaller number ; required the two numbers. ' Am. 24 ; 96. 7. A man bought a span of horses and a carriage for flOOO, paying three times as much for the horses as the carriage; required the cost of each. Am. Horses, $750; carriage, $250. 8. The salary of a clerk for a year was $1500, and he spent five times as much of it as he saved ; how much did he save ? Am. $250. LESSON III. 1. The difference of two numbers is 24, and the larger equals 4 times the smaller ; required the numbers. SoLiTTiOK. Let X equal the smaller num- operation. ber; then 4x will equal the larger number. And sir.?e the difference of the two numbers is 24, Ax—x will equal 24, or, subtracting, 3a; will equal 24. If Zx equals 24, x will equal one-thirdof 24, which Is 8, the smaller number; _ o *>. and Ax will equal 4 times 8, or 32, the larger ^^Zoo' ti,o lo,.™, number. Let X = the smaller. 4a; = the larger. 4a; — a; = 24 3a; =24 4a; =32, the larger. 2. The difference of two numbers is 28; and 5 times the smaller equals the greater ; what are the numbers ? Ans. Smaller, 7 ; greater, 35. 3. A has 28 cents more than B, and 3 times B's numbei equals A's ; how many has each ? Ans. A, 42 ; B, 14. 12 INTEODUCTION. 4. Mary gathered 21 flowers more than her sister ; how many did each gather if Mary gathered 4 times as many as her sister f Ans. Mary, 28 ; sister, 7. 5. Seven times a number, diminished by 3 times the number, equals 48 ; what is the number ? Am. 12. 6. Marie has 40 cherries more than Jane, and 5 times Jane's number equals Marie's number ; how many has each ? Ans. Marie, 50 ; Jane, 10. 7. A bought a house and lot, paying 6 times as much for the liouse as the lot ; what did he pay for each if the house cost $2660 more than the lot? Ans. House, $3200 ; lot, $640. 8. A and B enter into a copartnership, in which A's interest is 6 times as great as B's : A's gain was $650 more than B's gain ; what was the gain of each ? Ans. A's, $780; B's, $130. LESSON IV. 1. Jiilia and Anna had 24 oranges, and Julia had one-half as many as Anna ; how many had each ? SoiiUTiou'. Let as equal Anna's number ; then will — equal Julia's number; and since they together have 24, z-\ — equals 24. Adding, x plus ^ of x, or f of x, equals 24. Ti f of a; equals 24, ^ of a; equals J of 24, wbich is 8, Julia's number ; and f of x, or x, equals 2 times 8, or 16, Anna'e number. Leta; = = Anna's .l.\JSi. number. X 2' = Julia's number. -r = 24 3x 2 = 24 X 2~ = 8, Julia's number. a; = 16, Anna's number. 2. A's money, increased by one-half of his money, equals $60 ; what is his money ? Am. $40. 3. What number is that to which if its one-third be added, the sum will be 36 ? Ans. 27. 4. What number is that to which if its two-thirds be added, the sum wQl be 45? ' Am. 27. INTEODUCTION. 13 5. "What number is that wMcli bdng dimmishea by its threft eighths, the remainder will be 30? Am. 48. 6. Mary's age, diminished by its three-fifths, equals 6 years ; how old is Mary ? Am. 15 years. 7. If one-half of my age be increased by one-third of my age, the sum -will be 40 years ; what is my age? Am. 48 years. 8. Four times the distance from Philadelphia to Lancaster, diminished by 2^ times the distance, equals 102 miles ; required the distance. Am. 68 miles. 9. Benton lost four-fifths of his money, and then found three-fourths as much as he lost, and then had $120 ; how much money had he at first? Ans. $150. 10. Bessie gave three-fourths of her money to the poor, and then found two-thirds as much as she gave away, and then had $30 ; how much had she at first? Am. $40. LESSON V. 1. A man bought a hat, vest and coat for $35 ; the vest cost twice as much as the hat, and the coat cost four times as much as the hat ; required the cost of each. Solution. Let x equal the cost of the hat ; operation. then will Ix equal the cost of the vest, and Let a; = cost of the hat. 4a; equal the cost .of the coat, and their sum, 2a; = cost of the vest, a; 4- 2a; + 4a;, will equal the cost of all, or $35. 4a; = cost of the c oat. Adding, we have Tx equals $35. If 7a; equals a; + 2a; + 4a; = 35 $35, X equals one-seventh of $35, or $5, the 7a; =35 cost of the hat; 2a; equals 2 times $5, or $10, a;= 5, hat; 'the cost of the vest ; and 4a; equals 4 times $5, 2a; = 10, vest ; or $20, the cost of the coat. 4a; = 20, coat. 2. Divide the number 106 into three such parts that the first shall be twice the second and the second twice the third. Am. 60 ; 30 ; 15. 3. Three men, A, B and C, earned $216 ; A earned twice aa much as B, and C earned as much as both A and B ; how much did each earn ? Am. A, $72 ; B, $36 ; C, $108. 4. The sum of three numbers is 63 ; the second is one-half of the first, and the third one-fourth of the first ; what are tha numbers? Am. 1st, 36 ; 2d, 18 ; 3d, 9. 14 INTEODTJCTION. 5. A man, with his wife and son, earned $22 in a «Teek, the man earned twice as much as his wife, and three times as much as his son; what did each earn? Am. Man, $12 ; wife, $6 ; son, |4. 6. A man bought a horse, a cow and a sheep for $315; the cow cost 5 times as much as the sheep, and the horse cost three times as much as the cow ; required the cost of each. Ans. Horse, $225 ; cow, $75 ; sheep, $15. 7. A tax of $450 is assessed upon three persons according to the relative value of their property ; A is worth two-thirds as much as B, and B is worth three-fourths as much as C ; what is each man's tax? Ana: A's, $100 ; B's, $160 ; C's, $200. LESSON. VI. 1. A being asked how much money he had, replied that three times his money increased by $8 equals $80 ; how much had he 1 Solution. Let x equal A's money; then, by operation. the condition of the problem we shall have Sa;+8 = 80. Now, if 3x increased by 8 equals 80, Let a; = A s money. Zx will equal 80 diminished by 8, which is 72; ^■*^'' °" if 3a; equals 72, a; will equal one-third of 72, which ^ ^ is 24. Hence A had 24 dollars. ^^^^ 2. If three times a number increased by 12 equals 57, what is that number ? Ans. 15. 3. If three-fourths of the distance from New York to Troj be diminished by 21 miles, the result will be 90 miles ; what ia the distance ? Ans. 148 miles. 4. If A's age be increased by its two-thirds and 7 years more, it will equal 32 years ; what is his age ? Ans. 15 years. 5. If 2f times the money a boy spent on the Fourth of July be diminished by 40 cents, the result will be $5.65 ; how much did he spend? Ans. $2.20. 6. One-half of my fortune, plus one-third of it and $380 more, equals $2580; what is my fortune ? ^»w. $2640. 7. One-third of the trees in an orchard bear apples, one-fourth bear peaches, and the remainder, which is 100, bear plums ; re- quired the number of trees in the orchard. Ans. 240. IJSTEODUCTIOJS, 15 8. If 4 times what Mr. Jones spent during a summer vaca- fion be diminished by three-fifths of the sum spent and $680, 4ie result -will be $5100 ; what did he spend? Ans. $1700. LESSON VII. 1. Anna has 8 oranges more than William, and they to _gether have 36 ; how many has each ? SoLUTioK. Let X equal William's opekation, number; then, since Anna has 8 Let a; = William's namber. oranges more than William, a; + 8 will x+8 = Anna's number, equal Anna's number ; and since they a;+a;+8 = 36 both haTe 36, x pluf> x+8 wiU equal 2a;+8 = 36 36. Adding, we have 2x+8 = 36. '2a; = 28 If 2a; iTiareased by 8 equals 36, 2a; will a; = 14 William's number, equal 36 dimimsAed by 8, or 28. If a;+8 = 22, Anna's number. 2a; equals 28, x equals one-half of 28, or 14, William's number; and a;+8 equals 14+8, or 22, Anna's number. 2. A and B together have $35, and A's money, plus $9 equals B's ; how much has each ? Ans. A, $13 ; B, $22. 3. The sum of two numbers is 100, and the smaller number equals the larger diminished by 16 ; what are the numbers ? Ans. 42; 58. 4. A watch and chain cost $220, and the chain cost $20 less than five-sevenths of the cost of the watch ; required the cost of each. Ans. Watch, $14Q ; chain, $80. 5. A house and lot cost $5800 ; required the cost of each- if the lot cost $300 more thail three-eighths as much as the house. Ans. House, $4000 ; lot, $1800. 6. Blanche, Lidie and Eate went a-shopping and spent $70. Lidie spent $4 more than Kate, and Blanche spent $6 more than Kate ; how much did each spend ? Ans. Blanche, $26; Lidie, $24; Kate, $20. 7. A, B and C contributed $125 to a Sabbath-school; A gave $10 less than twice as much as C, and B gave $10 more than twice as much as C ; what did each contribute ? Am. A, $40; B,$60; C, $25. 1 6 INTRODUCTION. 8. A la;iy bought a hat, cloak and shawl for $78'; what did she pay for each, supposing that the cloak cost twice as much as the hat, plus $4, and the shawl twice as much as the cloak, lacking $4 ? Ans. Hat, $10 ; cloak, $24 ; shawl, $44. LESSON VIII. 1. If the height of a tree be increased by its two-thirds and 10 feet more, the sum will be twice the height; what is the height of the tree ? SoLTTTlON. Let X equal the height of the _ _♦>, i ■ i,< tree ; then, by the condition of the problem, ^x we have the equation a;+fa; + 10 = 2a;. Add- a; + — + 10 = 2z ing, we have f»;+10='2.'i;. If fz + lO equals fa; + 10 = 22; OPEBATIOSr. X X 2x, then 10 will equal 2x minus f«, or — ; or 10 = — 3 3 — will equal 10; hence x equals, three times. — = 10 3 ., 3 10. orSO. a: = 30, height. 2. If twice the length of a pole be increased by two-thirds of its length and 8 feet more, the sum will equal three times its length ; what is its length ? Ans. 24 feet. 3. Three-fourths of Nelson's age, increased by 6 years, equals four-fifths of his age, increased by 5 years ; how old is he ? Ans. 20 years. 4. Five times the money Jennie paid for her bracelets, dimin- ished by $8, equals three times the money she paid, increased by $8 ; what did she pay ? Ans. $8. 5. Emma bought a fan, shawl and cloak ; the fan cost $4 ; the shawl cost $4 more than two-thirds of the cost of the cloak ; and the cloak cost $4 more than the fan and shawl ; required the cost of the shawl and the cloak. Ans. Shawl, $28 ; cloak, $36. 6. A's money, plus $12, equals B's, and B's, plus $6, equals C's, and the sum of their moneys equals four and one-half times A's money ; how much money has each ? Ans. A, $20 ; B, $32 ; C, $38. 7. Three tinres what it cost Harry to attend college a year INTEODUCTION. 17 increased liy $50 equals twice the sum obtained by increasing the amount by $1 50 ; what did it cost him ? Am. $250. 8. A man driving his geese to market was miet by another, wh; said, " Good-morning, master, with your hundred geese." He replied, "I have not a hundred,' hnt if I had as many more, and half as many more, and two geese and a half, I would have a hundred." How many geese had he ? Am. 39. LESSON IX. In Problem 1, Lesson I., we supposed Henry's number, plus three times his number, to be equal to twenty-four. Suppose now, instead of representing the number twenty-four by the figures 2 and 4, we use one of the first letters of the alphabet, as a, to represent it. The problem will then become — 1. Henry's number of apples, plus three times his number, equals a ; how many apples has he ? OPERATION. Solution. Let x equal Henry's number; Let a; = Henry's number, then, by the condition of the problem, we will x+3x = a have x+Sx = a, or 4x = a; hence x equals a Ax = a divided by 4, which we express thus, ^. x = — 4 4 Known Numbers. — Now, it is evident that a may represent any other number, as 12, 16, etc. Hence, we see we may repre- sent known numbers by letters as well as by figures. A number represented by figures expresses a definite number of units, and may therefore be called a definite number. A number represented by a letter does not express a definite number of units, and may therefore be called an indefiniU. number. Substitution. — Since a represents any number, let us sup- pose, as at first, its value to be 24; if we then use 24 for a- 24 a ill the result « = -> we will have x = —-, or 6, which is the 4 4 same result as we obtained by using 24 in the solution in Lesson I. This using some particular value of a general quantity in an expression containing the quantity is called Substitution. 18 INTEOPUCTION. PKOBLEMS 2. Mary'a age is twice Sarah's, and the sum of their ages ia a years ; how old is each ? Am. Sarah, - ; Mary, — • 3. Find the age of each -when a = 36, by substituting the value of a in the result. Am. 12 years ; 24 years. 4. Divide the number m into two parts, such that the larger parfr will be 5 times the smaller. Ans. — ; 6 6 5. Find the value of each part when m = 144, by substitut- ing the value of m in the results. Am. 24 ; 120. 6. The difference of two numbers is a, and 5 times the smaller equals the larger; what are the numbers? Am. -; — 7. Find the value of each part when a = 24, by substituting the value of a in the results. Aiis. 6 ; 30. 8. What number is that to which if its one-third be added the sum will be 6 ? Ans. — . 4 9. Divide the number c into three such parts that the first shall be twice the second, and the second twice the third. Am. 1st, — ; 2d, — : 3d, -• '7 7 '7 10. If three times a number increased by n equals a, what is the number ? Ans. . 3 11. The sum of two numbers is a, and the smaller equals the larger diminished by e ; what are the numbers ? . a+e a—c Ans. ; •■ 2 2 12. One-half the length of a pole is in the mud, one-third in the water, and h feet in the air ; what is the length of the pole? Am. Gh. Note. — Let the pupil give special values to the general quantitiea L each of the above problems, and find the results by Substitution. ELEMENTARY ALGEBRA. SEOTIOIT I. DEFmiTIONS AND EXPLANATIONS. 1. Mathematics is the science of quantity. It treats of the properties and relations of quantity. 3. Quantity is anything that can be measured. It is of two kinds, Number and Extension. 3. Arithmetic is the science of Number; Gemnetry is the science of Extension, 4r. Algebra is a method of investigating quantity by means of general characters called symbols. 5. Algebraic Symbols are the characters used to represent quantities, their relations and the operations performed upon them. 6. The Symbols of Algebra are of three kinds, namely — 1. Symbols of Quantity ; 2. Symbols of Operation ; •3. Symbols of Relation. Notes. — 1. With beginners we regard Algebra as restricted to numberi, or as a kind of general Arithmetic. They may afterward be led to see how general symbols introduce ideas not found in Arithmetic; and eventually, that Algebra is a general method of investigation that may be applied to all kinds of quantity. 2. Some writers divide Algebra into Ariihmeiieal Algebra and Symbol- ical Algebra. Newton called it Universal Arithmetic, and many writers Hpeak of it as General Arithmetic. IVAlembert divides Arithmetic inU Numerigue, Special Arithmetic, and Algebra, General Arithmetic. 20 DEFINITIONS AND EXPL ANA 1 lOiSrH. SYMBOLS OF QUANTITY. 7. A Sjmibol of Quantity is a character used to represent a quantity. 8. The Ssrmbols of Quantity generally used are the figures of arithmetic and the letters of the alphabet. O. Known Quantities are represented by figures and the first letters of the alphabet, as 1, 2, 3, etc., and a, b, c, etc. 10. Unknown Quantities are usually represented by th, read is greater than, and <, read is less than. Thus, a>b and ab is read a is greate/r than h, and a- EXERCISES IN NOTATION. Express in algebraic language — 1. The sum of a and b. Am. a+b. 2. Three times b subtracted from a. Am. a- 36. 3. The sum of a and b, minus c. Am. a+b-c. 4. The product of a and b, minus c squared. Ans. ab - c\ 5. The sum of a and b, multiplied by c. Am. (a+6)c. 6. The square of m, minus m into n. Am. nv' - mn. 7. The sum of a. and b, into the difference of a and b. Am. (a+b) (a-b). 8. The square of a, plus the square root of a. Am. a'+i/a. 9. The square of the sum of a and b. Atis. (a+by. DEFINITIONS AND EXl'LANATIONS. 26 10. Four times a square into b, minus three times c square into u; cube. Ans. 4a'6-3ey. 11. The square of a plus b, divided by a minus b, plus four , times a into b square. Am. ~+iab'. a — b 1 2. The sum of a times x, and the square of b, divided -by a minus x. Ans. ?£±^'. a — X 1!]. The sum of the squares of b and c, divided by the dif " b' + d' rereuce of three times a and twice c. Ans. 3a — 2e 14. The cube of a~x, diminished by tlie square root of a plus*. Ans. (a-xy--i/a+x. 15. The cube of a, minus x, diminished by the sum of a and the square rootofa;. Ans. a'-x-(a+-[/x). 16. A trinomial with its second term negative, and twice the; product of the other two terms. 17. A homogeneous trinomial of the fifth degree, with the second term negative. NUMERICAL VALUES. 47'. The Numerical Value of an algebraic expression is the result obtained by substituting for its letters definite numer- ical values, and then performing the operations indicated. 1. Find the numerical value of (a' — ab)c when a = 5, 6 ■= 4, and c = 3. • SoLTJTiON. Substituting for a, b and opekation. B their assigned values, we have (5^-5 (a'-ab)o={5^-5x4r)x3 x4)x3; performing a part of the opera- =f25 — 20i x3= 5>rison gives rise to the equation. The equation is therefore the fundamental idea in Algebra; it is the basis oi all its investigations: Comparison is controlled ,by certain laws called axioms, and gives rise to certain operations called processes. 53. Algebraic Reasoning is employed in the solution of problems and the demonstration of theorems. 53. A Problem is a question to be solved. A solution of a problem is the process of obtaining a required result. 54. A Theorem is a'truth to be demonstrated. A demon- 28 DEFINITIONS AND EXPLANATIONS. stration of a theorem is a course of reasoning employed in estal> listing its truth. 55. An Axiom is a self-evident truth. Axioms are the laws which control the reasoning processes. AXIOMS. 1. If equals be added to equals, the glims will be equal. 2. If equals be subtracted from equals, the remainders will oe equal. 3. If equals be multiplied by equals, the products -will be equal 4. If equals be divided by equals, the quotients will be equal. 6. If a quantity be both increased and diminished by another, the value of the former will not be changed. 6. If a quantity be both multiplied and divided by another, the value of the former will not be changed. 7. Quantities which are equal to the same quantity are equal to each other. 8. Like powers of equal quantities are equal. Like roots of equal quantities are equal. REVIEW QUESTIONS. Note. — These Keview Questions are simply suggestive to the teaeher, who can extend them as fully as is deemed desirable. Define Mathematics. Quantity. Arithmetic. Geometry. Algebra. Symbols of Algebra. State the classes of syrobols. Define a Symbol of Quantity. Name Symbols of Quantity. Of Known Quantities. Of Unknown Quantities. Use of ; of oo . Of Accents. Of Subscript figures. The sign of continuation. Define a Symbol of Operation. Explain the sign of Addition, Sub- traction, etc. Define Coefficient. Power. Exponent. Index. Define a Symbol of Relation. Explain the sign of Equality, etc. Define an Algebraic Expression, the Terms, etc. Define Algebraic language. Numeration. Notation. Numerical Value. State jrinciples of positive and negative quantities. Define Reasoning. A Problem. A Theorem. An Axiom. Enunciate thfl Axioms. SEOTIOIfr II. FUNDAMENTAL OPERATIONS. ADDITION. 56. Auddition is the process of finding the sum of two -i more algebraic quantities. 57. The Sum of several algebraic quantities is a single quantity equal in value to the several quantities united. Note. — The symbol + was introduced by Stifelius, a "German math© matician, in a work published in 1544. CASE I. 58. To add when tbe terms are similar. Class I. When the terms have the same sign. 1. Find the sum of 2a, 3a and 5a. operation 2a Solution. 5a, plus 3a, are 8a ; and 8rt, plus 2a, are ^* 10a. Hence the sum of 2a, 3a and 5a is 10a. 5a 10a Rule. — Add the coefficients, and prefix the mm with its proper sign to the common literal part. EXAMPIiES. (2.) (3.) (4) C5.) (6.) 3a 5x -box ^a'c -da^b^e 4a 3x -lax 5a'o -ISa'b'd 5a 7x -6ax 12a'c -aVe 7a _85 -8aa; ISa'e - 6a'6°e i9a 23^ -26aa! 36a''e -34a^6'c 7. Find the sum of 4a, 6a and ,7a. Ans. Via. 8. Find the sum of -2a, -3a and -5a. Ans. -10a. 9. Find the sum of 3a6, 5a6, 6a6 and 8a6. A'ls. 22ab. 10. Find the sum of — 3ae, — 4ac, -7ac and -9ac. Aiu. -2yaft 30 FUNDAMENTAL OPBRATIDNS. 11. Find the sum of 5aW, la?V, 9a'6' and lOaV. Ans. 31a''5'. 12. Find the sum of &xY, l^f, 'd^f and 12a;y. Ans. 34ar'2/*. 13. Find the sum of -5a6c, -4a6e, -7a6e and -%ahe. Ans. —25abc. 50. Class II. When the terms have different signs. 1. Find the sum of 7a and - 4a. Solution, la is equal to 3a+4a. Now, -4a united opeeation with +4a, a part of 7a is equal to nothing, Prin. 3, Art. 7a 50; therefore —4a added to 3a+4a equals 3a. Hence, ~4« ~ 4a added to 7a equals 3a. 3a Solution 2d. Plus 7a may indicate some quantity inareased hy 7a, and —4a may indicate some quantity diminished by 4a. A quantity increased by 7a and then diminished by 4a is evidently increased by Sa; hence the sum of la and —4a is plvs 3a. * 2. Find the sum of —6a, 4a, —7a and 9a. Solution. The sum of the positive quantities, 9a and operation. 4a, is 13a; and the sum of the negative quantities, —7a —8a and -8a, is -15a. Now, -15a= -13a-2a; +13a 4a united to —13a is equal to nothing, and there remains —7a —2a. Hence the sum is — 2a. 9tt Solution 2d. The latter part may be given thus: —2a Any quantity increased by 13a and then diminished by 15a is evidently diminished by .2a ; hence the sum is minus 2a. Rule. — I. Find the sum oj the coefficients of the positive and negative terms separately. II. Take the difference of these sums, and prefix it, with the sign of the greater, to the common literal part. EXAMFIiES. (3.) (4.) (6.) (6.) (7.) + 7aa; -5aV -Iz' + 27 xy -2x'y\ -%ax +3aV -z= -ZAxy - 12a;y? + %ax — 9aV + 9a' -Ibdxy +/-!fz -3aj +4aV -bz' + 21 mj +28x'fa + Zax -7aV -iz' Find the sum — 8. Of 6a6, -5a6 + 8a6 and — 3a6. Am. Gab, ADDITION. 31 Am. - 6cd. Ans. 5xy. Am. —ian. Am. -lOa'b. Ans. — aV. 9. Of Serf, - Ged, -led, +8cd and -Acd. 10. Of Ixy, +dixy,-2xy, + Zxy and -Axy. 11. Of ban, +7an, -12an, +15an, -19a». 12. Of 1a% - Qa% + l{)a% + \2a% -30a^6. 13. Of 12aV, -6aV, + aV,-15aV, +7aV. U. Of 153^=^8, -19a;y''3, +12x2/%, -Waij/'s, -15a«/% and 15. Of 5ac'i\ +6ac'6^ -7ac'6^+8ac'6^ -17ac'6^ -4ac'6', 16. Of 21am''m;', - lQaw?m?, - 2\amhix', + 25am'nx', + 19am=n«', -25a?n'Ma^. 4^. Q, CASE II. 60. To add when the terms are dissimilar. 1. Find the sum of 3a, 46 and —oib. Solution. Since the quantities are dissimilar, we can- not unite them into one sum by adding their coefficients ; we therefore indicate the addition by writing them one after another with their respective signs. We thus have 3a+46-a6. 3a +46 2. Find the sum of 2a+3a6, 3a — 4a6 + 56 and Sab — 7b. SoLUTloir. We write the similar terms in the same column for convenience in adding, and begin at the left to add: 3a and 2a are 5a, which we write under the column added; 5a6, — 4a6, +3ab are +4a&, which we write under the column added; —76, +56 equals —26, which we write under the column added. Hence the sum is 5a+4a6 — 26. Rule. — ^I. Write similar terms, with their proper signs, in the same column. II. Add each column separately, and connect the results with their proper signs. EXAMFI^BS. OPERATION. 3a 46 — a6 -a6 OPEKATION. 2a +3a6 3a-4a6+56 5a6-76 _ 5a+4a6-26 (3) (4.) (5.1 2a +36 3x — 5xy 12a6^+28car' 5a -7b 7x+8xy -06^^+250;' a +96 a- -9x-6xy 24a6'-23ca;' 3a -86 4a- -5x+7xy -35a6'-17ca^ 11a-36 6a- -Ax+Aayy + 13ca;' 32 FUNDAMENTAL OPEEATIONS. 6. Find the sum of Zae-bax, 7ac + Gax, 5aG—12ax ana 9ac+ I5ax. Ans. 24ac+4aa;. 7. Find the sum of 5ab + 12bc-7ed, 9ab-18bc+lled and nab-15bo+lBcd. Ans. Slab -21bc+ lied. 8. Find the sum of 3ax-2b''c, 5ax + 7e\ 96'c-12c*, $ax + 15^ mid IWc - 18e'. • Ans. Wax + 21b'e - 8c\ 9. Find the sum of m + Bn^-5mn, 3m— 8n^, 7ra'— Swn, \9m+27mn and lQn^-17mn. Ans. 23m — 3m»j+18»il 10. Fmd the sum of a+26+-3c, *2a- 6,— 2c, b — a — e and c — a — b. Ans. a+ 6 + c. 11. Find the sum of 3a-4p + g, 7p + 3g-6, 9a-7 + 3p and 9g-12+llj9. ^ns.l2«;+ 17^9 + 13^-25. 12. Find the sum ofa+6 — c, a-b + c, a+c+b and b — a + c. Ans. 2a +26 + 2c. 13. Find the sum of 4a+ 7a'c - 8m', 7a+ 16m', 15a'c - 20m' + 17 and 12m' - 5 - 22a^c. Ans. 11a + 12. 14. Add 34aa;' - 18ay\ - 25aa;' - IZaf + Uaf, 16 + lbay\ \baf-l& and 22aa;' + 7a2/'-lla2/', Ans. 31aa:'+20ay''-9ajr'. 15. Add 12a; + %y- %z, ba~12y + 13a;, 7y - 16a; + lOs and 10a;-5a+122i. Ans. I'dx + Ay+lQz. 16. Add a;"-aa;^ + 36, 2,ax'-2b+y''^, 5a;"+46-32/^ 76-4a:" +f"-Bax\ Ans. 2x"-ax''-y'"+12b. 17. Add 5a-96 + 5c+3-d, a-36-8-c?, 3a+26-3c + 4 + 5ci, 2a+5c-6-3d. ^?is. lla-10J + 7c-7. 18. Add a;*-4a;'2/ + 6a;y-4a;j^ + 2/*, 4a;'2/-12a;y+12xy-42/*, 6a-y - 1 2a;?/' + Gy* and 4ay - 4y*. Ans. x*- y*. 19. Add a^+ab' + ae'-a^b-abc-a\ a'b + b" + be' - ab^ - b^c - abc and a'e + b''e + c' — a6c — bc^ — ao^. Ans. o' + 6' + c' — 3a6c. 20. Add 4a6 - 3inn f 10am — 6an, 7m?i — 7am + 4an, 3a6 + 7 an + 9, (3 - 4»JMi - 3am — bn' and An' — 15 — 2m'. Ans. 7ab — 2m^ — n^+5an. FACTORED FOEMS. 61. Similar quantities in any form may be added by taking the algebraic sum of their coefficients. ADDITION. Jja EXAMPLES, (1-) (2.) ' (30_ (4.) 6/7 8(a-6) -12/a + 6 5(m-^+2) V7 -5(«-6) 15iAr+b 7(m-M+2) _5j/7 6(a-6) -ISy'aTl. -9(m-w+2) l5i/7 9(a-6) -15T/aT6 3(m-M+2) 5. What is the sum of 5(x-y), - 12{x-y), Six-y), 10(a;-iJ and -14(a;-y)? ^ns. -8(a;-y). 6. What is the sum of 7(,a-b)\ -9(a-6)', +12(a-&/, + 16(a-6)', -18(a-6/? Ans. Sja-by. 7. Find the sum of S-f/a+gf, 5|/a+a;, -7-i/a+x, +8-[/a+x, - 5i/a+x and 12y'a+x. Ans. 16j/a. + a;. 8. Add 4cM + 7 (a^ - 6'), 6aa; - 5 (a^ - 6'), + 3(a^ - h^) - bax. 12(a'^ - 6') - 7aa;, IBCa" - 6^) + Qax, - %%{a^ - V). Ans. 7 ax. 9. Add 2a'-3(a+a;), 5a?+&{x-y)\ ia^-Jix-yy, 9{a+x)- 60,', 3(a+x) ^ 9(x - yf, a' - (a+x) + (x- yf. Ans. 6a'+8(a+x)-9{x-yy. 62. Dissimilar Terms having a common factor may Ix" added by taking the algebraic sum of the dissimilar parts, ep closing it in a parenthesis, and affixing the common fiictor. 1. Find the sum of ax+bx — cx. OPEEAT'ON. Solution, a times x, +b times x, — c times x, equals '"^ (a+6 — c) times x; hence the sum is (a i-b — c)x. — ex (a+b—cix. EXA9IFL.ES. 2. Find the sum of aa;'-6»' + ca^. Ans. (a-6+c)a^. 3. Find the sum of as'-mz^+n/ — gz*. Ans. (a — m+n — q)^. 4. Find the sum of 2(w;-26a;+(a — &)». Ans. 3(a — b)x. 5. Find the sum of 4Lax+3x+2ax—5x+bx — 5ax+2x — 2bx. Ans. (a — b)x. 6. Find the sum of Saj/— 252/+ (a +26 + c)?/. ^'**- (4«^ c)y- 7. Find the sum of San- 5am+2an — Zbn + '6am — 5m^C)bn-> 2am-3bn+6nn on. ' Ans. (5aic)n. 34 FUNDAMENTAL OPEKATIONS. SUBTRACTION. ©a. Subtraction is the process of finding the difference ol two algebraic quantities. 64. The Subtrahend is the quantity to be subtracted. 65. The Minuend is the quantity from which the subtrar hend is to be subtracted. 66. The Difference or Remainder is a quantity which, added to the subtrahend, will equal the minuend. Note.— The symbol - was introduced by Stifelius, a German mathe- matician, in a work published in 1544. CASE I. 67. To subtract when all the terms are positive. 1. Subtract 4a from la. * OPERATION Solution. 4 times a quantity subtracted from 7 times 7(j the quantity equals 3 times the quantity ; hence, 4a sub- 4.(f tracted from 7a equals 3a. 3^ 2. Subtract 7a from 4a. Solution. 4a equals 7a— 3a; 7a subtracted from opekation. 7a — 3a leaves— 3a; hence 7a subtracted from 4a 4a = 7a — 3a equals - 3a. _Ta = 7a -3a -3a Solution 2d. Plus 4a may indicate some quantity inereased by 4a, and +7a may indicate some quantity increased by 7a. A quantity inereased by 4a is evidently 3a less than the quantity inereased by la j hence, 7a subtracted from 4a equails minus 3a. 3. Subtract h+c from a. Solution. Subtracting b from a, we have the remain- operation. der a — 6 ; but we wish to subtract b increased by c from a, a hence the true remainder will be a— 6 diminished by e, or 6 +c a~b — c. (i—b — o Rule. — Change the, signs of the subtrahend and proceed as in uddition. Note. — Signs of terms are said to be changed when, being plus, they *re changed to minus, or tedng — ,"they are changed to +. SUBTRACTION, 35 EXAMPLBS. (4-) (5.) (6.) (7.) (8.) 12a Idny)/ 21mV 5a^+S6 3a+46 _9a _92^ 28mV 3a' +75 96 +2(! 3a 6a;'y -7mV 2a' -46 3a-56-2e 9. From 19a6 take 12a6. _lns. lab. 10. From 21ac' take 16acl Am. bac\ 11. From lOoay take 17aa^. Am. - laxy. 12. From 12mV take ISmVi Am. - GmV. 13. From 4a'+ 66 take 126. Am. 4a''-Qb. 14. From 7a + 5c take 10c. Am. la -5c. 15. From 3a;'+2y' take Ax^+f. Am. -3?+y\ 16. From 2a+36 take a+26. Am. a+h. 17. From 4a +26 take 2a +36. J.7M. 2a-6. 18. From a' +406+6" subtract a' +2a6 + 6'. Am. 2ab. 1 9. From a' + 6' subtract a' + 2a6 + b\ Am. - 2a6. 20. From 4a + 26 subtract 3a + 46 + 2c. Am. a - 26 - 2c. 21. From 7a'6 + 3ac subtract 5ac+4a'6. Am. 3a'b-2ac. 22. From 06 + 6e+c(i subtract 6c +2c(i+e. Am. ab-cd-c. CASE II. 68. To subtract irhen one or more terms are negative. 1. Subtract — c from +a. Solution, a equals a +e—e, since increasing and operation. diminishing a quantity by the same quantity does not +a = a+e — change its value. Now, — c subtracted from a +e—e, — e= — e leaves a + c. Hence, — c subtracted from +a leaves a+c a+o a + c. Solution 2d. The difference between any quantity operation. increased by a and diminished by C is evidently the sum + a of a and c; hence, — e subtracted from +a equals — c a + c. a+c 2. Subtract 6 — c from a. Solution. Subtracting 6 from a, we have the re- opebation mainder a — b; but we wish to subtract 6 diminished a by e from a ; we have therefore subtracted c too much, b — c aonsequently the remainder, a — 6, is c too small ; hence ^ZjZf^ the true remainder is o b ineresased by e, or a — 6 + c. 36 FUNDAMENTAL OPERATIONS. Bule. — I. Write the subtrahend under the minuend, placing mnilar terms one under another. II. Conceive the signs of the subtrahend to he changed, and then, proceed as in addition. £:xAMPi 6'. Ans. 2ab - 26', 19. From 3a+c+d-/-8 takec+3a-d Ans. 2d-f-8, 20. From 4a6 + 36' - 2e take 4a6 - 26' - 3d Ans. 56'- 2ef3d, 21. From 7am - 36c - c' take 5am — 2c' — 36c - 5a;'. Ans. 2am+c'+5a;'. 22. From 2a + 26 - 3c - 8 take 3c + 46 - 3a - 5. .<4?w. 5a-26-6c-3, 23. From a^+ 3a'6 + 3a6' + 6' take a' - 3a'6 + 3a6' - 6'. Ans. 6a'6 + 26». 24. ■ From a' - ■ 3a6 - 6' + 6c - 2c' take a' - 5a6 + 56c - 36' - 2c', Ans. 2a6+26'-46c. FACTORED FOEMS. 37 FACTOEED FORMS. 69. Similar quantities in any form may be subtracted by taking tlie algebraic difference of their coefficients. (!•) (2.) (3.) (4.) V6 12(a-6) 15iAT6 -7(a-J+4) V6 7(a-b) -l-iAT+b -12(a-6 + 4) V6 5(a-6) 22iAT5 5(a-6 + 4) 5. From diaP-y^) take -lix'-y'). Am. 12(x'-y'), 6. From - 6(a' - 6') take 12(a= - 6''). 4ns. - 18(a' - 6')- 7. From 6aXa - 6) take - 4aXa - 6). ^ns. lOa^a - 6). 8. Find the value of 5t/2- 7i/2+6i/2. ^ns. 4^/2. 9. From -5.cXc-(^) take -UxXc-dy Ans. 7xXc-d). 10. Find the value of IcXm -n)- 13^(m — n) + 12c\m - n). Ans. Ge'(m — n). 7©. Dissimilar terms having a common factor may be sub- tracted by taking the algebraic difference of the dissimilar parts, enclosing it in a parenthesis and affixing the common part. 1. From ax subtract ex. OPERATION SoivUTioN. a times x minus c times x is evidently equal ox tc (a — e) times 2;,-which is expressed thus (« — c^a;.' ex {a~c)x EXAiapi.x:s. (2:) (3.) (4.) (5.) a^ mz' avni as ba» -nz' cxy (a-c)xy z (a-b)x' _(m+n)z^ (a-l> 6. From 5az take baz. Ans. (5-6)a2. 7. From 1 mx take -Zax. Ans. . (c+S)ax. 38 FUNDAMENTAL OPERATIONS. 8 From az take bz — 82. Ans. (a — h + 2i)z. 9. From 4n^c+3e take Ic — iao. Ans. (w''-l+a)4c. 10. Fionx an + en +dn take n+ an idn. Ans. (e — l)n. 11. From (Qa+2x)6d take 4acd+2cdx. Ans. 2acd. 12. From 5a,^+lQb' take -3aV25^ Am. 8(aJ'+¥). 13. From 6ay — Swii/ subtract — 5»i2/+6c2/. Ans. {Sa-Se+in)2y. 14. From 6|/c — a|/c + J^/e subtract 2ai/c + 6|/c - 2i,/i; — (KCl/c J.M. (8-3a+aa;)i/c. USE OF THE PARENTHESIS. 71. The Parenthesis is frequently used in Algebra : we will therefore now explain its use in Addition and Subtraction. The plus sign before a parenthesis indicates that the quantity withia the parenthesis is to be added, and the minus sign indi- cates that it is to be subtracted. m Pein. 1. A parenthesis with the plus sign before it may be re- moved from a quantity without changing the signs of its terms. Thus, a + (b — e+d,) is equal to a+b — c+d. Prin. 2. A quantity may be enclosed in a parenthesis preceded by a plus sign without changing the signs of its terms. Thus, a+b — c+d — e is equal to a + {b — e + d—e), or to a+6 + ( — + d — e), etc. Pein. 3. A parenthesis preceded by the minus sign may be re- moved from a quantity if the signs of all its terms be changed. Tiiis is evident from the rule for subtraction. Thus, a—{b — o+d)ie equal to a—b + o—d. Pein. 4. A quantity may be enclosed in a parenthesis preceded by the minus sign if the signs of all its terms be changed. This is evident from the principles of subtraction, and, also from the previous principle^ Thus, a — 6 + c — d is equal to a-{b — G-i d); or tx. a ~b — ( — c+d]; etc. THE PARENTHESIS. 39 EXAMPIiEIS. Find the value— 1. Of —( — a') and x—(a — h). Arts, a'; « — a■^6. 2. Of +( — a6) and a — (fi-c+d). Ans. —ab; a-b + e — d. 3. Of - (6' - a') and 3e - (2o - 5). Ans. a' - 6' ; c + 5. 4. Of 4a - 56 - (a - 56 + 3e). ^m. 3(a - c). 5. Of 5a-26-3c-(-5e+2a-26). Ans. Ba+2c. 6. Put in a parenthesis preceded by a plus sign the last three terms of a+26-3c+d-4. Am. a+26 + (-3e+d-4). 7. Put in a parenthesis preceded by a minus sign the last three terms of Ba-Ab + 5e-7d. Ans. 3a- (46 - 5e + Id.) 8. Find the value of 2a — (b + e-d+e-f) plus 26 — (a-c + d — e+g). Ans. a+b~g+J. 72. Expressions sometimes occur containing iaore than one pair of brackets, as a— {b — (e — d)]. Such brackets may be removed in succession, beginning^ for convenience, with the inside pair. Note. — Brackets may alao be removed by beginning with the ouJ« pair, or with any 'pair. Find the value — 1. Of a-{6 + (e-d)}. Solution, a- {b+{o-d)\=a- \b+o-d] =a-b-c+d. 2. Of a-{b-(c-d)\. Ans.a-b+c-d. 8. Of a- {6-c-(d-e)}. Ans. a-b+c+d-e. 4. Of 2a-{6-(a-26)}. i9M.3a-36. 5. Of 8a-{6 + (2a-6)-(a-6){. Ans.2a-b. 6. 7a - [3a - { 4a - (5a - 2a) } ]. Ans. 5a. 7. Of 6a - [46 - { 4a - (6a - 46) | ]. Ans. 4a. 8. Of a-[26+{3c-3a-(a+6)} + {2a-(6+c)|]. Ans. 3a -2c. to FUNDAMENTAL OPERATIONS. EEJliABKS UPON ADDITION AND SUBTRACTION. 1. Addition and Subtraction ma,y also be explained by regarding the positive and negative quantities as representing, respectively, gain and loss in business, distance north and soidh, etc. But these illustrations, though they may a,id the beginner, are not sufficiently general to be embodied in 1 solutioi.. 2. The problem, subtract 7a from 4a, may be explained by the follow- ing method : la equals 4a+3a ; subtracting 4a fiom 4a, nothing remains, and there is still 3a to be subtracted, which we may represent by writing — 3a. This method, however, is not general ; it will not explain several cases, such as — 7a from 3a, nor the general problem, subtract — C from a. 3. Special attention is invited to the method of explaining Addition and Subtraction given in the " Solution 2d" of Articles 67 and 68. The pecu- liarity of the method consists in regarding a positive term as indicating that some quantity is increased by the term, and a negative term as indicating that some quantity is, diminished by that term, or in using an auxiliary quantity. Thus, to subtract —2a from +3a, we regard +3a as indicating that some quantity is to be increased by 3a, and — 2a that som,e quantity is to be diminished by 2a ; then since a •quantity increased, by 3a is greater than the quantity diminished by 2a, by the sum of 3a and 2a, or 5a, we infer that —2a taken from +3a leaves +5a. Hence we use ''a quantity" as auxiliary. The same idea is presented in the following form of statement: The difference between a quantity increased by 3a and diminished by 2a is evi- Viently the sum of 3a and 2a, or 5a ; hence — 2a subtracted from 3a leaves -i 5a. The plus sign before the remainder will show that the minuend is greater than the subtrahend ; the minus sign before the remainder will show that the minuend is less than the subtrahend. 4. This method enables us to give a simple explanation to each of the eight possible cases in the subtraction of monomials. It will be well to have the pupiW explain each of the cases given below: do- (2.) (3.) (4.) (5.) (6.) (7.) (8.) 7a 4a -7a -4a -7a 7a 4a -4a ia 7a -4a -7ffl 4a -4a -7a 7a 3a -3a -3a + 3a -11a 11a 11a -11a irCLTIPLICATION. 41 MULTIPLICATION. 73. Multiplication is the process of taking me quantity as lany times as there are units in another. 74. The Multiplicand is the quantity to be multiplied. 75. The Multiplier is the quantity by which we multiply. 76. The Product is the result obtained by multiplying. 77. The Multiplicand and Multiplier are called factors of lie product. Note. — The symbol x was introduced by Wm. Oughtred, an English lathematician, born in 1574. PRINCIPLES. 1. The product of two or more quantities is the same in whatever rder the factors are arranged. Thus, a times 6 is the same as b times a, as may be seen by assigning aec'ni values to tile letters; and the same is true of any number of uantities. 2. Multiplying any factor of a quantity multiplies the quantity. Thus, 4 times the quantity 2x3 equals 4x2x3, which is 8 x 8, or x3x4, which is 2x12. Thus, also, 3 times 2a is 6a; 4 times 3a6 i]2a&. 3. The exponent of a quantity in the product is equal to the \im of its exponents in the two factors. Thus, a" X a? equals a*, since a used as a factor imice, multiplied by a 3ed as. a factor three times, equals a used as a factor /»e times. It may [so be seen thus : a' x a' = aa x aaa, which equals aaaaa, which }ual9 a'. 4. The product of two factors having like signs is positive, and ie product of two factors having unlike signs is riegative. To prove this, multiply 6 by a ; - 6 by a ; 6 by - a, and - b J -a. First, +6, taken any number of times, ! a times, is evidently + ab. Second, — 6 taken once is — 6 ; taken rice, is —2b, etc.; hence, —6, taken ly number of limes, as a times, is -ab. -+«* +6 UJ.'J3.X6 -b +6 ' -b +a + a — a -a + aj -ab -ab -i~ab) 42 FUNDAMENTAL OPERATIONS. Thitd, 6 multiplied by —a means that h is to be talcen subtractivdy a times ; b taken a times is ab, and taken subtractivdy is — ab. Fourth, —6 multiplied by - a means that —b is to be taken sub- tractivdy a times; —b taken a times is — a6, and used subtractivdy ii -( — ab), which by the principles of subtraction is + ab. Hence we infer that the product of qvxmtities having like signs is plus, and having twlike signs is minus. CASE I. 78. To multiply a nionomial by a monomial. 1. Multiply 36 by 2a. Solution. To multiply 36 by 2a, we multiply by 2 operation and by a. 3& multiplied by 2 and by a equals 36 x 2 x a, 36 which, since the product is the same in whatever order the 2a factors are placed, equals 3x2xax6, which equals Gab. 6a6 Therefore, 36 multiplied by 2a is 6a6. 2. Multiply W by Sa,\ Solution. To multiply 4a' by 3a', we may multiply operation one factor by 3 and the other factor by a' (Prin. 2). 3 times 4a' 4 are 12, and a' times a' is a* (Prin. 3). Therefore 4a' 3a^ multiplied by 3a' equals 12a*. 12a' Rule. — I. Multiply the coefficients of the two factors together. IL To this product annex all the letters of both factors, giving each letter an exponent equal to the sum of its exponents in the two factors. III. Mahe the product, positive when the factors have like tigns, and negative when they have unlike signs. EXADIPIiES. (3.) (4.) (5.) 5a -6a' 5a' 26 3a 4a'' lOab -18a» 20a' (7.) (8.) (9.) 8ab'' 12aa; Idrn'n Sa'b 4cx 48aca;» ~5an 24a'6' - 75amV (6.) -6c'6 -3c + 18e'6 (10.) — 5a'c Gc'd BOaVd MULTIPLICATION. 43 11. Multiply 7mV by -SmV Am. -35mVx. 12. Multiply 12aV2/'' by la^c'xy. Am. Sia'c'xy. 13. Multiply -9a'6Vby -la'bV Am. 63a^J'eV 14. Multiply Aia+by by 2a. Am. 8a(a+6)'. 15. Multiply -a(ia-x) by 6. Am. -ab{a-ai) 16. Multiply (a+6)' by (a+bf. Am. (a+6)'. 17. Multiply a(a;- 2/)^ by 2(a;-y). Am. 2a{x-y)K 18. Multiply -5a;(m-n)^by -Zx(m- n)\ Am. 15a^(m — ny 19. Multiply a" by a". Am. a"-^". 20. Multiply b^ by b". Am. 6", 21. Multiply c"* by cl ^m. c" 22. Multiply d*" by d» + l ^ns. d*"+', , 23. Multiply (a - a;)*" by (a - a;) "". Am. (a - x)" 24. Multiply -3a\f'-m)» by -2a'(P-m)-'. CASE II. 79. To multiply a polynomial by a monomial. 1.' Multiply a — 6 by c. OPERATION. Solution. To inultiply a — 6 by e we must multiply each, a—b term by c. c times a is ae, and c times — 6 is — &c. Hence, ^ 3— 6 multiplied by c is ac — 6c. . ac—be Rule. — Multiply each term of the multiplicand by the multiplier. and connect the products by their proper signs. EXAMP1.BS. (2.) (3.) (4.) 7a' -36 &ax-5e\ 3m»-4w'+7 3a 3ae — Imn 21a' -9a6 ISa^'ca; - 15ac'y -6»i'»+8mw*-14jn» (5.) (6.) (7.) 4a» - 3a6" Se*" - Abe + 5d^ 4a;» - 5xf 2a* Scd ' 3afy-' 8a*'-6a''+'6" 9e"+'(i-12Jc''d+15cd"+' 12a;''+V"'-15a!V"' 44 FUNDAMENTAL OPERATIONS. 8. Multiply 5ax'-3a^yhj -Ga'x. Am. -30aV + 18a*a;V 9. Multiply llwi' - 3 by - 5. Am. - 55m' + 15. 10. Multiply a^-^ah\V by ah. Am. a?h- 1aW-^ah\ 11 Multiply 3a" -46" by a'"6''".- ^ws. 3a*'6'"'-4a*6*". 12. Multiply a»-'6-6»-'c by a6l Am. al'b" - ah'c. 13. Multiply 5a;'-7m'a; + 3J by Amx. Ans. 20OTa;*-28mV+14wi«. CASE III. 80. To multiply a polynomial by a polynomial. 1. Multiply 2a -6 by a +26. OPEEATION. Solution. a+ 26 times 2a— & equaL? a times 2a— b 2a— b plus 26 times 2a — 6. a times 2a — 6 equals 2a'' — a6; « +26 26 times 2a- 6 equals 4a& — 26^ Adding the partial 2a^— a6 products, we liave 2a»+3a6-26'i. Therefore, etc. +4a6 - 25' 2a«+3a6-26S Rule. — Multiply each term of the multiplicand by each term of the multiplier, and add the partial products. EXAMPLKS. (2.) , (3.) 2a- 36 a+6 a— 6 a+6 2a'— 3a6 a'+a6 -2a6 + 36' +a6 + 6' 2a'-5a6 + 36' a'+2a6 + 6'' (5.) a'+a6H 6' a— 6 a^ + a^b +ab' -a'b-ab'-b^ 7. Multiply 3a- 26 by 2a - 36. 8. Multiply a'- 6' by a' + 6'. (4.) a — 6 a +6 a'-a6 + a6-6" a' -6' (6.) a" -6" a'-6' ! a"+'- a'6» - a"6'+6"+' a"-''- a"6'-a'6''+6»+' Ans. 1 Ba'-13a6-66'. Am. a* — b'. MULTIPI.KJATION, 4fi 9. Multiply 3a; - 6y by 2x + 4y. Am. Gx' - 24y-. 10. Multiply o^+cd+(Phj c- d. Ans. &-d?. 11. Multiply ar'+y' by ar'-^/*. Ans.x^—^. 12. Multiply Ax - Zy by Ax + By. Ans. 16a;' - 9y\ 1 3. Multiply X* - x^2 + x'z^ — xi 4 s* by a; + z. Aiis: x" + /. 14. Multiplya"-1-6»-''bya'+6^ ^ns.a"-a%"-'+a"-W-6»- 15. Multiply aVH-a;y by aV — a;y. J.»m. uV — a;^. 16. Multiply Xi —yi by a;^ +y^. Aits. x-yr 17. Multiply 3|aH5ic' by 2a' +40*. _.4m. 7a* + 2diantity into its factors. 100. The Factors of a composite quantity are the quan- tities which multiplied together will produce it. FACTORING. 59 101. A Prime Quantity is one that canntt be produced by the multiplication of other quantities ; as, 11, a'+b^. 103. Quantities are prime to each other when they have nc common factor ; as 5 and 9, Ga'b and lied". 103. The Square Root of a quantity is one of its two equal factors ; thus, 2ab is the square root of 4a*6l Note. — Composition and Factoring are the converse of each other ; com- position is synthetic; factoring is analytic; one is the putting together of the factors to make the numher, and the other is the separating of the number into its factors. CASE I. 104. To resolTe a monomial into its prime factors. 1. Find the prime factors of 6a?b'. Solution. The factors of. 6 are 2 and 3; a'' = aa and 6' = 666 ; -hence, 6oW equals 2 x Saabbb. Hence we have the following rule : Rule. — Resolve the numerical coefficient into its prime factors, and annex to it each letter written as many times as there are units in the exponent. EXAMFIiKS. What are the prime factors — 2. Of 12a'6. Ans. 2x2x3aaa6. 3. Of 15a'6'e. Ans. 3 x baabbe. 4. Of 183? fz. Ans. 2 x 3 x Zxxxyyz. 5. Of 27a'»6'"c. Ans. 3 x 3 x 3a"a"6»6»6»c. 6. Of 105a"+'6"-'c. Ans. 3 x 5 x 7a"aa6"6-^6-'c. 7. Of 72a"5'V", when w ■= 1. Ans. 2x2x2x3x Babhccc. 8. Of 84a'"c^d*', when w = 2. Ans. 2 X 2 X 3 X 7aaaaccceccdddddddd. 6 ' 6': =2x3 =6 xbxb &a,^b^- =2x3aa666 60 COMPOSITION AND FACTOBING. CASE II. 105. To find one of tlie two equal factors of m ■uonomlal. 1. Find one of the two equal factors of 25a'6*. Solution. Kesolving ZSa'b* into ita factors, we have oPEHATroif. 5 X badbbbb. Since there are two 5's, one of the equal fac- 25a%* = tors will contain one 5 ; since there are two a's, the equal . 5 x 5aabbbb factor will contain &ne a ; since there are four b's, the equal 5abb ■= 5a6' factor will contain two b's; hence the factor is 6abb, or 5ab'. This, we see, is equivalent to extracting the square root of the coefficient, and dividing the exponents of the letters by 2. Rule. — Extract the square root of the coefficient and diidat, the exponents of the letters by 2. EXAMPLES. Find one of the two equal factors — 2. Of 4:aV. Am. 2a'&. 3. Of 9a*5». Am. da'b'. 4. Of IQxy. Am. 4x'y. 5. Of 36cW. Am. 6c*d\ 6. Of tT«y»". • Am. yy'A 7. Of 25a*"6^. Am. ba'"h\ 8. Of ^a;«V^ Am. ^x*'^. 9. Of 81(a+6)*. Am. %a+h)\ 10. Of 1764(a-a;)». Am. 42(a-a;)'. CASE IIL 106. To resolve a polynomial into two factors one of wliicli is a monomial. 1. Find the factors of 2ac - Aah. o iiT 1 i o • .. „ OPERATION. SOLTTTION. We see that 2a is a factor common to all 9 19 _a i ihe terms; hence, dividing 2ac — 4«6 by 2a we find the other factor to be o— 26: henCe, 2ac— iab = 2a(G—2b). o — zo ' 2a(c-26) FACTORING. 61 Rule. — Divide the polynomial by the greatest factor common to all the terms, enclose the quotient in a parenthesis, and prefix the divisor as a coefficient. kxa9ifi4e:s. Find the factors — 2. Of Qa'b+Qa'c. Ai^. 3a\2b + Bac). 3. Of Sa'b' - 12ab*. Ans. 4:abX2d' - 36"). 4. Of Uax'z+b&abx^. Am. 7axX2e+8b). 5. Of ae' - 6aV + adcK Ans. ae\l - abe + dc^). 6. Oi ax'y — a'xif^+dx'y. Ans. xy{ax - a^y' + dx'). CASE IV. 107. To resolve a trinomial into two eqnal bino- mial factors. 1. Factor a' + 2a6+6'. SoLTJTioN. a^ is the square of a, and 6' is the square operation. of b, and since 2ab is twice the product of a and b, the a'+iab + b'^ trinomial is the square of (a +6), (Theo. I.) ; hence, a+b (a+b)(a+b) is one of the two equal factors of. a'+2ab + b\ Rule. — Extract the square root of the terms which are squares, and if twice the product of these roots equals the other term, these roots, connected by the sign of this other term, will be one of the equal factors. EXAMPI-BS. Find one of the two equal factors — 2. Of a? - 2ah + b\ Ans. a-h. 8. Oi 3?+2xy+y'^. Ans. x+y. 4. Of A' - 2AB+B'. Ans. A-B. 5. Of 4a'' - 12ae + 9el Ans. 2a - 3c. 6. Of 9m'+12mn+in\ Ans. 3m + 2w. 7. Of l-2c' + c*. Ans. l-c\ 8 Of Ua^ i-40a"c"+25e*. Aiis. 4a»4 5c". 62 COMPOSITION AND FACTORING. CASE V. 108. To resolve a binomial consisting of the dif. ference of two squares Into its binomial factors. 1. Find the factors of a? — b^. SoIjUtion. The difference of the squares of two quan- operation. titira equals the product of their sum and difference, a'' — 6'= (Theo. III.) ; hence, a^-V = (a + 6) multiplied by (a-6). (a + 6)(a-6) Rule. — Take the square root of each term, and make their sum one factor and their difference the other factor. E:xAin:PL.E:s. 2. Factor a' — c''. Ans. {a+c)(a — c). 3. Factor c' - Ad\ Ans. (c+ 2d)(c - 2d). 4. Factor 4a;»-9y'. Ans. (2x+By)(2x-3y). 5. Factor aV- J V. Ans. (ax+bz)(ax-bz). 6. Factor \x'-\yh'. Ans. {:^x+^^{\x-\y^). 7. Factor ga'"- 16c*". ^ns. (3a»+4c'^)(3a"-4c'^). 8. Factor a*-c*. Ans. (a^+e^)(a+c)(a-c). 9. Factor x'^y'' — y^. Ans. y^{x +!)(»- 1). 10. Factor a'-6^ ^ms. (a*+60Ca'+J')(a+6)((i-6). CASE VI. 109. To resolTe a trinomial into two unequal binomial factors. 1. Resolve a'+5ae+6e^ into its factors. SoitTTioN. The first term of each factor is evidently operation. a;, the second terms must be 2c and 3e, since their a^ + 5ac+6e'= product will be 6c^ and their sum 5c. (Theo. IV. ) (a+2c)(a+3e) Rule. — Take the sqaare root of one term for the first term of each factor, and for the second term take such quantities that their product will equal the third term of the trinomial, and their sum into tJie first term of the factor viUl equal the second term of iha 'rinomial. FACTORING. 63 ISXAMPL.£:S. 2. Factor x' + Bx+2. Am. (x^lXx-> 2). 3. Factor a»+ 5a +6. Am. >+2)(a+3). 4. Factor a;" -a; -2. Am. (x+lXx-2'). 5. Factor a* — a — 2. Am. (a — 2) (a + 1). 6. Factor a;' + 7a; + 12. Am. {x +3)(x + 4). 7. Factor a'' - 3a - 10. Am. (a - 5)(a+ 2). 8. Factor x* - 9x' - 36. Am. (a;V 3)(a^ - 12). 9. Factor 4a;'' - 6a! - 40. Am. {2x+5')(2x-8). 10. Factor o'+4ac-21c'. ^ns. (a-3c)(a+7c). 11. Factor a^+5a"- 84. Am. (a"-7)(a"+12). CASE VII. 110. To resolve a qnadrinomial into two bino- mial factors. 111. When two binomials contain a common term their product will be a trinomial ; when the terms are dissimilar the product will be a qnadrinomial. 1. Factor ac+bc+ad+bd. OPEKATION'. Solution, ac+bo eqaah {a+b)c, snC ad+bd -, ,j+hlivide A and B. Therefore, since the last divisor divides A and B, and is a number of times the a. o. D., it must be once the G. c. D. Hence, the remainder which exactly divides the previous divisor is the G. c. D. of A and B. Note. — The latter part of the solution may be given as follows : Since the last divisor divides A and B, it cannot be greater than the G. c. D., ann since it is a number of times the Q. c. D., it cannot be less than the Q. c. D. ; hence. Since it is neither greater nor less than the G. C. D., it must be th« G.C.D. Rule. — Divide the greater quantity by the less, the divisor by the remainder, and thus continue to divide the last^divisor by the last remainder until there is no reiwinder ; the last divisor will be the greatest common divisor. Notes. — 1. When the highest power of the leading letter is the same in both quantities, either quantity may be made the dividend. 2. If both quantities contain a common factor, it may be set aside, and afterward inserted in the greatest common divisor of the other parts. 3. If either quantity contains a factor not found in the other, it may be canceled before beginning the operation. (Prin. 3.) 4. When necessary, the dividend may be multiplied by any quantity not a fe,ctor of the divisor, which will render the first term divisible by the first term of the divisor. (Prin. 3.) 6. If we obtain a remainder which does not contain the leading letter, there is no common divisor. 6. When there are more than two quantities, find the G.C.D. of two, then of that divisor and the third quantity, etc. The last divisor will be the greatest common divisor. 2. Find the greatest common divisor of a" — b' and a' - 2ai t b' SoiiTJTioN. Dividing a' — lab , J.5 1. 2 A2 u *• i OPERATION. I + b^ by a' — o^, we have a quotient of 1, and a remainder oi -2a& a^-l>^)a^^~2ab + b\l + 2b\ Eejecting the factor -26, ^ 5! which does not affect the greatest -2ab + 2b'' Rejecting tli« common divisor (Prin. 3), we have lactor a-b. Dividing a2_ 62 ty Qj- 6, a-b)a'-b^ i^a + b we have a quotient a + b, with no a^—ab remainder. Hence the last divisor, ab—b' a — 6, is the greatest common divi- ab — 6' «or of a' - V' and d? - lab + 6*. THE LEAST COMMON MULTIPLE. 6S EXAMPLES. 3. Find the greatest common divisor of a' — t'^ and a'- 2a6 + 6' Am. a+b. 4. Find the greatest common divisor of 2a?^-5a;^-i-3» and fx^-2x-2.- Ans. x-1. 5. Find the greatest common divisor of a? - »' and a^ — a;^ Alls. h — x. 6. Find the greatest common divisor of ab-^bc+ad + de and a^-c'. Am. a+e. 7. Find the greatest common divisor of a'+a;' and a^ — x^. Ans. a+x. 8. Find the greatest common divisor of 2otx'' — 2ay' and 4aa^ .+ 4a2/^. Am. 2a(x+y). 9. Find the greatest common divisor of a* -6* and a'+a^b -aV — V. Ans. a'-V. 10. Find the greatest common divisor of a? — x^ — 12x and jr- Ax -21. Am.x+^. THE LEAST COMMON MULTIPLE. 117. A Multiple of a quantity is any quantity of which it -is a factor. 118. A Common Multiple of two or more quantities is a quantity which is a multiple of each of them. 119. The Least Common Multiple of two or more quan- tities is the least quantity which is a multij)le of each of them, Note. — The primary idea of a multiple is that of a number of times a quantity. The above definition is based upon a derivative truth. An- other derivative definition is, a multiple of a quantity is any quantity which it will exactly divide. Pein. 1. A multiple of a quantity contains all the factors of the qiuxntity. Thus, it -is evident that a multiple of ab, as 4a'6, must contain all the factors of ab. 70 C mPOSITION AND FACTORING. Pris 2. A common multiple of two or more quantities cm- tains all the factors of each quantity. Thus, il is evident that lia^bc, which is a common multiple of 2ab and Sac, contains all the factors of 2ab and Sao. Prin. 3. The least common multiple of two or more quantities contains all the factors of each quantity, and no other factors. Thup, it~is evident that the least common multiple o('2ab and Zac, which is 6abo, must contain all the factors of 2ab and Sao, oi it would not contain the quantities ; and it must contain no other factors, or else it would not be the least common multiple. CASE I. 130. To find tbe least common multiple by fac- toring. 1. Fiud the least common multiple of 12a'bG' and ISa'bd'. Solution. We first resolve the quan- opebation. titles into their prime factors. I'la^bd' —2x2x 3a'. b . (? The least common multiple must ISa'^Sd'^ = 2 x 3 x 3a^ 6 . d' contain all the different prime factor.?, i,.c.M = 2x2x3x Sc?. b . e^. d? and no others (Prin. 3). All the dif- =36a'6e^d^ ferent prime factors are 2, 2, 3, 3, a', b, c^, ?). 6. Of a{h - c) and 6(6' -^ c^. Ans. ab(,¥ - c'). 7 Of (a'-60 and a'-'2ab+b\ Ans. a'-a'b-ab'+b'. TJIK LKAST COMMON MULTIPLE. 71 8. Of aXa - 2) and x\a' - z") . Ana. a?x\a' - z% 9. Of ZxXIa - 1) and Axy{^a^ - 1). Ans. lIx'yiAi? - 1). 10. Of a? - 'f and a;' - f. Ans. x*-xf+ afy - y\ 11. Of 3a(a-6), 4ac(a^-6') and 6cXa+J). Ans. 12acXa'-&'). 12. Of m' + imn+n^ and m'+w\ Ans. (m?+n^)(m.+n'). Note. — Young pupils can omit the next case of Least Common Multi- pit until review. • CASE II. 131. To find tlie least common multiple when ttae quantities are not readily factored. The niethod will be readily understood from the following principle : Principle. — -The least common multiple of two quantities eqiuils either ■ quantity multiplied by the quotient of the . other quantity divided by their greatest common divisor. For, let A and S be any two quantities, and let operation. their greatest common divisor be represented by e, A = ax e and the other factors by a and 6, respectively; then B = b x. c we shall have the L.c.M. = ax6xc, Case L; but i,cM=ax6xc A A 6xc — £, and a=- ; hence, L.c.M. = - xB, There- — _ v 7? c e '^ -° fore, etc. c Note. — The least common multiple of two quantities equals their prod A AxJB uct divided by their greatest common divisor, for —xB = 1. Find the least common multiple of a' — b' and a^ — Bab + 26'. SoiiUTioir. We first find the great- operation. estj common divisor to be a — 6. Then G. c. D. = a — 6 the L. c. M. equals a'— b^ multiplied by ^ c m = (a''—b^) x 5[!i:.^a6 + 26' the quotient of a'' — 3a6 + 26^ divided " ' ' a~b by a-b, or (a2-6'')(a- 26), which -'{a'-b'){a-2b) equaL. a» - 2a'b - ab^ + 26'. = a» - 2a'b -ab' + 26» 72 COMPOSITION AND FACTORING. Rule. — I. Find the greatest common divisor of the two qymi/ tities; divide one quantity by it, and multiply the other quantity bij the quotient. II. When there are more than two quantities, find the least- eom- man multiple of two of them, then of this multiple and the third quantity, etc. EXAMPIiKS. 2. Find the least common multiple of a;' — a — 12 and x^-ix- 21. Ans. «' - Sa!" - 5a;+ 84. 3 Find the least common multiple of x^ +• 5a; + 6 and a?+%x+ 8. Ans. a;'+9iBV26a;+24. 4 Find the least common multiple of a''+4a6 + 36' and a' -51 Aris.a^+Za^h-aV-%h\ 5. Find the least common multiple of a'' + 3a6 + 26' and e-ah- &h\ Ans. a''- laV - 66^ 6. Find the least common multiple of a:'' — aa; + 3a; — 3a arid a;^ — 3a; — ax + 3a. Ans. x^ — ax^.— Qx+%a. 7. Find the least common multiple of a;' — a; — 2, x^+Sx+2 and 3;'' + 5a; +4. J.ws. a;*+5a;'-20a;-16. REVIEW QUESTIONS. Define Composition. A Compopite Quantity. State the relation of lomposition to Factoring. State the four theorems of Composition. Define Factoring. Factors. A Prime Quantity. Quantities prime te each other. State each case. Give the rule for each case. State tho three theorems of Factoring. Define Common Divisor. Greatest Common Divisor. State the cases. Tie principles. The rules. Define a Multiple. A Common Multiple, ^ilie Least Common Multiple. State the cases. The principles. Tho rules. SEOTIOIiJ' IV. FRACTIONS. 133. A Fraction is a number of the equal parte of a unit 133. A Fraction in Algebra is expressed by two quantities, one above the other, with a straight line between them; as, a a — x — or e a — z 134. The Denominator of a fraction denotes the number of equal parts into 'which the unit is divided. It is written below the line. 135. The Numerator of a fraction denotes the number of equal parts taken. It is written above the line. 136. An Entire Quantity is one that has no fractional part ; as, 2a'', a + b, etc. 137. A Mixed Quantity is one that has both an entire and a fractional part ; as, a+-> ax c m+n 138. An Algebraic Fraction is usually regarded as the expression of one quantity divided by another ; thus, - means a divided by c, etc. PRINCIPLES OF FRACTIONS. 139. The Principles of Fractions are general laws show- ing the relation of the value of the fraction to its numerator and denominator. Peis'. 1. Multiplying the numerator or dividing the denomi- nator of a fraction by a'ly quantity multiplies the value of the fraction by that quantity. 7 1* 74 FRACTIONS. If. we multiply the numerator of a fraction by n, there will be n time« as many parts taken, each of them the same size as before ; hence, the value will be n times as great. If we divide the denominator by n, the unit will be divided into l-»itb as many parts ; hence, each part will be n times as great ; and the same number of parts being taken, the value of the fraction will be n times'as E;roat. Therefore, etc. Pkin. 2. Dividing the numerator or multiplying the denomi- nator of a fraction by any quantity divides the value of the frao- tion by that quantity. If we divide the numerator of a fraction by n, there will be only 1-nth as many parts taken, each of the same size as before ; hence, the value of the fraction will be 1-nth as great. If we multiply the denominator of a fraction by n, the unit will be divided into n times as many parts ; hence, each part will be 1-nth as great as before ; and the same number of parts being taken, the .value of the frac- tion will be 1-nth as great. Therefore, etc. Pbin. 3. Multiplying or dividing both numerator and denomi- nator of a fraction by the same quantity does not change the value of the fraction. For, since multiplying the numerator multiplies the fraction, and multi- plying the denominator divides the fraction, multiplying both numerator and denominator by the same number both multiplies and divides the , value of the fraction by that number, and hence does not change the value. In the same way it may be shown that dividing both terms does not change the value of the fraction. Therefore, etc. Pein 4. The value of a fraction is equal to the quotient of ife nnmerator divided by its denominator. SIGNS OF THE FRACTION. 130. The Sign of a fraction is the sign written before the dividing-line, showing whether the fraction is to be added or subtracted. Note. — This sign is sometimes called the apparent sign of the fraction and the sign of its value, the real sign. KEDtrCTlON. 76 Prin. 1. Changing the sign of the numerator or denominator changes the sign of the fraction. For, this is the same as multiplying or dividing the fraction by —1, which will change the sign of the quantity (Prin. 1, Art. 129). Thus, M— X , —a+x a—x , H equals > or . etc. C + 3 C + Z —C — Z Prin. 2. Changing the sign of both numerator and denomi- nator does not change the sign of the fraction. For, this is the same as multiplying both numerator and denominator by — 1, which wUl not change the value of the fraction. (Prin. 3, Art. 129.) EEDUCTION. 131. Reduction of Fractions is the process of changing their form without changing their value. CASE I. 133. To reduce a fraction to its lowest terms. A Fraction is in its lowest terms when the numerator and denominator are prime to each other. 1. Reduce ■■ — to its lowest terms. 12a'bV Solution. Dividing both terms of the fraction by opebation. the common factors, 4,, a', b' and c, we have it equal ■ 8a?b''o 2a to — (Prin. 3, Art. 129) ; and this is the lowest term of ISa^&'c* 3c oO the fraction, since the terms are prime to each other. Rule. — Divide both numerator and denominator by their eonvmon factors; Or, Divide both terms by their greatest common divisor. e:xampi.es. 15aV 3.1; 2. Eeduce to its lowest terms. Ans. — - 2ba'x 5a' ■ 54a;V 32* 3. Reduce :— to its lowest terms. Ans. ——■ assy's' 2a^ 76 FRACTIONS. 4. Reduce — ■■ to its lowest terms. Am. 66a*a;¥ 7W a' — b^ 1 5. Reduce to its lowest terms. „ ^ , 2a+2& , ., , ^^ . 2 6 Reduce to its lowest terms. Am. a'-b^ a-b 7. Reduce to its lowest terms. .Ins. 2(a6 - 6) 26 2^'»+< 2a' 8. Reauce to its lowest terms. Ans. — Bar"-' 3 4£j2g»+i 2c' 9. Reduce to its lowest terms. Ans. — 6aV»-' 3a a;" — 9 x — S 10. Reduce to its lowest terms. Am. 2a;' + 10a;+12 2a;+4 11. Reduce to its lowest terms. Ans. • d'+2ab+b'' a+b 12. Reduce . to its lowest terms. Ans. x'+2ax-8a' x+ia 2.2" _ 952™ 18. Reduce — -— to its lowest terms. x'"-6b''x'' + %^ a;" +36" Ans. a;" -36" CASE II. 133. To reduce a fraction to an entire or mixed quantity. 1. Reduce to a mis;ed quantity. Solution. The value of a fraction ia equal to the operatiok. q[uotient of the numerator divided by the denominator, ao+b b Diiriding ao + bhy e, we have a for the entire part, and c e — for the fractional part. Therefore, etc. c Rule. — 1. Divide the numerator by the denominator for the entire part, continuing the division as far as necessary. n. Write the denominator under the remainder, and annex the result to the entire part -with the proper sign. EEDUCTION. 77 EXAMFIiES. I. Kediice to a mixed quantity. J.j!s. ae+ — 3. Reduce to a mixed quantity. Ans. 2x — ~ a+x %+x 4. Reduce to a mixed quantity. Ans. a+c — - a — c 3a' — 3»' 5. Reduce to an entire quantity. «-* Ans. Bia'+ax+aT). 6. Reduce to a mixed quantity. Ans. x+2z-\ • Qc-£f ^ ^ x-z 7. Reduce to a mixed quantity. a+b 26' Ans. a^ — ah+V- a+b 8. Reduce to a mixed quantity. Ans. 1 + (x^-z'y x'-s 9. Reduce ^ to a mixed quantity. Ans. 1 — (x+zf Qc+z)* CASE III. 134. To rednce a mixed quantity to a fraction. 1. Reduce a+- to a fraction. X Soi/unosr. It is evident that 1 =—; hence a equals X o a times —.or — . which, added to — equals 1- -> or ^^ XX X ^ ^ ax , c ax+e ax+3 r- X XX X Rule. — Multiply the entire part by the denominator of the fraction; add the numerator when the sign of the fraction is plus, and subtract it when the sign is minus, and write the denominator under the resuM. Note. — When the sign of the fraction is minus, remember tri change the ligns of all the terms. 78 FRACTIONS. EXAMPLES. 2. Reduce Ba+ — '— to a fraction. Ana, 2 2 3. Reduce 2z+ to a fractioik Ans. 3 3 4. Reduce x-\ to a fraction. Ans. ■ a+x a+x 5. Reduce a+e to a fraction. Ans ■ a—B a—e 6. Reduce 4a; to a fraction. Ans. • 4 4 _ _ . 2ae — e' , » ,. i («-c)' 7. Reduce a to a traction. Ans. -• a a 8. Reduce 4a+a;+ to a fraction. Ans. • a—x ■ a—x 9. Reduce 2a; — 5 ^ to a fraction. Ans. — ^^ -■ x-B x-B /j2 ~2 ~2 g2 10. Reduce a + a; to a fraction. Ans. a—x a—x CASE IV. 135. To reduce fractions by changing factors from one term to anotlier. Peinciple. — Any factor may be transferred from one term of a fraction to the other if the sign of the exponent be changed. This principle is readily proved by applying the principles of Article 91 a' 1. Change the terms in the fraction — (^ a -nr- 1 «' a 1 OPERATION. SoLTjTiON. First, -^ =- a* X — r ; but a^''-=- (Prin. 3, Art. 94); ' C e» «-» c» a-V hence.a'x— =— X— =-!-. 2d. ^=a'xi- = aVc '-o'o-' REDUCTION. 79 tt^ 11 1 Second, — ~a'x— -; but — = c-»; hence, a' x --= a' x c"', or a'o-*. Rule. — In changing a fachyrfrom one term of a fraction to the other, change the sign of its exponent. „5 a 2. Change the terms iii ■• Ans. a^c'x''. eV 3 Change the terms in — - — Ans. a'x'y*z~\ 4. Change — — to an entire quantity. Ana. Ba'^b'^if, 5. Change to an entire quantity. Ans. (a — a;)(a+a;)~'. a+x 6. Change to an entire quantity. ^"'^ Ans. 3x5-\a+ey. 7. Change ^ _ to a simpler form. Ans. Ke* — eV). 8. Change ^— ^ to an entire quantity. ^ ~y Ans. (x — y')(jc+y)~\ 9. Change -^^ ^ to an entire quantity. Ans. a(b^ — -— and — ■ Ans. 4a 6c 3w' 12acrv' 12am' 12acM' . 3 46' , 2ac . IScz IGaS'a 6a'e o. — — J - — - and Ans. 2a'e Sac' Ach 12aVs 12aV3 12aVs g a-b a+b 6e(a-6) 2a(a+6) 33aV _ a'e ' Sac" ^^ ' ^' 6aV ' 6aV ' 6aV '' „ ab be J cd . a6(a+6) beCa-b) cd 7. -> — - and Ans. — ^^ -> — -, a-b a+b a'-b' a'-b' a'-b' a'-i' „ 2aa; 3aa; , 4aa! o. -> — and x—1 x+1 «'— 1 . 2ax^+2ax Sax' — Sax Aax Ans. x'-l x'-l a^-1 a'e — ae' a' — ae e(a — b) , f -^ c. e{a — c) c(a — e) e(a — o) , (a-cy (a + c)' Ans. — ^—' — ^— • (a'-o'y ia'-cy ^^ ^ . ^ ^ ^ ., w ,>. . (a + c)' (a-eV a'+c* 11. » and . Ans. ^ — -> -> a' — c" a' — e" a' — c^ Ans. —. —J — rr ' c a — c 10. 12. C a- ■c a — c , rand (fl + C)^ o+c ia-cy a+c a — c a — c ■f a+c and a^+^ a^-e a a' -> and a* a»+l a'-l o*-l , a(a'-l) aW+l) a' Ans. -^ -t -^ -y a*-l a*-l ci'-\ 13, ^. -and (a+6)(6-e) (a+6)(e-6) Ans. {a+bXb-c) (a+bXb-e) OPEKATION. 82 FEACTIONS. ADDITION. 139. Addition of Fractions is the process of finding the simplest expression for the sum of two or more fractions. Principle. — To be added, fractions m/ust have a common de- nominator. « For, they then express similar fractional units, and only similar unitu can be united into one sum. 1. Find the sum of - and -• n ' n Solution, a divided by n, plus b divided by n, equals (a+b) divided by n. 2 + £ = £±° . n n n 2. Find the sum of - and — n m SoiiTJTioN. Since the denominators are not alike, we must first reduce the fractions to a common , , denominator. The common denominator is mn. —-\ = H „™ f. ?,»; n m mn mn a , am , o , on _ j- -j j - equals , and — equals . am divided am+bn n mn m, mm, = by mn, plus bri divided by mn, equals [am+bn) '^^ divided by Mj»i. Rule. — ^I. Beduee the fractions, when necessary, to their least common denominator. II. Add the nwmerators, and write the comm^on denominator under their sum. Notes. — 1. Eeduce each fraction to its lowest terms before adding, and also the result after addition. 2. Mixed quantities may be added by adding the integral parts and uniting the sum with the sum of the fractions. EXAMPLES. Find the sum — 3. Of -+-■ Am. — ry- b d bd ^„a b d . ade + Vc+bd* 4 0f-+-+— Ans. — b d c ode ADDITION. ' 83 6, Of 2^+-?£-+i«L. ^^, 2cm'4 3m+46 6. Ofi+^^i. ^,^.^±^£±^. X y z xyz 7. Ot — +0--— • J.n«. o aoc 3ac 8. Of — - — V — - — An». a. 9. Of -^^— Ans. ^™ rt m+M m— n m — n 10. Of -+ ■■ Am. a + b a—b a' — b^ 11. Of — ^5_ + — * ^^. 1 2a-26 26-2a 12. Of-, ^^ and ^^. ^«a.-^. 2c ac oc * 2c ^„ ^„ a" aj-a*" , a^a; - 2is' . x^-s" 13. Of — , and Am. ^■ 2a; dl'x 2aV aV ^A r>pl+« A l-« J 2(l+aO 14. Of :; and Am. — ^ -~ • 1 — a 1 +a 1— a' 15. Of£::2!,l^and^::^. Am.^. xy yz xz 16. Of 3aV^^ and 4a'+-?^. im. 7a'+?^^ 3 2a 6a 17. OfAandlii'. Am}-^. a+1 a'+s s 1— a;' 1 +a;' 1 — ar 1 19 Of ■ and Am. , (d~bXb-c) ia-bXe-b) b-e a+b b+c c+a . . 20 Of -— + + Am.Oi (b-cXc-a) (e-a)(o^6) (a-6)(6-e) 84 FRACTIONS. SUBTEACTION. 140. Subtraction of Fractions is the process of findiig the simplest expression for the difference of two fractions. Principle. — To be mihtraded, fractions must have a common denominator. For, they then express similar fractional units, and only similar units can be subtracted. 1. Subtract - from -• n n „ 1- ■ 1 1 , . 1 ,. . , , , OPEEATION. SOLUTION, a divided by n, minus o divided by ra, , . 1 / i,\ J- -J J 1. •' ' a b a-b equals (a—o) divided by n., = "^ ^ , n n n 2. Subtract - from — n m Soi/TJTION. Since the denominators are not alike, we must first reduce the fractions to a com- opekation. mon denominator. The common denominator is « o an om ^ a an /ft bm ,. . , , ^ m n mn mn mn. — — and- = ; an divided by ?wn, nn — hm m mn n rnn "" "'"• minus bm divided by win, equals {an—bm) '''''* divided by m/n. Rule. — I. Reduce the fractions, when necessary, to their leaA common denominator. II. Subtract the numerator of the mbtrahend from the numerator of the minuend, and write the common denoininator under the result, Notes. — 1. Reduce each fraction lo its lowest terms before subtractiag, and also the result after subtraction. 2. Mixed quantities may be subtracted by subtracting the integral parts, and uniting the difference with the difference of the fractions. EXAnPIiES. __a^,e . ad-be 3. f rom - take - • Ans. • b d bd . _, 2a ^ T 2c . ICam-c) 4. h rom — take • Ans. —^ — n mn mn _ _, 3aa; ^ , 4» . Qa\-^cx 5. Irom take — ■• Ans. 2c^ 3ac 6c«!' SUBTE ACTION. 85 „ T, + 6 ^ , a — h 6. From ——- take 7. From "^ take ^. a a 3a, 8. From 5a' + — take 3a' + ae 9. From 4c take 2c — ^^^-^^ J.ms. 2c — ^ns. 6. 4w«. — a 2a ax . „ , 3a;' - 2a<' Ans. 2a' + OCX a + 3 . „ 9 a — 3 a a' — 3a lAT-L a + 6^,6 — a . a' + 6' 10. From take Ans. a b ab 11. From take -• -4ns. a — b a + b a' — b^ 12. From take Ans, a-b a'-b' a'-b' 13. From take • Ans. a 1 + a 1-a' ^._ a;+l^,a;-l , ' 4x 14. From take Ans. x-1 x+1 sd'-l 1 +3' 1 — a' As' 15. From take Ans. ■ ■ 1-3' l+«' it> -n- J ^ p « 3a 2aa; . 4a 16. Fmd value of 1- -• Ans. a + x 17. Find value of -+-: — r- — ;; — -• Ans. a + b a'-b'' a' + b^ a*-b* 18. Fmd value of :: — - — ;: — — +- — -—• Ans. l-2a; l + 2a; l-4a;' 19. Find value of - — :; — ; - : — r— — r-- Ans. (a+b)(b-e) (a+6)(c-6) b-c 20. Find value of ( — +- |(a+6) -( )• Ans. — \m n) \ m ' n j n Bf FRACTIONS. MULTIPLICATION. 141. Multiplication of Fractions is the process of finding a product when one or both factors are fractions. CASE I. 148. To multiply a fraction by an entire quan- tity. 1. Multiply - by c. SoLTJTioN. Since multiplying the numerator of a frao- operation. tion multiplies the value of the fraction (Prin. 1, Art. 129), £ „ -, ^ £2 ^. a , ao b b c timea — equals o o 2. Multiply 4 by h. SoLxrraoN. Since dividing the denominator of a frac- operation tion multiplies the fraction (Prin. 1, Art. 129), 6 times a , a a , a b' b -equals-. Rule. — Multiply the numerator or divide the denominator of the fraction by the multiplier. Notes. — 1. The second method is preferred when the denominator ii divisible by the multiplier. 2. It is often convenient to indicate the multiplication, and cancel equal factors in numerators and denominators. KXAMPtiES. 3. Multiply - by n. Ans. — c c 4. Multiply by cd. Ans. c^d 5. Multiply - — - by 4*". Am. MXJLTIPLICATION. 87 6. Multiply -— by Sac'. Ans. e'(a — x) c(a ~ x) _ ,. ,^. , Tna; , „, ^ . 2'mx 7. Multiply — by 2(m — x). Ans. • (m — x) m — x 8. Multiply by a" - 3?. Am. ^^ -• (a — xy a — x 9. Multiply , t"^^ , by a;* - y*. Am. -^^^^ • a?-2xy+f x-y 10. Multiply by a'- 1. Am. 5a'x+5a'x. a—1 11. Multiply — by 2a(x-l). Am. — — — •« CASE IL 143. To multiply an entire or fractional quantity by a fraction. 1. Multiply a by-' c Solution, a multiplied by 6 is 06; hence, a mul- opebation. tiplied by 6 divided by c is a& divided by c, or ax-= — c c c 2. Multiply 7 by V o a Solution. - multiplied by c is — ; hence, - multi- opEBAnon. plied by c divided by d must be — - divided by d, which TX-;=r-: - ■^ •' 6 o a oa iB^i. (Prin. 2, Art. 129.) Solution 2d. Prom Art. 135, Prin., - equals opeeation. a e a6-» and - equals cd"', and ah- » multipUed by ^ >« ^ = «^"^>< "d- - cd-i equals ae&-'d-', which, by Art. 135, Prin., ae6->d-' = — , . ac fid !S equal to — - • od - 88 FRACTIONS. Bllle. — Multiply the numerators together for the numerator, and the denominators together for the denominator, canceling conmon factors. Notes. — 1. When there are common factors in the numerators and denominators, indicate the multiplication and then cancel the common factors. 2. If either factor is a mixed quantity, reduce it to a fraction before multiplying. EXAniFl.ES. 3. Multiply — by -• Ans. m n mn . ,^ ,^. - 2ax , 3a;' j «* 4. Multiply by — — • \ Ans. — 3c 2a e Sa^e . 4cn . 6c' by — r- Ans. • 2n' ^ 5a' 5an . ,^ ,^. , fflfi'", oar" , a'b" 6. Multiply — - by Ans. ar mo" mx' n TIT 1^- 1 a-x, aV . ax\a-x) 7. Multiply -^ by --• Av^. \ ' a 6b do 8. Multiply by • Am. c' a + c e' 9. Multiply - — by Ans. 4ax a + X 4ax 10. Multiply^ by i^. ^«.. Mll^. ^ •' 6a' •' 1 - a Sa" 11. Multiply ^^ by ^^'. Ans. a^-b\ a + b a — b 12. Multiply a-« by ^. Am. '^'-^^ c 3a 3 13. Multiply a + - by a Ans, a? XX a? 14. Multiply — -— by — - • Ans. — i— -^ 3m' n + z TO* DIVISION. 89 15. Multiply by -• Ana. {l + hf ' a+b "1 + 6 16. Multiply -^ by ^. Ans. ^-^^^. (.a -by e' cia-b) 17. Multiply 1+^ by ^+1 Ans. 2+%^. a a ' ad be 18. Multiply -^, ^^^ and .-^^ together. Ans. - • m+w. m' m — n m 19. Multiply IZ^ by ^^\ Ans. (ln^Xl+f). l-a' , l-rc' X + X^ X 20. Multiply <^~^> by <''^^^' ■ ^m. **' 21. Multiply ■ by ^ — - • Ans. a^-2ab + b^ a?+ab a 22. Eequired the value^f | 1 x ( )■ ^ ^ \b-e b-cj Y-V a?-V) Ans. 1. 23. Eequired the valu6 of {^\ (a- J) 4^-^(-^- Ans. 1. DIVISION. 144. Divlfiion of Fractions is the process • of dividing when one or both terms are fractional. CASE I. 145. To divide a fraction by an entire qnantity. 1. Divide — by b. e SoLTJTioir. Since dividing the numerator of a fraction a5 OPERATION. divides tlie fraction (Prin. 2, Art. 129), to divide — by 6 ^ ^ e — (-{)=— we divide the numerator by 6, and have — • c 90 PEACTIONS. 2. Divide ~hy b. e I SoiiTTTios'. Since multiplying the denominator of a fraction divides the fraction, to divide — by 6 we multi- e ^ ■ h ~ ply the denominator by 6, and have " ^' ^' •' ' fie Rule. — Divide the numerator or multiply the denominator oj the fraction, by the divisor. Note. — It is often convenient to indicate the division and then cancel common factors. E'KJUtrPl.ES, 3. Divide by 2ax. Ans. ba bo 4. Divide — — by 46c'. Am. ad ad f. _.. . , abed 1 „ , . alicd 5. Divide by 2a;\ Ans. mn ■ 2mnx* b. Divide — ; — by oz*e. Ans. ■• a'b' aVc 7. Divide ^(^Zf) by 2aV. , Am. ^^. 3c 6aV 8. Divide ^^ by o(a; +1). Am. — • ab' aW 9. Divide ^^^ by oXa - b). Am. ^^^^. a-c e\a-c) i(\ T>- -A a'a;-a^ I, „ „ i ax(a-x) 10. Divide by ac^+e^x. Am. ^ ■' . c" e'*' CASE II. 146. To dlTide an entire or a fractional qnantitj by a fraction. 1. Divide a by - • SoLTmoN. a divided by 6 equals ^; but since the ™™.^t„„ divisor is b divided by c, the quotient must be c times as b ao a ao «-!--■=-- great, or -xc = e 6 6 6 DIVISIOK. 91 2, Divide 7 by -■ a SoiCTiON. - divided by e equals (Prin. 2, Art. o oxc 129) ; but since the divisor is e divided by d, the quo- tient must be d times as great, or x d, which equals 6xc ay.d ad bxc' be OPERA riosr. a c ax d b d b xo ad OPERATION. a c , -^-^ab- b d ab-^ cd- ad "be Solution 2d. From Art. 135j Prin., - = ab-\ b and also — = ed~*; a6-' divided by cd~^ equals €L ab-^ ,. , .ad . which equals ■; — ed-i be By inspection, we see that the same result can be obtained by inverting the terms of the divisor and multiplying ; hence the following Rule. — Invert the terms of the divisor and proceed as in m/uMiplieation. Notes. — 1. When there are common factors in numerators and denomi- nators, indicate the operation and then cancel common factors. 2. If either term is a mixed quantity, reduce it to a fraction before ■ i'viding. XJXAMPIiES. 3. Divide — by ^• e ^ d 4 Divide — by — -• cy xf -_,.., A whicb equals — x— or o a be CM ic CPEEATIOK a h a G c b d d a d ad b be COMPLEX FRACTIONS. 93 SoL.tiTioN 2d Sir.ce multiplying both terms of a frac- 0PEii\Ti0N. don by the same quantity does not change its value (Prin. a , , 3, Art. 129), if we multiply both terms of the complex. 6 _ad fraction by the least common multiple of their denomina- ~ , , J)C tors, 6 and d, we will have the complex fraction equal to d ad to' Rule. — Divide the numerator by the denominator, as in division; Or, Multiply both terms oj the complex fraction by the lead eommon multiple of their denominators. EXAMPLES. 2. Keduce — to a simple fraction. Ans. — 3 2a' ex ab 3. Reduce to a simple fraction. Ans. 4ffla; 2g^x a b 4. Reduce — to a simple fraction. Ans. — — - , 1 ^ 2a+2 1+- a i4 5. Reduce to a simple fraction. Ans. , , ^- a+- a ' 1 «(e+l) c-1 . e+1 6. Reduce to a simple fraction. Ans. . _ i ' a+b /p _L ly ^ ^ fj . 7. Reduce to a simple fraction. Ans. r 94 FKACnONS. 8. Reduce 1 - — — r to a simple fraction. 1+- a 9. Reduce to a simple fraction. n Ans. 1 a+1 Ans, n+1. a — 1+- 10. Reduce — to a simple fraction. Ans, • a — 6 VANISHING FEACTIONS. 148. A Vanishing Fraction is one which reduces to the form - when certain suppositions are made. Thus, — . when x = l, becomes . or — So, also, . when x-\ 1-10 ' a-x a = a;, becomes equal to — • Principle. — Vanishing fractions contain a ■ common factor in the numerator and denominator, which reduces to zero when a epedal supposition is made. Thus, by factoring becomes ^^ — , in which the factor, x—\ x—1 x— 1, is common to both terms, and is equal to when x = l. 1. Find the value of when ai=x. a — x Solution." J{ v,e substitute a for x, we opeeation. will have - ; but factoring the numerator and S^ZZ^ = (a+x)(a-x) ^ " a—x a-x dividing by the denominator, we have the a+x = 2a fraction equal to a+x. Substituting The value of X, we have a+a, or 2a. Hence, the value of the gi?en fra© lion when a =a; is 2a. VANISHING FKACTIONS. 95 Rule. — Cancel the common factor which reduces to zero, and tiien make the supposition which reduced the fraction to - • EXAMPLES. 2. Find the value of -when a; = l. " An». 2. x—1 x?— 1 3. Find the value of when a; = 1. Ana. 3. x — 1 of— a' 4. Find the value of when x = a. Ans. 3a'. X— a «3 _3 Q 5. Find the value of when x-^a. Ans. —■ x'-a' 2 x* — a* 6. Find the value of when x = a. Ans. 4a'. x — a (x — aV 7. Find the value of when x = a. Ans. 0. of -or 8. Find the value of when a; = 1. Ans. 5. 1-x 9. Find the value of when x = B. Ans. t- x^+Ax-^1 5 1 — a;"* 10. Find the value of when a; = 1. Ans. m. 1 — x /p** ^« 11. Find the value of when x = a. Ans. na""'. x — a REVIEW QUESTIONS. Define a Fraction. The Terms. Numerator. Denominator. A Mixed Quantity. State the principles. What is the sign of a fraction ? State "the principles of the signs. Define Reduction of Fractions. State the cases. The rule for each case. Define Addition. Subtraction. Multiplication. Divison. How many cases in each ? Give the rule for each case. How is a quantity changed from one term of a fraction to another ? Define a Complex Fraction. Give the rule for the reduction of complex fractions to simple fractions. Define a Vanishing Fraction. When does a fraction become a vanishing fraction ? How do we find the value of a Tanishing fraction ? SEOTIOI^ Y. SIMPLE EQUATIONS. 149. An Equation is an expression of equality betweei two equal quantities. Thus, 2a; + 4 = 20 is an equation. 150. The First Member of an equation is the quantity on the left of the sign of equality. 151. The Second Member of an equation is the quantity on the right of the sign of equality. 153. A Numerical Equation is one in which the known quantities are expressed by figures ; as, 3a; — 4 = 17. 153. A Literal Equation is one in which some or all of the known quantities are expressed by letters; as, 2x — a = b, 01 x — 2 = b. 154. An Identical Equation is one in which the two mem- bers are the same, or will become the same by performing the operations indicated ; as, 3a; + 1 = 3a; + 1 ; or, 3a; + a = 2a; + 2a + a; - a. 155. The Degree of an equation containing but one un- known quantity is determined by the highest power of the unknown quantity. 156. A Simple Equation, or an equation of the first degree, is one in which the first power is the highest power of the unknown quantity ; as, a; + a = 6. 157. A Quadratic Equation, or an equation of the second degree, is one in which the second power is the highest power of the unknown quantity ; as, a;'-l- aa; = b, or a;" = a. • 158. A Cubic Equation, or an equation of the third degree, is one in which the third power is the highest power of the unknown quantity ; as, 3? + 4x''+Sx = 8. Note.— The symbol = was introduced by Robert Eecorde, who gave as hia reasMi for it that "npe 2 thynges can be moare equalle" than two parallel lines. 96 TRANSFORMATION OF EQUATIONS. 97 TEANSFOEMATION OF EQUATIONS. 159. The Transformation of an equation is the process of changing the terms without affecting the equality of the uiemhers. 160. Equations may be transformed by means of the fiJ- lowing axiomatic principles : PRKVCIPLBS. 1. The same or equal quantities may be added to both members if an equation. 2. The same or equal quantities may be subtracted from both memiers of an equation. 3. Both members of an equation may be multiplied by the same or equal quantities. -, 4. Both members of an equation may be divided by the same w equal quantities. 5. Both members of an equation may be raised to the same power. 6. Both members of an equation may have the same root extracted. 161. In the transformation of equations there are two prin- cipal cases — 1. Clearing of fractions ; 2. Transposition of the terms, CASE I. 163. To clear an equation of fractions. 1. Clear — ■ =- of fractions. 4 3 6 Solution. The least common multiple of the denomi- operation. natora is 12 ; multiplying both members of the equation by Zx 2x 5 12, we have to — 8a; = 10; hence, multiplying both mem- "^ 3"~6 bers of the equation by the least common multiple of the qy_ aj.„ in denominators clears it of fractions and does not change the squality of the members. (Prin. 3.) 9 98 SIMPLE EQUATIONS. Rule. — Multiply both members of the equation by the leasl common multiple of the denominators, reducing fractional terms to integei's. Notes. — 1. An equation may be cleared of fractions by multiplying eacb term by all the denominators. 2. If a fraction has the minus sign before it, the signs of all the teruu ■)( the numerator must be changed when the denominator is removed. EXAMPLES. Clear the following equations of fractions : Ans. 3a; + 2a; = 10. Ans. 8a;+9a; = 10. Ans. 18a; -20a; = 21. Ans. 3x+x = 2i-2x, Ans. 2a; — 6a+36 = a+a;, Ans. 2x-2ab + 2b' = ac-bc. 8. = Ans. Aax - 36a; = 2ca; — 2dx. 3 4 6 9, _££, — ^^=4. Ans. 2ax-2bx-3ax-Sbx = 4a'-i¥. a + b a-^ b 10 2:^— ^ = 2- 3a-'. Ans. Sx -A = 2a- B. a U. ^-?-l = _^. Ans. (x + iy-(x-iy-3a. r-l a; + l x^ — Tl 12. - = ^—i. Ans. 8- 2x-2x-x. ? I 3 2. a; a; 6 2"^3~3' 3. 2a; 3a; 5 3 "^4 "e" 4. 3» 5a; 7 4 6 8 5. a; a; . a; _+_=4 2 6 3 6. X „ a + x — a+6 = — — 3 G 7. ^ b '- a-b 2 TRANSFOEMATION OF EQUATIONS. x-3 x+B 13. ■ ^^ - ^^ = 6f . ^ms. 7(a; + 3y-7(x-By = 48(a;' - & ). CASE II. 163. To transpose the terms of an equation. 164:. Transposition is the process of changing a term from me member of an equation to the other without affecting the equality of the members. 1. In a;+6 = a, transpose b to the second member. Solution. Subtracting 6 from both members, which , _ does not affect the equality of the members (Prin. 2, t,~ h Art. 160), we have a; = a — &.• ^— x = a—b 2. Inax — a = bx+e, transpose a and bx. ^ OPERATION. Solution. Adding a to both members and sub- ax — a — bx+o tracting bx from both members, which, according to a = a Prin. 1 and 2, will not affect the equality of the ax = bx + a+o members, we have ax — bx = a+c. bx = bx ax--bx = a + G In both of these examples we see that in changing a quantity from one member to the other, the sign of the quantity is changed ; hence the follow- ing rule. Rule. — A term may be transposed from one. member of an equor Hon to the other, if, at the sams time, the sign be changed. EXAllIFIiXlS. In the following examples transpose the known terms to the second member and the unknown terms to the first member : 3. 2x+e = a. Ans. 2x = a — e. 4. 3a; - 2 = a;+4. Ans. 3a; - a; =■ 44 2. 5. 5x Tb = Bx+a. Ans. 5x - 3x = a - b. 6. 3a - 26 = 6a; — ax. Ans. ax — Qx = 2b- 3a, 7. a' — a'x — an = Bx\ Ans. — Sx' — a'x ^an — a'. 8. oa;* + 5ac - 2a = cx. Ans. ax' -cx = 2a - 5ae. „ „, a-b _ 6a;'-a; . _. 6a;'-a; a-6 9.2a;' = 5a + Ans. 2x' — ^oai -— -■ 2 3 3 2 100 SIMPLE EQUATIONS. SOLUTION OF SIMPLE EQUATIONS, CONTAINING ONE UNKNOWN QUANTITY. 163. The Solution of an equation is tlie process of finding the value of the unknown quantity. 166. The Root of an equation is the value of the unknown q^uantity, 167. To Verify the root of an equation, we substitute its value for the unknown quantity and reduce the members to identity. The equation is then said to be satisfied. Note. — The solution of an equation is often called the reduction of the eqiiaiion. To reduce an equation is therefore to solve it. CASE I. NUMEEICAL EQUATIONS. 1. Find the value of x in the equation 3a; — 4 = 12 — a;. Solution. Transposing the unknown terms to the opeeation. first member and the known terms to the second mem- 3x — 4: = t2 — x her, we have 3a;+a; = 12 + 4; uniting the terms, we 3a;+a! = 12+4 have 4a; = 16 ; dividing by the coeflB.cient of x, we have 4a; = 16 a; = 4. a; = 4 Vebification. Substituting for x its value in the vebification. given equation, we have 3 x 4— 4=12 —4 ; reducing, we 3 x 4— 4 = 12-4 have 8 = 8; and since this is an identical equation, the g^g roo* is ijerified. 2. Find the value of a; in the equation - — - = - + -• 2 b 2 6 OPEEATtON. Solution. Clearing the equation of fractions by mul- . £ _ ^ _ i + * Jiplying by 6, we have 3a; — 5 = 3+a;; transposing the 2 6; 2 6 terms, we have 3a; — a; = 3 + 5; uniting the terms, we 3a: — 5=3 + a! have 2a; = 8; dividing by the coeiEcient of x, we have 3a; — a; =3+5 a; = 4. 2a;=8 Verification. — Substituting the value of a; in the equation, we have f—|= j + l; reducing, we have 1 = 1; 5_ij.4 and since this is an identical equation, the root is '_? verified. SOLUTION OF SIMPLE EQUATIONS. 101 Rule.' — I. Clear the equaiion of fractions, if necessary. 11. Transpose tlie unknown terms to the first member of the equation and the known terms to the second m,ember. III. Reduce each member to its simplest form, and divide both members by the cffeffident of the unknown quantity. To verify the result : Substitute the value of the unknown quavr tity in 'the equation, and if the Tuembers are identical the result is correct. Notes. — 1. It is sometimes advantageous to transppse and make some reductions before clearing of fractions. 2. When the coefficient of the unknown quantity is negative we may multiply both members by —1, or divide by the negative coefficient. e:xampi.e:s. 3. Given4a;+6 = 2a; + 10, tofinda;. Ans.x = 2. 4. Given 5a;+4-2a; = 10+a;, to find x. Ans. a; = 3. 5. Given 18 - 3x = 5a; + 2, to find x. Ans. x = 2. 6. Given 7a;+6 = 5.T+14, to find a;. J.ns. a; = 4. . 7. Given 4a; - 17 = 4a; + 13 - 6a;, to find x. Ans. x = 5. 8. Given 2a; - 1 2 = 5a; - 30, to find x. Ans. a; = 6. 9. Given 6a;+ 16 = 9a; - 5, to find x. Ans. x = 7. 10. Given -+- = 51+U, to find x. Ans. x = 12. 2 3 11 Given ---+- = 2' to find a;. Ans.x = l^. 3 4 5 12. Given --- +3 = -+4|, to find x. Ans. x^\2. 1 3. Given ?^ + ^ = ^+ 1^, to find x. Ans. a; - 2. 5 6 5 14. Given ^-^J-^-U^, to find x. • Am. a; = 7. 7 4 6 15. Given 4(a; + 1) = 3(a; + 2), to find x. Ans. a; = 2. 16. Given '^+'^ = ^^ - 10, to find x. Ans. ar - 14. 3 6 2 9» 102 SIMPLE EQUATIONS. 17. Gi^eii^i^ + - = 4-^^.tofinda;. Am.X'-S^ 2 3 4 18 Given a; +=^^- = 12--^-^^' to find a;. Ans.x^5. 3 2 19. Given ^^+- = 12 - ^^^ . to find x. Ans. x = 14f 3 2 3 20. Given ^^ - ^11^ = ?^^ + i, to find x. Am., a; = 28. 2 o l to find x and g.. 4ns. x = &; 3 = 5. ff-f=-M 8. Given -\ 3^. 2?/ r to ^^^ ^ ^iid y. . iT"T^"M Am.x-n;y = \b r2^+32,=16| 9. Given -; 2x-Sv ( *" ^'^'^ * *°*^ ^* (2a; g ^■^^ J J-ws. a; = 5; y=-5 Sx — Sjf 5y 10. Given ' ^ ^ = 13 5a; 7^.-42/ , f to find a; and y. :^ = 7 I ^ns. a; = 6; y = 12. CASE III. ELIMINATION BY ADDITION AND SUBTRACTION. 185. Elimination by addition and subtraction consists in adding or subtracting the equations when the coefficients of one of the unknown quantities are alike , or are made alike. 11 • OPEKATION. 2x+Sy = = 12 (1.) 3a;+22/= = 13 (2) 6a; + 9y = = 36 (3) Gx+^ = = 26 (4) 52/ = = 10 (5) 2/ = = 2 ;6) 2z+6 = = 12 ;7) x = =3 (8) 126 SIMPLE EQUATIONS. 1- ^^^"^ { StS-ls } to find :. and 3,. Solution. Multiplying equation (1) by 3, and equation (2) by 2, to make the coefficients of x alike, we have equations (3) and (4). Subtracting (4) from (3), we have 5y = 10, from which y = 2. Substituting the value of 2/ in equation (1), we have (7), from wliich we find a; = 3. Rule. — I. Multiply or divide one or both equations.if neces- sary, so that the coefficients of one unknown quantity shall be the tame in both. II. Add these equations when ilie signs of the equal coeffidenta are unlike, and subtract the equations when they are alike. Notes. — 1. If the coefficients to be made alike are prime to each other, multiply each equation by the coefficient of that quantity in tlie other equation. 2. This method is generally preferred in finding the value of one quan- tity, the value of the other quantity being found by substitution. BXAMFIiES. 2. Given ■! , „ „ >■ to find a; and «. Ans. x = l: i/ = l. l4x+2y = 6 ) " " 3. Given ■! „ „ , „ ?■ to find x and y. Ans. a; = 4 : 1/ = 3, ( 2x+3y=17 J " ' " . 4 Given ■] „ / , „ r to find x and y. Ans. x = 5; u = 2. (2a;+4^ = 18 J ^ ' 5. Given \ „ „ „ [-to find a; and y. Ans. a; = 7; y = 5. ( 6x-Sy = 2 j " . -^ 6. Given \ „ ' _,„^ [■ to find a; and u. Ans. x = S;y = Q. ( 9^+3x = 105j ■^ 7. Given ] _ a r ^ find x and y. (-^ ^~^} , a + b a-b Ans. X-——; y = —r- ax+by = ab '\ i ^ ( ax+by = ab '\ *• ^^^^^ j 2ax+3&8/ = ^ [to find x and y. .4»w. a; = -;y--- .MI8UELLA.NEOUS EXAMPLES. 127 9. Given ^ I V to find a; and y. 3 ■ 2x- 8 3x-iy = 2 ^ lO' G^'^i 3X-2. 2/+6 ^, l^tofind.andy MISCELLANEOUS EXAMPLES. 186. In the following examples the pupil will exercise his • judgment which method to use : '■«>™ {:::::.-"}■ ^-{i^i ■6. Given i"**-""}'. ^~{r? 7,Given{f:}^=n- ^- l^l'^ 8, Given {rf=;n. ^- 1::^' 128 SIMPI-E EQUATIONS. r x+i x-i .6 11. Given \y — 1 y y \,x-y = \ J 2 GJven- r£±2/+a.=i5 1 . 1 13. Given^^^ l- Am. \ «!'' 14. Given j*^+«^=2fn. 15. Given {"^+^f = 2f{. ^« ( ax —by = 2n ) 16. Given}«^+^^=«1. ^n.. ( ox+ay=d ) 17. Given \a b _bx- ay = .8. Given}f+'^=«|. Ans. ( oa;— ay = b ) & a J V^ a + 6' C a b_ \ r bo -ad 20. Given J ^ ^''^ L ^m T" f " ^^^ I V- = n \ y J 21. Given {(«+«^^-'?'='n. A^. e d ( I 6c — ad + - = n I I 3/= -■ ' ' wie-na fa;=6 PEACTIOAL PROBLEMS. 129 PRACTICAL PROBLEMS. 1^7. Several of the problems in Art. 176 contained more than cme unknown quantity, but the conditions were such that they were readily solved with one unknown quantity. 188. In the following problems the solution is most readily (!tf'ected by using a separate symbol for each unknown quantity. 189. In such problems the conditions must give as many independent equations as there are unknown quantities. CASE I. 1. A drover sold 6 lambs and 7 sheep for $71, and at the same price 4 lambs and 8 sheep for $64 ; what was the price of each? OPERATION. Solution. Let a;=the price of the Let a; = the price of lambs, lambs, and 2^= the price of the sheep; and y = the price of sheep, then, by the first condition we have eqiiSr &x+ly — 'Jl (1) tion (1), and by the second condition we 4a;+8j/ = 64 (2) haveeq. (2). Multiplying eq. (1) by 2, X2x + Uy = lA2 (3) we have (3). Multiplying (2) by 3, 12a;+2% = 192 (4) we have (4). Subtracting (3) from (4), 10w = 50 (5) we have (5). Dividing by 10, we have y = h (6) (6). Prom which we find « = 6. 6a; + 35 = 71 (7) a: = 6 (8) 2. A farmer hired 8 men and 6 boys one day for $36, and at the same rate next day he hired 6 men and 11 boys for $40; what did he pay each per day ? Ans. Men, $3 ; boys, $2. 3. A man paid $1.14 for 12 oranges and 13 lemons; but the piice falHng \, he r.eceived only 62 cents for 9 oranges and 11 lemons ; what was the price paid for each ? Ans. Oranges, 3 cts. ; lemons, 6 cts. 4. A man hired a men and b boys for |m, and at the same price e men and d boys for %n ; what price did he pay each ? 1 tr ^md — nh . ^an — me Ans. Men, $ — ; — ;— ; bays, $— — — - . ad- OB ad — 00 130 SIMPLE EQUATIONS. CASE II. 1. There is a fraction such that 1 added to its aumerator will make it ^, and 1 added to its denominator will make it ^ ; what U the fraction ? Let - = the fraction. x + 1 1 2/ 2' (1) X 1 (2) y + 1 3" 2x + 2=y (3) 32;=3/ + l. (4) Solution. Then, by the 1st condition, and by the 2d condition. Clearing (1) of fractions, clearing (2) of fractions, Subtracting (3) from (4), x-2 = l transposing, x = S substituting in (3) and reducing, y = 8 hence the fraction is ■ — =— . Ans. y 8 2. Find a fraction such that if 2 be subtracted from the numerator the fraction wUl equal \, or if 2 be subtracted from the denominator the fraction will equal \. Ans. ■^^. 3. If -4 be subtracted from both terms of a fraction, the value will be \, and if 5 be added to both terms the value will be | ; what is the fraction ? Ans. -f . 4. If 1 be added to both terms of a fraction, its value will be ^ ; and if the denominator be increased by the numerator, and the numerator be diminished by 1, the value will be -^ ; re- quired the fraction. • Ans. ■^. 6. Required the fraction such that if the numerator be in- creased by a the result will equal — > and if the denominator be n 'ncreased ])y a the result will be — Ana. — ^^ -. in- an (w - - n') PKACTICAL PROBLEMS. 131 CASE III. 1. Raquiied the number of two figures which, added to the number obtained by changing the place of the digits, gi ves 77. and subtracted from it leaves 27. SoLUTioir. Let a; = the tens' digit, and y = the units' digit ; then 1 Oa; + 2/ = the number, and lOy + a; = the number with digits inverted. Then, by 1st condition, V^x-^y -vV^y +x = n , (1) and by 2d condition, lOy+a: — (lOai+y) = 27 ; (2) uniting terms in (1), lla;+ll^ = 77; (3) dividing by 11, x+2/ = 7; (4) uniting terms of (2),' 9^ — 9a; = 27; (5) dividing by 9, y—x = 3-'~' (6) from (4) and (6), x = 2, a,nd y = 5; hence the number is 2 tens and 5 units, or 25. 2. Required a number of two figures which, increased by the number obtained by inverting the figures, will equal 132, and which diminished by that number will equal 18. Ans. 75. 3. A number consisting of two places being divided by th? sum of its digi1;p, the quotient is 4 ; and if 36 be added to it, the digits will be inverted ; required the uumber. Ans. 48. 4. There is a number which equals 5 times the sum of its two digits ; and if three times the sum of the digits, plus 9, be sub- tracted from twice the number, the digits will be inverted ; re- quired the number. Ans. 45. CASE IV. 1. The amount of a certain principal at simple interest, for a certain time, at 5%, is $260 ; and the amount for the same time, at 8^, is $296 ; required the princij)al and time. 132 SIMPLE EQUATIONS SoitTTION. Let K^the prin'-ipal, . and y = the time. 5 Then, by the 1st condition, xii x \-x = 260 f 1) ' ' "^ 100 g By the2d ccndltion, xy-K +a; = 296 (2) ' -^ 100 Whence, a; = 200, the principal, and 2/ = 6,' the time. 2. A certain sum of money on simple interest amounts in a certain time, at 6%, to 1310, and at 10%, for tlie same time, to $350 ; required the time and principal. Ans. Prin., $250 ; time, 4 yrs. 3. The amount of a certain principal for 7 years, at a certain rate per cent., is $810, and for 12 years, at the same rate, is $960; required the principal and rate. Ans. Prin., $600; rate, 5%. 4. A certain sum of money, put out at simple interest, amounta in m years to a dollars, and at the same rate, in n years, to h dol- lars ; required the sum and rate per cent. . Ti . an—hm 100(5 — a) Ans. Prm., ; rate, ; — • n — m an — bm CASE v." 1. There are two numbers whose sum equals twice theii product, and whose difference equals once their product; re quired the numbers. Solution. Let X = the greater number, and y = the less number. Then, by the 1st condition, x +y = 2xy (1) and by the 2d condition, x—y — xy (2) Adding (1) and (2), tx^Zxy Dividing bv x, 2 = Zy Whenoe, y = \ and x = %. MISCELLANEOUS PEOBLEMS. 133 2. Tliere are two numbers whose sum equals thr^-e times theii product, and whose difference equals once their product ; what are the uumbers ? Am. 1 and |. 3. There are two numbers such that twice their sum equals 3 times their product, and twice their difference equals once their product ; what are the numbers ? Ans. 2 and 1. 4. There are two numbers such that their sum equals 4 timea their quotient, and their difference equals twice their quotient ; what are the numbers ? Ans. 9 and 3. 5. There are two numbers whose sum equals a times their product, and whose difference equals b times their product ; • 2 2 what are their values ? Ans. ; a — b a + b MISCELLANEO US PR 0BLEM8. 1. A person buys 8 lbs. of tea and 3 lbs. of sugar for S2.64, and at another time, at the same rates, 5 lbs. of tea and 4 lbs. of sugar for $1.82; find the price per pound of the tea and sugar. " Ans. Tea, 30 cts. ; sugar, 8 cts. 2. If the greater of two numbers be added to ^ of the less, the sum will be 30, but if the less be divided by \ of the greater, the quotient will be 3 ; what are the numbers ? Jns. 25 ; 15. 3. A said to B, " Give me $200 and I shall have 3 times as much as you ;" but B replied, " Give me $200 and I shall have twice as much as you ;" how much had each ? ^M. A, $520; B, $440. 4. A lady having two watches bought a chain for $20 : if the uhain be put on the silver watch, their value will equal \ of the gold watch, but if it be put on the gold watch, they will be iforth 7 times as much as the silver watch ; required the value of each watch. Ans. Gold, $120 ; silver, $20. 5. Find the fraction which becomes equal to \ when . the numerator is increased by 1, and equal to \ when the denomi- nator is increased by 1. Ans. ^. 6. Mary and Jane have a certain number of plums : if Mary had 4 more she would have 3 times as many as Jane, but if she 134 SIMPLE EQUATIONS. had 4 less she would have ^ as many as Jane ; how many haa each ? ■ Ans. Mary, 5 ; Jane, 3. 7. A bill of £120 was paid in guineas (21s.) and moidores (27s.), and the number of pieces of both kinds was just 100; required the number of pieces of each kind. Ans. 50. 8. A boy expends . 80 cents for apples and pears, buying his apples at 4 and his pears at 5 for a cent, and afterward accom- modates his friend with half his apples and one-third of his pears for 13 cents at the price paid ; how many did he buy of each ? Ans. Apples, 72 ; pears, 60, 9. A person rents 25 acres of land for £7 12s. per annum : for the better part* he receives 8s. an acre, and for the poorer part, 5s. an acre ; required the numbe:^ of acres of each sort. Am. 9 ; 16. 10. If 9 be added to a number consisting of two digits, the two digits will change places, and the sum of the two numbers will be 33 ; what i-s the number ? Ans. 12. 11. Said A to B, " If you will give me $20 of your money, I will have twice as much as you have left ;" but says B to A, " If you will give me $20 of your money, I will have thrice as much as you have left ;" how much had each ? Am. A, $44 ; B, $52. 12. A and B laid a wager of $20 : if A loses, he will have as much as B will then have ; if B loses, he will have half of what A will then have ; find the money of each. 4ms. A, $140; B, $100. 13. There is a number consisting of two figures which is double the sum of its digits ; and if 9 be subtracted from 5 times the number, the digits will be inverted ; what is the number ? Am. 18. It. Several persons engaged a boat for sailing : if there had oeen 3 more, they would have paid $1 each less than they did . if there had been 2 less, they would have paid $1 each more than they did ; required the number of persons and what each paid. Ans. 12 persons ; $5 each. 15. A and B ran a race which lasted five minutes : B had a start of 20 yards, but A ran 3 yards while B was running 2, THREK OE MOKE UNKNOWN QUANTITIES. 13a id won by 30 yards ; find the length of the course and the )eed of each. Ans. 150 yds. ; A 30 yds. and B 20 yds- a min. 16. If a certain rectangular field were 10 feet longer and 5 ,et broader, it would contain 400 square feet more ; but if it ere 5 feet longer and 10 feet broader, it would contain 450 [uare feet more ; required its length and breadth. Am. 30 ft. ; 20 ft. 17. A market-woman bought eggs — some at 3 for 7 cents, ad some at two for 5 cents, paying $2.62 for the whole ; she fterward sold them at 36 cents a dozen, clearing 62 cents ; how lany of each kind did she buy ? Ans. 48 ; 60. 18. A and B ran a mile, A giving B a start of 20 yards at le first heat, and beating him 30 seconds ; at the second heat . gives B a start of 32 seconds, and beats him 9-j-\ yards ; at hat rate per hour does A run ? Ans. 12 miles. SIMPLE EQUATIONS, CONTAINING THREE OE MOEE UNKNOWN QUANTITIES. 190. Equations containing three or more unknown quan- ties may be solved by either of the methods of eliinination Iready explained. C x+ y+ 2=6 1. Given \ x+2y+3z=14: (^2x+Si/¥ «=11 to find X, y and a. Solution. Subtracting equation (1) from inition (2), we have equation (4). Multiply- 'g (1) ^7 2, and subtracting from (3), we have i) ; subtracting (5) from (4), we have (6) ; di- iding (6) by 3, we have z = 3; substituting the ilue of » in (4), we have 8. Transposing, we wey = 2; substituting the values of z and y in .), we ^ave (10^ ; from which we have a; = l. OPERATIOK. x+ y+ z=6 x+2y+Sz = li 2x+Sy+ z = ll 2/ + 22 = 8 ^1 83 = 9 z = 3 2/ + 6 = 8 y = 2 2;-l-2 + 3 = 6 (1) % (3) (4) (5) («) (7) (8). (9) (10) (in 136 SIMPLE EQUATIONS. Rule. — I. Eliminate suecessiwly the same unknown qvxmtitji from each of the equations ; this will give a number of equations and of unknown quantities one less than the given number. II. In the same way eliminate one of the unknown quantities from each of the derived equations, and thus continue until an equation is found containing one unknown quantity, and then find the value of the unknown quantity in this last equation. III. Substitute this value in one of the equations containing two unknown quantities, and find the value of a second; substitute these values in an equation containing three unknown quantities, and find the value of a third, and thus continue until all an found. Notes. — 1. The pupil will exercise his judgment which quantity to eliminate first, and aiso what method of elimination to use. 2. When one qf the equations contains but one or two of the unknr>wn quantities, the method of substitution wiU often be found the shortest. EXAMPLES. 2. Given 3. Given a;+2?/+ z=13 x+ y + Zz=l5 2a; + 3y+23=22 2a; + 42/ + 32 =35 a;+3;y+23=23 Zx-by+Az'= 17 Ana. An^. a;=2, z=3. ^ = 3, z=5. 4. Given 2a;+42/-3z = 22 ix-2y+5z = \8 Qx+ly- z = 63 x = Z, z=4. 5. Given 6. Given 7. Given 2a; + 32/-3 = 27 3a; -42/ + 32 = 12 4a; +22/ -52 = 15 x + \y + lz = Z1 ^x+\y+\z=\b \x¥^y+\z = l'i x->-y + z = 24: x—y+z=8 X \y—z = Q Ans. Ans. Ans. x=7, 2—5. a;=12, y=20, z =30. a; -7, y-n. 2 = 9 THREE OE MOEE UNKNOWN QUANTITIES. 13? Cx+y\-e=.a^ Cx=i{b+e), 8, Given i x+y-z=b [■ Ans. \ y==Ua-c), \_x-y~z = o} {,z^i{a-b). Cx+y = a^ Cx^i\a+b-c), 9. Given l x+z =b [■ Am. ] 2/ = i(a-l-e-6), ^y+z=cj (_3=K6+c-a). x y 10. Given ^-+i = 6\. Am. X z z x=i, ^+y=\\ I- « x=- a b \ -"""2 11. Given ^^+i = ik. Am. ^=2 2 12. Given ^ a;+z-2/T6 [■ • ^ns. ■] 3^=i(a+e), ■a;+2/+2r=12" 13. Given ^^r^r"=jH. ^m. ' x+z+u = li ' .y+z+u = l5. '^+^+^ = 3 a; w z hi \ ^=' 14. Given / +- = 1 >• Ans. les w- ' 7 oranges for 34 cents ; what was the price of each? Ans. Apples, 1 ct.; peaches, 2 cts.; fears, 3 cts.'i_ oranges, 4 cts. PKOBLEMS. 135; 8. Divide the number 27 into 4 sucK parts that if the first art 'is increased by 2, the second diminished by 2, the third lultiplied by 2, and the fourth divided by 2, the results will be qual. Ans. 4 ; 8 ; 3 ; 12. 9. A and B can do a piece of work in 12 days, A and C in 5 days, and B and C in 20 days ; how many days wUl It take ach person to perform the same work alone? Ans. A, 20; B, 30; C, 60. 10. The sum of 3 fractions is 2\ : the sum of the first and hird equals twice the second fraction, and the difierence between he first and third is ^ of the third fraction ; what are 4he frac- ions ? Ans. f , |, f . 11. In a naval engagement the number of ships captured was ' more, and the number burned was 2 less, than the number unk. Fifteen escaped, and the fleet consisted of 8 times the lumber sunk. Of how many ships did the fleet consist ? Ans. 32. 12. A cistern has 3 pipes opening into it. If the first be ilosed, the cistern may be filled in 20 minutes ; if the second be ilosed, in 25 minutes; if the third be closed, in 30 minutes, low long would it take each pipe alone to fill it ? Ans. 1st, 85f min. ; 2d, 46^ min. ; 3d, 35i^ min. 13. I have 3 watches, and a chain which is worth $60. The irst watch and chain are worth ^ as much as the second and hird watches ; the second watch and chain are worth -f as much IS the first and third watches ; and the third watch and chain .re worth 3 times as much as the first and second watches. iYliat is the value of each watch ? Ans. 1st, S20 ; 2d, $60 ; 3d, $180. 14. There is a number consisting of 3 digits : the sum of the ligits is 9; the digit in the tens' place is half the sum of the ther 2 digits ; and if 198 be added to the number, the result fill be expressed by the figures of the number reversed • re- [uired the number. Ans. 2b4. 15. A sum of money consists of quarter dollars, dimes and i&lf dimes. It is worth as many dimes as there are pieces of Lioney ; it is worth as many quarters as there are dimes ; and 140 SIMPLE EQUATIONS. the number of half dimes is one more than ths number oi dimes. What is the number of each ? Ans. 3 quarters ; 8 dimes ; 9 half dimes. 16. Three boys, A, B and C, were playing marbles. Firsl; A loses to B and C as many as each of them has ; next, B losea to A and C as many as each of them now has ; lastly, C losea to A and B as many as each of them now has ; and it is then , found that each of them has 16 marbles. How many had each at first ? Ans. A, 26 ; B, 14 ; C, 8. 17. Some smugglers discovered a cave which would exactly hold the, cargo of their boat, which consisted of 13 bales of cotton and 33 casks of liquor. While they were unloading, a custom-house cutter hove in sight, and they sailed away with 9 casks and 5 bales, leaving the cave f full ; how many bales oi casks would it hold ? Am. 24 bales ; 72 casks. 18. A person has 2 horses and 2 saddles : the better saddle cost $50, and the other $15. If he puts the better saddle upon the first horse, and the worse saddle upon the second, then the latter is worth $50 more than the former ; but if he puts the worse saddle upon the first, and the better saddle upon the secona horse, the latter is worth If times as much as the former ; what is the value of each horse ? Ans. 1st, $165 ; 2d, $250. KEVIEW QUESTIONS. Define Independent Equations. Give an example of them. Define Simultaneous Equations. Give an example of them. Define Elimina- tion. What is the use of Elimination ? How many methods are there ? State the methods of elimination. Define Elimination by Substitution. State the rule for it. Define Elimination by Comparison. State the rule for it. Define Elimmation by Addition and Subtraction. State the rule for it. Is there any other method of elimiiiation ? Which method of elimination is preferred ? What effect has transposition upon the signs of terms ? Why does kransposition change the sign of a term? Why does changing the signs oi all the terms of an equation not affect the equality of the members? In solving problems when should » separate symbol be used for eacit unknown quantity? SUPPLEMENT TO SIMPLE EQUATIONS. 191. This Supplement to Simple Equations embraces the following subjects: Zero and Infinity, Generalization, Negativt SolvMons, Discussion of Problems, and Indeterminate Problems. Note. — Teachers desiring a short course may omit this Supplement to gimple equations. ZEEO AND INFINITY. 193. Zero and Iniillity often occur in algebraic expres- sions. Such expressions may be interpreted by the following principles : Prin. 1. 0x^ = 0; that is, if zero be multiplied by a finite quantity the product is zero. • For, multiplied by 2 is 0, multiplied by 3 is 0, etc. ; hence, mul- tiplied by any number is 0, or x .4 is 0. Pein. 2. - =0 ; that is, if zero be divided by a finite quantity A the quotient is zero. For, divided by 2 is 0, divided by 3 is 0, etc. ; hence, divided by any nvmher is 0, or -s- .4 ia 0. Pein. 3 -=-A; that is, if zero be divided by zero the quotient ie any finite quantity, or is indeterminate. For, if in Prin. 1 we divide both members by 0, we have - ="-^> in wMcL A is any finite quantity. Pein. 4. — = oo ; that is, if a finite quantity^ be divided by zero the quotient is infinity. To prove this; puppose we have the fractiDn - • Now, if a remains con 142 SUPPLEMENT TO SIMPLE EQUATIONS. etant, the smaller b is, the greater will be the quotient, hence, it h becomes infinitely small, the quotient will become infinitely large; henca, when 6 is zero, the quotient is infinity. Pein. 5. X 00 = J. ; that is, if zero he multiplied by infinity, the product is a finite quantUy. For, clearing the equation in Prin. 4 of fractions, we hav(! x oo —^ ii wl'.ich A is any finite quantity. Pein. 6. ^ = ; that is, if a finite quantity he divided by infinity, the quotient is zero. For, dividing both members of the equation in Prin. 5 by oo , we have ( equals A divided by infinity, which proves the principle. GENERALIZATION. 103. Generalization is the process of solving general prob- lems, and interpreting the results. 19.4. A General Problem is one in which the quantities are represented by letters. 195. A Formula is a general expression for the solution of a problem. A formula expressed in ordinary language gives a rule by which all the problems of a class may be solved. CASE I. 1. The difference between a times a number and b times the number is c ; required the number. Let X = the number. Then, ax — bx = c; whence, x = a — b Expressing this formula in ordinary language, we have the following rule : Rule. — Divide the difference of the products by the difference oj the mvltipliers. GENERALIZATION. 143 Apply this formula to the following problems : 2. The difference between 5 times a number and 3 times the number is 12 ; required the number. 3. The difference between 9 times a number and 6 times a number is 162 ; what is the number ? CASE II. 1. Find a number which being divided by two given num- bers, a and b, the sum of the quotients may be e. Let X = the number. Then,-+-=c; a b , abc whence, x = a + b This formula, expressed in ordinary language, gives the following rule : Rule. — Divide the product of the three given quantities by the sum of the divisors when the sum of the quotients is given, and by the difference of the divisors when the difference of the quotients is given. Apply this formula to the following problems : 2. Find a number which being divided by 4 and by 6, the difference of the quotients is 4. 3. Find a number which being divided by 3 and by 7, the difference of the quotients is 16^ CASE III. " L A can do a piece of work in a days, and B in 6 days ; in n'(\at time can they both do it ? Let x = the number of days in which both can do it. Then, -+\^-- a b X ab wbence, x - a + b 144 SUPPLEMENT TO SIMPLE EQUATIONS. This formula, expressed in ordinary language, gives the fol lowing rule : Bule. — Divide the product of the numbers expressing the time in which each can perform the work by their sum. Apply this formula to the following problems : 2. A can do a piece of work in 4 days, and B in 8 days ; iu A hat time will they together do it ? 3. A can reap a field in 6 days, and B in 9 days ; in what time can tTiey together reap it ? 4. A pound of tea would last a man 12 months, and his wife ■ 6 months ; how long would it last them both ? CASE IV. 1. The sum of two numbers is a, and their difference is b ; what are the numbers ? Let X = the greater number, and y = the smaller numbei. Then, x+y = a, and x—y=b. a + b Whence x = and 1/ = 2 a—b TheSe formulas, expressed in ordinary language, will give the following rules : Rule. — ^I. To find the greater number, add half the difference to half the sum. II. To find the less number, subtract half the difference from half the sum. Apply the formulas to the following problems : Required the numbers whose — 2. Sum is 20 ; difference is 4. 3. Sum is 62 ; difference is 14. 4. Sum is 221 ; difference is 29. MISCELLANEOUS PROBLEMS. 145 MISCELLANEOUS PROBLEMS. 196. The pupils -will obtain the formulas in the following problems, and derive rules from them : 1. Divide the number a into two parts, so that one part shall be n times the other part. Ana. ; w+1 n+1 2. The sum of two numbers is a, and n times one number equals m times the other ; what are the numbers ? . ma na Ans. ; m+n m+n 3. The difference of two numbers is a, and n times one num- ber equals m times the other ; what are the numbers ? m — n m — n 4. The sum of two numbers is a, and n times their sum equals m times their difference ; what are the numbers ? . (m,+n)a ^ (jn — n)a 2m 2m 5. Divide the number a into two such parts that one part increased by h shall be equal to m times the other part. . m& — b a+b Ans. ; m+1 m+1 6. A is m times as old as B, and in a years he will be n times as old ; what is the age of each at present ? ^^.A's,=^^^^;B's,-^^^^^- m—n m—n 7. A courier starts from a place and travels a miles a day ; n days after he is followed by another, who travels b miles a iny ; in what time will the second overtake the first ? . na , Ans.. days. b — a 8. I bc'Jght two kinds of sugar — one at a cents a pound, and the other at b cents a pound ; how much of each kind must I take to make a mixture of m pounds worth e cents a pound? . m(e — b) m{a — c) Ans. -^; — a—b a—b 146 SUPPLEMENT TO SIMPLE EQUATIONS. NEGATIVE SOLUTIONS. 197. In tlie solution of a problem the value of the unknown quantity is sometimes a negative quantity. 198. A solution of a problem which gives a negative qua» tity is .called a Negative Solution. 1. What number must be added to the number 12 that the result shall be 8 ? SoLTTTiON. Let X equal the number ; then, operation. l2+x = S0TX=-i. Now, the result - 4 is Let x = the number, said to satisfy the question in an algebraic sense, Then, 12 + a; = 8 since — 4 added to 12 equals 8 ; but the prob- a; = — 4 lem is evidently impossible in an arithmetical sense. Since, however, adding —4 gives the same result as subtracting + 4, the negative result indicates that the problem should be. What num- ber must be subtracted from 12 that the result shall be 8 ? 2. A is 45 years old, and B is 15 years old ; how many years hence will A be 4 times as old as B ? Solution. Let a; = the number operation. of years; then, solving the prob- Let a: = the number of years, lem, we have a; =-5. This re- Then, (45+a;) = 4(154-a;); suit, — 5, indicates a reckoning of whence, x= —5 time backward instead of forward ; hence, it was 5 years ago instead of 5 years hence. The problem should, therefore, be modified to read, "How many years since," instead of "How many years hence." 3. A is 45 years old, and B is 15 years old ; at what tiraa from the present is A 4 times as old as B ? SoiiUTiON. Solving this problem by supposing the date to be in the future, we find x= -5. Had the result been -1-5, it would have indicated 5 years hence. The problem is stated in general terms, so as to admit either result, and need not be modified in its statement. This result, --5, therefore indicates 5 years since, or 5 years ago, which, by examining the problem, we see is the correct time. 4. There is a fraction such that if 1 be added to its numer* NEGATIVE SOLUTIONS. 147 tor the result is ^, and if 1 be added to the deaominator the re- sult is ^ ; what is the fraction ? _4 Soi-TJTION. Solving this problem, we find the fraction to be This can be verified algebraically, but is absurd arithmetically considered. 1 he problem can be made consistent, however, by changing the word " added " to ■' sul)tracted." 199. From these examples and illustrations we derive the foJbwiug conclusions: 1. The neffotive solviion indicates some inconsistency or ab- surdity in the statement of the problem, or a wrong supposition respeding some condition of it. 2. A problem which gives a negative solution can usually be so modified that it will become consistent, and the result will be positive. 3. The negative solution somHimes merely indicates the direction in which the result is to be reckoned. SOO. The pupils will solve, interpret, and modify the enun- ciation of the following problems : 5. What number must be added to 18 that the result may be 15? • Ans. -3. 6. What number must be subtracted from 12 that the result may be 15 ? Ans. — 3. 7. Eequired a number such that f of it shall exceed f of it by 2. Am. -24. 8. A man was born in 1825, and his son in 1855 ; find at what time the father's age is 4 times the son's age, dating from 1870. Ans. -5. 9. A man labored 10 days, his little son being with him 8 days, and received $18; at another time he labored 14 days, his son being with him 12 days, and received $25 ; required the wag(! of each. Ans. Father, $2; son, —25 ets. 148 SUPPLEMENT TO SIMPLE EQUATIONS. DISCUSSION OF PROBLEMS. 301. The Discussion of a problem is the process vjf asMgtt ing different conditions to^a problem, and interpreting the results which thus arise. ]. If we subtract b from a, by what number must the result be multiplied to give a product of c ? OPERATION. Discussion. Let x represent the number; Let 2; = the number, then we shall have (a— 6)cB = c, from which we {a — b)x = c find the value of x equal to c divided by a — 6. x = —^— a—b S03. This result may have five different forms depending on the values of a, h and e. I. When a is greater than h. In this case, a— 6 is positive; and c being positive, the quotient is positive; , hence, the required number is positive. This is as it should be, for the problem then is. By what shall we midtiply a positive quantity to produae a positive quantity? Evidently the multiplier should be positive. II. When a is less than b. In this case a—b is negative, and consequently c divided by a— 6 is negative; hence the required number is negative. This is as it should be, for the problem then is. By what shall we midtiply a negative quantity to pro- duce a positive quantity f Evidently the multiplier should be negative. III. When a is equal to b. In this case a — b = and a; = — = 00 [Prin. 4, Art. 192) ; that is, no finiJ.e quantity will answer the conditions. This is correct, since the problem then becomes. By what must we mviliply 0, nothing, to produce 0, somethingf Evidently by no finite quantity. IV. When c is 0, and a is either greater or less than b. In this case we have x = , which equals (Prin. 2, Art. 192) ; thai a —b Is, the multiplier is zero. This is correct, for the problem then is. By what must we multiply a — b, something, to produce nothing f Evidently the mul- tiplier should be tero. PliOBLEM OP THE COUEIEES. 149 V. Wlien c = and a = b. In this case we have a; = — . or any jiumtity whatever (Prin. 3, Art. 192). This is correct, for the problem then is, By what shall we multiply io pro- duce 0? Evidently the multiplier may be any quantity. Note. — Let the pupils illustrate each form by using particular values fcr a, b and c. PKOBLEM OF THE COUEIEES. 203. The Problem of the Couriers is a fine illustration )f the subject under consideration. It was originally proposed by Clairaut, an eminent French mathematician. 1. Two couriers start at the same time from two places, A and B, a miles apart, the former traveling m miles an hour, and the latter n miles an hour; where will they meet? There are evidently two cases of the problem. Case 1. When the couriers travel toward each other. Discussion. Let A and B represent the two opebation. places, and P the point at which they meet; AB = a. Let x = AP, the distance which the first travels; then a—x = PB, the distance which the second travels. Then — = the number of hours '-^^ m the first travels, and =the number of hours n the second travels. They both travel the same X a—x qwt time; hence, we have — = — . from which we ■'' »« j.n m n m-t-n find the values of a; and a— a;. ax "" 1st. Suppose m = rt,; then, by substituting, we m+n find x = — , and a-x = -; that is, if they travel at the same rate, each will 2 2 . ' , travel half the distance. This is evident from the nature of the problem. 2d. Suppose n = I then a; = a and a — x = 0; that is, if the second does not travel, the first travels the whole distance, a,nd the second no distance. This is evident &om the conditions of the problem. 3d. Suppose w =0; then x = and a — x = a; that is, the first travels 1 p 1 a = = AB Let X- ^AP; then a — X- = PB. X a—x m n X x—a m n X- am m-n -a- an m—n 150 SUPPLEMENT TO SIMPLE EQUATIONS. no distance, and the second travels the whole distance. This is evident from the conditions of the problem. Note. — Let the pupil make the suppositions m = 2n, m = \n, etc., and interpret the results. Case II. When the couriers travel in the same direction. Discussion. Let A and B represent the two operation. places, and P the point where they meet, each A B f traveling in the direction of P. AB = a. Let I— — — J^— — I X = AP, the distance the first travels ; then a = AS. :c — a = £P, the distance the second travels. The Let ,x=AP; limes they travel are equal; hance we have then x—a = BP. — = . from which we find the values of x m n and x—a. 1st. Suppose ?n>n; then x and a; — a will be 'positive. That is, they will meet at the right of B. This is as it should be by the conditions of the problem. 2d. Suppose n>m; then x and x — a are both negative. That is, the point of meeting must be at the left of A, instead of at the right. Hence, if the second courier travel faster than the first, that they may meet, the direction in which they travel must be changed. This is evident from the conditions of the problem. 3d. Suppose r>i=n; thena; = .or oo.and a;— a = — .oroo. That is, if the couriers travel at the same rate, they can meet at ho finiUt distance from A ; in other words, the one can never overtake the other. This is evident from the circumstances of the problem. 4th. Suppose a = 0; then x = . or 0, and a; — a= . or 0. m—n m—n That is, if the couriers are no distance apart, they will have to travel no distance to be together. This is evident from the circumstances of the prollem. 5th. Suppose m = n and a = 0; then x = - and a;— a = -. or anything. That is, if the couriers are no distance apart, and travel at the tame rate, they will always be together. This is evident from the nature of the problem. 6th. Suppose n = 0; thena; = = aanda;— a = 0. m, That is, if the rate at which the second travels is zero, the first couriei PEOBLEMS FOR DISCUSSION, 161 travelb the ■who.e distance, and the second no distance. This is evident from the circumstances of the problem. 7th. Suppose «i=2ra; thena; = = 2a and x — a— a. m That is, if the rate at which the first travels is twice the rate at irhich the second travels, the first courier travels ivdce the distance from A io B, overtaking the second a miles from B. This is evident from the oircum stances of the problem. 8th. Suppose m is plus and n is minus ; then am , — an x = anda— a = 7n+n m+n Here the value of a; is posiiiiie, and the value of a;— a is negative; hence, they meet at the right of A and at the left of B, or between A and B. This is evident, since minus n indicates that the second courier travels toward A ; hence they must meet between A and B. Changing the signs of the last expression, we have a—x = . which m+n is the same as the expression for the distance the second courier travels, sbtained in the first case of this problem. Case I. is therefore but a special ase of the general problem just discussed. PROBLEMS FOE DISCUSSION. 1. Required a number that, being successively multiplied by m and n, the difference of the products shall equal a. Ans. m — n When will the result be negative? When indeterminate? Wher infinite ? Illustrate with numbers. 2 B is a years old, and A is m times as old ; at what time • II . . . , 1 T-.n 4 (™- n)a will A be n. times as old as B ? Ans. x = ^^ —• n-1 Interpret the result when m>n; m,^ whence, 7a;- 36 = = 7a;+60, and xx = = 96, or x= 9^ r« = - = 00. This value of x is infinite, which shows that no finite number wiU andwei the conditions of the problem ; it is therefore impossible. Note. — ^When a problem contains more conditions than imknown quan- tities, the conditions which are unnecessary are said to be redundant. EXAMPI-ES. 3a; + 5 3a;- 6 1. Find the value of a; in the equation Solution. Reducing the second member, we have equation (2) ; clearing of fractions, we have (3) ; transposing terms and factoring, we have (4) ; dividing by the coefficient of x, we have a; =— > or oo , which indicates that no finite value will answer the conditions. x+2 x-2 operation 1st. 3a;+5 3a; -6 a;+2 a;-2 (1) 3a;+5 a;+2 (2) 3a;+5 = 3x + 6 (3) (3-3)a; = 6-5 (4) Oxa;=l .=i-» (5) 154 SUPPLEMENI TO SIMPLE EQUAriONS. „„,•.. ^, ,. - OPERATION 2d. Solution 2i>. Clearing the equation ol q j.f; q _fi fractions and reducing partly, we have (2); = (1) transposing and uniting, we have (3) I whence ^^^ ^~ we have (4). This apparent value of x can- ^a;'- a; -10 = 3a; -12 (2, not be verified, a? the pupil may see by sub- — a;= — 2 (3) etitution. . a; = 2 (4) 2. Required a number such that its \, increased by its \, is equal to its W, diminished by its 3=3-. 2 Ana. {x = -\. Indeterminate. ••('-?)• 3 Required a number whose f , diminished by 4, is equal to the sum of its ^ and \, diminished by 3. Ans. (x = Qo). Impossible. 4. A and B dug a ditch for $20, A receiving $2, and B $3, a dg.y ; how many days did each labor, if they did not labor the same number of days ? Ans. Indeterminate. 5. Twenty years ago, A was 40 years old, and his son was only \ as old ; now the son is ^ as old as the father ; gaining thus, when will the son be as old as the father ? Ans. (a; = 00 ). 6. Find a fraction such that if 2 be subtracted from the "numerator, or if 3 be added to the denominator, the resulting fractions will equal -f. Ans. Indeterminate. 7. Required a nuniber such that 4 times the number, dimin- ished by 12, divided by the number minus 3, may equal 4 times the number, plus 9, divided by the number plus 3. Ans^ Impossible. Note. — The 6th reduces to the indeterminate form, altliough f, |, ^|, etc., will answer the conditions of the problem. REVIEW QUESTIONS. State the principles of Zero and Infinity. Define Generalization. A General Problem. A Formula. A Negative Solution. State the prin- ciples of negative solutions. Define the Discussion of a Problem. An Indeteminate Problem. An Impossible Problem. When is a problem Indeterminate ? When in possible 1 SECTIOI5' YI. INYOLUTION, EVOLUTION AND RADICALS INVOLUTION. S06. Involution is the process of raising a quantity to any given power. 207. A Power of a quantity is the jproduct obtained by using the quantity as a factor any number of times. 308. An Exponent of a quantity is a number which indi- cates the powcfr to which the quantity is to be raised. Thus, let a represent any quantity ; then, a = a}, is the first power of a. aa = a^, is the second power of a. aaa = a', is the third power of a. aaaa = a*, is the fourth power of o. When the exponent is n, as a", it indicates the nth power of a. S09. The Exponent (called also the Index) indicates how many times the quantity is used as a factor. The First Power of a quantity is the quantity itself. The Second Power of a quantity is called its square. The Third Power of a quantity is called its cube. PRIBTCIPIiES. 1. The square of a quantity is the product obtained by udrug the guantity as a factor twice. 2. The cube of a quantity is the product obtained by using the quantity as a factor three times. 3. All the powers of a positive quantity are positive. For, the square of a positive quantity is positive, since it is the product of two positive quantities ; and its cube is positive, since it is also the product of two posiliv*. quantities, etc. 1S& 15(} INVOLUTION. 4. The EVEN powers of a negative quantity are positive, and the odd powers are negative. The square is positive, since it ia the product of two negative quantitia; the athe is negative, since it is the square, which is pofiiim, multiplied by the quantity, which is negative; the fourth power is positive, since it is the cube, a negative, multiplied by the quantity, a negative; etc. CASE I. 310. To raise a monomial to a given power. 1. Raise ia'b to the third power. OPERATION • Solution. Multiplying 4a% by itself, we have 16a*6', ^ 21 and multiplying the square by 4a%, we have 64a%'. Ex- < ji amining the result, we see we have the cube of the coefficient, and the letters of the given quantity with three times the exponents which they have in the root. 4a% 64a66' Rule. — I. Raise the coefficient to the required power, and mulr tiply the exponent of each letter by the index of the power. II. When the quantity ia positive, the power will he positive; when the quantity is negative, the even powers will be positive and the odd powers negative. BXAMPLiES. 2. Square of Sab\ Ana. 9a'b\ ' 3. Square of 5aV. Ans. 25aV. 4 Square of -6aV. Ans. 36aV. 5. Cube of 3a'x. Ans. 27aV. 6. Cubeof -4aV. Ans. -6UV. 7. Fourth power of Ba"b\ Ans. 81a*»6". 8. Fourth power of - 2a!*"c". Ans. IGa'^c*". 9. Fifth power of -a'iV. Ana. -a^'^if. 10. Fifth power of 2a-'b^c\ Ans. 32a-'Wc^. 11. Sixth power of 2aV. Ans. 64a"c". 12. Seventh power of - 2aV. Ans. -128aV*. 13. Eighth power of -a^¥e-^d". Am. a^^'b'*c~^^d''" FRACTIONS. 167 14. nth power of a^6V. Ans. a^b'"e*^. 15. nth power of -2a; V'^*- -^ns. =i=2r'x^y-'"z\ 16. Valueof (-2a'6'/. - J.n«. -32a"6". 17. Value of (-a'6"c)*. Ans. a«6V. 18. Value of (-26-%-')'. Ans. -1286-''m-"'. 19. Valueof (-a"6^)». Ans. ^a"''b^'. 20. Value of (-a;^6-*'c'»)". 4m ±a;'^''6-'»'c"". CASE II. 311. To raise a fraction to a giren power. a' 1. Find the third power of — Solution. Using the quantity three times as a operation. , , , a^ a^ a' a' a' a' a' a* actor, wehave — x — x — = — x — x — = — c° c° c° c' c' (T c» e* Rule. — Raise both numerator and denominator to the required power. EXAMPLES. 2. Square of — Ans. ^ Be 9c' 3 Square of . Ans. ^ V 16e* 4. Cube of Ans. cd' eW 2ael 5. Cube of - 6. Square of . SaV 4aV 7. Fourth power of — - 8. Fifth power of — 9. Sixth power of — 2e? g'c" a"6c .47JS. 8aV 27a;« 9e^-* .ilMS. 16a**-* .Ans. ■ 16c" Ans. ■ 32a;y aV» A«, c**-' 158 INVOLUTION. 10. Second power of Ans. . 11. Third power of — Ans. 12. Find value of / — '^^\ ■ Ans. \^2aVJ \2cdj' 16a*"-' 13. Find value of | — ^^^^^^ ) . Ans. — - 14. Find value of ( — — - 1 • Ans. 2"c''d" 15. Find value of | | , when m is even. Ans. 16. Find value of ( 1 , when m is odd Ans. 17. Find the square of the fraction — f :^ ■ Ans.^^^^^^. (x-3)' 18. Find the cube of the fraction -j-^r — "^ ■ ar-5d'+6a Ans.^^^1^. (a -3/ CASE III 313. To raise a polynomial to any given power. 1. Ti find the square of a+b. SoLTjTiON. {a + by = (a + b)(a + b), which hy multiplying we find to De equal to d' + 2ab + b\ Rule. — Find the product of the quantity taJcen as a factor at tnany times as there are units in the eapponent of the power. POLYNOMIALS. 169 2. Square of a;-l. Ane. x' ~2ai+l. 3. Cube cf a-c. Am. a? - 3a''e + 3ac* - e*. 4. Square of •2a'' - 3e. . Ans. 4a* -12a'c+9c\ 5. Cube of 1 - a;. Am. l-Zx+Bx'- a?. 6. Cube of 2a-6'. Am. 8a'-12a'6'+, etc. 7. Fourtb power of a — b. Ans. a* — 4a'b + 6a^b^ - , etc. 8. Fourth pbwer of 2a — c*. Ans. 16a' — 32aV+, etc. 9. Square of a — 6 + e. Ans. a? — 2ab + ¥+ 2ao — 2bo + c'. 10. Square of a'- 26 +e'. Am.a*-4a'b+W+2aV-4bc^+c*. 11. Cube of a+b+c. ^M.a'+3a'6 + 3a6^+6' + 3a'c+6a6c+36'e+3acV36c'+c'. 12. Cubeof 2a-36^+c». Ans. 8a'- SGa'b'+Mah*- 276° + 12aV- 36a6V + 276V + 6ac°-96V + e». CASE IV. 313. Special methods of squaring a polynomial. 1. Find the square of a+6+c+d. OPEKATIOK. Solution. Squaring the polynomial by actual multiplication, and arranging the c^Tr+^+^xj^ terms, we shall have the square as written a'+b^+o^+d'+2ab+2ac+ m the margin Examining this, we see a 2ad+2be+2bd+2cd. certam law which may be stated as fol- lows : Rule. — The sqimre of a polynomial equals the square of each term and twice the product of the terms taken two and two. Notes. — 1. The rule may be briefly stated thus : The square of every one, twice the product of every two. 2. The pupils will readily solve the given problems mentally vy meant of this rule. 160 INVOLUTION. 2. Square of a+b+c. Am. a'+F+e^+2ab + 2ac+2bfi. 3. Square of a+b — c+d. Am. a'+b'+e'+d^+2ab-2ae+2ad, etc. 4. Square of u — x+y — s. Am. v?+a?+y'+^ — 2vm+ 2uy — 2uz, etc. 5. Square of a+6+c+d+e. Am. a^^b''+e^+d^+e^+2ab + 2ac, etc. 6. Square of a + b+c+d+e+f+g. Am. a^+V^+e +d^+^+p+g'', etc. 7. Square of 2a +36 + 4c +d Am. 4a'+96''+16c'+d'+12a6+16ac, etc. 8. Square of Z—m+n — o+p — g+r. Am.P+rn?+n^+ o' +p^ +q'+r'- 2lm, etc. 9. Square of a+b — e — d+e+f—g — h+i+j. Am. a'+b'+e'+d^+e'+f'+g^, etc. 10. Square a polynomial consisting of the letters of the alpha Det to m. Am. a'+b^+(^+d'+ ^ +p + g', etc. 314. Second Method. — Arranging the terms in anothei order and factoring, we have the following : (a+b + e+dy = a'-i^2ab+b'+2(a + b)e+(^+2(_a+b+c)d+d'. Stating this in ordinary language, we have the following rule: Rule. — The square of a polynomial equals the square of the first term, plus twice the product of the first term into the second, plus the square of the second, plus twiee the sum of the first tv)0 terms into the third, plus the square of the third, etc. S13. Third Method. — Still another form is the following which pupils may translate into common language : (ai-6+c+d)''=aV(2a+i)6 + [2(a+6)+c]e+[2(a+6+c)+s_ last expressioil, we perceive a law in its formation which may be expressed as follows : Rule. — The cube of a polynomial equals the cube of the fiest term, plus theee times the squaee of the fiest itito the second, plus THEEE times the fiest into the squaee of the second, plus the CUBE of the second ; phis theee times the squaee of the SUM of the FIEST and second into the thied, plus theee times the SUM of the first and second into the squAee of the third, ■plus the cube of the thied, etc. EXAMPLES. 2. Cube6+c+d. Ans. 6'+36'c+36c'+e'+3(6 + c)'(i+,ete. 3. Cube x+y+z. Ans. a? + , etc., + 3(a; + y^z + 3(a; + y)z' + z'. 4. Cube a+b+c+d. Ans. a' + ,etc., +3(a+6+c)'d+3(a+6 + e)(Z'+d'. 5. Cube a+b+c+d+e. Ans. a'+.,etc., +3(a+6+c+d)'e+3((H-6 + e+d)eVe'. 6. Cubea+6 — c+d-e. Ans. (j^+,eic.,+Z(ar+b-^e+d)^ — (?. 7. Cubea;'-2y+2'. Ans. a;«-6a;*y+12a;y-82/' + 3(a;'- 22/)V+,e{c. Note. — In solving the 7th, expand a + 6 + c, and then substitute u^ for a, -22/ for 6, and z* for c, or involve it directly. ' 317. Anothee Form. — The formula for cubing a polyno- mial may be put in another form, which is sometimes more con- venient, as follows : (a+6+c)' = a»4(3a'+3a6 + 6'')6 + [3(a + 6y+3(a+6)c+c']& 162 INVOLUTION. THE BINOMIAL THEOREM. 218. Tlie Binomial Theorem expresses a general method of raising a binomial to any power. 319. This theorem affords a much shorter method of raising oinomials to required powers than the tedious process of multi- plication. Note.— The Theorem was discovered by Sir Isaac Newtnn. It was con- sidered of so much importance that the formula expressing it was engraved upon his monument in Westminster Abbey. 3S0. To derive the binomial theorem, we will raise two binomials to different powers by actual multiplication, and then examine these powers to discover the law of their formation. Let us raise a + 6 to the 2d, 3d, 4th and 5th powers. a +b a +b a^+ab ab + b^ 2d power, a" + 2ab + V a +b a' + 2a'b + ab' a'b+2ab'+b' 3d power, a? + Sa''b + Sab''+¥ a +b a*+Sa'b + ^a'b'+ab' a'b + Sa'b' + 3ab' + b* 4th power, a* + ia'b + Qa'b'' + 4a6' + b* a +b a' + ia'b + 6aW+ Wb' + ab* a*b+ 4aW + &a'b^ + 4a6* + ¥ 6th power, a^+5a*b + 10a*b' + 10aW + 5ab*+b* TUE BIXOMIAL THEOKEM. 163 Raising a - 6 to the 2d, 3d, 4th and 5th powers, we have — (a-6)' = a'-3a^6 + 3a6'-fi'i (a - 6)* = a* - ^a% + 6a^6^ - 4a5' + S* ; (a - hf ■^a?- 5a'b + lOa'b^ - 10a'6'+ §a6* - b\ In deriving a law from these examples there are five things to he considered : 1st. The number of terms ; 2d. The signs of the terms ; 3d. The letters in the terms ; 4th. The exponents of the letters ; 5th. The coefficients of the terms. I. NuMBEE OF Teems. — Examining the results in the given examples, we see that the second power has three terms, the third power four terms, the fourth power five terms, the ffth power six terms ; hence we infer that The number of terms in any power of a binomial is one greater than the exponent of the power. II. Signs of the Teems. — By examining the signs of the terms in the different powers, we infer the following principles : 1. When both terms of the binomial are positive, all the tenns will be positive. 2. When the first term is positive and the second negative, all the ODD terms, counting from the left, will be positive, and all the EVEN term^ will be negative. III. The Lettees. — By examining the letters in the differ- ent powers we infer the following principle : The first letter of the binomial appears in all the terms excejA lie lad; the second letter appears in all except the first; and theii product appears in all except the first and the last. IV. The Exponents. — By examining the exponents of the terms in the different powers, we infer the following principles : 164 INVOLUTION. 1. The exponent of the leading letter or quantity in the firgl term is the same as the exponent of the power, and decreases by- unity in each successive term toward the right. 2. The exponent of the second letter or quantity in the second term is one, and increases by unity in each successive term toward the right, until m the last term it is the same as the exponent of fht power, 3- The sum of the~ exponents in any term is equal to the expo- nent of the power. V. The Coefficients. — By examining the coefficients of the terms in the different powers, we infer the following prin- ciples : 1. The coefficient of the first and the last term is 1. 2. The coefficient of the second term is the exponent oj the power. Thus, in the second power it is 2 ; in the third power, 3 ; in the fonrth power, 4 ; and in the fifth power, 5. 3. The coefficient of any term, multiplied by the exponent of the leading letter in that term, and divided by the number of the term, equals the coefficient of the next term. Thus, in" examining the powers of a + 6 or a — 6, we see that in the 4th power the coefficient of the second term, 4, multiplied by 3, the exponent of a in that term, and divided by 2, the number of the term, equals 6, the coefficient of the following term ; in the 5th power we have 5, the coefficient of the 2d term, multiplied by 4, the exponent of a in that term, and divided by 2, the number of the term, equals 10, the coefficient of the 3d term ; also 10, multiplied by 3 and divided by 3, equals 10, the (X)efficient of the 4th term, etc. Notes. — 1. We see that the coefficients of the last half of the terms when even, or the terms after the middle when odd, are the same as the coefficients of the preceding terms, inversely ; hence we may write the coefficients of the terms after the middle term without actual cal- culation. 2. The above method of deriving the Binomial Theorem is -by observa- tion and induction from particular cases. For a more rigid demonstration see Supplement, page 330. THE BINOMIAL THEOREM. 165 1. Raifre x-y to the fourth power. Solution. Letters and exponents, x* x^y x^y^ xy^ y* Coefficients and signs, 1 —4 +6—4+1 Combining, a;* — 4a;^^ + 6a;V — 4a;2/'+^* Note. — In practice, we first write the literal part of the development, Mid then, commencing at the first term, insert the coefiicients with their signs. 2. DeTelop'(o+ar)'. Ams. a' + 3a''a;+3aa;'+a;'. 3. Develop (a + c)*. Am. a* + 4a'c + 6aV+4ac'+c*. 4. Develop (a - Vf. Ans. a" - 5a*b + lOa'b' - lOa'b^ + 5ab* - b\ 5. Develop (x — yf. Am. x' - 6x'y + 15a;y - 20a^2/' + 15a;y - 6xf+y\ "6. Develop (a +6)'. Am. a' +7a'b + 21a'b' + 35a'b' + 35a»6* + 21a'b' + 7ai' + b\ 7. Develop {a-xf. Am. a^ - Sa'x + 28aV - 56a=6' + TQaV - 56aV + 28aV-8«a;'+a;l 8. Develop (1- a;)'. Am.' l-6x + 15a;'' - 20a;' + 15a;* - 6a^ + a;'. 9. Develop (a - of. Am. a"- 9a'o + 36aV - 84aV + 126aV - 126aV + 84aV-36fflV + 9ac'-e^ 10. Develop (a+e)". Am. a'" + lOa'e + 45aV + 120a'c' -^ 210aV + 252aV + 210aV + 120a'c'' + 45aV + lOtfc' + e'". 11. Develop (a+x)\ n(w-l) „ , „ w(7i-l)(n-2) ..^ Am. tt"+ri(t"-^a!+ ^ V -V+-'^ ^^ ^ a" V + naaf~^+af*. 16P INVOLUTION. BINOMIALS, WITH COEFFICIENTS AND EXPONENTS. SSI. The Binomial Theorem can also be applied to bino mials when one or both terms have coefficients and exponents. 1. Raise 2a'' + 36 to the fourth power. Solution. Let 2a' = m and 36 = n; then m+n will equal 2a''+3b. We have {m + n)* = m*+ irn^n + GmV + 4mn? + n*. Substituting, {2.a''y + 4{2a'y3b + 6{2a'')\Sby+'i[2a'')(,3by+i3bY. Keducing, IQa^ + 96a% + 216a*b^ + 2Wa^b' ^81 b*. Note. — It can also be solved by writing the second expression direotlT »fld then reducing, without making the substitution. 2. Develop (2a - 36)'. Ans. 8a' - 36a^6 + .54a6' - 276'. 3. Develop (a' + 3a;) ^ ^ns. a«+12a«a;+54aV + 108aV+81a;*. 4. Develop (4a -3aV)'. Ans. 64a' - 144aV + 108aV - 27'»,V. ^ „ , / aV . . 4aa;' 6aV 4a'a; a* 0. Develop a; — 1. Ans. a; 1- + — \ cj c e' r2ad-o^. ' 8QUAEE ROOT OF NUMBKRS. 173 SQUAEE BOOT OF NUMBEKS. S3o. The method of extracting the Square Boot of Mint- bers is most satisfactorily explained by Algebra. PKINClPIiES. 1. The square of a number consists of twice u-s many figures cu Ihe number, or of twice as many less one. Any integral number between 1 and 10 consists of 1" = 1 one figure, and any number between their squares, 1 10' = 100 and 100, consists of one or fwo figures ; hence the square iqqh .-iq qoO of a number of one figure is a. number of one or teo -.nnra — i nnn 000 figures. Any number between 10 and 100 consists of ' ' two figures, and any number between their respective squares, 100 and 10,000, consists of three or four figures ; hence, the square of a number of two figures is a number of three or four figures, etc. Therefore, etc. 1, If a number be pointed off into periods of two figures each, beginning at units' place, the number of full periods, together with the partial period at the left, if there be one, will equal the number of places in the square root. This is evident from Prin. 1, since the square of a number contains (trace as many places as the number, or twice as many, less one. 3. If we represent the units by u, the tens by t, ihe hundreds by h, the thousands by T, we mil have the following formulas : {t+uf^f+2tu+u'; (h-rt+uy = K'+2ht+f+2(h+t)u+u'; {T+h+t+uy^T'+2Th+h''+2{T+h)t+f+2iT+h+t)u+u\ 1. Extract the square root of 2025. SoLUTioisr. Separating the number into periods of two figures e^h, we find there are two figures in the root (Prin. 2) ; hence the foot consists of tens and units, and the number equals the square of the tens, plus tvm the tens into the units, plus the square of the units. (2i+-u)u =85 x 5 = 425 The greatest number of tens whose square ii contained in 2025 is 4 tens; squaring the tens and subtracting, we bare 15 • OPERATION. P + 2tu+u' = 20-25 C 40 i2 = 40« 1600 6 425 45 2< = 40x2 = 80 u= 5 174 EVOLUTION. 425, wLich equals turice Ihe tens into the tmits, plvs the square of the units Now, since twice the tens into the units is generally greater than the unitt into the vnits, 425 consists principally of turice the tens into the units ; hence if we divide by twice ihe tens, we can ascertain the units. Timce the tent equals 40 x 2, or 80 ; dividing 425 by 80, we find the units to be 5 ; then {2t+u)u, which equals ifu + u^, equals (80+5) x 5, or 425; subtracting, nothing remains. Hence the square root of 2025 is 4 tens and 5 units, or 45. 2. Extract the square root of 104976. Solution. Separating the number into periods of two figures each, "beginning at the right, we find there are three figures in the root, and the toot consists of hundreds, tens and units, and the number equals h' + 2ht + P + {2h+t)u+u''. OPERATION. ■h' + 2M+e' + 2{h + t)u+u''^ 10-49-76(^300 h''= 300'= 9 00 00 20 2ht+f + 2{h+t)u+u' 14976 4 2A= 300x2 = 600 324 t= 20 {2h+t)t= 620 X 20= 12400 " 2(A+<)m+m'= I 2576 2{h+t)= 320x2 = 640 M= 4 [2(A + «)+m]m= (644) X 4 2576 The greatest number of hundreds whose square is contained in 104976 Is 3 hundreds ; squaring and subtracting, we have 14976 remaining, which equals 2ht+f+, etc. Now, since twice the hundreds into the iois is gene- rally much greater than fi +, etc., 14976 must consist principally of tudct the himdreds into the tens ; hence, if we divide by twice the hundreds, we can ascertain the tens. Twice the hundreds equals 300 x 2 = 600 ; dividing by 600, we find the tens to be 2, etc. Note. — In practice we abbreviate the operation as in PRACTict work by omitting the ciphers and con- 10'49'76C324 densing the other parts, preserving merely the trial and true divisors, as is indicated in the margin. Here 6 is the first trial divisor and 62 the first true divisor ; 64 is the second trial, and 644 the second true umsor. From these explanations we derive the following rule : 3 9 62 149 644 124 2576 2576 SQUARE ROOT OF NUMBERS. 175 Rule. — I. Commence at units' place and separate the number into periods of two figures each. n. Find the greatest number whose square is contained in the left-hand period, place it at the right- as a quotient and its square wider the left-hand period, subtract and annex the next period to the remainder for a dividend. III. Double the root found, and place it at the left for a trial divisor, divide the dividend exclusive of the right-hand figure by U;' the quotient will be the second term of the root. IV. Annex the second term of the root to the trial divisor for the true divisor, multiply the result by the second term of the root, sub- tract and bring down the next period for the next dividend. V. Double the root now found, find the third term of the root as before, and thus proceed until all the periods have been used. Notes. — 1. If the product of any true divisor by the corresponding term of the root exceeds the dividend, the term of the root must he dimin- ished by a unit. 2. When a dividend, exclusive of the right-hand figure, will not contain the trial divisor, place a cipher in the root and at the right of the trial divisor, then bring down the next period and proceed as before. 3. To extract the square root of a decimal, we point off the decimal into •periods of two figures each, counting from ■amis' place, amd proceed as with whole numbers; the reason of which may be easily seen. When a number is not a perfect square, annex ciphers and find the root on to decimals. 4. To find the square root of a common fraction, it is evident that we extract the square root of both numerator and denominator. When these terms are not perfect squares, the shortest way is to reduce the fraction tc a decimal and extract the root. EXAMPIiES . Extract the square root — 3. '576. Ans. 24. 11. 5503716. Ans. 2346 4. 1296. Ans. 36. 12. 4137156. Ans. 2034 5. 2401. Ans. 49. " 13. 11594025. Ans. 3405 6'. 56644. Ans. 238. 14. m- Ans.^ 7. 119025. Ans. 345. 15. 164ff. Ans. 12| 8. 207936. Ans. 456. 16. 12.96. Ans. 3 6 9. 321489. Ans. 567. 17. .0064. ^m.-.08 10 6421156. Ans. 2534. 18. 10.6929. Ans. 3.27 3a%+3a62+6' 176 EVOLUTION. CASE in. 336. To extract the cube root of polynomials. S37. The method of extracting the cube root of a polyno inial is derived from the law for the cube of a polynomial. I. Find the cube root of a'+Ba'b +3a6'+6'. Solution. The first term, a', opebation. is the cube of the first term of a^ + Sa'b + Sab^+b' {^a+b the root (Art. 220) ; hence, the a' first term of the root is the cube g^^s root of a', or a. Cubing a and g^j + 3ab subtracting it from the polyno- mial, we have Sa'b + 3a6^ + &', which equals three limes the square of tht first term of the root into the second, plus, etc. ; hence, if we divide the first term of the remainder by three times the first term of the root, we can ascertain the second term. Three times a squared equals Za', the trial divisor ; dividing Za^b by Za'', we obtain 6, the . second term of the root. Adding to the trial divisor three times the product of the first term of the root by the last term, and the square of the last term, we have for a complete divisor 3a^ + 3a& + 6^; multiplying this by 5, we have Sa'b + Sab^+b^ subtracting this, nothing remains. Hence the cube root of the given polynomial is a +6. Rule. — I. Arrange the terms of the polynomial with reference to the powers of some letter. II. Extract the cube root of the first term; write the remli as the first term of the root ; subtract its cube from the given polynomial by bringing down the next three terms for a divi- dend,. III. Take three times the square of the root found for a trial divisor ; divide the first term of the dividend by it, and write the qvxitient for the next term of the root. IV. Add to the trial divisor three times the product of the first and last terms of the root, and the square of the last term, for a complete divisor. Multiply the complete divisor by the Msl term of the root, and subtract the product from the divi- dend. CUBE BOOT OF NUMBEES. 177 V. If there are other terms of the polynomial remaining, pro- ceed in a similar manner until the work is completed. Notes. — 1. Arrange the terms of each new dividend, when necessary, with reference to the powers of the leading letter of the root. 2" When there are three terms in the root, to obtain the third, we use the first two terms as we did the first term in obtaining the second. EXAlUPIiES. Find the cube root — 2. Of a^ — Sa^x+Bax^ — s?. • Ans. a — x. 3. Of a'+6a'6 + 12a6»+8&'. Ans. a+26. 4. Of 3^ - 6x* + Vh?- 8. Ans. s^-2. 5. Of a' + Za^b + 3o6^ + 6' + Za^c + 6aJc + We + 3ae' + 36e? + e'. Ans. a+b + e. 6. Of a'-3a'+5a'-3a-l. Ans. a^-a-1. 7. Of a'-6a^ + 15a*-20a' + 15(i''-6a+l. Ans. a'-2a+l. 8. Of.m«+6m'-40m'+96m-64. Ans. m'+2m-4. 9. Of a? - Baaf + 5aV - 3a^x - a\ Ans. x'-ax-a\ CUBE EOOT OF NUMBEES. 338. The method of extracting the Cube Root of Number* is most satisfactorily explained by means of Algebra. ' PRINCIPIiES. 1. The Guhe of a number consists of three times as many figures us the number, or of three times as many less one or two. , Any integral number between 1 and 10 consists of one 1=1 figure, and any integral number between their cubes, 10' = 1000 1 aEdJ.000, consists of one, two or three figures; hence 100' = 1,000,000 fhe mhe of a number of one figure is a number of one, lv!0 or three figures. Any number between 10 and 100 consists of two figures, and any number between their cubes, 1000 and 1,000,000, consisti oi fova-.five or six figures; hence the cube of a number of two figures eon- sistB of three times two figures, or three times two, less one or two fignr«>» Therefore, etc. 178 EVOLTTTION. 1. Ij a number be pointed off into periodi of three Jiguret each, beginning at units' place, the number of full periods, together with the partial period at the left, if there be one, will equal the number of figures in the root. This is evident from Prin. 1, since the cube of a number contains tHree times as many places as the number, or three times as many, less one or two, 3. ]ff we represent the units by u, the tens by t, the hundreds by h, etc., 'we will have the following formulas: 1. (t+uy = f+Zfu+Uu^+u^ ; 2. lh+t+uy = W + Wt+Zhf+f + Z(h+tyu+Z{Ji,+t)u^+u\ 1. Extract the cube root of 15625. SoiiUTiON. Separating the number into periods of three figures each, we find there are two figures in the root (Prin. 2) ; hence the root consists of tens and units, and the number equals the te?is' + 3 x tens' x «mu+3tu^+u^ = St'= 3x20= = 1200 3tu= 3x20x5= 300 m'= 5"= 25 (3t'+3tu+u^')u 1525x5= 7625 The greatest number of tens whose cube is contained in 15625 is 2 tern; cubing the tens and subtracting, we have 7625, which equals 3 x temr' x uniti + 3 X tens x units'' + uni's^. Now, since 3 x tens^ x units is generally greater than 3xtensxunits^ + , etc., 7625 consists principally of 3 x tens' x units; hence, if -^^ ™ practice. vre divide 7625 hy3xtens'i we can ascer- 15-625 (.25 tain the units. 3 x teru^ = 20' x 3 = 1200 ; ^' "" ^ dividing 7625 by 1200, we find the units to ^ ^ ^fl^ = 1200 r7625 be 5. We then find 3 x tejis x units equals 3 x 20 x 5 = 300 3x20x5 = 300, and units' equals 5^ = 25; ^'"° 25 taking their sum, we have 3t' + 3tu + u' 1525 = 1525 ; and multiplying by the units, we have {3t'+ 3tu+u')u = 1525 x 5, or 7625; and subtracting, nothing re- mains. Hence the cube root of 15625 is 25. iMOTE— In practice we omit the naughts and abbreviate js is seen in the margin. 7625 25 7625 CUBE ROOT OF NUMBERS. 179 2. Estiact the cube root of 14J06125. SHOWN BY LETTEES. AS IN PRACTICE. 14-706-125 (.245 W = 2008= 8000000 14-706-1251.24S 3W=3x 2002 -=120000 3ftt = 3x 200x4= 24000 «!= 40^= 1600 145600 tih + Cjhi, + 3(A + «)«2 + u>= ' 3(7, + «)a = 3 X 2402 = 1728OO 3(ft + «)« = 3x240x5= 3600 ««= 52 = 25 176425 6706125 5824000 882125 882125 2» = 8 201x3 = 1200 S706 20x4x3= 240 48= 16 1456 5824 882125 2405x3 = 172800 240x5x3= 3600 5^= 25 176425 882125 Note. — In practice, abbreviate by omitting ciphers and using periods instead of tbe wliole number each time. Rule. — I. Separate the number into periods of three figures each, beginning at units' place. II. Find the greatest number whose cube is contained in the left- hand period ; place it at the right and subtract its cube from the period, and annex the next period to the remainder for a dividend. III. Tftke 3 times the square of the first term of the root re- garded as tens for a trial divisor ; divide the dividend by it, and place the quotient as the second term of the root. IV. Take 3 times the last term of the root multiplied by the pre- ceding part regarded as tens; write the result under the trial divisor, and under this write the square of the last term of the root ; their sum will be the complete divisor. V. Multiply the complete divisor by the last term of the root ; subtract the product from the dividend, and to the remain- der annex the next period for a new dividend. Take 3 times the square of the root now found, regarded as tens, for a trial divisor, and find the third term of the root as before ; and thus continue until all the periods have been used. Notes. — i. If the product of the complete divisor by any term of the root exceeds the dividend, that term must be diminished by a unit. 2. When a dividend will not contain a trial divisor, place a cipher in the root and two ciphers at tiie right of the trial divisor, and proceed as 3. The cube root of a fraction equals the cube root of the numeratoi 180 EVOLUTION. and dei.ominator. When these are not perfect cubies, reduce t A Radical is an indicated root of a quantity ; !>a |/a, ail/24, 3v/(a- a;), etc. 343. The Coefficient of a radical is the quantity which in- dicates the number of times it is taken. Thus, in 3y^(a'a;) and m(o*— a;?)*, 3 and m are the coefficients. 343. The Degree of a radical is indicated by the index of the radical sign, or by the denominator of the fractional expo- nent. Thus, l/a ; 6^ ; (2aV)' are radicals of the second degree. y^a' ; 6^ ; {a^a^y are radicals of the third degree. Li l/a'; (26) » ; (a — 6)" are radicals of the wth degree. 344. Similar Radicals are those which have the same q[uantity under the same radical sign ; as i/(a^e) and 4i/(a''e) ; also a-{yc and be". Note. — A quantity whose root cannot be expressed without a radical sign or fractional exponent is called an irrational quantity or a sard ; when the root expressed can be obtained exactly, it is called a raiionai quantity. EEDUCTION OF RADICALS. 345. Reduction of Radicals is the process of changing their forms without altering their values. 346. The reduction of radicals depends upon the following principles : Pein. 1. Any root of the product of two or more quantities eguak the product of the same roots of those quantities. Thus, 1/(06) = ya x -yb. For, i/'{ab) equals a^b^ (Art. 231), and a^6^ eqnals ^/a x f/b. In the same way we may prove that i/(ab) = i^a x y'b. Pein. 2. Fractional exponents may he added, subtracted, multi- plied, and divided the same as integral exponents. This may be readily inferred from the relation of integers and iiactions ; it will be rigidly demonstrated in the artinle on the Theory of Exponents. 184 EADICALS. CASE I. 347. To reduce a radical to Its sliuf>lest form. 948. A radical is in its simplest form when the radical parj- is integral, and contains no factor of which the given root can be extracted. 1. Eeduce i/(48<»'a;) to its simplest form. Solution. Besolving the quantity operation. under the radical sign into two factors, v/{48a';i:) = T/(16a^x3ax) one of which, 16a^, is a, perfect square, and =T/{16a^)l/(3aa;) the other, Zax, not a perfect square, =4a|/(3aa;) we find i/(4&a?x) equals ■/ (16a' x Sax), which (Prin., Art. 246) equals i/(16a') x y'{Zax) ; whicli, extracting the square root of 16a', equals Aa-\/{Zax). Rule. — I. Separate the quantity under the radical into two factors, one of which shall contain all the perfect powers of the same degree as the radical. II. JExtract the root of the rational part, and prefix it as a coefficient to the other part placed under the radical sign. EXAMPLES. Eeduce the following radicals to their simplest form : 2. y^iia'x). Ans. 2a^x. 3. v^(9a'c). Ans. 3ai/(aB). 4. i/CSaV). Ans. 2axyi2x). 5. 3i/(12a'). Ans. 6aY(3a). ■ 6. a/(48a'6). Ans. AaY(3ab). 7. -i/(32a«6y). Ans. 4a'b'eY(2e). 8. v^J4aX«-&)}. Ans.2a(a-b)K 9. 2i/ { 9(a' - a'o) ] . Ans. Ga^i^a - c). 10. a^ix' - A). Ans. aa;iX(l - s). 11. 2(ax' -bx')^. Ans. 2x(a-bx)K 12. ba^idiaWe'). Ans. 15a6ci'/(2a'c). REDUCTION OP EADICAIjSr 185 13. (75a'a^y)^. 14. (24a^6V)i 15. ]yim2aXb'-b*c)i\. 16. (a+6)(o'-2a'6+a5')i 17. (m-n)(2a7rv'+4:amn+2an')^. ■18 2ai/(8xfy+WxY+8xi/'). Ans. 5ax\3a,xyy' . Am. 2a6(3a6V)^. Ans.3a-^l2b[b-e)il. Ans. (a' - b^)-\/a. Ans. (m^-n')-i/2a. Ans. 4a{x+y)-\/(2xy). WHEN THE BADICAL IS FRACTIONAL. 349. A Fractional Badical is reduced to its simplest form by changing it to an entire quantity. 1. Keduee 2]/f to its simplest form. Solution. Multiplying both terms of the radical by 3 to make the denominator a perfect square, we have 2^f ; factoring, we have 2^/^x6, which equals %^\y.-\/^ (Art. 246), which, extracting the square root of \ and multiplying, equals |i/6. Rule. — I. Multiply both terms of the fraction by such a quantity as will render its denominator a perfect power of the degree OPEEATION = 2/y^= 2/^x1/6 XL Resolve the quantity under the radical into two factors, one of which is a fraction and a perfect power of the degree indicated, md proceed as before. Reduce the following to their simplest forms : Ans. f t/5. 2. l/f 3.1/1. Ans. ii/14. Ans. ^j/5. Ans. — -j/Sa. 5 6. 2v/|. 7. 2 '14a* «• «^'S Ans. ii/10. Am. — y'12a. o A w. 2ay'6ab. 9. 6z \2707 Ans. 2xs(2Axy'z) i 1 86 RADICALS. CASE II. 350. To reduce a rational quantity to the form of a radical. 1. Keduce 2a'x to the form of the cube root. SoiiUTiON. Since any quantity is equal to the operation. cube root of its cube, 2a''x is equal to the cube root 2a''x = f^ (2a'x}''. of (2a^a;)', which equals the cube root of Sa'x'. = ^(8aW) Rule. — liaise the quantity to the power indicated by the given root, and place the result under the corresponding radical sign. Note. — The coefficient of a radical or any factor of a coefficient may be placed under the radical sign by raising it to the power indicated by the radical, and multiplying the quantity under the sign by the result. EXAMPLES. 2. Keduce 2a'c to the form of the square root. Ans. |/(4a*e'). 3. Reduce Za%^ to the form of flie cube root. Ans. ^(27a'b'). 4. Reduce a + 26 to the form of the square root. Ans. ■|/(a'+4oA+46'). ■ 5. Reduce 2a^b^e^ to the form of the fourth root. Ans. ^(IGa^bh). ■ 6. Express 2a-y/b without a coefficient. Ans. -f/^ia'b). 7. Express Za'-^(2ae'} without a coefficient. Ans. •i^(54aV). 8. Express {x+y)i/z without a coefficient. Ans. ■\/z(x+yf. 9. Express ^y/{Zax) without a coefficient. Ans. •»/ — ^• 10. Express 6a|/(ca;) with a coefficient of 2; Am. 2i/(9a''ca;). 11. Express — •(/(2af') without a literal coefficient. * Ans. 2v^(2a»J). 12. Express -r\ -~ without a coefficient. Ans. ^/-^-^• i y 9c^ \ 16 ADDITION OF RADICALS. 187 CASE III. 351. To rednce radicals of different degrees to a common radical Index. 1. Eedupe a^ and b^ to a common index. o 1 1 » 1 II OPERATION. SoLTJTioN. J equals f; hence a^=as, which equals (a')t, which equals f a». i a* = a^ = (os')^, or ^'a* equals f; hence 6^=6^, which equals (6')^, fis =6^ ={6'')*, or f^b' which equals ^V. Rule. — ^I. Seduce the exponents to a common denominator. II. Raise each quantity to the power indicated by the numerator of its reduced exponent, and indicate the root denoted' by the com- mon denominator. 'EX.ATHBL.Wia. 2. Eeduce a* and e^ to a common index. Ans. i^d' ; -^c*. 3. Keduce 4' and 6* to a cgmmon index, Ans.-tyW; 1X2X6. 4. Eeduce «» and b"" to a common index. Ana. '^a'";'y'h^. 5. Eeduce |/a, ■^b' and -^^ to a common index. Ans.^a^; ^b'; ^/ (8a'a ) = 2ai/ (2aa;) ; 2aT/ (2aa;) . Za^/ (^ax) plus a|/'(2aa;) equals 3ai/(2aa;). Rule. — I. Reduce the radicals to their simplest form. II. If the radicals are then similar, add their coefficients and annex the common radical. III. If the radicals are not similar, indicate their sum hjf the proper signs. EXAMPLES. Find the sum — 3. Of 1/12 and i/27. Ans. 5i/3. 4. Of ^20 and y'^b. Ans. 5/5: 5. Of by'ia^b) and a^bK Ans. 2abi/b. 6. Of i/((f6) and a^b\ ' A.ns. {a+ab)-y/b. 7. Of 3i/(3a'a;) and ai/(27aa;). 4ns. 6ai/(3aa;)., 8. Of ay'ClStMO and x-y/{2,2a'). Ans. 7axi/(2a). 9. Of a-i/(^ac') and ei/(a'c). 4m. 2ac|/(ac). 10. Of 2i>'(16a) and 3,V(54a). Ans. IS-^C^a). 11. Of i/50, i/72 and i/128. 4ns. 1 V2. 12. Of i/(28aV) and cy (112a''c). Ans. 6ao-i/(7c). 13. Of y'2 and 2,/'|. 4ns. 2v'2. 14. Of 2|/3 and 3y^^. 4ns. 3^/3. 15. Of 4;^^ and 6^^\. Ans. \^%. 16. Of |/(2(/a;) and v^(26'a;). 4»m. (a+6)y^2a;. 17. Of i/(a^m) and -\/{(^n^. Ans. a^j/m+^n}. 18. Of i/(a*c), 2i/'(«'6'c) and i/(6*c). 4ns. (a+b)ya. 19. Find the sum of 2x^/i50a'c), 3i^(24aV), ■^V(72acar') itnd 2.E^(8H*)- 4ns. 12aa;(y^2Sc+v°'"3a). SUBTRACTION OF RADICALS. 189 SUBTEACTION OF KADICALS. 233. Subtraction of Radicals is the process of finding the difference between two radicals. 1. Subtract S^/S from 2i/32. Solution. Beducing the radicals to their operation. Bimplest form, we have 2t/32 = 8i/2, and 2v/32 = 2>/{16x2) = 8i/2 3i/8 = 6/2. 6t/2 subtracted from 8/2 a^/S =3;/ (4 x 2) =6j£2 leaves 2/2. ' 2/2 Rule. — I. Seduce the radicah to their simplest form. II. If the radicals are then similar, subtract the coefficient of the subtrahend from the coefficient of the minuend, and annex the common radical. III. If the radicals are not similar, indicate their difference by the j)roper sign. KXAMPIiBS. 2. From y(20a) take i/(5a). Ans. v^(5a). 3. From •/'(49aa;') take y'(25aa;'). Ans. 2x-^/{ax). 4. From 3i/(12a') take ay'(27a). Ans. 3ai/(3a). 5. From ^(Uda'ftake ■^{Sa''). Ant. S^a'. 6. From 2|/(a^e) take a-y/^. Ans. (2a — ac)y'c. 7. From v^(12a) take 2J— ■ Ans. |/3a. 8. From -^(jidOa'x) take :,X(54a*a;). Ans. 2a^(2ax). 9. From 3i/| take 2i/|. Ans. f i/3. 10. From 4t>'32 take A-fX^. Am. Q^2. 11. From i/{a'+ii'x) take i/iQab' + Qb'x). Ans. (a - Bb)i/(a + x), 12. From |/(2a'+4a'6 + 2a6') take i/(2a'-4a^6 + 2rt/.=). Ans. 2b-[/(2a). 13. Find tie value of 7b-i/(a'x) - a-i/(9ab'x) -^ 5i^(a'ove ; hence the entire quotient is =6i/2k 3y'8x, which, reduced, equals &^2x. Rule.^I. Divide the coefficient of the dividend by the coef- ficient of the divisor, and the radical part of the dividend by the radical part of the divisor. II. Annex the latter qi^otient to the former, and reduce tht result to its simplest form. EXAMFIiElS. 2. Divide 6i/(aa;) by 2y^a. Ans. Sy x. 3. Divide V27 by 2i/3. Ans. 6. 4. Divide 5v/(27ae) by 3i/(3a). Ans. 5/c. 5. Divide 6]/54a by 3i/27. Ans. 2i/(2a). 6. Divide 3i/(72a6) by 2i/(66). Ans. 3i/(3o). 7. Divide 6v/(20a) by 2i/(30a). Ans. t/6. DIVISION OF BADlCALS. 193 8. Divide 2-/ J by 1/2J. Am. ax/2. 9. Divide 15(a'6^)i by 3(ai')^. Ans. bah^b. 10. Divide S^J^ClGaV) by 2^{2ax). Ans. bx^a. 11. Divide i^l by i^|. Ar..lv^. 12. Divide (1 - a?)^ by (1 - a)^ . Ans. (1 + a) ' . CASE II. 361. To divide radicals of different degrees. 1. Divide 4v^(aa;) by 2i^a. Solution. We reduce the radicals operation. in a common index, and then divide iy/^ax) 4-^(aV) _„ ,,, .^ i-f/iax) equals ^^{a'x^); %■} a equak 2fa 2^a^ 2f a'; ^-^{aV) divided by 2^.a« equals 2y'(a2^), according to Case I. Rule.— I. Reduce tJie radical parts of the dividend and divisor to a common index, and divide as in Case I. ; or, n. Reduce the radicals to the form, of fractional exponents, and divide as in simple division. x:xAiHFL.x;s. 2. Divide d^{a'c) by i/a. Ans. Z-^(ac^). 3. Divide Bi^TaV) by 2i/(aa;). Ans. S/Coa^). 4. Divide 2^^(06) by 2-i^Xah). Ans. iTCfl*). 5. Divide 8|/(a«) by 4:x/{ax'). Ans. 2y'x-\ 6. Divide &y'Z by S^TS. • Ans. 2^Z. 7. Divide 12 by V^- ^'**- ^V^^- 8. Divide 4v'/(c2) by 6i/(ac). Ans. f^— • e 9. Divide av>'a; by ey^a;. Avs- ^ T'"' 10. Divide!-^- by 1^^- ^««. I-J^- -17 194 RADICALS. INVOLUTION OF RADICALS. 2G2. Involution of Radicals is the process of raising radical quantities to any required power. 1. Find the cube of 2-\/a. Soi/UTION. Expressing the radical with a frac- operation. tlonal exponent, we have (2a4)'; raising it to the (2i/a)'=(2xai}* third power, we have 2' X fli ; which reduced gives =2'xai = 8^a^ 8a^a. =8ay'a Rule. — Raise the rational part to the required power, and mul- tiply the fractional exponent by the index of the power ; or, Raise the rational and radical parts to the required power, and reduce the result to its simplest form. Note. — Dividing the index of the radical by any number raises the radieai to a power denoted by the number. Thus, the square of y^ a is y^ a. Find the EXAMPIiBS. 2. Square of 3iX(ac'). Am. 9ci7(a'c). 3. Cube of 2i/(aa;). Ans. 8aa;i/(aa;). 4. Cube of 3i/(2a;). Ans. bixi/{2z). 5. Square of - i/(2a). Ans. — • 2 6. Cube of 4 J-^- Ans. lQx\* (Aa'x). 7. Fourth power of 3-»/-- Ans. 9a', 8 Fourth power of 2a-*/-- Am. l&a^x^{a'x). 9. wth power of ay^x. Ans. a"|^af. 10. Third power of -,r2a' x ^(aa^). Ans. 2a^x\/(ax\ IL Square of f/a -j/a;. Ans. a — 2-]/(ax) +x. 12. Square of •i/2+a|/2. Am. 2i ia+2a'. EVOLUTION OF RADICALS. 196 EVOLUTION OF EADICALS. 263. Evolution of Radicals is the process of exti acting any required root of radical quantities. 1. Find the cube root of a'-^e'. BoiiUTlON. Reducing the radical to the form operation. of a fractional exponent, we have a*e? ; extractr ■^a^^cfl = ^ (a'ct) — ing the cube root hy dividing the exponents by 3, i_ •, J ■ etc* — Ctiy G we have ao^, which equals a-^o. Rule. — Extract the required root of the rational part, and divide the fractional exponent by the index of the root ; or, Extract the required root of the rational and radical parts, and reduce the result to its simplest form. Note. — Multiplying the index of J,he radical by a number extracts a root of the radical denoted by the number. Thus, the square root of y^a is y^a. EXAMPLES. Find the— 2. Square root of 4y^a\ Ans. ±2T^a. 3. Square root of V(4a;'). Ans. ^Z^(2x). 4. Square root of IQy'XBx). Ans. ±4^(3a;). 5. Cube root of 2\/(ax). Ans. -^{Aax'). 6. Cube root of 2ay'(2a). Ans. y'(2a). 7. Fourth root of 3a;X(3a). Ans. ^(Za\ 8. Fourth root of ii/(2a). Ans. i7(ia). 9. Fifth root of 4cV(2c). Ans. 1/(20). 10 Fifth root of 4a;/ (4a;). Ans. i^'(4a;). 11 Cube root of --v|-- -^Jw. \-y/(2a^. 2i \ Ji 3' 12. Square root of ^J^- Ans. \^iZa% 13. Fourth root of t-Jt" ■^'^- il^'(2«). 1 96 EADICAiS. RATIONALIZATION . S64. Rationalization is the process of removing the radi «!al sign from a quantity. CASE I. 365. To rationalize any monomial surd. 1. Rationalize i/a, also a' . SoLTTTioif. Multiplying -^/a by ■\/a, we have a, a opebations. .•ational quantity. l/a x ^/a = a SoLTiriON. Multiplying a' by a*, we have a, a oli •/. o^ ^a rational quantity. Rule.- -Multiply the surd by the same quantity v/ith a frae- limal exponent which added to tf^ given exponent shall equal unity. EXAMFIiES. 2. What :&ictor mU rationalize a* ? Ans. a*. 3. What factoi will rationalize y''(a'c) ? Ans. y^ai^. 4 What factor will rationalize 2]/(a5)? Ans. y^{c?V). 5. What factor will rationalize 3i7(a'6)? Ans. ^(aV). CASE II. 366. To rationalize a binomial surd of the second degree. 1. Rationalize -j/a — -y/h. Solution'. Since the product of the sum and difference oPEEAiioy, of two quantities equals the difference of their squares, if j/a — i/^ we multiply -\/a- yb hy^/a+i/b, we obtain a — b, a i/a + t/b rational quantity. a — b Rule. — -Multiply the given hinomial by the same binomial, ivith the signs of one of the terms changed. RATIONALIZATION. 197 EXAMPIiES. What factDr will — 2. Rationalize i/a+^a? Aiis. ya — \/x. 3. ■ Eationalize 1/2 - ^/S ? Am. i/2 + v/3. 4. Rationalize 2|/a+ ^5? An8. 2|/a-|/5. 5. Rationalize a +|/6? Atis. a — yb. CASE III. 367. To rationalize eitlier of the terms of a frac> tioual surd. 1. Rationalize the denominator of ——' yx SoLTTTiON. Multiplying both terms of the fraction operation. by -\/x, we have — '- — . in which the denominator is ^ y V ^ _ ^V^ ^ ' |/a; y'a; x rational, and the value of the fraction is not changed. Rule. — Multiply both terms of tJie fraetion by a factor that will render either term rational which may be required. EXAMPLiES. 2. Rationalize the denominator of Ans. iT/'2. l/2 ^^ 3. Rationalize the denominator of ■• Ans. ■ 2-i/d 6 1 /O -I 4. Rationalize the denominator of • Ans. — l+l/3 2 5. Rationalize the denominator of —^ Ans, — ■ — 3-t/3 2 6. Rationalize the denominator of 7. Rationalize the denominator of ya- ye Ans. /^ ; or, putting a = i/n, we have ±av/ — 1. In all higher powers the form will be i a^ ~ !• REDUCTION OF IMAGINARY QUANTITIES. Keduce to simplest form — 1. i/ — n'. Ans. =t»i|/ — 1. 2. i/^. Ans. ±2i/^. 3. v-'^^W- Ans. ^Za^y-1. 4. iZ-iea". ^?is.±4aV^- 5. i/-n°. , ^»M. ± n^i/ - 1. 6. v/"-4aV. ^w. ± 2ac'v/^l. 7. ,iy-iQd'b\ Ans. ^^ah-^/'-i. 8. i/ - 8aV. ^ns.±2acV"^- ADDITION AND SUBTRACTION OF IMAGINARY QUANTITIES. 1. Add -/"^V and y^'-b^. OPERATION. SoLtmoN. \/ -a^ = a\/ — 1 ; y' — 6" = 6i/— 1; ^— a" = ai/^l &ddire; these two quantities, we have (a+6)j/ — 1. i/ — V —fir'^] (w + ft)^"^ EXAMPr.ES. 2. Add v^ a' and ;/ - c'. 3. Add i/"=4 and 1/"=^. Ans. (a+ci-j/ — 1. IMAGINARY QUANTITIES. 199 4. Add i/'^ and |/'"^^T8. Ans. Qj/^. 5. Subtract /^l from i/"^^16. Ans. 2|/^. 6 Subtract -[/ -2m' from |/ - 8m^ ^ns. m|/-2. MULTIPLICATION OF IMAGINARY QUANTITIES. 1. Multiply 3^/"=^ by t/^. OPERATION. Solution. 3/^2 = 3T/2xi/^and /"^ 3/-2 = 3y'Fxi/^l = y'S'x i/ — 1 ; multiplying, we have 3i/6 y' — 3= y^Fx^/ — 1 '<(/^)'. ■which equals 3i/6x -l,or -3v/6. 3/6x (v/'^)2= - 3v^ Note. — Had we multiplied the quantities under the radical sign -at 5rst, we could not have determined the sign of the product. EXAMPLES. 2. Multiply i/^ by 2t/^. Ans. -2y^6. 3. Multiply av/'^=7' by 2v'''^P. ^ns. -2a6\ 4. Multiply 1 + 1/"=^ by 1 - 1/^. -^^js. 2. 5. Multiply l+i/^^ by l + i/~^. Am. 2^/'=^. DIVISION OF IMAGINARY QUANTITIES. 1. Divide 4^/"^ by 2|/^3. Solution. 4v'^F=4t/6x /^T, and opebation. 2/^ = 2/3xi/^^; dividing the divi- 4/^ = 4/6x/^^ „^^ dend by the divisor and canceling the 2-/~^ = 2-i/3x^/^^-l common factors, we have 2y2. NoTB. -In dividing, it is not necessary to reduce to the general form though it is sometimes convenient. EXAinPIiES. 2. Divide 6^/"^ by 2^/"^. ^««- f I ' ^• 3. Divide4i/^=^byai/=2. ^ws. 2/2. 4 Divide ai/^6c by ■|/":^^2ac. Ans. i/3a. 5. Divide 2v/"=l by \ + V~^- ^'*^' 1 + t/-^- 200 RADICALS. PRINCIPLES. 1 . The PEODtrcT of two imaginary quantUies is real, with the sign before the radical the eeveebe of that given by the common rule. Thua, i/^^a^x/ — 6'=-a6; and —i/-a''x—i/ — b'=-ab; bul — ■\/ — a'^xy' — b''=+ ab, etc, 2. The QUOTIENT of two imaginary quantities is real, with the sign before the radical the same as that given by the common rule. Thus, /"^a' -^ t/^:^ =+-;- /^=^ H- - -v/"^^ =+-; also- /^a"' ==- c c + /^F=--, etc. e SQUAKE EOOT OF BINOMIAL SURDS. Sy©. Some binomials cohtaining a radical quantity are squares of binomials, and will thus admit of the extraction of the square root. Thus, (2+ l/3y= 4 + 41/3 + 3, or 7+41/3; hence the square root of 7+41/3 is 2+ VS. 371. Since the second term in the square of a binomial is twice the product of the other two terms, if we reduce the bi- nomial surd to the form a + 2 Vb, a will Jse the sum and b fhe product of two numbers. 1. Find the square root of ll + 6l/2. Solution. — 11 + 61/2 = 11 + 21/] 8 ; now find two numbers whose sum is 11 and product 18. These numbers, we see by inspection, are 9 and 2; then ll + 21/18 = 9 + 2l/r8 + 2, the square root of which is 3 + 1/2. Note. — The numbers 9 and 2 can be obtained by letting a; +^ = 11 and a^ = 18, and finding x 3,-aAy. BXAMPLEIS. Find the 2. Square root of 14 + 61/5. Ans. 3+ Vb. 3. Square root of 28 + 10 l/g. A.ns. 5 + 1/3. 4. Square root of 11 - 4 VI. Ans. 2 - Vl. 5. Square root of 15 + 6 1/6. Ans. 3 + 1/6. 6. Square root of 5 + 2l/6. Ans. 1/2+1/3. EADICAIi EQUATIONS. 201 7. Square root of 8-2l/15. Ans. V3- V5. 8. Square root of 9 + 2vTA. Am. V2+V1. 9. Square root of 14 - 4 V&. Aiis. 2VB- V^. 10. Square root of 30 + 12 1/6. Ans. 2 T/3 + 3 1/2. 11. Square root of 3a — 2aV2. Ans. Va- V2a. 12. Square root of 2a+2l/a^-6^ Ans. Va + h+ Va - 6. 13. Square root of m' — 2n Vm? — n'. Ans. Vm' -n' — n. 14. Square root of 7 + 30 V~^. Ans. 5 + 3 V^^. 1 5. Square root of 1 2 V~^ - 14. Ans. 2 + 8 V^^. 16. Square root of 24 V^^ --23. Atis. 3 + 41/"=^. 17. Square root of 40l/"^- 23. Ans. 5 + 4l/^. 18. Square root of 1^18 - 4. Ans. 1^2( V2 - 1). 19. Square root of 4 1/3 - 6. Ans. l^3( t/3 - 1). 20. Square root of 6 t/5 - 10. Ans. j^'6( T/5 - 1). ■Note.— In thelSth, leta;+2/ = 'l/18 and^^4; inl9th,let a:+2/ = 4l/3, and, since 6 = 21/9, let xy = 9, etc. EADlCAL EQUATIONS. SOLVED AS SIMPLE EQUATIONS. d73. Radical Equations are those wMch contain the un- known quantity in the form of a radical. 373. Some radical equations may be solved by the princi- ples of simple equations, examples of which will now be presented. 1. Given i/x — 3 = 2, to find x. Solution. — Given, T/a; — 3 = 2, transposing, Va; = 5, squaring, a; = 25. Ans. 2. Given l/(5 + 2 l^cc) = 3, to find a;. Solution.— Given, 1/(5 + 2v'a;) = 3, squaring, 5+2V'x = 9, transposing and reducing, 2T/a; = 4, dividing and cubing, x = 8. 202 EADICALS. 3. Given i/5"+7+vS^ = 6. Solution. Given, /a; + 7 + v^a;-5 = 6, transposing, ■/a;+7 = 6 — i/a; — 5, squaring, a;+7 = 36 — 12y'c(;— 54 a; 5, transposing and reducing, i/a; — 5 = 2, squaring, cc — 5 = 4, transposing, a; = 9. EXAMFIiES. 4. Given i/2x+ 5 = 9, to find x. Ans. a; = 8. 5. Given |/(a; - 3) + 2 = 5, to find x. Ans. a; = 12. 6. Given |/(a;+5) =v^a;+l, to find a;. Ans.x = 4. 7. Given 3 + 1/(23; + 4) = 7, to find x. Ans. x = 6. 8. Given 8 - -^^x = ^/(a; - 16), to find x. Ans. x = 25. 9. Given / (6 + y^SS) + 5 = 8, to find x. Ans. a; =^ 9. 10. Given -^/x ~2 = \/{x- 24), to find x. Ans. a; = 49 5 11 . Given |/(a; + 2) = — — > to find a;. Ans. a; = 3. l/(a;+2) 12. Given i/(a;-9) + i/(a;+ll) = 10, to find a;. ^?m. a; = 25, 25a 13. Given -(/(a; — a) = -j/a; — -^i/a, to find x. Ans. x = — — 14 Given = » to find a;. j4ns. a; = 6, ■\/x 3 1 5. Given -\,'{x+. 4a6) = 2a - 1/ a;, to find x. Ans. x = {a-bf, 1 6. Given x+-i/{a — x)= — ; . to find x. Ans. x = a-l l/(« - x) 17. Given i,'(a; - a) +-i/(a; - 6) - \/(^a - b), to find x. Ans. a; = ft. 18. Given — - — = ^^, to find x. Ans.x-=-- — -y/x SEOTIOI^ YII QUADRATIC EQUATIONS. 274. A Quadratic Equation is one in which the second power is the highest power of the unknown quantity ; as, a;' = 4, and 2a;'' -+30; = 5. 275. There are two classes of quadratic equations — Pure or Incomplete quadratics, and Affected or Complete quadratics. 370. The term which does not contain the unknown quan- tity is called the absolute term, or the term independent of the unknown quantity. Notes. — 1. A quadratic equation is also called an equation of the secmd degree. An equation of the fourth degree is called a bi-quadratie equation, 2. When there are two unknown quantities, a quadratic is defined aa tn equation in which the greatest sum of the exponents in any term la wo. PURE QUADRATICS. 377. A Pure Quadratic Equation is one which contains the second power only of the unknown quantity ; as a" = 16. 378. The General Form of a pure quadratic equation, is xa;' = 6. 1. Given 2a;'+6 = 24, to find x. SoiiUTION. Given, 2a;»+6 = 24, transposing an d reducing, x' = 9, extracting the square root, a; — ± 3. 70X 204 QUADRATIC EQUATIONS. 2. Given ax''+n = cx'+m, to find x. Solution. Given, ax' + n = ox^ + m, transposing, ax'' — cx'' = m—n, factoring, (a— c)a;' = >w— M, m — n reducing, a;^ = a— e , ,. , , Im—n extracting square root, a; = ± -»/ ' a—o Rule —I. Reduce the equation to the form ao^ = h. II. Divide by the coefficient of x^, and extract the square root oj both members. EXAMPLES, 3. Given a;' + 5 = 3a;' - 13, to find x. Ana. a; = ± 3. 4. Given {x - 6)(a;+ 6) = - 11, to find x. Ans. a; = ± 5. 5. Given x^+ah = ba?, to find x. Ans. x= ^ \\/(ab). 6. Given 2a;'' - 3 = -^ + 12, to find x. Ans. a; = ± 3. 7. Given x^+a'+b''^ 2ab + 2a;^ to find x. Ans. a; = ± (a - 6). 8. Given 4a; + 8 = (a; + 2)^ to find x. Ans. x=^2. 9. Given ; +- ; = 3, to find x. Ans. x = =t^i/3. 1-a; 1+a; \ ""^ 4 4 1 10. Given =-, to find x. An^. x= ±9. a;-3 a; + 3 3 L. Given 2a;+5a;-' = 3a;-lla;-', to find x. Ans. x= ±4. x + Z a;— 3 12. Given -+ -=^, to find x. Ans. a;- ±6. x-Z x+Z 13. Given + = 1, to find x. Ans. a; = ± 2, a; + 1 a; + 4 CC Qi X Ji 14. Given -+- = -+-, to find x. Am. a;= ^v^(ab\ a X b X r \ ' 1 5. Given (n - a;)' + (w 4 a;)' = 3»i', to find x. Arn.x-^ J-j/g! PEINCIPLES OP PURE QUADRATICS. 205 379. Radical Equations sometimes become pure quad- ratics when cleared of radicals. 1. Given x/(a;'+9) = v'(2a;'-7), to find x. Am. x= ±4. 2. Given |/(a^-4) = 2v^(a-l),tofinda;. Am. x= ±2-y/a. X. Ans. X— ±^. 3. Given a/(— ^- — )=i/a;,to find 4. Given {x+ay = -, to find x. Am. x- ±av/2. (x - ay 5. Given i/(a;+?n) = |/fa;+y^(w'+a;')], to find «. Ans. a! = |/(m' — w'J. 6. Given -1/(0;+ a) =— . to find x. l/(x - a) ^^_ ^^^ l/(a»+t»). 7. Given y'{a?+2ax+^{x'-A)\ ~a+x, to find x. Am. x= ±i/(a*+4). 8. Given -[/\se'+-i/(x*-n*)] =n, to find x. Am. X" ±n. 9. Given ;/ (a;+m) =-iy(a;V»i''), to find x. Am. x 2m '2a? 10. Given x+y'(a?-vx'^= , to find x. PRINCIPLES OF PUEE QUADEATICS. 380. The Principles of Pure Quadratics relate to the form of the equation and tlie relative value of its roots. PRINCIFIiES. 1. Every pure quadratic equation may be reduced to the form ax'^b. For, it is evident we can reduce all the terms containing x' to one term, as nx^, and all the known terms to one term, as b ; hence the form will become ax^ = b. 2. Every pure quadratic equation has two roots, equal in numerical value, but of opposite signs. 206 QUADRATIC EQUATIONS. Dem. 1st. The general form of a pure quad- operation. ratic is ax^ = b; dividing by a, we have x^ = b ax'= b divided by a; representing the quotient by m\ a_^_ i wehave a;^ = m'; extracting the square root, we a have x= ^m. Therefore, etc. a:= zfcni Dem. 2d. From, the equation x^ = 7n\ by a;^ = m» transposition, we have x^ — rn' = 0; factoring, we x'—m'' = have (a; + m){x — in) = 0. This equation can be (a; + m) (a; — m) = satisfied by making x — m equal 0, or by making x — m = x-Hn equal 0, and in no other way (Art. 192, x=+m Prin. 1). Making a; — »n = 0, we have a;= +»l; x+Tn=0 making x+m = 0, we have x= —m. Therefore, x= —m etc PBOBLEMS PRODUCING PURE QUADRATICS. 1. The product of two numbers is 48, and the greater is 3 times the smaller ; required the numbers. Solution. Let X = the smaller number, and 3x = the larger number ; then 3a;xa; = 3a;»^=48; whence, a;^ = 16 ; then a;= ±4, and 3a; =±12. 2. The sum of two numbers is 10, and their product is 24 ; required the numbers. SoLiTTiON. Let 5 + a; represent operation. the greater number, and 5 — a; the 5+a; = the greater number; emalltr number ; their sum will be 5 — a; = the less number. 10, which satisfies the first con- (5 + a;)(5— a;)=24 diti -,n of the problem . By the sec- 25 — a;'' = 24 ond condition, (5+a;)(5 — a;) = 24, a;'' = l or 25 — a;' = 24. Eeducing,wehave 3;^ -tj a;=±l. Using the positive value, 5 + a; = 6or4 we ha ve 6 or 4 ; using the negative 5 — a; = 4 or 6 value, we have 4 or 6. 3. The sum of two numbers is 15, and their product is 54 ; required the numbers. Am. 9 j 6. PROBLEMS. 207 4. Tlie sum of two numbers is 12, and the sum of their squares is 74 ; required the numbers. Am. 7 ; 5. 5. The difference of two numbers is 5, and their prodi ct is 84; required the numbers. Ans. 12; 7. • 6. Divide the number 24 into two such parts that their product shall be 140. Ans. 14; 10. 7. The difference of two numbers is 4 and the sum of their squares is 208 ; what are the numbers ? Ans. 12; 8. 8. Kequired a number whose square is 432 more than the square of ^ the number. Ans. 24. 9. What number is that which, if multiplied by J of itself, is 72 greater than the square of its f ? Ans. 36. 10. What two numbers are to each other as 4 to 5, and the difference of whose squares is 81 ? Ans. 12 ; 15. 11. Eequired two numbers whose product is 48, and the quo- tient of the greater divided by the less, 3. Ans 12 ; 4. 12. There is a rectangular field containing 4 acres whose length is to its breadth as 5 to 2 ; required its dimensions. Ans. Length, 40 rods ; breadth, 16 rods. 13. There is a number whose third part squared and sub- tracted from 170 leaves a remainder of 26 ; what is the number ? Ans. 36. 14. I of the square of twice a number is equal to -f of the square of | of the number, increased by 28 ; what is the number ? Ans. 6. 15. A man has two cubical bundles of hay, one of which contains 999 cubic feet more than the other; what are the dimensions of each, if the smaller is f as long as the larger ? Ans. 12 ft. ; 9 ft. ' 16. A merchant bought a piece of cloth for $216, and the number of dollars he paid for a yard was to the number of yards as 2 to 3; required the price per yard and the number of yards. Ans. 18 yards at $12 a yard. 17. A man bought a field whose length was to its breadth as 5 to 4 ; the price per acre was equal to the number of rods in the length ci the field, and 5 times the distance around the 208 QUADEATIC EQUATIONS.. field equaled tlie number of dollars that it cost ; i-equired the length and breadth of the field. Ans. Length, 60 rods ; breadth, 48 rods. 18. From a cask containing 81 gallons of wine a vintner draws ofi" a certain quantity, and then, filling the cask with water, draws ofi" the same quantity again, and then there re- mains only 36 gallons of pure wine ; how much wine did he draw ofi" each time? Ans. First, 27 gals. ; second, 18 gak; AFFECTED QUADRATICS. 381. An Affected Quadratic Equation is one that con- tains both the second and first powers of the unknown quantity; as, a? + 4m = 12. 282. The General Forms of an afiected quadratic are ax^ +bx = c, and x^ + 2px = q. 1. Given a;'+6a; = 16, to find the value of x. SoLTJTiON. If the first member of this equation operation. were a perfect square, we could extract the square x^+6x =16 root of both members, and find the value of x from x^+6x+9 = 25 the resulting equation. Let us, therefore, see if we a;+3 = ± 5 can make the first member a perfect square. »= — 3±5 JTie square of a binomial is equal to the square of a; = + 2 t}ie first term, plus twice the product of the first term, into a; = — 8 the second, plus the square of the second. If now we consider x^+6x as the first two terms of the square of a binomial, the first term of this binomial will be the square root of x'' or x ; and 6a; will be twice the first term of the binomial into the second ; hence, if we divide 6a; by twice the first term, 2x, the quotient, 3, must be the second term of the binomial, and its square, 9, added to the first member of the equation, will render it a perfect square. Adding 9 to the first member to complete the square, and to the second member to preserve the equality, we have a;'+6a;+9 = 25. Extracting the square root, we have a;+3 = ± 5 ; from which, using the positive value tf 5, we have a; = 2 ■ and using the negative value, we have a;= -8. Rule.- — I. Reduce the given equation to the general form i^+2px = q. AFFECTED QUADRATICS. 209 II. ^ Add to both members the square of one-half the coefficient o/x. III. Extract the square root of both members, and solve the resulting simple equation. 2. Given a;' - 8a; = — 7, to find x. Solution. Completing the square by adding operation. the square of half the ooefficient of x to both x' — Sx= —7 - (1) members, we have i;'' - 8a; + 1 6 = 9. Extracting a;' -8a; +16 = 9 (2) the square root of both members, we have a; — 4=±3 (3) x—4= ±3. Using the positive value of 3, we a; = 7 (4) hav,e a;=7; using the negative value, we have x — 1 (5) VEEinCATION. 7a-8x7=-7; or49-56=-7; also, P-8xl=-7; or 1- 8= -7. 3. Given ar'+4a; = 32, to find x. SOLTTTION. * Given the equation, a;' + 4a; = 32, completing the square, a;' + 4a; + 4 = 36, extracting the root, a; + 2 = ±6 whence, a;= —2 +6= +4, and • a:- -2-6= -8. Both of which will verify the equation. 4. Given a;'' - 12a; = - 20, to find x. Solution. Given the equation, x^ — V2,x—— 20, completing the square, a' — 1 2a; + 36 = 16, , extracting the root, a; — 6 = i 4 ; whence, a; = 6+4 = 10, and a; = 6-4 = 2. • 5 Given a;' - 5a; = 6, to find x. Solution. Given the equation. completing the square, which reduced, equals ^ extracting the root, whence, and a;' — 5a; = 6, a;»-5a;+(f)^ = 6 + (i)', a;'-5a;+(|)' = 6 + y = V, a: = f+J= T-e, x = \-l^ \. 210 QUADRATIC EQUATIONS. " 6. Given Be' -7a; = 66, to find x. Solution. Given the equation, 3x'~ 7a; = 66, dividing by the coefficient of x', x^- — = 22, completing the square, a;» - |a; + (|)^ = 22 + , »)», which reduced, equals a;' — |x+(|)'>= ^, extracting the root, x — ^= =t V i whence, a; = |+V = +6, and a: = f-V=-V- Reduqe the following equations :' 7. a;'+2a; = 24. Ans. a; = 4, or -6. 8. a;'+2a; = 35. Ans. a; = 6, or - 7. 9. a;' + 4a; = 32. - ^m. a; = 4, or - 8. 10. a;'' - 6a; = 16. Ans. a; = 8, or - 2. 11. a;' - 3a; = 18. Ans. a; = 6, or - 3. 12. a;^ - 5a; = 84. Ans. x = 12, or - 7. 13. a;^- 11a; = 60. Ans. a; = 15, or -4., 14. x' - ISx = 140. Am. x = 20, or - 7. 15. a;^ - 14a; = - 24. Ans. x = 12, or 2. 16. a;'- 4a; = 1. ^w^. a; = 2 ± i/5 17. 3x'-4a; = 39. Ans. a; = 4|, or -3 18. (a;-l)(a;-2) = 2a ^ns. a; = 6,' or - 3. 19. a;'' -6a; =-13. 4ns. a; = 3 ± / ^. 20. 4(&'-l) =4a;- 1 Ans. a; = f, or -^. 21. (2a; - 3)' = 8a;. Ans. a; = 4^, or f 22. a;'-3 = ^. Ans.x = n,OT-U. 6 '^t> 23. a;'+2pa; = q. Ans. a; = -p ± i/C^ +2)- 24. o^-ax- b. Ans. a; = ^(a i /a' +46). 25. a;" - 2na = 7>i' - w'. Ans. x = n^m. 26. a;' - aa; - Ja; = - a5. Am, x = a,or b. AFFECTED QUADRATICS. 21] 27. a<- = 5. Am. x~4. x-3 5 h— 4x 5 28. » = 2+— Am. x = ^, or -i. 29. "^—^ + 2a; = 12. ilns. x = 5, or 3|. 30. J_+^±^ = 12. ^^. 3 or - 2|. a; + 3 2 3 . -^ J.ns. a; = 2, or — 3. a; = l, or — 4. .32. X a;+l 13 a: + 1 a; 6 33, a; + 2 a;+l 13 a; + l a:+2 6 34. a;+l a!-2 9 a;-l a; + 2 5 35. a;''+2aa; = al 36. 3a'a;"'-a;= -2a. 37. a;'-2aa; = 6'-a'. 38. x^ - (a- b+e)x = (h - -a)e. Arts, a; = 3, or — f . ^M. a; = a(-l±|/2). :S. X = 3a, or — a. .Ans. x = a + b, or a — b. x = a — b, or +c. S83. The roots of a complete quadratic may be directly written by the following rule : Rule. — I. Reduee the quadratic to the form x''+2px = q. II. Take, with a contrary sign, one-half of the coefficient of x, plui or minus the square root of tlie sum obtained by adding the vpimi of half the coeffi/siei\t of x to the second member. The reason for this rule may be shown as follows: Solving the generil quadratic, x''+1px = q, we find that a; = —p i Vi.l'^P') ; comparing the root with the general equation, we see that —p is \ of 2p with the sigil changed, and y" {q+p^) is the square root of the second member plus the square of half the coefficient of x. Note.— PupUs sbguld be required to solve some of the previous exam- ples by this rule 212 QUADEATIC EQUATIONS. SECONB METHOD OF COMPLETING THE SQUAKE. 284-. A SECOND METHOD of Completing the square enables us to avoid fractions in the. solutions of all forms of quadratics. 285. The method already given involves fractions when the coefficient of a; is an odd number or a fraction. Note. — This method is Bometimes called the Hindoo Method of solving quadratics. I. Given ax'+bx = c, to find the value of x. Solution. To complete operation. the square without using ax'+bx — c (1) fractions, it is evident that ia'x'' + iabx = iao (2) the first term must be a ia''x''+4tabx+b'' = 4:ao + b^ (3) perfect square and the see- 2ax+b= ±y'(4ae+6'') (4) ond term niust be divisible 2ax= —b± ■/{•iac + b^) {5) oy2.. ^ -&=fcT/(4ac+5') -^ . To make the first term a 2a perfect square and the second term divisible by 2, we multiply both members by 4a, which gives eq. (2), If now we consider 4:aV + iabx as the first two terms of the square of a binomial, the first term of this binomial will be the square root of 4aW, or 2ax ; ^abx will be twice_ the product of the first term by the second ; hence, if we divide ^abx by two times 2ax, or ^ax, the quotient 6 will be the second term of the binomial, and its square, W, added to both mem- bers of the equation, will complete the square. Extracting the square root and reducing, we have the values of x ex- pressed in equation (6). Rule. — I. Seduce the equation to theformax''+bx = e, in nhich the three 'terms are integral and prime to each other. II. Multiply the equation by 4 times the coefficient of x' ; add the square of the coefficient of x-fo both members. III. Extract the square root, and find the value of x in the re- sulting simple equation. Notes. — 1. A quadratic may also be solved by multiplying by thi coefficient of x', and adding the square of one-half the coefficient oix. 2. Let pupils make a rule from the formula which expresses the rosi .>! the general qusdiutic, ax^ + bx = c.. COMPLETING THE SQUARE. Reduce 3a;^+5a; = 22. Solution. Given the equation, 3x^ + 5a; = 22, multiplying by 4x3 or 12, 36a;2H-60a; = 264, completing the square, 36a;* + 60a; + 25 = 264 + 25, 213 extracting the root, 6a; +5= ±17, transposing and reducing. 6a;=-5±17; whence, \ a;= +2, and a;=-3f. \ 3. Reduce 2a;' -3a; = 9. Ans. a; = 3, or - lA. 4. Reduce 3a;' + 8a; = 28. • Ans. x = 2, or - 4|. 5. Reduce 4a;''+ 7a; = 11. Ans. a; = 1, or - 2|. 6. Reduce 53;=^+ 2a; = 88. Am. a; = 4, or — 4|. 7. Reduce Sar"- 4a; = 156. Ans. a; = 6, or — 5^. 8. Reduce 4ar'- 45a; = 36. Ans. a; = 12, or — f. 9. Reduce 8a;' -7a; ^165. Jims, a; = 5, or —i^. 10. Reduce 9a;' -7a; = 116. Ans. a; = 4, or - 3|. 11. Reduce 2a;='+aa; = 6. Ans. x = ^\ -a±^(a'+86)}. 12. Reduce "-^''=^^*-«^. b X J.?M. a; = 36, or 3(a-6). 13. Reducear'+3a; = 5. ^JM. 1.1925 + , or -4.1925+. 14. Reduce 3?"+ 2a; = 5. Ans. 1.449 + , or -3.449+^ 15. Reduces;'- 8a; = -8. Ans.&.82S + , or 1.172+. 16. Reduce 5a;' — 4a; = 2. ^M, 1.148, or - 0.348. THIED METHOD OF COMPLETING THE SQUAEE. 386. A THIED METHOD of Completing the square is stated b the following rule : Rule. — I. Make the coefficient of ike first term of the quadratic a positive square. II. Divide the second term by twice the square root of the first, and add the square of the quotient to both members. 214 QUADRATIC! EQUATIONS. EQUATIONS IN THE QUADRATIC FORM. 387. Any equation which is in, or may be put in, the quad- ratic form, may be solved by the following methods. 388. An equation is in the quadratic form when it con tains but two powers of an unknown qua* dtj, and the index of one power is twice that of the other a?, x'"'+ax" = b, or CASE I. 389. When the unknown term of the quadratic is a monomial. 1. Given x*-6x'' = -8, to find x. SOLUTIOS-. Given the equation, a;*— 6a;' = — 8, completing the square, x^ — 6a;' +9 = 1, extracting the square root, x' — 3— ±1, transposing and reducing, a;' = 4, or 2, extracting the square root, z — ± 2, or ± y'2. 2. Given a^ - 4a;' = 32, to find x. Solution. Given the equation, . a;*— 4a;' = 32, completing the square, x^ — 4a;' + 4 = 36, extracting the square root, a;*— 2= ±6, transposing and reducing, a;' = 8, or —4, extracting the cube root, a; = 2, and a; = f(-4). Note. — Since the odd roots of a negative quantity are real, by extra< t- ing the cube root of 4 and prefixing the minus sign we find the approii- mate value of the second root of the above equation. EXAMPLES. 3. Given a;*+4a;' = 32, to find x. Ar^. a;- ±2, or±2i/(-2). 4. Given a;*- 5a;' = 36, to find x. Ans.x=±B, 01 t2v'(-l). EQUATIONS IN THE QUADRATIC FORM. 215 5. Given x^-1a? = 8, to find x. Am. a; = 2, or - 1. 6. Given x'+ix'" = 5, to' find x. Am. a; = ± 2, or == 1. 7. Given x+Z-i/x = 18, to find x. Am. a; = 9, or 36. 8. Given a;' + ix' = 6, to find x. Am. x = l, or - 125. . 1 9. Given x^ - aa;" = b, to find x. Am. x = (^a ± -\/b+^ar) » . 10. Given x'" + 4a;" = 12, to find x. Am. x = 1^2, or 1/ - 6. 11. Given a;'+ 7a;^ = 3|, to find x. Am. x = \-^2, or 11^^450. 1 2. Given y^x^ 2^' 3? = |, to find x. Am. a; = ^, or - 1|. f} 2 1 3. Given a;" — ax^ = b, to find x. Ans. x = (^a =1= yb+\c?)». 14. Given — = — — — > to find x. Ans. x = 64, or 4. i + l/x \/x CASE II. 390. When the unknown term of the quadratic Is a polynomial. 391. When a polynomial becomes the basis of the quad- ratic form, we may consider it as a single quantity, and proceed as in the previous case. 1. Given (a;'+2a;)V4(a;'' + 2a;) = 96, to find x. Solution. Given the equation, (a;*+2a;)=i + 4(x2 + 2a;) = 96 ; completmg the square, (a;H 2a;) ' + 4(:^'' + 2a;) + 4 = 100 ; extracting the sq. root, a;2+2a;+2=±10; transposing, a;^ + 2a; = 8, or — 12 ; completing the square, a;* + 2a; + 1 = 9, or — 11 ; extracting the sq. root, a;+l= ±3, or ± i/( — 11); whence, a; = 2, or— 4, and a;=-l±y'(-ll)- Note.— This problem may also be solved by placing x'^ + 2x=y, giving the equatior^ 2/^+4^/ = 96, which may be solved and the value of y thus found placed equal to a;' + 2a;, from which the value of x can readilj be found. 216 QUADRATIC EQUATIONS. 2 Given (x' - 2)» + 2x'' = 67, to find x. Solution. Given the equation,' {x' -2)' +2x^ = 67; subtracting 4, ' {x'-2)^ + 2x^-i = 63; factoring, (a;'' - 2)H 2(x« - 2) = 63 ; completing the square, (x'-2y + 2{x''-2) + l = 6^; extracting the sq. root, (x^ — 2) + 1 = i 8 ; whence, aj" = 9, or — 7, and a;= ±3, or ±^-7. Note. — The 4 was subtracted to put the equation in the quadratic form. Sometinaes this can be done by transposing a term or by adding oi Bubtracting some quantity. EXAMPLES. 3. Given (x" - 3) V 4(a;' - 3) = 5, to find x. Am. a; = ± 2. 4. Given (x' + Bxy -6(x'+ Bx) = 216, to find x. Ans. a; = 3, or - 6. 5. Given (a;' - 4a;)^ + {x^ - Ax) = 3f , to find x. Ans. a; = 41, or - ^. 6. Given i/(5+a;) + /(5+a;) = 6, to find x. Ans.- X = 11, or 76. 7. Given (-+«) +6 (-+«;) =40, to find x. V J V J ^ns. a; = 2, or-5±i/(21). 8. Given x — i/(a;+ 5) = 1, to find x. Ans. a; = 4, or - 1. 9. Given {x - if - 6i/(a; - 4) = -i^. to find x. ^'^ ^ns. a; = 8,or4+i?'4 10. Given (a;V2a;-3)V7(a;V2a;-3) =60, to find x. Ans. x = 2, -4, or -l±2i/.(-2). 11. Given (a;' -9)' -11a;' +40 = 21, to find x. Ans. a; = ± 5, or =t 2. ' 12. Given (a;' - 4a; +5)'' +43;"- 16a; = -8, to find x. ^m. a; = 3, or 1, or 2 ± i/( - 7). Note.— Add 59 to both members of Ex. 11 ; add 20 to both members ■ ■ of Ex. 12. Several of the problems in Art. 290 have other results than 1 1 those given. V' PROBLEMS. 217 PBOBLEMS' PRODUCING AFFECTED QUADIUlTICS. 1. Find two numbers such that their sum is 16 and their product is 60. Solution. Let X = one number ; then 16 — a; = the other number, by the conditions, (16 — a;)a! = 60 ; which gives a;' — 16a; = — 60 ; whence, a;=10, or6, and 16-a; = 6, orlO. Hence the two numbers are 10 and 6. 2. A man sold a watch for $24, and lost as much per cent, as the watch cost him ; what did the watch cost him ? Let a; = the cost of the watch; then a; = the loss per cent, and xx = = theloss; 100 100 ' a;*' . therefore, x = 24 : ' * 100 whence, a; = 60, or 24. Both of these values will satisfy the conditions of the problem. The pupil win show this by verification. 3. A lady divided $144 equally among some poor persons; if there had been two more, each would have received $1 less ; required the number of persons. Solution. 144 Let a;=the number of persons; then = what each received; X 144 and =whai each would have received if there had been two inore; 55+2 then = - 1; whence a; = 16, or —18. The number of persona a;+2 X was therefore 16 ; the negative result will not satisfy the problem in au arithmetical sense. KoTE. — I^ in the problem, 2 more be changed to 2 lesk, and |1 leas to $1 more, the ccrrect result will be 18. 19 21 S QUADRATIC EQUATIONS. 4. A' gentleman divided |50 between his tv\o sons in such a manner that the product of their shares was 600 ; what was the share of each ? Ans. $30 ; $20. 6. The wall which encloses a rectangular garden is 128 yards long, and the area of the garden is 1008 square yards ; what is its length and breadth ? Ans. 36 yds. ; 28 yds. 6. An officer wishes to arrange 1600 men in a solid body, so that each rank may exceed each file by 60 men ; how many must be placed in rank and file ? Ans. 20 ; 80. 7. A merchant bought a number of Bibles for $50, which he sold for $5.50 a piece, and thus gained as much as one Bible cost; how many Bibles did he buy? Ans. 10. 8. The perimeter of a room is 48 feet, and the area of the floor equals 35 times the difference of its length and breadth ; what are the dimensions of the room? Ans. 14 ft. ; 10 ft. 9. Two boys, A and B, bought 10 oranges for 24 cents, each paying 12 cents ; if A paid 1 cent more apiece than B, how many oranges did each buy ? Ans. A, 4 ; B, 6. 10. A lot of sheep cost $180, but on 2 of them being stolen, the rest averaged $1 more a head than at first ; find the number of sheep. 11. A man walked 48 miles in a certain time: if he had gone 4 miles more per hour, he would have gone the distance in 6 hours' less time ; how many miles did he travel per hour ? Ans. 4 miles. 12. In an orchard containing 180 trees there are 3 more trees in a row than there are rows ; required the number of rows and the number of trees in a row. Ans. 12 rows, and 15 trees in a row. 13. In a purse containing 52 coins of silver and copper, each silver coin is worth as many cents as there are coyyer coins, and each coppesr coin is worth as many cents as there are silver coins, and the whole is worth $2 ; how m.iny are there of each ? . Ans. 2 silver ; 50 copper. 14. A person distributed $6 equally among a number of PEOBLEMS. 219 paupei-s; and as there were 5 less than he suiiposed, they each received 10 cents apiece more than they otherwise would ; how . many paupers were there? Ans. 15. 15. The expenses of a party amount to $10 ; and if each pays 30 cents more than there are persons, the bill will be set- tled ; how many persons are there ? Ans. 20. 16. There is a number consisting of two digits whose sum is 10 and the sum of whose' squares is 62 ; it is required to find the number. Ans. 46, or 64. 17. Mr. Leslie sold his horse for $171, and gained as much per cent, as the horse cost him ; what was the first cost of the horse? Ans. ^90. 18. A person laid out a certain sum of money for goods, which he sold again for $24, and lost as much per cent, as the goods cost him ; what was the first cost ? Ans. $40, or $60. , 19. A. yacht sails 90 miles down a river whose current moves 3 miles an hour, and is gone 16 hours ; required the rate of sailing. Ans. 12 miles an hour. 20. A farmer bought a number of sheep for $80 ; if he had bought. 4 more for the same money, he would have paid $1 less for each ; how many did he buy ? Ans. 16 sheep. 21. A man bought a quantity of meat for $2.16. If meat were to rise in price 1 cent per pound, he would get 3 pounds less for the same sum. How much meat did he buy ? Ans. 27 lbs. 22. The plate of a mirror, 18 inches by 12, is to be set in a frame of uniform width, whose surface is to be equal to the surface of the glass ; required the width of the frame. Ans. 3 inches. 23. Todhunter gives the following beautiful little problem : Find the price of eggs per dozen when two less for 12 cents raises the price 1 cent per dozen. Ans. 8 cts. 24. A and B start at the same time to travel 90 miles ; A liavels 1 mile an hour faster than B, and arrives 1 hour earlier ; at what rate per hour did each travel ? Ans. A, 10 mi. ; B, 9 mi. 25. A person bought cloth for $72, which he sold again at 86 1- a yard, and gained by the bargain as much as one yard cost liim; required the number of yards, .^rur 12. 220 QUADRATIC EQUATIONS QUADRATIC EQUATIONS CONTAINING TWO UNKNOWN QUANTI'nES. Si>S. The Degree of an equation containing two or more uukuowii quantities is determined by the greatest sum of the exponents of the unknown quantities contained in any term. Thus, 2ax + Sxy = 4a is an equation of the 2d degree ; Zxy + Ax'y = 12 is an equation of the 3d degree. 393. A Homogeneous Equation is one in which the sum of the exponents of che unknown quantities in each term which contains them is the same. Thus, and x' + Zx^y+'Zxy'+f^'n are each homogeneous equations. 394. A Symmetrical Equation is one in which the un- known quantities are similarly involved, or one in which they can change places without destroying the equation. Thus, a;'+/ = 13, and-+^ = 2i y X and a^+2/'-a!2/+3a; + 3^ = 22 are each symmetrical. 395. Quadratic Equations containing two unknown quan- tities can generally be solved by the rules for quadratics if they come under one of the following cases : I. When one of the equations is simple and the otliei , quadratic. II. When each equation is homogeneous and quadratic. III. When each equation is symmetrical. Note. — Two quadratics containing two unknown quantities usually pro- duce a biquadratic in elimination ; hence all quadratics containing tw« unknown Quantities cannot be solved by the rules for quadratics. CONTAINING TWO UNKNOWN QUANTITIES. 221 CASE I. 296. When one of the equations is simple and th« other quadratic. 397. Equations of this case can generally be solved by sui stituting in the quadratic equation an expression for the valu of one unknown quantity found from the simple equation. I. Given | J;J;^, = 13 } ' *° ^^ ^ ^^'i 2/- » Solution. 3a; +3/ =9, (1) a;2+3/»=13. (2) From(l), 2/ = 9-3a;; (3] squaring (3), ^^ = 81 - 542: + 9a;^ (4) substituting in (2), k' + 81 - 54a; + 9a;« = 13 ; (5) reducing (5), a;^ - Va; = - V ; completing tlie square, x' — ^^x+ (H)' = 'Ars i ' extracting sq. root, *"~H=='=A; * whence, a; = 2, or 3|, and ^ = 3, or — 1^. EXAMPIiES. Find the values of x and y in the following equations : 2.Given{^^='\l. Am. |=«=J-J |.a;+2/ = 8j (.3/ = 3, or5. f a; — y =3 I < f a;-5, 1 ^'-3,^ = 21 r ^'"- U = 2. 3. Given i. Given or 2, or 4. 5. Given I ^^=' |. Ar^. \''=t'"' ~l\' "^ l4a;-32/ = 10j (2^ = 2,or-5J, or +41, or -71. IP' 222 QUADRATIC EQUATIONS &. Given 9 Giren 10. Given or J X +y =6 -4ns, '}■ >=±7 or±5i/(-l) 2/- ±5, ori7i/(-l), (S^ = 2, or4 CASE 11. - 398. When each equation is homogeneons and quadratic. 309. Equations in this case are usually most conveniently solved by substituting for one unknown quantity the product of the other by a third unknown. ^- ^'"^^'^ { it v=24 } ' *" fi°*^ ^ ^°^ y- Let Bub^ituling in (1), substituting in (2), from (4), from (5), equating (6) and (7), clearing of fractions, etc., solving (9), substituting in (6), and wlienc«, and Solution. x^+xy =10, ay+2y' = 24 . y = vx, x^+vx'' = 10, x' = - — , 1+v 24 X' 10 24 1+v v+2v' 10v'-'iv = 12, t) = f, or- 1+1 .., 10 = 4, 50; (1) (2) (3) (4) (5) (6) CO (8) (9) (10) 1-^ a;= ±2, or ±5v'2, 1/=^ ±3, or ±4j/2. CONTAINING TWO UNKNOWN QUANTITIES. 223 ^■«'-{:;ir.2:::}-^~{;::::;i:;;:;:i (a;»+ 2/» = 13j (3/= ±3, or =p J/i _ „. (a:^-^' = 3 ) . ra;=±2, or ±V5, 7- ^^^^^ U-2.^ W = 2 }• ^'^- L= ±1; or ^11/5: CASE III. 300. Wben eacb equation is symmetrical. 301. There is no general method for the solution of equa- tions in this case. The various expedients employed depend upon the powers of binomials and principles of factoring. 1 . Given \ , „ r > to 6nd x and y- Solution. xy=2^, (I) x+y=lQ. (2) Squaring (2), x" + ixy +y^ = 100, (3) multiplying (1) by (4), 4xy =-96 , (4) subtracting (4) from (3), x^--2xy+y'' = ^, (5) extracting square root, x—y=^% (6) uniting (6) and (2), x=&, or 4, aud 2/ = 4, or6. 224 QUADRATIC EQUATIONS 2. Given \ ^^^ . ,„ L to find x and y. Solution. x''y''—ixy=\% Completing the square, extracting square root, whence, Bquaripg (1), subtracting 4 times (5), extracting square root, whence, and x'y^ — ixy+4: = xy-l-- 16, ±4; (1) (2) (3) (4) (5) 3. Givenl",^''' xy = Q, or — 2, x''-'2.xy+y* = l, or 33, 2;-2/=±l, or ±v'33; a; = 3, or2, or ^5 =1= i/33), 2/ = 2, or3, or i(5±i/33). 3 [• , to find X and y. Solution. x+y='I, Dividing (2) by (1), squaring (1), subtracting (3) from (4), dividing by 3, subtracting (6) from (3), extracting square root, whence, and x'^ — xy'^■y'^=\Z, a:' + 2a;y+y' = 49, 3a;^ = 36, a;^ = 12, a;2-2xy+2/' = l, x-y= ±1; a;=4, or 3, 2/ = 3, or4. (1) (2). (3) (4) (5) (6) Notes. — 1. Let the pupils see that the values of x and y in these equa- tions are not equal to each other; for when x = A,y = ^; and when x=Z, 2/ = 4. Their values are interchangeable. 2. The signs i and =f are equivalent when used independently ; but when taken in connection they are the reverse of each other. Thus, il r= =ta and y== =]=6, then when x=+a, y— —b; and when x=- a, y-^b. lx-y=l i 5. Givenj^^fr'' !• i:xAMFL.i:s. Am. 1"=!'°^-!' iy=4,0T-5. Am. \ = 4, orl, = 1, or 4. CONTAINING TWO UNKNOWN QUANTITIES. 225 6. Given ^ x» ^f 4a: .A. Am. \ ""^t' °' ^V ^•<^^^«M?+2/')-C(:.+^) = 16}- U = 3,5, , ,or-l=F/(-14). 8.Given!t/(7);^ = n. ^n«. j^=J 9. Given {7^;^, I. Jin.. {"=''•"•-'' (a:'— y' = 162J ^ (y = 4, or— 6. 10. Given j^-^'=;«l. ^-{'1' (.a;^-a3/» = 6j ' {y = % = 3, or -2, or- 3. a;^ x+y=6 ■n i-i- ) — +— =9 f -( fa;=4, or2, 11. Given { y X )■• Am. ^ '■ \7X or 4. 303. Equations which are not symmetrical may sometime* be so combined as to produce a symmetrical equation. 303. Equations that are not symmetrical with respect to the unknown quantities themselves may be symmetrical with respect to some multiple or power of these quantities. 1. Given \ , „„ [^ > to find x and u. (3^'+a^=36j ' Solution. ^-^xy=^; (1) y''+xy=^&. (2) Adding, x' + Ixy +^^ = 81 ; (3) evolving, a;+2/=±9; ^ (4) subtracting (2) from (1), a;»-2/'=9; (5) dividing (5) by (4), x-y=:^\; (6) -adding (4) and (6), 2a;=±10; (7) whence a:=±5; (8) subtracting (6) from (4), 22/= ±8; (9) whence 2/= ±4.^ (10) 226 QUADEATIC EQUATIONS, 2. Gi '''^° { ^i V -^109 } • *° ^''^ "^ ^""^ »• SOLUTIOH. a;4 2y=13 (Ij a;'+4y2=109 (2) Squaring (1), a;''+4a;3/ + 42/= = 169; (3) subtracting (2) from (3), 4xy = 60 ; (4) subtracting (4) from (2), x^ — 4xy + 4y^ = i9 ; (5) evolving, x — 2y= ^7; " v6) adding (1) and (6), 2a; = 20, or 6;' (7) ■ wlignce , x = 10, or3; (8) subtracting (6) from (1), 43/ = 6, or 20; (9) wlience 2/ = l)Or5. (10) Note.— In Example 1, the sum of the two equations gives a symmet rical equation. Example 2 is symmetrical with respect to x and 2y. eixampi.e:s. 3. Given \xy+y^8 I ■ (2/ = 4, or2, (,a;' + V = 45j (3^-1, or -2. xy=ab a b 5. Given -J »+3^^2 [■• ^"«- j^^Zj' 6.Givenf^'+^^=^° I. Ar^A^^'^^' {xy-x-yi) (2/ = 2, or4. 7.Giveni«,\ I- ^m{^=^'°\°' i £!+i^ = i I U = 0, or6. 8.Given{^^"^:=;;« ^ ^l- ^n«. } ^=«' °^ " ^' (.a;-2/ + >/(a;-2/) = e j (^ = 5, or- 9. 9. Given } t'T"''. 1 ^m- I "'"\r "'; 10. Given { "f'tK^^ ]■ Ans. \ * = J •"■ -^' l.a:'-^=»19 j (^-2, or -3. PROBLEMS. 227 or |/&, or i/of. 12. Given { '^ ' 2a4«.4 ) . ^,ja. j ^= V«, 13. Given j^^-^i'^n. ^n.. {^^J'"'-;' 14. Given .}^=^«, 1. 4n.. (""J'^'^J 15. Given \^Zl^^y^.^^^y.\ ^^- {"yl^CX 16. Given [^'+<=^^(^+^n. ^n«. }^=f °=-;' 17. Given j^-;r^-;=jn. ^n«. f^^^'^--^' 18. Given Pt^r^M. ^-. j^^JJ or 2. 19. Given JX=^ , J . ^m. f^^^tj or ±4, or ±3. or ±2, or ±3. 20. Given \x^y = h j |a;=4,orl, PRODUCING QUADRATICS WITH TWO UNKNOWN QUANTITIES. 1. The sum of two numbers is 7, and the sum of their squares is 25 ; required the numbers. Arm. 4 and 3. 2. The difference of two numbers is 2, and the difference of their squares is 20 ; required the numbers. Ans. 6 and 4. 3. Divide 97 into two such parts that the sum of the square roots of those parts may equal 13. Ans. 81 and 16. 4. The difference of two numbers is a, and the difference of their square roots is j\/2a; required the numbers. Ans. ^a and §. 5. Find two numbers whose product is 3 times, their sum', and the sum of their squares is 160. Ans. 4 and 1 2. 228 QUADRATIC EQUATIONS. 6. Divide the number 10 into two such parts that the sum of the cubes of the parts may be 280. Ans. 6 and 4. 7. The difference of two numbers is 3, and the difference of their cubes is 117; required the numbers. Ans. 5 and 2. 8. Find two numbers whose product is 6 times thBir difference, and the bum of their squares is 13. A^. 3 and 2. 9. The sum of two numbers is a, and the sum of their cubes is 4a''; required the, numbers. J.?is. f (1 =t i/5) ; f(l=Fy'5). 10. Two men, A and B, can together do a piece of work in 12 days ; in how many days can each do it if it takes B 10 days longer than A ? Ans. A, 20 days ; B, 30 days. 11. A colonel forms his regiment of 1025 men into two squares, one of which has 5 men more in a side than the other; required the number of men in a side of each. Ans. 20 ; 25. 12. A farmer sold 7 calves and 12 sheep for $50 ; and the price received for each was such that 3 more calves were sold for $10 than sheep for $6 ; what was the price of each ? Ans. Calves, $2 ; sheep, $3. 13. Find two numbers such that their difference added to the difference of their squares shall equal 6, and their sum added to the sum of their squares shall equal 18. Ans. 3 and 2. 14. The expense of a sociable was $70, but. before the bill was paid, 4 of the young men sneaked off, in consequence of which each of the others had to pay $2 more than his proper share ; how many young men were there ? Ans. 14. 15. A merchant sold some cloth for $24, and some silk at $1 less a yard for the same sum ; required the number of yards of each, provided there were 2 yards of silk more than of cloth. Am. Cloth, 6 ; silk, 8. 16. A and B run a race ; B, wlio runs slower than A by a mile in 2 hours, starts first by 2 minutes, and they get to the 4-mile stone together • required their rates of running. Ans. A, 8 mi. ; B, 7|- mi. an hour. 17. A certain rectangle contains 300 square feet ; a second rectangle is 8 feet shorter and 10 feet broader, and also con- tains 300 square feet ; find the length . and breadrh of the first rectangle. Ans. Length, 20 ft. ; bres^dth. 15 ft PROBLEMS. 229 18. A bought two pieces of cloth of different sorts ; the finer cost 1 dollar more a yard than the coarser, and there were 10 yards more of the coarser than the finer ; find how many yards there were in each piece, provided the coarser cost $80 and the finer $90. Am. 30 yds. ; 40 yds. 19. The area of a rectangular field is 2275 square rods; anr] if the length of each side is diminished by 5 rods, the area will be 1800 rods ; required the dimensions of the field. Ans. 65 rods ; 35 rods. 20. There is a certain number, of two digits ; the sum of the squares of the digits is equal to the number increased by the product of the digits, and if 36 be added to the number, the digits wUl be reversed ; what is the number ? Am. 4(3. 21. A person bought two cubical stacks of hay for £41, paeh of which cost as many shillings per cubic yard as there weie yards in the side of the other ; and the greater stood on more ground than the less by 9 square yards; what was the price of each? Am. £25 and £16. 22. A laborer dug two trenches for £17 16s., one of which was 6 yards longer than the other, and the digging of each trench cost as many shillings a yard as it was yards in length ; what was the length of each ? Ans. 10 yds. ; 16 yds. 23. Required two numbers such that their sum, their product and the difference of their squares shall be equal to one another. 24. Two partners, A and B, gained $18 by trade : A's money was jn trade 12 months, and he received for his principal and gam $26 ; B's money, which was $30, was in trade 16 months. How much did A put in trade? ^Jia. $20. 25. The fore wheels of a carriage make 5 revolutions more than the hind wheels in going 60 yards ; but if the circumfer- ence of each should be increased one yard, the fore wheels will make only 3 more revolutions than the hind wheels in the same liifetance ; required the circumference of each. Ans. 3 and 4 yards. 20. An English landholder received £7 4s. for a ceitaiu quantity "^ ^^00+ -"ri an o o*^ 1 ^rice less by In. 6d 20 230 QUADRATIC EQUATIONS. per bushel, for a quantity of barley which exceeded the quan- tity of wheat by 16 bushels ; how many bushels were there of each ? Am. 32 bu. wheat ; 48 bu. barley. 27. A and B run a race around a two-mile course. In the first heat B reaches the winning-post 2 minutes before A ; in the second heat A increases his speed 2 miles per hour, and B diminishes his as much, and A then arrives at the winning-post 2 minutes before B. Find at what rate each man ran in the first heat. Ans. 10 mi. per hour ; 12 mi. per hour. PKINCIPLES OF QUADRATIC EQUATIONS. 304. The Peinciples of Quadratics are the relations which exist between a quadratic and its roots. " Note. — This subject may be omitted by young pupils, and even bv older pupils until review, if the teacher prefers. PRIWCIPILB I. Every quadratic equation has two roots, and only two. F'lEST. The general form of the operation. complete quadratic is a;'' +2pa; = g'. x''+2r>x = o (1) Completing the square of the gen- rci+2px+p^ = Q+p^ (2) eral quadratic, and finding the l" f 11' value of X, we have two values, x=-p + y q+p' (3) -p + -/q+p^ani -p~y'^+^, x^^-p-y/q+p^ (4) which proves the principle. Second. This proposition can also be demonstrated in another way, m follows : Assume that m^ = q +p^, oim = i/q+p^ then we have x^+2px +p^ = rri? ; or, in another form, . {x +2>)^ = w? ; transposing, {x-\-pY—m^ = ^; factoring, {x->rp-\-m){x-\-p—m) = ' = 5' Let p^ = q x^+2px+p' = x=—p+0=-p {x+py = x= —p — (>= —p , {x+p){x+p)=Q x= —p x= —p Making the same substitution in the .equation and reducing, we have {^+p){x+p) = 0; dividing by the first factor, we have a;+p = 0, oi x= —p; dividing by the second factor, we have x+p = 0, or x= —p. In the first and second forms the results will be different. Fourth. — Suppose q to be greater than p' when q is negative. If in the third or fourth form we assume q operation. numerically greater than p', the quantity x^+2px^—q Under the radical becomes a negative quan- x= —p^-\/p'' — q tity, and the value of x is therefore imaginary. Hence, the root of an equation in the third and fourth forms is imaginary when q is numerically greater than p'. EXA9IFI.EIS. 1. Find a number such that its square Jncreased by four times the number equals zero. Aks. — 4. 2. Eequired the number such that its square, plus 6 times that number, shall equal minus 9. Aws. — 3. 3. Div'de 8 into two sucli parts that their product shall be equal to 20. Am. 4 ± 2t/^1. Why does the root in the last equation become imaginary? Which supposition does Ex. 2 Ulustrate? What does Ex. 1 illustrate? 236 QUADRATIC fiQUATlONS. IMAGINARY ROOT. 300. An Imaginary Root of a quadratic is a root which contains an imaginary quantity. 310. The Imaginary Root occurs in the third and fourtk forms of a quadratic upon a certain supposition. 311. We shall now discuss the imaginary root under three distinct heads : First. — AVhen does a quadratic give an imaginary root ? Prin. 1. A quadratic gives an imaginary root when the known term is negative and numerically greater than the square of half the coefficient oj the first power of x. For, if q ia negative and numerically operation. greater than p'', the quantity under the radi- x' + Ipx = -q cal is negative, and we shall have the square a;= —p ± Vip' — q) root of a negative quantity, which is imagin- ary. Therefore, etc. ^ Second. — What is assumed by a quadratic which gives an imaginary root ? Prin. 2. A quadratic which gives an imaginary root assumes that the product of two quantities is greater than the square of lialf their sum. For, since 2p is the sum of the two roots operation. with its sign changed, p^ is the square of half %'■ + Ipx = —q the sum of two quantities, and q is the prod- a; = — p ± ^Z {p' - q) uct of the two roots, with its sign changed ; hence, when q is negative and greater than p', the quadratic assumes that the product of two quantities is greater than the square of half their sum. Third. — Prove tKat this assumption is fe.lse. Prin. 3. — The product of two quantities can never be greater than the eq^'iart of half their sum ; hence, the above assumption is fake. OPERATION. 2p = (p+z) + (p- -2), Product ;. {p+z){p-z)=p^-z'; Sum, {p + z) + (p-z) = 1p; « /SumV I2 j' &)'-'■■ Now, p^>p^-z^; hence, P"T> Product. V 2 / IMAGINARY EOOT. 237 Let 2p rtpiesent any num- ber, and let it be divided into two parts, p + z and p-z; the product of the two parts is p^ — z^; the si.m of the parts is 2p, and the square of half their sum is p^. Now, p* is greater than p^—z'; hence the product of two numbers can i^ever be greater than the square of half their sum From the above discussion we see that a quadratic of the form x'zk2px= —9, in which q is greater than p'',- assumes that the product of two quantities is greater than the square of half their sum, which is absurd. When a problem furnishes such an equation, the problem is impossible. EXAMPIillS. 1. Divide the number 12 into two parts such that their product shall be 40. Ans. 6 ^ 2y^^^. 2. A farmer thought to enclose 40 square rods in rectangular form by a fence whose entire length shall be 20 rods ; required its length and breadth. Ans. 5 ± y' — 15; 5 =f t/ — 15. Why do these problems give an imaginary result ? What is incorrect in the first? What is incorrect in the second? REVIEW QUESTIONS. Define a Quadratic Equation. State the two classes of Quadratics. De- fine a Pure Quadratic. An Affected Quadratic. Give examples of each. How do we solve a pure quadratic ? How solve an affected quadratic T State each method of completing the square. Explain each method. Define a Quadratic of two Unknown Quantities. A Homogeneous Equa- tion. A Symmetrical Equation. What ca-ses of quadratics of t*o un- known quantities can be solved? Give examples. Show the method of solution. Define principles of Quadratics. State the principles of Pure Quad- ratics. State the fiist four principles of Incomplete Quadratics. State thr four forms of a quadratic. State the principles of the forms. Define at Imaginary Root. State the prinpiplgs of an Imaginary Root SEOTIOIT YllJ. RATIO AND PROPORTION. EATIO. 31 S. Ratio is the measure of the relation of two siiuilai quantities. Thus, the ratio "of 8 to 4 is 2. 313. The Symbol of ratio is the colon, : , read to, or is to. Thus, a : c indicates the ratio of a to c. 314. The Terms of a ratio are the two quantities com- pared. The first term is the Antecedent ; the second term is the Consequent. The two terms together are called a Couplet. 315. A ratio is expressed by writing the two quantities with the symbol between them (Art. 27), or by writing the conas quent under the antecedent in the form of a fraction. Thus, the ratio of a to c is a : e, or -• 316. A Simple Ratio is the ratio of two quantities. A Compound Ratio is the product of two or more simple ratios. Thus, (a : J)(c : rf), or r^^' 317. A Compound Ratio is usually expressed by writifig the simple ratios one under another. Thus, \ „'. J \ expresses the ratio compounded of a : 6 and e : d. 318. A Duplicate Ratio of two quantities is the ratio of their squares ; as, a? : (?. A Triplicate Ratio of two quanti- ties is the ratio of their cubes ; as, d : &. 319. A Ratio of Equality exists when the two terms are equal. When the antecedent is the greater, it is called a ratio of greater inequality when less, a ra.tio of less ineqiiality, 23S RATIO. 239 Notes. — 1. The symbol oi ratio, : , is supposed to be a modification of the symbol of division. 2. Batio is usually defined as the relation of two numbers. This is indefinite, however, for the ratio is the measure of the relation. 3. A few authors divide the second term by the first, calling it tbe French Method. This is wrong in method and name, as nearly all the French mathematicians, like the German, English, etc., divide the first ta m by the second. PRI1VCIPI.EK. 1. The ratio equals the quotient of the antecedent divided by the cm-sequent. OPEBATIOII. Thus, if r represents the ratio of a to e, we have r = a:o r=a:c, or r equals a divided by c (Art. 315). There- _a fore, etc. c 2. The antecedent is equal to the product of the consequent and ratio. OPEKATtON. For, if r = a divided by c, clearing of fractions, we have _ a a = r.c. Therefore, etc. e 3. The consequent is equal to the quotient of the antecedent divided by the ratio. OPEBATION. a For, if r =- a divided by a, clearing of fractions, we have c r.c = a; and dividing by r, we have c — a divided by r. r.c=a Therefore, etc. • a r 4. Multiplying the antecedent or dividing the consequent multi- plies the ratio. For, a ratio is expressed by a fraction whose numerator is the antece- dent and denominator the consequent ; and multiplying the numerator or dividing the denominator multiplies the fraction (Art. 129, Prin. 1), Therefore, etc. 5. Dividing the antecedent, or multiplying the consequent, divides the ratio. For, a ratio is expressed by a fraction whose numerator is the antecf- lent and denominator the consequent; and dividing thp numerator OJ 240 RATIO AND PROPORTION. multiplying the denominator divides the fraction (Art. 129; Prin. Therefore, etc. 6. Multiplying or dividing both terms of a ratio by any num does not change the ratio. For, a ratio is expressed by a fraction whose numerator is the anti dent and denominator the consequent ; and multiplying or dividing b terms of a fraction does not change its value (Art. 129, Prin. 3). Th( fore, etc. Note. — These principles are restricted to simple ratio. Similar prii pies may be proved of compound ratio. CASE I. 3S0. Problems whicb arise in simple ratio. 1. Find the ratio of ^a^ to 2fl5. Ans. 2a 2. Find the ratio of 3 bufehels to 2 pecks. Ans. 6 3. Find the ratio of a' — x' to a+x. Ans. a — x 4. The ratio is 2a and the consequent 3a& ; required the an -cedent. - Atis. Wb 6. The antecedent is 6aV and ratio 2a' ; required the cor quent. Ans. Sac" a c' 6. The ratio is - and the antecedent is - ; required the o sequent. Ans. —r. or 7. If the ratio of a to 6 is -j^, what is the ratio of 5a to 4^ Ans. I 8. If the ratio of 3a to 26 is f , what is the ratio of a to 6 ! Ans. f 9. If the ratio of 2m to 5n is f , what is the ratio of 5m to 5 Ans. 5 10. If the ratio of a to c is f, what is the-ratio of a+e a-c? Ans. -9 11. The ratio of two numbers is a+b, and the consequent •i — b; required the antecedent. Ans. a" - 6' RATIO. 241 CASE II. 3S1. Problems wliicli arise in compound ratio. 1. Find lie ratio compounded of 8 : 15 and 21 : 24. Solution. The ratio of 8 to 15 is ^ ; the ratio of operation. 21 to 24 is fj; compounding them by taking their '■=-TV''li'°A product, we have A x fj=^. 2. Find the ratio compounded ofa:b and b' : Sax. Ans. - — ox 3. Eequired the value of — \8:15J' \e:di Xe'tdi W bd' cd 4. Given the compound ratio of a; : 8 and 6 : 9 equals -j^, to find the first antecedent. Ans. a; = &§•. f 9 • 12 ") 5. Given the compound ratio I '..^>=\,to find the second antecedent. Ans. 6. 6. Given the compound ratio ] „■< .' f =7,to find the second consequent. , Ans. 3^. 7. The ratio of 2 : 3^ equals the compo'ind ratio \ -.o\ [ > required the second consequent. Ans. 32. 8. The duplicate of the ratio x : 1 equals the ratio 27 : a; ; required the value of x. Ans. 3. 9. The ratio a-x:b-x is the duplicate of the ratio a:b; required the value of x. Ans. ■ -• a + b 10. The duplicate ratio of a; : a equals the compound ratio \ u. \'> i^equired the value of x. Ana. x - 6'. 21 242 KATIO AND PROPORTION. PROPORTION. 333. A Proportion is an expression of equality betwec e^jual ratios. Thus, a formal comparison of the equal rati 8 to 4 and 12 to 6, as 8 : 4 = 12 : 6, is a proportion. 333. The Symbol of proportion is the double colon, : Thus, a : b :: e : d ia read the ratio of a to 6 equals the rat of G to d; or, a is to 6 as c is to d. 334:. The Terms of a proportion are the four quantiti (compared. The first and fourth terms are the extremes, and t] second and third are the means. 333. The OoupletS are the two ratios compared. The fii couplet consists of the first and second terms ; the second coupi _ consists of the third and fourth terms. 33®. A Mean Proportional of two quantities is a quantr which may be made the means of a proportion in which tl two quantities are the extremes ; as, a : b : : b : c. 337. A Continued Proportion is one in which each cons quent is the same as the next antecedent; as, a : 6 : : 6 : e : : c : 338. Quantities are in proportion by Alternation when ant cedent is compared with antecedent and consequent with cons quent. Thus, if a : b : : c : d, hj alternation, a : c :: b : d. 339. Quantities are in proportion by Inversion when tl antecedents are made consequents and the consequents antec dents. , Thus, ii a : b : : e : d, hy inversion, b : a : : d : e. 330. Quantities are in proportion by Composition when tl sum of antecedent and consequent is compared with either ant cedent or consequent. Thus, if a : b : : c : d, hj compositio a : a+6 : ; c : e + d. 331. Quantities are in proportion by Division when tl difference of antecedent and consequent is compared with anf cedent or consequent. Thus, if a : b : : c : d, by divisio a : a — b : : c : e -d. Note. — Eatio arises from the comparLson of two qucailUies; proportii from the comparison of two ratios. A proportion is therefore a compai son of the results^ of two previous comparisons. S'iMPLE PROPORTION. 243 SIMPLE PROPORTION. 333. A Simple Proportion is an expression of equality between simple ratios ; as, a : b :: o : d. 333. A Proportion may be written in the form of an equa- Hun. Thus, a : b :: c: d becomes - = -. d 334. This Equation is called the fundamental equation of the proportion. It lies at the basis of the principles of pro- portion. 333. The Principles of simple proportion are expressed in the following theorems : THEOEEM I. Tn every proportion the product of the extremes is equal to the produet of the means. Let a:b::c:d; .then (Art. 333), ? = ^; o a clearing of fractions ad = bo. Therefore, etc. THEOEEM II. Either extreme is equal to the product of the means divided by Oie other extreme. Let a: b :: : d; then (Theo. I.), ad=-bc; , be , , bo hence, a — — ; and a = a a Therefore, etc. COE. — Either mean equals the produet of the extremes divided 6y the other mean. 244 - EATIO AND PKOPOETION. THEOREM III. If the product of two quantities equah the product of two oL quantities, two of them may be made the extremes, and the ot two the means, of a proportion. Let ad=bc; a e dividing by bd, ( Therefore, etc, 6 d' or (Art. 333), a -.b :: c: d. THEOEEM IV. A mean proportional between two quantities equals the sqtu root of their product. Let a : 6 : : 6 : c ; then (Theo. I.), b'' = ac, and 6 =l/ac7 Therefore, etc, THEOEEM V. If fofjtr quantities are in proportion, they will be in proporti by ALTEENATION. Let a: b :: c: d; then, ad = bc; dividing by dc, — =— ; e d whence, a: c •.:b : d, Thereforef, etc. THEOEEM VI. If four quantities are in proportion, they will be in prop'orti by INVEESION. Let a : b :: c: d; then (Theo. I.), bo = ad; dividing by ac, — "= — ; a c whence, b : a :: d: c Therefore, etc. SIMPLE PROPORTION. 245 THEOEEM VII. Ij Jour quantities are in proportion, they will he in proportion by COMPOSITION. Let . a : b :: : d; then will a+b : 6 : : o+d : d. For, b~d' adding 1 to each aide, T'^^^Zj'^^' a+b c+d b d ' whence, a+b : b :: c + d: d. reducing, ■V Therefore, etc. THEOREM VIII. If four quantities are in proportion, they will be in proportion by DIVISION. Let a: b :: c: d; then will a -b:b::e-d: :d. For, " a _ b'd' subtracting 1, 2_1 = «_1. b d ' reducing, a—b G—d b d ' whence, a- -b:b::o-d: d. Therefore, etc, THEOREM IX. If four quantities are iri proportion, like powers or roots of Omse quantities will be proportional. Let a:b::c:d; then a e_ raising to nth power. a" c" _ hence, a": &» :: c": d»; similarly, Therefore, etc. lilt o" : 6" : : en : d». 246 RATIO AND PEOPOETION, THEOREM X. Equimultiples of two quantities are proportional to the qua ties themselves. Let a and 6 be any two quantities. t-p ,,. , . , ma a mu.'iplymg by m, — - = - ; mo whence, ma : mb :: a : b. Therefore, etc. THEOREM XI. If four quantities are in proportion, any equimultiples of first couplet will he proportional to any equimultiples of the sea couplet. Let a:b::c:d; then T = ";' b d ma no and "~r = ~3i mb nd whence, ma : rrib :: nc: nd. Therefore, etc. THEOREM XIL If two proportions have a couplet in each tht same, the ot eoupleU win form a proportion. Let and then, hence, or, Therefore, etc. a :6i :: c: d, a :6; :: e : ■ f; a b~ c , a d b ', e d e : id: : e: /• SIMPLE PEOPORTICN. 247 THEOBEM XIII. The uroduds of the corresponding terms of two proportions art 'proportional. Let a:b :: o: d, and m:n::p:q\ then, a h" d' and m p n-q' multiplying, am cp _ bn dq' whence. am -.bn-.-.cp: dq. Therefore, etc. THEOEEM XIV. If any number of quantities are in proportion, any antectifd wi'l he to its consequent as the sum of all the antecedents is to the sum of all the consequents. Let a:b :: g: d:: e -.f, etc.; then will a ib :: a+o+e:b + d-\-f. For Theo. I., ad = bc, and . af =be; also, ab = ba. Adding, ab+ad+af=ba+bc+be, factoring, a{b +d+f) = 6(a+ e + e) ; whence, a : b :•. a+c + e : b + d-¥f. Therefore, etc. ADDITIONAL THEOEEMS. 336. These theorems will afford pupils an opportunity to exercise original thought in applying the principles of propor- tion. Part of them may be omitted until review. 1. 1{ a:b::c:d, prove that am : bn : : cm : dn. 2. n a:b::e:d, prove that a^ : b'' : : ac : bd. 248 BATIO AND PROPORTION. 3. Jf a -.h-.-.c id, prove that a? : TD PROPORTION. 249 Bq'iares is to the square of their sum as 13 to 25 ; what are the numbera ? Solution. Let X and jr represent the numbers. Then, xy =24, (1) and x'+y^ : x'+2xy+y' : : 13 : 25. . By division, Theo. VHI., 2xy : {x+yY : : 12 : 25, substituting for 2xy, 48 : (x+y)^ : : 12 : 25, dividing antecedents by 12, 4 : (x+yY :: 1 : 25 ; extracting the square root, 2 : x+y :: 1 : 5 ; from Theo. I., x+y = 10. From which the values of x and y can readily be found. 3. What is the ratio of 6a iuches to b .yards ? Ans. — . 00 4. Two numbers are in the ratio of 2 to 3, and if 3 be added to. each, the ratio is that of 5 to 7; find the numbers. Ans. 12 and 18. 5. Two numbers are ia the ratio of 4 to 5, and if 6 be taken , from each, the ratio is that of 3 to 4 ; find the numbers. Ans. 24 aind 30. 6. Two numbers are in the ratio of 3 to 5, and if 2 be takei, from the less and 5 be added to the greater, the ratio is that of 2:5; find the numbers. Ans. 12 and 20. 7. Find the number which added to each term of the ratio 5 : 3 makes it f of what it would have been if the same number had been taken from each term. Ans. 1. 8. Find two numbers in the ratio of 2 to 3, such that their difference bears the same relation to the difference of their squares as 1 to 25. Ans. 10 and 15. 9. Find two numbers in the ratio of 3 to 4, such that their sum has to the sum of their squares the ratio of 7 to 50. Ans. 6 and 8. 10. Find two numbers in the ratio of 5 to 6, such that their sum has to thp difference of their squares tiie ratio of 1 to 7. Ans. 35 and 42. 250 RATIO AND PROPORTION. 11. The sum of two numbers is 10, and the sum of their squares is to the difference of their squares as 13 to 5 ; required the numbers. Aiis. 6 and 4. - 12. The difference of two numbers is 6, and their product ia to the sum of their squares as 2 to 5 ; what are the numbers ? Am. 12 and 6. 13. Two numbers are to each other as 3 to 2, and if 6 be added to the greater and subtracted from the less, the results will be as 3 to 1 ; what are the numbers? Ans. 24 and 16. 14. The product of two numbers is 12, and the difference of their cubes is to the sum of their cubes as 13 to 14 ; required the numbers. Ans. 6 and 2. 15. There are three numbers in continued proportion : the middle number is 60, and the sum of the others is 125 ; what are the numbers ? Ans. 45 ; 60 ; 80. 16. A quantity of milk is increased by water in the ratio of 7 : 6, and then 8 gallons are sold ; the remainder, when mixed with 8 gallons of water, is increased in the ratio of 7 to 5 ; how much milk was there at first ? Ans. 24 gallons. REVIEW QUESTIONS. 1. Define Eatio. The Terms. A Simple Ratio. A Compound Ratia A Duplicate Ratio. A Triplicate Eatio. A Ratio of Equality. Of In- iquality. State the Principles of Ratio. 2. Define a Proportion. The Terms of a Proportion; Extremes; Means; Couplets. A Mean Proportional. A Continued Proportion. Proportion by Alternation. By Inversion. By Composition. By Division. State the Funlamental Equation of Proportion. Enunsiate Ihe Theorems. SEOTIOI^ IX. PROGRESSIONS. 338. A Progression is a series of quantities in which the terms vary according to some fixed law. 339. The Terms of a progression are the quantities of which it is composed. 340. The Extremes of a progression are the first and last terms ; the Means are the terms between the extremes. NoTE.-^The general term for Progression is series. There are many different kinds of series ; the only two appropriate for an elementary algebra are arithmetical and geometrical progression. ARITHMETICAL PEOGRESSION. 341. An Arithmetical Progression is a series of quan- tities which vary by a common difierence. 34S. The Common Difference is the quantity which, added to any term, will give the following term : thus, in 1, 3, 5, 7, the common difference is 2. 343. An Ascending Progression is one in which the quantities increase from left to right; as, 1, 3, 5, 7, 9, etc. 344. A Descen,ding Progression is one in which the terms decrease frpm left to right ; as, 12, 10, 8, 6, etc. 343. The Terins considered in Arithmetical Progression are five, any three of which being given the other two may be found. THE FIVE TERMS. 1. The first term, a ; 3. The number of terms, n ; 2. The last term, I; 4. The common difierence, d; 5. The sum of the terms, S. 261 252 PEOGBESSIONS. * 346. Prin. — The common difference is positive in an ascendr ing series, and 'legative in a descending series. CASE I. 347. Given the first term, the common ditferk ence and the number of terms, to find the last term. 1. Given a tte first term, d the common difference, n the number of terms, to find the expression for I, the last term. SoLTTTlON. The 1st term is a; the 2d term iaa + d; the 3d term is a+d plus d, operation. or a-F2d; the 4th. term is a+3d, and so 1st term = a on. By examining these terms, we see that 2d term = a+d any term equals a, pliis the product of d taken 3d term = a+2a as many times as the number of terms less one; 4th term = a + 3a hence the reth term equals a+(n-\]d; .•. wth term = a+(n-l)rf or, representing the nth term by I, we have or, l = a+{n—\)d l = a+{n—l)d. Rule. — To the first term add the product of the common diffes/"- ence multiplied by the number of terms less one. Note. — An ascending series of n terms may he written as follows : a, a+d, a+2d, a+3d .... a+(n—l)d. A descending series of n terms may be written as follows : a, a—d, a— 2d, a—5d .... a— (n— l)d!. EXAMPIiES. 2. Find the 12th term of the series 2, 5, 8, 11, etc. SoLtTTiON. In this problem a = 2, d = 3 and w = 12. operation. The formula forthe lasttermis Z = a+(n — l)d; substi- l'^a+(n—l)Cl tutingthe values of a, d and n, we have Z = 2 + (12 — 1)3; ^=-^2+ (12 — 1)3 and reducing, we have 1 = 35. i ^ 2 + 33 = 35 Note. — The prob'«m may also be solved by the rule instead of substi tating in the formula. 3. Find the 18th term of the series 1, 4, 7, 10, etc. Av~ -52. 4. Find the 17th teTm of the series 3, 7, 11, 15, etc. ' Am 67. AEITHMETIOAL PROGRESSION. 253 5. rind the 20th term of the series 1, 2|, 3f , 5, etc. • Am. 26^, 6. Find the 14th term of the series 29, 27, 25, 23, etc. Am. 3. 7. Find the 40th term of the series 1, 2|, 4|, 6, etc. Aris. 66. 8. Find the 15th term of the series |, A|, f, i|, etc. * Am. 0. 9. Find the 30th term of the series a, 3a, 5a, 7a, etc. Am. 59a. 10. Find the nth term of the series 2, 4, 6, 8, etc. Am. 2n. 11. Find the nth term of the series 26, 46, 66, etc. Am. 2bn. 12. Find the nth term of the series 1, 3, 5, 7, etc. Am. In - 1. 13. Find the nth term of the series, 2, 2|-, 2f, etc. Ans. ^(n+5), 14. If a body fall 16^ feet the 1st second, 3 times as fat the 2d second, 5 times as far the 3d second, and so on, how far will it fall the 20th second? Am. 627^ ft. 15. If a body fall n feet the 1st second, 3n feet the 2d, 5n feet the 3d, and so on, how far will it fall the f th second ? Am. (2i-l)n. CASE II. 348. GiTen the first term, tbe last term, and the number of terms, to find the sum of tlie terms. 1. Given a, the first term ; I, the last term ; and n, the numbei of tei-ms, to find the expression for 8, the sum of the terms. Solution. We have the series, 5'=a + (a + d) + (a+2d) + («+3d) . . . +Z. inverting the series, 8= l+{l- d) + {l~ 2d)+ (i!-3ci) . . . +a. Adding the series, 2,S'=(a+;) + (a+0 + (a+0 + («+0+ ...+(a+i!) That is, (a+Z) taken as many times as there are terms, or n times. Hence, 2>S={a+Z)»»; whence, 5=/-— W or 9-(a+0. 254 PROGRESSIONS. Rule. — Multiply (lie sum of the extremes by one-half of the number of terms. 2. Find the sum of the arithmetical series whose first term is 2, last term 35, and number of terms 12. « OPERATION. Solution. Inthiaproblem, a = 2, n = 12, and S=^{a+l) i = 35. Substituting the values of these terms in gf= '^^(2 + 35) the formula, S=^(a + l), we have S= y (2 + 35), o^ 6 x 37 = 222 Am. or 6 X §7, which is 222. Find the sum — 3. When a = 3, i = 40 and w = 16. Ans. 344. 4. Of 12 terms of the series 2 + 6 + 10 + 14, etc. Ans. 288. 5. Of 16 terms of the series 3 + 7 + 11 + 15, etc' Ans. 528. 6. Of 12 of the odd numbers 1 + 3 + 5 + 7, etc. Ans. 144. 7. Of 12 of the even numbers 2+4 + 6 + 8, etc. Ans. 156. 8. Of 18 terms of the series ■|^+l-^+2^, etc. An^. 159. 9. Of 25 terms of the series ^+l + l| + 2, etc. Ans. 162^. 10. Of 12 terms of the series 20 + 18 + 16, etc. Ans. 108. 11. Of 17 terms of the series .2 + .25 + .3 + .35, etc. Ans. 10.2. 12. Of n terms of the series 1 + 8 + 5 + 7, etc. Ans. n\ 13. Of w terms of the series 2+4+6 + 8, etc. Ans. n^+n. 14. Of n. terms of a+3a+5a+7a+, etc. Ans. an'. 15. Of 6 terms of (a - 56) + (a - 36) + (a - 6) +, etc. Ans. 6a. 16. If a body fall 16315- feet the 1st second, 3 times as far the yd second, 5 times as far the 3d second, and so on, how far will it fall in 20 seconds ? Ans. 6433| ft. 17. If a body fall n feet the 1st second, 3n feet the 2d, 5m feet the 3d, and so on, how far wiU it fall in t seconds ? Am. t\ ft. ARITHMETICAL PEOQEESSIOK. 255 CASE III. 349. Oiren any three of tlie five quantitie^s to find cither of the others. , 350. The Fundamental Formulas of arithmetical pro- gression are — 1. l^a+(n-l)d; 2. S=i(a+l). These are called Fundamental because by means of them we can solve all Ifce problems which may arise. S51. There are three classes of problems, since the four quantities may all be in the 1st formula, or all in the 2d for- mula, or part in the first and part in the 2d. 352. Class I. When the four quantities are all contained in the first fundamental formula. 1. Find the formula for d, having given a, n and I. SoLTjTiON. The formula l = a+(n—l)d operation. contains all the four quantities; we will 7 = n + (n — l\d (l\ therefore find the value of d from this for- 7 — n = (n—l)d (2) mula in terms of the other quantities. Trans- I — a posing, we have equation (2) ; dividing by the "= _ ^ W coefiicient of d and transposing, we have equation (3), which is the value of d required. 2. Find the value of a, having given n, d and I. 3. Find the value of n, having given a, d and I. 4. The first term is 90, the last term 34, and the common difference 4 ; required the number of terms. Ans. 15. 6, What term of the series 2, 5, 8, etc., is 35? What term of the series 29, 27, 25, etc., is 3 ? Ans. 12th ; 14th. 6. The mth term of a series whose common difference is 2, is 2n ; what" is the first term? Ans. 2. 7. The wth term of a series whose first term is 1, is 2n - 1 , required the first four terms of the series. Ans. 1, 3, 5, 7. 256 PROGRESSIONS. S59. Class II. "When the four quantities are all contaiaed in the second fundamental formula. 1. Given a, I and S, to find n. Solution. The formula S=^a+l) coittains aU operation the four quantities ; we can therefore find the value S=^(a+l) (1) of n from this formula in terms of a, I and S. 2S=n(a+l) (2) Clearing of fractions, we have equation (2) ; dividing 215 bj a + I and transposing, we have equation (3), ** °^ a+ I which is the value of n required. BXAMPIiES. 2. Find a, having given n, I and S. 3. Find I, having given a, n and S. 4. The first term is 2, the last term 35, and the sum of the terms 222 ; required the number of terms. Ans. 12. 5. The last term is 27, the number of terms 12, and the sum of the terms 180 ; required the first term. Ans. 3. 6. Required the last term of the series whose first term is 1, number of terms n, and sum of terms n'. Ans. 2n - 1. 7. Required the first four terms of the series whose first term is 1, number of terms n, and sum n'. Ans. 1, 3, 5, 7. 8. Required the last term of the series whose first term is a, number of terms n, and sum of the terms an\ Ans. a(2n — 1). 354:, Class III. When part of the quantities are in the first and part in the second fundamental formula. 1. Given d, n, I, to find S. Solution. The two fundamental for- opeeation. mulas contain one quantity, namely, a, not l = a+{n—l)d (1) involved in this problem; hence we may S=''(a+l) (2) combine these formulas by eliminating a, ■ ■ and obtain an equation containing the four <^='' — {n—ijCt [o] quantities, d, n, I and . ' ( = ai'''' GEOMETRICAL PKOGEESSION. 263 Rule. — Multiply the first term by the ratio raised to a power whose index is one less than the nvmber of terms. Note.— An ascending aeries of n terms may be written as follows:, a, ar, ar\ ar^ . . . . ar^"^, ar'^-^. EXAMFIiES. 2. Find the 8tli term of the series 2, 4, 8, etc. Ans. 256. 3. The first term is 3 and ratio 4 ; what is the 7th term ? Ans. 12288. 4. The first term is 729, the ratio \ ; required the 12th term. Ans. ^. 5. Find the wth term of the series 1, 2, 4, 8, etc. Ans. 2""^ 6. Find the nth term of the series 2a, 4a^ 8a', etc. Ans. (2a)". 7. Find the nth term of the series 2, 4a, 8a'', etc. Ans. 2"a"-'. 8. If a merchant doubles his capital every 4 years, and be- gins with $4000, how much has he at the end of 20 years ? Ans. 1128000. CASE II. 370. GlTen tlie first term, tbe last term and the number of terms, to find the sum of the terms. 1. Given a, the first term ; I, the last term ; and .n, the number .of terms, to find an expression for S, the sum of the terms. SoiiTJTION. We have S=a + ar+ar^+ar^+ ar"-'; (1) multiplying (1) by r, rS = ar + ar^-\- ar^ + ar"-^+ar«. (2) Subtracting (1) from (2), rS- S=ar"-a; (3) feetoring, 5(r-l) = ar»-a; (4) wlieacf «^= ; — (5) 264 PROGRESSIONS. This may be put in another form by substituting a value for ar\ From Case I. we have I — ar"~^ ; multiplying by r, rl = ar^ ; substituting in (5), S= r-\ Rule. — Multiply the last term by the ratio, siibtrad -the firs ierm, and divide the remainder by the ratio less one. EXAMPLES. Find the sum of the series — 2. When a = 2,1 = 256, and r = 2. Ans. 510. 3. When a = 3,1 = 12288, and r = 4. Ans. 16383. 4. Of 9 terms of the series 2, 4, 8, 16, etc. Ans. 1022. 5. Of 12 terms of the series 1, 2, 4, 8, etc. Ans. 4095. 6. Of 10 terms of the series 1, 3, 9, 27, etc. Ans. 29524. 7. Of n terms of the series 1 + 2+4+8, etc. Am. 2"- 1. 8. Of n terms of the series 1, 3, 9, 27, etc. Ans. ^(3"- 1). 9. Of n terms of the series a+2a+4(i+8a, etc. Ans. a(2»-l). 2" — 1 10. Of m terms of the series l+i+l+i, etc. Ans. ■• 11. Of n terms of the series 1 +^ + ^ + ■^, etc. Ans. 12. Of n terms of the series 1 — -j+i — i-, etc. On-l /' "I 2""' / 13. A laborer agreed to work one year at the rate of $1 fo! January, $2 for February, $4 for March, and so on ; how muct did he receive in the year ? Ans. $4095. 14. A servant-girl saved $160 one year. Now, if it were possible for her to save half as much again every year as the previous year for 8 years, how much would she save ? Ans. $11981.87f INFli^ITE SERIES. 265 INFINITE SEEIES. 371. An Infinite Series is a series in which the number of terms is infinite ; as,l,\,\, ^, etc. 37S. The Sum of a decreasing geometrical series to infinity is the limit toward which the series approaches as the number ")f terms increases. 1. Find the sum of a decreasing geometrical series to infinity. OPERATION. Solution. In a decreasing series, r is less than 1 ; „ rl — a ... hence, for a decreasing series we change formula (1) to r — l ' formula (2), that the denominator may be positive. a — rl ,_. Now, as the number of terms increases, the value of ^ i_^ ^ I I decreases ; hence, when the number of terms is infinite, When ri = I must become infinitely small ; that is, ; hence, ri = 0, and the formula for Ans. 2, 6, 18. 11- The sum of the iirst and third of three numbers in geo- metrical progression is 10, and the sum of the cubes of the first and third is 520 ; required the numbers. ' Ans. 2, 4, 8. 12 There are three numbers in geometrical progression ; the «um of the first and second is 32, and the sum of the second and third is 96 ; what are the numbers ? An^. 8, 24, 72. PROBLEMS. 27 i- 13. The product of three numbers in geometrical progression is 216, and the sum of the squares of the extremes is 153 ; re- quired the numbers. Ans. 3, 6, 12. 14. The sum of three numbers in geometrical progression ia 39, and the sum of the extremes multiplied by the mean is 270 ; what are the numbers ? Ans. 3, 9, 27. 15. There are three numbers in geometrical progression whose Bum is 52, and the sum of their squares is 1456 ; what are the numbers? Ans. 4, 12, 36. 16. It is required to find three numbers in geometrical pro- gression such that the sum of the first and last is 30, and the square of the mean is 144. Ans. 6, 12, 24. 17. Of four numbers in geometrical progression the sum of the first' and third is 20, and the sum of the second and fourth is 60 ; what are the numbers ? Ans. 2, 6, 18, 54. 18. Eequired to find four numbers in geometrical progression such that the sum of the first two is 10, and of the last two is 160. " Am. 2, 8, 32, 128. In the 15th Example divide the 2d equation by the first. In the 18th, let X, xy, xy^, xy^ represent the numbers. REVIEW QUESTIONS. Define Progression. Arithmetical Progression. The Terms. Ex- tismes. Means. Ascending Progression. Descending Progression. How many terms? State the four cases. The rule for each case. The formula for each case. How many ca-ses are possible ? Define Geometrical Progression. What is the value of the ratioin an ascending progression? In a descending progression? State the three cases. Give the rule and formula for Case I. and Case II. Define an Infinite' Series. State the rule for the sum of the terms of an infinite series. Hew many cases are possible? 272 MISCEI-LANEOUS EXAMPLES. MISCELLANEOUS EXAMPLES. 1. Add (ffl + 26)a;" and (2a -fi^". At^. (3a+6)«" 2. Subtract 3(m — an) from 3a(m — n). Ans.- 3m(a — 1), 3. Multiply a"-' by SaT*''. Am. 3a'»+". 4. Multiply a^ - b^ by a* +6*. Am. a^ - 6^. 5. Divide a"6"'-" by a''-'"6~". ^n«. a^i". 6. Divide a' — 2^ by a^ — s'. J.W. a^ + a*z' +2^. 7. Divide a''+6" by a^+bT Am. a^ - a'6J+6». 8. Multiply a' + 2aa; - a^ by a" + 2aa; + a;^ JLns. a*+4a'a;+4aV-a;*. 9. Divide 2a* + 27a6' - 81 6* by a + 3 J. Ans. 2a' - 6a'6 + ISaS'' - 27^>^ 10. Divide a^-2a'+l by a''-2a+l. Am. a*+2a' + 3a'+2a+l. 11. Value of a- {2b-'(3e + 2b-a)\. Ans. 3c. 12. Valueof 16-{5-2a;-[l-C3-a;)]}. Am.9+3x. 13. Valueof 15a;- [4- [3-5a;-(3a;-7)]}. ^ra. 7a;+6. 14. Valueof 2a; -[32/-^ 4a; -(52/ -6a; + 72/) I ]. Am. 12x-15y. 15. Value of a - [5b - {a - (5e - 2e^^6 - 46) + 2a - (a - S+c)!]. Am. 3a -2c. 16. Prove (a-6)' + 6'-a' = 3a6(6-a). 17. Prove (a'+ab + by-(a'-ab + b'y = iabia'+V). 18. Prove (a+J+c)'-(a' + 6' + e') =3(a + 6)(6 + c)(a+c). 19. Expand (a"-2)(a"+2)(a''+3)(a»-3). A?isa*"-13a'" + 36. 20. Factor a* +6*; a^-6'; a? + b^; a*^-b^. 21. Factor »«-6»; a^6«; a'-J«; a'+6'; a^-J**. 22. Factor a' +9a6 + 206'; oTif - b" when a; = 3. .4»w. - 3. a;''-7a; + 12 42. Value of ; when x = a. Am. oo. (a -a;)' Ao -ir 1 „'aa;^ + ac,''- 2aea; , . a 4d. Value of > whena; = c. J.w». -• bx^ - 26ca; + 6c'' 6 ..^. a;+l, 3a;— 1 „ „, . 44. Given — =a;-2, to find x. Ans. x = 2. 3 5 .p. ^. 5a;- 7 2a; + 7 „ ^ , „ , 45. (jiven — ~ r ^ 3a; - 14, to find x. Ans. x-7. .„ „. 3a;- 4 6a;- 5 3a;- 1 ^ „ , , 46. Given — — = — — — , to find a;. Ans. x = 2\. 2 8 16 47. Given 5a; - [8a; - 3{16 - 6a; - (4 - 5a;)}] -=6. Ans. x = 5. .. ^. a; + 3 a;- 2 3a;- 5 1 , „ , 48. Given —- = + 7' to find x. Am. x - 28 ^ o Iz 4 ,„ „. 3 + a; 2 + x 1 + x ^ , „ , . 49. Given = 1, to find x. Ans. x^U 6— X 2— X 1— X en /T a;'-a;+l a;'+a; + l „ „ , 50 Given -— + — = 2a;, to find x. Ans. x- 0. a;-l ai+l MISCELLANEOUS EXAMrLES. 27£ _^ _. a(a-x) b(b+x) ^ « j . j, 51. Given ^ ^ - -^^ = x, to find x. Am. x^a-b. b a 52. Given ?^£^1^ + *^^— ^ = a;, to find x. Ans.x^a+b. b a 1 1 a- b „ -, . 2ab 53. Given ■- > to find x. Ans. x = X— ax— bx^ — ab a + b 54. Given J J_ ^ I- ^'^■{^ = 7. 55- Given J ^^^^^ I. , ^-. j^I^'^ aVo 59. Given J « " I- ^ns. ^ ,j,,. 60. Given { l"""^^,:'?:?;!^' 1 " ^-- '^^^' la{x-y) + b{x+y)^\^ (2/ = 0. 1 61. Given ^ 7x-ny+'2z=9 V- Am. j 3/ = l, x+y+Zz=12 ) U=3, -2/+z=va-) ' . (x = Ub + o-a), 62. Given .^ x+z-^b [■ Ans. ] y = \[a-b + c) x+y=o) (,2=i(a+6-c). ry+z-x = a^ ^x = i{b+c), 63. Given \ x+z-y= b [ . Am. j y = i{a+c), 276 MISCELLANEOUS EXAMPLES. '5+y+i=l a b o abo 64. Given (^+^+|=l ). Am. ] y , ,^, ^„„ ^ a c b I I \ ab+oc+ao \,b a c X y z I h I \^°°"' 65. Giveu n+-~-==l \. Am. ]y = b, \^ y ^ ( (0=a 2a_6_c _ . a; 2/ z 66. Value Vf^)' ^^^^ ^°°* °^ ^\/^- ^'^- *" ^V^^"" 67. Value (aa; + 6i/y + (aa;- fey)'. Ans. 2aV + 6aa;6y. 68. Value (a+6 + c+d)'' - (a-6+c-d)l J.n«. 4(a+c)(6 + d). 69. Square root of a«-12a^+60a*-160a' + 240a'-192a+64. AnB. a'-6a'+12a-8. 70. Fourth root of 16a*-96a'2/+216ay-216a2/' + 8l2/*. Ans. 2a -Sy. 71. Cube root of x'+3r' + Gx*+73?+Qx'+3x+l. Ans. x^+x+1. 72. Sixth root of l + 12a+60a'+160a'+240a*-i-192a^+64a^ Ans. 1 + 2(1. 73. Simplify J - (a^yi-^ x | - ( - a;)-'|*. 74. If x = ^ . and y -=^ > find the value of thj l/3-l * -1/3 + 1 jxpression x'+xy-^'f. Ans. 15. 75. Given i/(a; — o) = > to find x. Ans. x= =>= ay 2. l/Or + a) a; — 4 Ix—X 76. Given = . to find x. Ans. * =• =^1/ (~^)• 2a; + 1 a;+4 77. Given J^^Ill = ]/^_I± , to find x. Ans. a; - 9. i^r + lO T/a; + 23 MISCfiLLANEOtrS EXAMPLES. 277 12- a- 78. Given 4x = 22, to find x. Ans. a; = 6, or 2 J. x—3 79. Given = 5 > to find x. Ans. a; = 11, or 3. 80 Given = — • to find x. Ans. x = 6. or 2i. x-B x-1 15 ' 81. Given = — -. to find x. Ans. a; = 7, or 21. a;-2 x-4 5 " ' 82. Given = — > to find x. Ans. a; = 8, or 2^4- x-4: x-2 12 . ' 11 £,„„. a;+la;-12a;-l . . „ 83. Given + = • Ans. x = 4, or 0. a; + 2 a;-2 a;-l -, ^. .r-2 a;+2 2(a; + 3) . -, , -, 84. Given + = -^^ -• Ans. a; = 11, or 0. a; + 2 a;-2 a;-3 ^ 85. Given ^ZV^^j. Ans. x= ±^^6 a+^d'-x' 1 + * = ±3, ±2. „„ „. a; + a x — a b+x b—x . , /r i^ 89. Given = ; 7 Ans. x=^ V («*)• x-a X + a 0— x b + X 90 Given {r/^i^.^^l} " ^^- {TX-t f 3a;+22/ = 5a;2/| , . (a;i.f,oiO, 91..Giyen|^5^_^_4^^^^- ^'^ |;!/~iorO. 278 MISCELLANEOUS PROBLEMS. 92. Given {"+^7^n. Ans. a + b x = , or(\ crO. 93. Given lab V- Am. \^x'^+y'' = az-^by ) 94 Given {-^+-2/ = 28) ^^_ |,= ±4, or ±1^2, 2/=±3, or ±V2. 95. Given {^^^.f ^='H. ^n.. f^"!'"-^'' 96. Given } -'+^: + 2-+2^ = 23 ) ^^ | . = 3, or 2, {xy = & ) . (.2/ = 2, or3. 97. Given p^-^; = ^-^ = = « = H. Ans. {"=!' 98. Given f^'-C^"-^^'"''^'!- ^-. {^^J 99. Given j-'^+^'-'-^z' = = 210 : 1U| ^^^_ |.=3, 100. Given j^^^^T^/^ , ,,,,}• Am. j^^^J!' MISCELLANEOUS PROBLEMS. 1. A cliild was born in November, and on the tenth day ot December he is as many days old as the month was on the day of his birth ; when was he. born ? ' Ans. Nov. 20th. 2. After A has received $10 from B he has as much money as B and $6 more ; and between them they have $40 ; how much money had each at first ? Ans. A, $13 ; B, $27. 3. A father has 6 sons, each of Vi^hom is 4 years older than his next younger brother, and the eldest is 3 times as old as the youngest ; find their respective ages. Alls. 10 yrs., 14 yrs., 18 yrs., 22 yrs., 26 yrs., 30 yrs. MISCELLANEOUS PROBLEMS. 279 4. Four men club to buy a set of ten-pins, but by cluobing with 2 more the expense of each is diminished $1.76 ; what did the set cost? ^^s. $21. 5. A pudding consists of 2 parts of flour, 3 parts- of rais-ins and 4 parts of suet ; flour costs 3d. a pound, raisins 6d., and suet 8d. ; find the cost of the several ingredients of the pudding when the whole cost is 2s. 4d. Ans. 3d., 9d., Is. 4d. 6. A market-woman being asked what she paid for eggs, replied, "Six dozen eggs cost as many pence as you can buy eggs for eightpence." What was the price a dozen ? Ans. Ad. 7. The tens' digit of a number is less by 2 than the units digit, and if the digits are inverted the new number is to the former as 7 is to 4 ; find the number. Am. 24. 8. In paying two bills, one of which exceeded the other by ^ of the less, the change out of a five-dollar note was ^ the dif- ference of the bills ; find the amount of each bill. Ans. $2; $2|. 9. Two persons, A and B, own together 175 shares in a railway company: they agree to divide, and A takes 85 shares, while B takes 90 shares and pays $250 to A ; find the value of a share. Ans. $100. 10. Find the fraction such that if you quadruple the nume- rator and add 3 to the denominator, the fraction is doubled ; but if you add 2 to the numerator and quadruple the denominator, the fraction is halved. • Ans. |. 11. How many sheep must a person buy at $35 each that, after paying $1.50 a score for folding them at night, he may |ain $394 by selling them at $40 each ? Ans. 80. 12. A colonel, on attempting to draw up his regiment in the form of a solid square, finds tha,t he has 31 men over, and that he would require 24 men more in his regiment in order to mcrease the side of the square by 1 man ; how many men werb vfiere in the regiment ? -A.ns. 760. i3 In a certain weight of gunpowder the saltpetre com- 280 MISCELLANEOUS PROBLEMS. posed 6 pounds more than ^ of the weight, the sulphtir 5 poun less than -I, and the charcoal 3 pounds less than ^ ; how lua; pounds were there of each of the three ingredients ? Ans. 18; 3; 3. 14. A cistern could be filled in 12 miLutes by 2 pipes whi run into it ; and it could be filled in 20 minutes by 1 alon in what time could it be filled by the other alone ? Ans. 30 min. 15. Divide the number 88 into 4 parts, such that the fii increased by 2, the second diminished by 3, the third multipli by 4, and the fourth divided by 5, may all be equal. Am. 10, 15, 3, 60. 16. A and B began to play together with equal sums money : A first won $100, but afterward lost ^ of all he th had, and then his money was i as much as that of B ; wh money had each at first ? Ans. $300. 17. A and B shoot by turns at a target: A puts 7 bulL put of 12 into the bull's-eye, and B puts 9 out of 12 ; betwe them they put in 32 bullets ; how many shots did each fire ? Ans. 24. 18. A person buys a piece of land at $150 an acre, a: by selling it in lots finds the.value increased threefold, so tl he clears $750, and retains 25 acres for himself; how ma: acres were there ? Ans. 40. 19. A and B play at a game, agreeing that the loser shi always pay to the winner $1 less than ^ the money the los has : they commence with equal sums of money, and after has lost the first game and won the second he has $2 kc than A ; how much had each at the commencement ? Ans. $6. 20. It is between 11 and 12 o'clock, and it is observed tl the number of minute-spaces between the hands is | of what was ten minutes previously ; find the time. Ans. 20 min. of 12 o'clock. 21. A clock has two hands turning on the same centre : t «»iftcr makes a revolution every 12 hours, and the n'ow^r eve MISCELLANEOUS PROBLEMS. 28i 16 hours ; in what time will the swifter gain one compleie revo- lution on the slower ? Ana. 48 hrs. 22. An officer can form his men into a hollow square 4 deep, and also into a hollow square 8 deep : the front in the latter formation contains 16 men fewer than in the former for- mation ; find the number of men. Ans. 640. 23. A certain number of 2 digits is equal to 4 times the sum of its digits ; and if 18 be added to the number the digits are reversed ; find the number. Ans. 24. 24. Prove that if 2 numbers difier by a, the diflTerence of their squares equals 2a times their arithmetical mean. 25. If two integers difier by 2, show that the difierence of their squares equals 4 times the integer between them. 26. The two digits which form a number change places on the addition of 9, and the sum of the original and the result- ing number is 33 ; find the digits. Ans. 1 and 2. 27. Which is the greater, and how much — the square of a number, or the product of the number a unit less than it by the number a unit greater than it ? Ans. The square is 1 greater. 28. If a certain rectangular fioor had been .2 feet broader and 3 feet longer, it would have been 64 square feet larger ; but if it had been 3 feet broader and 2 feet longer, it would have been 68 square feet larger ; find the length and breadth. Ans. Length, 14 ft. ; breadth, 10 ft. 29. When a certain number of 2 digits is doubled and in- creased by 36, the result is the same as if the number had been reversed and doubled, and then diminished by 36; also the number itself exceeds 4 times the sum of its digits by 3 ; find the number. ^ns. 59. 30. A sets out from M to N, going 3^ miles an hour ; forty minutes afterward R sets out from N to M, going 4J- miles an DOur, and he goes half a mile beyond the middle point before be meets A ; find the distance between M and N. Ans. 29 mi. 31. A person walked a certain distance at the rate of 3i| 282 MISCELLANEOUS PROBLEMS. miles an hour, and then ran part of the way back at the : of 7 miles an hour, walking the remaining distance in 5 n utes : he was gone 25 minutes ; how far did he run ? Ans. ^ m: 32. Two trains, 92 feet long and 84 feet long respectiv are moving with uniform velocities on parallel rails : when t move in opposite directions they are observed to pass each ot in 1^ seconds, but when they move in the same direction faster train is observed to pass the other in 6 seconds ; find rate at which each train moves. Ans. 30 mi. and 50 mi. per houi 33. If the sum of 2 fractions is unityj show that the fi together with the square of the second, is equal to the seco together with the square of the first. 34. A man has a rectangular field containing 4 acres wh length is to its breadth as 8 : 5 ; required the length and brta of the field. Ans. 32 rods ; 20 rods 35. The sum of two numbers is 17, and the less divided by greater is to the greater divided by the less as 64 : 81 ; what the numbers ? Ans. 8 and S 36. There are two quantities whose product is a and quoti b; required the quantities. Ans. ^\/ab and =*=«/- 37. A father gave to each of his children on New Yei day as many books as he had children : for each book he g 12 times as many cents as there were children, and the cost the whole was $15 ; how many children had he? Ans. 5 38. A man bought a field whose length was to its brea as 3 to 2 ; the price per acre was equal to the number of r m the length of the field, and 4^ times the distance around field equaled the number of dollars that it cost ; what was length and breadth of the field ? Ans. 60 rods ; 40 rods 39. A and B carried 100 eggs to market, and each recei' ihe same sum : if A had carried as many as B, he would hi received 36 cents for them, and if B had carried oidy as ms MISCELLANEOUS PKOBLEMS. 283 as A, L( would have received only 16 cents for them , ho\i many eggs had each ? Ans. A, 40 ; B, 60. 40. Several gentlemen made an excursion, each taking $484 : each had as many servants as there were gentlemen, and the Dumber of dollars which each had was 4 times the number of all the servants ; how many gentlemen were there ? Ans, 11. 41. There is a rectangular field whose breadth is -f of the length. After laying out \ of the wholeT ground for a garden, it was found that there were left 400 square rods for mowing ; required the length and breadth of the field. Ans. 25 rods ; 20 rods. 42. There are two square grass-plats, a side of one of which :g 10 yards longer than a side of the other, and their areas are as 25 to 9 ; what are the lengths of the sides ? Ans. 25 ydf. , 15 yds. 43. A farmer bought a number of sheep for $80 : if he had Hbught 4 more for the same money, he would have paid $1 less tor each; how many did he buy? Ans. 16 sheep. 44. A man divided 110 bushels of coal among a number of poor persons : if each had received 1 bushel more, he would have received as many bushels as there were persons ; find the number of persons. Am. 11. 45. A person bought a certain number of yards of cloth for !{36, which he sold again at $4 per yard, and gained as much on the whole a? 4 yards cost ; find the number of yards. Ans. 12. 46. A cistern can be supplied with water by two pipes, one i)f which would fill it 6 hours sooner than the other, and they both would Pll it in 4 hours ; how long will it take each pipe alone to fill it? Ans. 6 hrs. : 12 hrs. 47. The side of a square is 110 inches long ; find the dimen- sions of a rectangle which shall have its perimeter 4 inches longer than that of the square, and its area 4 square inches \^ss than that of the square. Ans. Length, 126 in. ; breadth, 96 in. 284 MISCELLANEOUS PROBLEMS. 48. The third term of an arithmetical progression is 4 ti the fii-st term, and the sixth term is 17 ; required the first terms of the series. Aiis. 2, 5, 8, et 49. A square tract of land contains | as many acres as tl are rods in the fence enclosing it ; required the length of fence. . Ans. 320 rod 60. An English woman sells eggs at Such a price thai piore in half a crown's worth lowers the price threepence score ; required the price per score. Ans. 15( 51. An English woman sells eggs at such a price that fewer in half a crown's worth raises the price threepence score ; required the price per score. Ans. 12( 52. Find the side of a cube which shall contain 4 timei many solid units as there are linear units in the distance twfedn its two opposite corners. Ans. 1-^\ 53. A grass-plat, 18 yards long and 12 wide, is surroun by a border of flowers of uniform width ; the areas of the gi plat and border are equal ; what is the width of the border ! Ans. 3 yd 54. A and B set out to meet each othei from two places miles apart : A traveled 8 miles a day more than B, and number of days in which they met was equal to half the num of miles B went in a day ; how far did each travel before t met ? Ans. A, 192 mi. ; B, 128 m 55. Two clerks, A and B, sent ventures in a ship bounc India : A gained $120, and at this rate he would have gai as many dollars on a hundred as B sent out. B gained \ which was but one-fourth as much per cent, as A gain how much money was sent out by each ? Ans. A, $100 ; B, |12( 56. In a collection containing 27 coins, each silver coii worth as many cents as there are popper coins ; and each cop "oin is worth as many cents as there are silver coins ; and whole is vorth $1 ; how many coins are there of each sort? Ans. 2 of silver ; 25 of eoppej ^MISCELLANEOUS PEOBLEMS. 285 57. The third term of an arithmetical . progression is 18, and the seventh term is 30 ; find the sum of 17 terms. Am. 612. 58. A man by selling a horse for 264 dollars gains as much per cent, as the horse cost him ; required the cost of the horse. Atis. $120, 59 Two numbers are in the ratio of 4 to 5, but if one is ii creased and the other diminished by 10, the ratio of the result- ing numbers is inverted ; required the numbers. Am. 40 and 60. 60. Show that the difference between the square of a number consisting of 2 digits, and the square of the number formed by changing the places of the digits, is divisible by 99. 61. A laborer, having built 105 rods of fence, found that, had he built 2 rods less a day, he would have been 6 days longer in completing the job ; how many rods did he build per day ? Am. 7. 62. The common difference in an arithmetical progression is equal to 2, and the number of terms is equal to the second term ; what is the first term if the sum is 35 ? Ans. 3. 63. If the sum of two fractions is unity, show that the first, together with the square of the second, is' equal to the second together with the square of the first. 64. A man having a garden 12 rods long and 10 rods wide, wishes to make a gravel-walk half-way around it ; what will be the width of the walk if it takes up ^ of the -garden ? Am. 9.234 ft. 65. If 8 gold coins and 9 silver ones are worth as m«ch as 6 gold coins and 19 silver ones, find the ratio of the value of a silver coin to that of a gold coin. Afis. 1 : 5. 66. A rectangular picture is surrounded by a narrow frame which measures altogether 10 linear feet, and costs, at 12 cents per foot, 20 times as many cents as there are square feet iu the picture ; requirt'd the length and breadth of the picture. Ans. Length, 3ft.; breadth, 2 ft. 286 MISCELLANEOUS PROBLEMS. 67. A person walked to the top of a moiintain at the rate oi 2J miles an hour, and down the same way at the rate of 3^ miles an hour, and was gone 5 hours ; how far did he walk alto- gether? Ans. 14 mi. 68. A and B hired a pasture, into which A put 4 horses, and B as many as cost him 18 shillings a week. Afterward, B put in 2 additional horses, and found that he must pay 20 shillinga a week ; at what rate was the pasture hired ? Ans. 30s. a week. 69. Two partners, A and B, gained $700 by trade : A's money was in trade 3 months, and his gain was $300 less than his stock ; and B's money, which was $250 more than A's, was in trade 5 months ; required A's stock. Ans.' $500. 70. A sets put for a certain place, and travels 1 mile the first day, 2 the second, 3 the third, and so on ; five days afterward B sets out from the same place, and travels 12 miles a day ; how long will A travel before he is overtaken by B? A')is. 8 or 15 days. 71. A certain number of students go on an excursion ; if there were five more, and each should pay $1 more, the expense would be $Q1^ more ; but if there were 3 less, and each should pay $1^ less, the expense would be $42 less ; required the number of students and the fare of each. Ana. Number, 14 ; fare, $8^. 72. A traveler set out from a certain place, and went 1 mile the first day, 3 the second, 5 the third, and so on ; aft^.r he had been gone three days a second traveler set out, and went 12 miles the first day, 18 the second, and so on; in how many days wiil the second overtake the first ? Ans. In 2 or 9 days. 73. A set out from C toward D, and traveled 7 miles a day: after he had gone 32 miles, B set out. from D toward C, and went every day y\- of the whole journey ; and after he had traveled as many days as he went miles in 1 day, he met A ; required the distance between C and D. Ans. 76 nii., or 152 mi. 74. Two trains start at the same time from two towns, and MISCELLANEOUS PROBLEMS. 287 each proceeds at a uniform rate toward the other town : when they meet it is found that one train has run 108 miles more than the other, and that if they continue to run at the same rate they will finish the journey in 9 and 16 hours respectively; required the distance between the towns and. the lates of the trains. Ans. Distance, 766 mi. ; rate, 1st, 36 mi. ; 2d, 27 mi. an hour. 75. A crimiaal having escaped from prison, ' traveled 10 hours before his escape was known ; he was then pursued so as to be gained upon 3 miles an hour ; after his pursuers had trav- eled 8 hours, they met an express going at the same rate as themselves, who had met the criminal 2 hours and 24 minutes before; in what time from the commencement of the pursuit will they overtake him ? Am. 20 hrs SUPPLEMENT. SECTION X. INEQUALITIES, INDETERMINATE AND HIGHER EQUATIONS. INEQUALITIES. 384. An Inequality is an expression signifying that one quantity is greater or less than another; as az — b>c. 385. The First JVEember of an inequality is the part on the left of the sign ; the second member is the j)art on the right. 386. In treating inequalities the terms greater and less must be understood in their algebraic sense ; thus, 1. A negative quantity is regarded as less than zero, 2. Of two negative quantities, the greater is the one which has the less number of units. .387. Two inequalities are said to exist in the same sense when the first member is greater in both or less in both ; thus, 4>3and6>5. 388. Two inequalities are said to exist in a contrary- sense ^ when the first member is greater in one and less in the other ; thus, 4 > 1 and 3 < 5. ■ 389. The following examples will be readily solved by the student. i:xaih:p£.es. 1. Given ~^.+ -^>-j + :^,to find a limit of x. SoLUTiON.-^Clearing of fractions, we have 6a; + 8a; > 9a; +20; transpos- ing, etc., we have 5x > 20 ; hence, x>4. 25 289 290 INEQUALITIES, INDETERMINATE, ETC. Sx 2. Given 6a;> — + 14; find the limit of x. Ans. x>4. 3. Given ,.- ~ o < 7 ~ i^ ! fiucl a limit of x. Ans. x> 5. 4. Given — — -r- > -r- - 2 ; find a limit of x. Ans. x 16 and 2x + y = 12; find the limits of x and y. J.ns. a;<5^; 2/>l. 7. Given 3a;-5<2a;+l and 4a;+l>13 + a;; find the value of X if integral. . Ans. x = 5. 8. Given x+2y>18 and 2a; + 3y = 34 ; find limits of x and y. Ans. a; < 14 ; y>2. ' 9. Twice an integer, plus 5, is less than 3 times the integer, plus 3, and 4 times the integer, less 4, is greater than 6 times the integer, minus 12 ; required the integer. Ans. 3. 10. Twice a number, plus 7, is not greater than 19 ; and three times the number, minus 5, is not less than 13 ; what is the number? Ans. 6. THEOEEMS IN INEQUALITIES. 1. Prove that the sum of the squares of two unequal quantities, a and h, is greater than twice their product. For, (a — 6)^ is positive whatever the values of a and 6 ; hence, (a — 6)^>0; or, a'-2a& + 6^>0. Hence, a^+Wyiab. 2. Prove that a* + J'' + c' > a6 + ac + 6c. By Theo. 1, 0^+ V>^>1ah, a^+ e'>2ac, fi^+ e2>2&c. Hence, adding, 2a' + 26'*+2c'>2a6 + 2ac + 26c. Whence, a?+ 6'+ c'> a6+ ac+ he. 3. Prove that a + 6 > 2 Vah, unless a = 6. 4. Prove that a^h + a¥ > 2a?lf, unless a = 6. 5. Prove that 3a''+6''>2a(a + 6), unless a = h. INDETERMINATE EQUATIONS. 291 6. Prove that a^ + l>a' + a, unless a = 1. . 7. Prove that the sum of any fraction and its reciprocal is greater than 2. 8. Prove that yr,+— „>-+£, unless a = J. 0^ a' a b , 9. Prove that a - 6 > ( Va - Vby, when o > 6. 10. Prove that the ratio of a' + b^ to a° + 6' is less than the ratio of a + 6 to a^ + 6^ 11. If n? = a^+¥, and y'' = ^ + d', which is greater, xy or ae+ hi, and xy or ad+bel Ans. xy, INDETERMINATE EQUATIONS. 390. An Indeterminate Equation is an equation in which the values of the unknown quantities are unlimited. 391. Thus, in the equation 2a; + 3y = 35, x and y may have different values ; and if any value be assigned to one of the quantities, a corresponding value may be found of the other. - 393. The solution of indeterminate equations, though the number of corresponding values is unlimited, is usually limited to finding positive integral values. 393. Of the several interesting cases that may arise we shall consider only two. Note.— The treatment of indeterminate equations is usually called Indeterminate Analysis. CASE I. 394. To find positive integral values of tUe un- known quantities in the equation. 1. Given 2a; +32/ = 35, to find positive integral values for a; and y. Solution. Given, 2a; + 3y = 35. (1) Transposing, 2a; = 35-32/. (2) Whence, ■ x = ^-^ = V1 -y + ^-^ . (3) Since y is an integer, Yl-y is an integer; and since also x is an integer, 1-2/. , ■ -r^ 18 also an integer. 292 INEQUALITIES, INDETERMINATE, ETC. Let m represent this integer. Then, 1-y 2 -"*' (5) and l-y = 2m. (6) Whence; y = \-1m. (7) Sub. in (1), Ix+Z -6m = 35. / (8) Whence, a: = 16 + 3TO. (9) In equation (7), for y to he integral and positive, m may be 0, or neg.ative, but cannot be positive. In equation (9), for x to be iutegral. and positive, m can be 0, or positive, or negative while less than 5. Hence m. may be 0, -1, -2, -3, or -4. Substituting these values of m in (7) and (9), we have a; = 16, 13, 10, 7, 4, 1. 2/= 1, 3, 5, 7, 9, 11. Note. — We shall use Int. to mean a» integer. 2. Given 7a; + % = 23, to find positive integral values for X and y. Solution. Here, x^'^'^^Z-y.^-^^. (1) 2 2?/ Now, — z— ^ must be an integer. 2 2?/ 2 7??? If we put — =-" = w, then y = — - — , a, fractional expression ; but we wished to obtain an integral expression for the value ofy. To avoid this 2 2?/ difficulty, it is necessary to operate on — =-^, so as to make the coefficient oi y a, unit. 2 2w . . 2 ~ 2?/ Since — z— ^ is integral, any multiple , of — --^ is integral. Multiply, then, by some number that will make the coefficient of y contain the denominator with a remainder of 1. Multiplying by 4, we have 2-2y . 8-8?/ , ,1-2/ ^^ 1 — 1/ Hence, —=^ = Int. = m, and 2/ = 1 — 7m. Sub. in (1), a; = 2 + 9m. Here, for x and y to be positive integers, m can be only 0. Substituting 'm = 0, we have x = 2 and 2/ = 1 • Note. — Other methods of reducing besides multiplying may be used, .as may be seen in the following solution, the object being to obtain an in- tegral form for the value of y. INDETEKMIlfATE EQUATIONS. 293 3. Given 19a; - 14y = 11, to find integral values of x and y. Solution. Subtracting, -^ - ^^g— = ^jg— = 7m«. Hence, ^^ =jw, and 2/ = 19m+6; and, a; = 14»i - 5. Taking m = 0, 1, 2, 3, etc., we have x = h, 19, 33, 47, etc., 2/ = 6, 25,. 44, 58, etc. Notes. — 1. If the equation is in the form ax + by = c, the number of answers will be always limited, and in some cases a solution is impossible. The form ax—by= ± c will admit of an infinite number of answers. 2. If in ax^by = c, a, and b have a common factor not common to c, there can be no integral solution. 4. How can 78 cents be paid with 5-cent and 3-cent pieces, and in how many ways ? Solution. Let a; = the number of 5-cent pieces, and 2/ = the number of 3-cent pieces ; then 5a; + .32/ = 78, from which, by the method explained, above, we find a; = 15, 12, 9, 6, 3, 0; and 2/ = !, 6, 11, 16, 21, 26. Hence it can be paid in 5 ways when both kinds of pieces are used. 5. Given 2x + 3y = 25, to find positive integral values for x and y. Ans. a; = 2, 5, 8, 11 ; y = 1, 5, 3, 1. 6. Given 3a; - 8^/ = - 16, to find positive integral values for x and y. Ang. a; = 8, 16, etc. ; y = ^, 8, etc. 7. Given 8a; + lit/ = 49, to find positive integral values for x and y. Ans. a; = 2 ; y = 3. 8. Given 14a; = 52^+17, to find the least positive integral values for X and y. ^ns. x = 3 ; y = 5. 9. Given 19^—132/ = 17, to find the least positive integral values for x and y. Ans. x = 5; 2/ = 6. 10. Divide 100 into two such parts that one may be divided by 7, and the other by 11. Ans. 56 and 44. 294 INEQUALITIIS, INDETERMINATE, ETC. 11. In how many different ways may I pay a debt of £20 in half-guineas and half-crowns ? Ans. 7 ways. 12. In how many ways can £100 be paid in guineas and crowns ? Ans. 1 9. 13. What is the simplest way for a' person who has only guineas to pay 10s. 6d. to another who has only half-crowns? Ans. In 3 guineas, receiving 21 half-crowns. Note. — The crown equals 5 shillings, and the guinea equals 21 shillings. CASE II. 395. To find the least integer which, dlTided by giTen numbers, shall leave given remainders. 1. Find the least integer which, being divided by 17', leaves a remainder of 7, and being divided by 26 leaves a remainder of 13. Solution. Let a; = the required integer. Then, -rz- and —^ = integers. x — 7 Let -r=- = m ; then, x = 17m + 7 ; substitute this in the second fraction, 17?w + 7-13 17m- 6 ., 26— = -^6~ = -^"*- „ 26m 17m- 6 9m + 6 ^ Hence, -^ ^g— , or -^^ = Int. , , 9m+6 „ 27m-i-18 ,m+18 ^, Whence, — r^ — = Int., which we represent by n. Tlien, — — — =n; hence m = 26ra — 18. Now, if »i = l, we shall havem = 8. Hence, - x = Vim + 7 = 17x8-1-7 = 143, the number. Anothek Solution. Let jN'=the number. mu -y-7 ,,,. ., N- 13 ,„, Then ^^ = a;(l), and— ^^=2/(2). Whence, N=llx + 7 (3), and JV=26y-H3 (4) ; and nx+1 = 1Qy + Vd, or, 17a; -26^ = 6. Then find x and y, as in Case I., and substitute in (3) and (4). INDETEEMISTATE EQUATIONS. 295 2. Find the least number which, being divided by 3, 4, and 6, shall leave respectively the remainders 2, 3, and 4. SoLXTTioN. Leta; = the-integer,then— -— = /»«. = w; whencea;='3m+2. o Also, — ^ =Int; by substitution, ==^~- = n; whence, m = ?i+^^. v>, . n+1 Placing -g- =p, we have w = 3p-l; m = 4p-l, and x = 12p-l. But,^^ = M. = ^^Pf^ = 2p-l + f. Now, f = Jn«.; hence, fx3 = 'f^^Int.=p+^. Put, ^ = q; then, p = 5q, and a; = 602-1. Now iiq = l,x = 59; if 2 = 2, a; = 119, etc. 3. Find the least integer which, being divided by 6, shall leave the remainder 2, and divided by 13 shall leave the re- mainder 3. Ans. 68. 4. FindJ;he least number which, being divided by 17 and 26, shall leave for remainders 7 and 13 respectively. Ans. 143. 5. What is the least integral number which, being divided by 3, 5, and 6, shall leave the respective remainders 1, 3, and 4 ? Am. 28. 6. A man buys cows and colts for $1000, giving $19 for each cow and $29 for each colt ; how many did he -buy of each ? , Ans. 45 cows and 5 colts, or 16 cows and 24 colts. 7. A farmer bought 100 animals for $100 : geese at 50 cents, pigs at $3, and calves at $10 ; how many were there of each kind? Ans. 94, 1, 5. 8. A farmer buys oxen, sheep, and ducks, 100 in all, for £100 ; required the number of each if the oxen cost £5, the sheep £1 , and the ducks 1 shilling each. Ans. 19, 1, 80. 9. A lady bought 10 books of three diiferent kinds for $30 ; the first kind cost $4^ each, the second $2^ each, the third $2 each ; required the number of each kind. Ans. 4, 2, 4. 10. A market-woman finds by counting her eggs by threes she has 2 over, and counting by fives has 4 over ; how many had she if the number is between 40 and 60 ? Ans. 44 or 59. 296 INEQUALITIES, INDETERMINATE, ETC. 11. A boy has between 100 and 200 marbles ; when he counts them by 12s, 10 remain, but when he counts them by 15s, 4 remain ; how "many marbles had he ? -Ans. 154. 12. A person wishes to purchase 20 animals for £20 : sheep at 31 shillings, pigs at 1 Is., and rabbits at Is. each ; how many of each kind can he buy ? r Sheep, 10, 11, 12. Ans. \ Pigs, 8, 5, 2. (Rabbits, 2, 4, 6. Note. — The solution of indeterminate equations of a higher degree is called Diophantme Analysis. HIGHER EQUATIONS. 396. A Cubic Equation is an equation in which the highest power of the unknown quantity is the third power; as ar'+4a;'' + 5a; = 10. 397. A Biquadratic Equation is an equation in which the highest power of the unknown quantity is the fourth power; as a;*+3a;' + 4a;V5a; = 13. 398 . The general form of a higher equation is a;" + aa;°~' + 5x°~" + . . . te+M = 0. 399. No general method of solving equations above the second degree, that is practicable, has yet been discovered. Note. — Cardan's method for cubics and Ferrari's method for biquad- ratics fail in so many cases as not to be practically general. Abel has shown that a general solution above the fourth degree is impossible. 400. The following Principles, which are demonstrated in higher algebra, may often be used in finding the roots of an equation. Pein. 1. if «■ is a root of an equation (the unknown quantity being x), the equation is divisible by x — a. Thus, if 2 is a root, the equation is divisible by a; — 2; if —2 is a root, the equation is divisible by a; — ( — 2), or x+2. Prin. 2. The coefficient of the second term, ajif~^, is the sum of the roots, with their signs changed. Pein. 3. The term independent of x, when in the first member, is the product of the roots, with their signs changed. HIGHER EQUATIONS. 297 Thus, a cubio equation, in which a, b, and e are the roots, is equivalent to {x~a){x-b){x-c) = 0; and when developed is 3?-(:a+b + c)x''+ (ab +aG + be)x—{abc) = 0. , Note.— Principles 2 and 3 will often enable us to conjecture the roots of an equation ; and Prin. 1 will enable us to test any supposed root SOLUTION OF CUBIC EQUATIONS. 401. Cubic Equations can often be solved by special arti- fices, a few of which will be given. CASE I. 403. Solution by inspection and the application of tlie aboTe principles. 1. Given a^ - 6^;" + 11a; = 6, to find x. Solution. The factors of 6 are 1, 2, and 3 ; and their sum equals the coefficient of the 2d term ; hence we may suppose one of these factors, as 3, to be a root of the equation. Transposing 6 to the first member, and divid- ing by a;— 3, we have a;' — 3a;+2 = 0; therefore, 3 is one root, and solving a;' — 3a; = —2, we find 1 and 2 to be the other roots. 2. Solve a;'-9a;^ + 26a; = 24. Ang. x = 2,S, and 4. 3. Solve a;* - 1 la;^ + 38a; = 40. Ans. a; = 2, 4, 5. 4. Solve a;" - 3a;' - 10a; = - 24. Ans. a; = 2, - 3, 4. 5. Solve a;' + 4a;'' + a; = 6. Ans.x = l, -2, -3. 6. Solve 3? - 4x' -7x= - 10. Ans. x = l, - 2, 5. 7. Solve a;' - 2a;' + 4a; = 8. Ans. a; = 2, 2V^, -2lA=T. 8. Solve ar- - 7a;' + 16a; = 10. ^n«. a; = l, 3+1/ -1, 3 - lA^l. CASE II. 403. Solution by making both members a perfect cube. 1. Givena;'+3a;'H-3a; = 7. Solution. Given ^af>+Sx'+3x = 7. (1) Adding 1, a?" + 3a;H 3a; + 1 = 8. (2) Whence a; + 1 = 2, or x = 1. Dividing Eq. 1 by a; -l,we have a;" + 4x + 7 = 0. Whence, a;= -2 + T/-3 and -2-1/-3. 298 INEQUALITIES, INDETERMINATE, ETC. ,2. Solve x» - Sa;" + 3* = 9. Ans. x = 3, + V^, - V=3. 3. Solvex' + 6a;' + 12x=-16. Am.x=-4, -1+ V"^, -l-V-5. 4. Solve 3^+9x' + 27x- -35. Ans.X" -5, -2+V^, -2- V^. 5. Solvea;'-9x^+27a; = 91. a; = 7, l + 2l/^, 1-21/^. CASE III. 404. Solution when tlie equation is readily fac- tored. 1. Given a;' - 6a; = - 4. Solution. (jiven a;' — 6a; = — 4. Whence a' — 4a; — 2a; — 4. And a;(a;2-4) = 2(a;-2). Since this is divisible by a; — 2, a; — 2 = 0, or a; = 2. Dividing by a; — 2, we have a;(a; + 2) = 2 or a;' + 2a; = 2, whence, a; = — 1 ± 1/3. 2. Solve a;'-3a;= -2. ^ns. a; = l, 1, -2. 3. Solve a;' - 3a; = 2. J.'ns.. a; = - 1, - 1, 2. _ 4. Solve a;'- 7a; = - 6. Jjis. a; = 1, 2, - 3. 5. Solve a;* - a'a; - a; = - a. .4ms. x = a, ^{-a^ l/o^+T). 6. Solve a;»-a;'-2a;= -2. ^m. a; = l, + l/2, - 1/2. 7. Solve a^ - 2a; = 1/3. ^ns. a; - l/3, |( - l/H ± V^). 8. Solve 3a? - 7«' - 7a; - - 3. Ans. a; - -J, 3, - 1. CASE IV. 405. Solution by reducing to a biquadratic, and then changing to make both sides squares. 1. Given 3?-lx= - 6, to find x. HIGHER EQUATIONS. 299 Given Whence Add 4x^ Complete squai'e, Hence Whence Solution. iK'-7a;= -6. X a^-3a;»+ (ir 2. Solve a;' + 3a; = 14. 3. Solve 3^-120; = 16. 4. Solve 2;' + 6a; = 6. 5. Solve ari- 4a; = 48. 6. Solve a;'-! 3a; = -12. 7. Solve a;' + 6a; = 20. (1) (2) •*-3a;2-4a;i'-6a;. =4a;»-6a;+(^|y ^^ 2 3 „ 3 „ 3 a;'-2 =2a;--, or -2x+^. x = 2, ora;'+2a;=3. .• . x = l, oi>-3. ^rw. a; = 2, -l=i=V^. An8.x = A, -2, -2. ^MS. a; = l, ^( - 1 =t 1/^^23); Ans. a; = 4, 2( - 1 ± V^). Ans. a; = 1, 3, - 4. ^ns. a; = 2, -l±3l/^^. SOLUTION OF BIQUADRATIC EQUATIONS. 406. Biquadratic Equations may often be solved by special artifices, a few of which we present. CASE I. . 407. SolntJon Ytj inspeeUon and applying the principles of equations. 1. Given a;* - lOa;' + 35a;' - 50a; = - 24, to find x. Solution. We notice that 24 = 1x2x3x4, and 10 = 1 + 2 + 3+4; hence, we presume that some of these factors are roots of the equation. , Dividing by a; — 1, we see that the equation is divisible by a;- 1 ; hence 1 is a root ; and in a similar way wa find that 2, 3, and 4 are roots. 2. Solve a;* - 5a;» + Sx'' + 5a; = 6. iws. a; = 1, - 1, 2, 3. 3. Solve a;* + a;' - 7a;' - a! = - 6. Ans. a; = 1, - 1, 2, - 3. 4.' Solve a;* - 6a;' - a;' + 54a; = 72. Ans. x^2,Q, -3, +4. 5. Solve a;* - 4ar' - 9x' + 16a; = - 20. Ans. x^ -1, 2, — 2, 5. 6. Solve a!*-4«'- 8a;' + 4a;= - 7. Ans. 1, - 1, 2± Vtt. 300 INEQUALITIES, INDETERMINATE, ETC. CASE II. 408. Solution by factoring when the factors can foe readily ofotained. 1 . Given a^ + 4a;'' - 8a: = 32, to find x. Solution. Given a;* + 4a? -8a; = 32. Transposing, a;* + 4a;' = 8a; + 32. , Factoring. a;'{a;+4) = 8(a;+4), or, (a;'-8)(a: + 4) = 0. . Whence, a;'- 8 = and a;+4 = 0, and a; = — 4 or 2. Dividing, a;' — 8 by a;— 2, we obtain a quadratic, from which »= — 1 ± V~—Z. 2. Solvea;*-2a;»-«= -2. Ans.x = l, 2, J(-1±T/^. 3. Solve a;* -3a;' -8a; =-24. AnS:X^2,Z, -lij/^. 4. Solve X* — ax* - n°x = - aw'. Ans, x=a^ n, -( - 1 =3= V - 8), 5. Solve a;* + 3a? -3a; = 9. ^jis. a;=-3; >/3, ^(-l=tT/-3), CASE III. 409. iSolntiou foy reducing to a quadratic form. 1. Given a;* + 2a;' - 3a;'' - 4a; = .5, to find x. Solution. Given a;*+2a;' — 3a;'— 4a; = 5. Whence, (a;'+a;)''-4(a;Ha;) + 4=9, of «Ha!-2=±3. From which x can be found. 2. Solve a;* - 4a;^ + ix' - 8a; = 1 2. Ans. 1 ± T/3, 1 ± V^7. 3. Solvex*+2a;'-7a;''-8a;=-12. Ans. 1, 2, -2, -3. 4. Solve a;* + 2a;'- 3a;^- 4a; =-4. Ans. \, 1, -2, -2. 5. Solve «♦ - gar" + 1 la;' - 6a; = 8. ^ns. i(3 ± V/I7), i(3 ± 1/'-^). KECIPEOCAL EQUATIONS. 301 CASE IV. 410. Solution by reducing both members to a binomial square. 1. Solve «* - 6a;' + 12a;' - 10a; = - 3. SOI/TTTION. Given a;*-6a;' + 122;''-10a;= -3. Whence, (a;'-3a;)H3a;2-10a;= -3, + 4a;'-12a; = a;2 + 2a;-3, +4:(x'-Sx)+4: = x''+2x+l; or {x^-3x)'+4{x'-3x) + i'=x'+2x+l. Whence, (a;*-3a;) + 2 = a;+l. Whence, x = l; 1; i; 3. 2. Solve a;*-4a;» - 19a;'+46a;= -120. Ans. x= -2, - 8, 4, 5. 3. Solve a;* + 42;=- a;' -16a; = 12. Ans.x-- -1, -2, 2, -3. 4. Solve a;* - 93;= + 30a;= - 46a; = - 24. Ans. x = \, 4, 2iT/^. 5. Solvex*+4a;»-6a;''+4a; = 7. ^ns. a; = ± l/^ - 2 =t l/lT. 6. Solve is*-12a;=+48K''-68a;= -lb. Ans. x = B, 5, 2i VS. Note. — Cubics and biquadratics, when any of their roots are integral, can usually be solved by artifices similar to those we have explained. The more general ioaethods of finding approximate roots of numerical equations are those of Double Position, Newton's Method of Apprommaiion, and Hamei's Method, for an explanation of which the student is referred to Works on Higher Algebra. RECIPROCAL EQUATIONS. 4it. A Reciiprocal Equation is one in vyhieh the recipro- cal of X may be substituted for x without altering the equation. 412. Thus-, in a;* - 3a!' + 4a;' - 3a; + 1 = 0, if we substitute - for ' X X, we shall obtain the same equation. l^foTtB.— 1. Such equations are also called fe&uMng eguations, because the coefficients recur in the same order. 2. It can be shown that a reciprocal equation of an odd degree is divis- '. ible by x-l or x + 1, according as the last term is positive or negative. 3. Also, a reciprocal equation of an even degree is divisible by x'-X When itslast term ie positiee. 302 INEQUALITIES, INDETERMINATE, ETC. EXAMFIiES. 1. Solve a;*-3a' + 4a;'-3a; + l = 0. Solution. Given a;4_3a;3+4a;J_3a;+i = o. Divide by x^, a;«-'3a;+4--+i;=0. X x^ Whence ^'"^^'"K*"''^)"'"*"'^' Or (,riy^,(li)=-. Complete the sq., (-i)"-("^)-|-i Extract the root / ,1\ 3 ,1 Whence X = l, \,l{\J^V-rZ. Note. — It is sometimes simpler to substitute some other quantity, as 0, for a; + -, and find the value of x from that of z. ' x' 2. Solve iB*+a;' + a;+ 1=0. Ans.x=-l, -1, |(1±t/^. 3. Solve a;* - 5x' + 6a;'' - 6a; + 1 = 0. Ans. a; = 2 ± t/3, ^(1 =t V^S). 4. Solve a;*-10a;' + 26a:'-10a;+l=0. ^ns. a; = 3±2l/2, 2±l/3. 5. Solve a;* - Sx* + 3a; + 1 = 0. Ans. a; = 1 ± l/2, ^(1 ± l/5). 6. Solvea;*-|a;' + 2ar'-fa;+l=0. ^ms. a; = 2, ^, ± l/^. 7. Solve a;* - 3a!» + 3a; - 1 - 0. Am. a; ■= =t 1, ^(3 ± 1/5). 8. Solvex' + a;~''+a;+a;~'-4 = 0. A71S. a; = l, 1, |(-3il/5). Notes. — 1. In equation 7 divide by a;^ — 1, and then reduce and find the values of x. 2. It may be readily shown that any two corresponding pair of roots are reciprocals of one another. SECTION XL EXPONENTS AND LOGARITHMS. THEORY OF EXPONENTS. 413. An Exponent denotes the power of a quantity or the number of times it is used as a factor. Thus a' means a,x ax a, or a used as a factor three times ; and o" means axaxax . . . . to n factors, or o used as a factor n times. 41'!:. By the original conception of a power the exponent n could be conceived only as a positive integer, and the rules for multiplication, division, etc. were all based on this conception. 4:15, Subsequently, it was seen that division gave rise to negative exponents and evolution to fractional exponents, and that these could be used the same as positive integral ex- ponents. Note. — We shall now give a complete logical discussion of the subject, assuming only the definition of an exponent and the rules of addition and subtraction. POSITIVE EXPONENTS Prin. 1. When m and n are positive integers, a" x a" = a"'^". For a'' = a'xaxax .... to »? factors. (Def) And a'^ =axaxa>i . . . . to n factors. (Def) Hence a'»xa" = axax .... xaxaxa. . . . to w + n factors, which by the definition equals a"+". Pein. 2. When m and n are positive integers, and m is greater thanw, a"'H-a»=-a"'-". 303 304 EXPONENTS AND LOGAEITHMS. (Prin. 1.) For gm-n xa"= a'»-"+''. Reducing, a'^-'^xa'^=a™. Dividing by a", a" a" Pkin. 3. When m and n are positive integers, (a")" equals a"^. For {a'')'^ == cT- X a'" X a'" X .... ton factors; But a'^ = axaxax .... to m factors. Hence (a™)" = axa>»> = a"-". Pein. 9. Prove that (a")" = a"" when one or both exponents are negative. First, suppose m is negative, and let m——r. Then, («"•)»= (a-')"= (ar)"=-Jn = «""'=«'"" Second, suppose n is negative, and let n =■- —p. Then, («")"= (a"')-^=(^=^ = «-'^^«'"" Third, suppose m and w are both negative, and let m=- r and n= —p. • Then, (a-)»=(a-)-' = (^ = ^ = «'' = «-'"''-" = «™'- 26* 306 EXPONENTS AND LOGARITHMS. Pein. 10. Prove that v^'^=a", when either m or n, or both, are negative. First, suppose m is negative, and let m= — r. n/T 1 _r •» Then v'a»= l/a-- = \ ~ = l-a " = a". n. . Second, suppose n is negative, and n=—T. Let x = ya^. Then aj-'-a", and aj-'^a", or , = a'»; ^^ = ^1 ^= '^='" ' =«"• Third, suppose hoth m and n are negative; let m= -J3, n= — r, and X = v'a"'. 11 - K Then a;" = a'", a;-' = a-J', — = — , ^ = aP, x = ar = a . Note. — No practical significance is attached to a negative index of a root ; but the form is a possible one, and the above demonstration proves the principle to be general. 417. Thus we see that whether m and n are positive or negative integers, we have the following: I. a™ xo" = a'"+». III. (a")" = «"*". m II. a" ^ a" = »"-". IV. Vdr^a'". FRACTIONAL EXPONENTS. 418. A Fractional Exponent arises from evolution by- dividing the exponent of the power by the index of the root, when the former is not a multiple of the latter. 419. We shall show the meaning of the fractional exponent and prove a few principles to be used in showing the univer- sality of the fundamental rules. ' Pein. 11. Prove that (a")'' = a" when — and^ are positive or negative. FEACTIOlfAL EXPONENTS. 307 First, suppose p is positive, — being either positive or negative. Eaising a" to thepih power, we have a'^xa" xa"x ... top fac- toi-s,-or a" " »'" • ■/ ^P^"-''^a"^^a'^_ Second, suppose ja is negative, and let p= —s. Then («")!'= (a")-=7^,=-4r.=a »=a "''-^=a-^. m Pkin. 12. Prove that a" = l/a", m and n being either positive or negative. First, suppose m is positive or negative, n being positive. By Prin. 11, (o")"=a"'; extract wth root, a" = V'a™. Second, suppose n is negative, and let n == — s.' Then, («")»= (a^) =— ^ = -L = a*.. Hence, (a^)» = a'», and a"= l/rf^. CoK. Hence, a" = 1/ a, or Va = a" ; also {cr) " = a , or a. Note. — This principle was derived under the previous article, but is here proved by another process of reasoning. Pein. 13. Prove that Va^=(a"')*. ,Let cr = x; then Vor= v^=a;". (Prin. 12, Cor.) But since K-o", a;" = («"•)"'; hence -|/a" = (a'»)". CoE. Hence a^ = (a")*. Pein. 14. Prove that a^x 6^ = (a6)^ Let x=x^xbK Then, ai^^{a^xby={a^)''x{b^)''^ab. 1, Hence, a?* = (a6) ; therefore, a; = (a6)". (Prin. 12, Cor.) 308 EXPONENTS AND JLiOGAElTHMS. Cor. 1. In the same way it may be shown that 1 i 1 1 CoK. 2. Hence also a" x 6" x e" = (a6c)". Pein. 15. Prove that a" x 6" = (o6)". Let x=c?xb^.' mm m m Then, a;» = (a" x6")»=(a'')"x (6")» = a"'x&"'=(a6)". Hence, X^ = {ab)^ ; and x = (afi) " ; therefore, a" x 6" = (a6)". Pein. 16. Prove that (a^y=={arf. By Prin. 11, (ft^)" = a^ ; let a; = a". 1 Then, x" = o», and x = (a^f. Prin. 12. CoE. In a similar way it may be shown that (a"')" = (a")", Pein. 17. Prove that (a""/ x (o^f = (a" x a")*. Let a; = (a*")" x (op)'' j then p:^ = a™ x a? ; 1 Hence, a; = (a"* x a**) '^. Prin. 18. Prove that a" = aT m Let a; = a" ; then a;** = a% and af^ = a"^. Hence, x = a'^; therefore, a" = a**". 430. *We shall now proceed to show that the rules for mul- tiplication, division, involution, and evolution apply to fractional exponents as well as integral. FRACTIONAL EXPONENTS. 309 Pein. 19. Prove that a'xo' =a^^'. £ J. jw «!■ o'xo'=a"xa'". Prin. 18. ='{aP'fx{a^)^. Prin. 13, Cor. = (a?" X a^f'. Prin. 17. = (ajp'+^f. =J^^S^J+r Cge. In -the same way it may be shown that a'-i-a-=a' PBm. 20. Prove that (a')' =a". Let «=(«')•; then.a?'= {ay=a«. Hence a;'' = o"^ ; and x = a?'. CoE. In the same way it may be shown that tl Z Zj.r £v« P^ 8 1 q q- a q^f . qf = a = a =a . ^' ScHOUirM:. When the exponents in Prin. 19 and 20 are negative, we can let m and n represent the fractions, and since the principles are true for m and n, they are true for negative fractions. Or we can prove them hj the methods used for negative integral exponents. 431. It is thus shown that the following relations are uni- versal, m and n being positive or negative, integral or fractional. I; a'»xa" = a'"+". III. (a")" = a"". II. «"• - a" = a"*-". IV. l/a" = a". Note. — The student will be interested in noticing that this general dis- cussion has introduced two forms of expression that are not usually em- ployed in algebra — viz. negative indices and fractional indices of roots. Thus, since n is general, ^\a gives the forms ^/a; */a; jla. 310 EXPONENTS AND LOGARITHMS. 1. Prove «"*-" = ■ EXAlIPIiES. 2. Prove (a) "^=-7. a" 1 3. Prove V^ar = -—. 4. Show the meaning of the negative exponent, a~". 5. Shaw the meaning of the fractional exponent, a' or a". 6. Show the meaning of a fractional index, ^/a or J^/a. 7. Show the meaning of a negative index, ya or v'a. 8. Show the meaning of a negative fractional index, ^/a or ^ /a. EXAMPLES IN EEDUCTION. 11. (4a-|)-i 1. 9-^. 2. 4-1 3. 64"l 4. 1 81-*' 5. («-)* 6. 1 7. (x-T'. 8. Va^. 9. (m-i)-i. 10. (x-^f. Ans. ^. Ans. \. Ans. \. Ans. 27. . 1 Am. —i- a'. .4ns. — . a' J.ns. m^. Ans. -. Ans. -. o 12. (64n-')-*, 13. (32a-'^)i 14. (diar'yi. 15. v^a^. 16. Ans. (1 16" 17. Il4a\ 18. Msa- Ans. -:. a* Ans. — -' 1/5 ^ns. -. a Ans. 4a. J-Jis. 8a'. Ans. •(1)' 19. -M. 64 20. 21. PKACTIONAL EXPONENTS. Ans.i 27 ■"^a Ans. Am. in if} 24. ~J^la-"'b'". Ans.a^%-"'\ 23. -^a 6' 25. ija "6". 26. 311 Ans. \bj ' ^ms. a"e°'. EXAMPLES m MULTIPLICATION, 1-1 1 1. a' X a ^. J.n«. a^. ^ /l\l 2. a ''x |/a. »-- I |(m-n) 3. a'xa^' '. Ans. VaT. 4. aVxa-'a"". Ans. e^. 11 .s 5. x^x |/a;'. Jjis. a;' 7. a" X a". Ans. a " Multiply the following : 8. y X i/y~^. Ans. y . 9. (4"'a)"-"+*x(4'"a)''-"^. Ans. 1. 10. a'"-"x3a'"^''x4a"+=c. Jlns. 12a<'"+'e. 11. («-"&? X a"6-' X a^+^fi*. 12. (a + 6)'"+"c" X (a + 6)''^c--». .Ans. a 6 • 13. a^6~-x a^^" X (a^b^y^-'^ 14. (8-Va;)'"+'"x(8'(i-V)"'+'". , ins. 8'"a"'a;'"+"*- 15. (a+6)-Vx(a+6)"+V"x(a + 6y+"c-l Ans. (a-by-*''e'-\ 16. (o* + &*) by a* - 6^- Ans. a* - 6*. 17. 0"+" + 6°^ by «"+" - 6""". ins. a^<'»+»> - 6«»-2). 18. a;" - 2/'™'"™ by xT'-^-f. . J.ns. cc'™'?/" - cc""-"?/"". 312 EXPONENTS AND LOGARITHMS. Ans. a" -6". 3 Ans. a;^ + 2x^H-a;-4. 19. a"+6''bya"-6". 20. x^x^+2 hy x+x^-2. 2 112 11 21 . a;^ + x^y^ + if by x^ - y^. 22. a^ + a'+l bya-'-a-Hl. 23. a~*+ffl"*+lby a~*-l. 24. J-2 + ar^hyJ-a'^. 25. -3a-'+2a-^6-^by -2a-'-3a-*6. Ans. x-y. Am. a*+l+a~*. Ans. a~' — 1. 1. a'-3a^+3a~^-ffl-i Ans. -4a-'6-'+9a-«6. i 26. a -a'+a -«''+» -a+a -1 by a^+1, Ans.a*--!. EXAMPLES IN DIVISION. 1. aKa-K 4ws. a . 13. (a-xy^-^ia-xf- 2. a '-5- a*. Ans. a ^- Ans. (a - a;)"' 3. a^-.aP-'. 4 ns. a". 14. 6ah^-^3a~h^.Ans.2ab~\ 4. 5. 6. Ans. b^. Ans. a"~\ 4ns. a"'". 15. 4a"V^2aV*. 7. 8. 1 2 4ns. 6^ 4ns. a^. 16. 4 ns. (a-b)"^. 9. 10. 11. n an 2 3 a -5-a . ^y, Ans. a". Ans. 1 — 1 . 17. 18. a''b~e^'^a--^b'e''-\ Ans. a^b-^c'-^. 12. Ans. I - [ah FRACTIONAL, EXPONENTS. 313 'Divide the following : 19. J-bi by a^-bK Ans. ahaHKb^. 20. J-b^ by J-bi Am. J + a¥ + a¥+bK 21. a - a; by a' - x^. Ans. J + Jx^ + a;l 22. a"-6'»bya''-6*. Ans. a' +a'b' + a'b' +b\ 23. a="- 6-"» by a= - 6 \ Ans. a' +a% ' +a'b'^+b * . 24. a'^W-^o"^ by a"^'6''+'c'^. Jms. a'^26-=^»c'^^ 25. a'6-' + 2 + a-'6' by ab-' + «-'&. ^ ??,«. aft-' + a'^b 26. 8a;-' + 272/-' by 2a;~^ + 32/^1 ^ns. 4x~*-6a;"V* + %"*- an __3w n g 27. a' -a 'hya-a', 4?ig, a"+l+a-^'. 28. X - xy^ + x^y — y^ by x^ - y^. Ans- ^+y- 29. aK aW + b^ by a^ + ah^ + 6*. ^ns. o* - aW + 6'. 30. x-^+x~''y~''+y~* by a;~Va;""'^/~M-2/~^ J.ns. x~'+x~^y~'-^y~''. 31. a* + 6* - c^+2a*6* by aKb^+cK Ans. a^+b^-eK 32. Find the value of f (a'"*)' x (a--)^| ''^, Jns. «=. 33. Find the value of [ [ (a - 6)"}""*]^. Ans. (a - 6)"-\ 34. Findthevalueof KS^-^r^xCS-"-^'")-'?^^. Ans. b". 35. Find the value of U(aTr7^i\(a'')TV- Ans. a'. 36. Find the value of [{(a-)"^}^]"'^ [{(a^)'}^]''.^«s.l. 314 EXPONENTS AND LOGARITHMS. LOGARITHMS. 433. The Logarithm of a number is the exponent denot- ing the power to which a fixed number must be raised to pro- duce the first number. Thus, if B" = N, then x is called the logarithm of N. 4:33. The Base of the system is the fixed number which is raised to the different powers to produce the numbers. Thus, in ^ = iV, a; is the logarithm of N to the base JB; so in 4' = 64, 3 is the logarithm of 64 to the base 4. 434. The term logarithm, for convenience, is usually written loff. The expressions above may be written logiV=a;and log 64 = 3. 435. In the Common System of logarithms the base is 10, and the nature of logarithms is readily seen with this base; thus, 10' =100; hence log 100 =2. 10' =1000; hence log 1000 =8. 10* =10,000; hence log 10,000 = 4. IQ^.zm ^ 234 ; hence -log 234 = 2.369. 436. We shall first derive the general principles of log- arithms, the base being any, number, and then explain the com- mon numerical system. PRINGIPIiES. Prin. 1. The logarithm of 1 is 0, whatever the base. For, let B represent any base, then jB" = 1 ; hence by the definition of a logarithm, is the log. of 1, or log 1 = 0. Pein. 2. The logarithfn of the base of a system of logarithms is unity. For, let B represent any l)ase, then B^ — B; hence X is the log. of B, or log B = l. LOGARITHMS. 315 Prin. 3. The logarithm of the product of two or more numbers is equal to the sum of the logarithms of those numbers. For, let w = log J/, and n = log iV. Then, B^ = M, £" = N. Multiplying, jB"+" = Jf x JV. Hence m\-n = log (MxN). Or, log (if X iV) =log M+log N. Prin. 4. The logarithm of the quotient of two numbers is equal to the logarithm of the dividend minus the logarithm of the divisor. For, let m = log M, a,nin = \og N. Then, B'^ = M, B^ = N. Dividing, B'''-^=M^N. Hence, log {M-i- N)=m — n. "Or log (M^N) = log M-log N. Prin. 5. The logarithm of any power of a number is equal to the logarithm of the number multiplied by the exponent of the power. For, let m=log M. Then, B^ = M. Eaising to nth power, ^'<'" = if''. Whence, log M" = n x w. Or log M^ = n x log Jf. Prin. 6. The logarithm of the root of a number is equal to the logarithm of the number divided by thelndex of the root. For, let m = log M. Then, B'^ = M. Taking nth root, B^ = M^. Whence, log J/ =-. Or logM^=l°S^. 71 427. These principles are illustrated by the following ex- amples, which the pupil will work. . 316 EXPONENTS AND LOGARITHMS. EXAMFI/ES. 1. Log (a.h.e) = log a + log 6 + log e. 2. Log! — )=log a + log &-log e. 3. Log a" = w log a. 4. Log (a'b") ■=% log a +2/ log h. aj'b'' 5. Log — -;- = X log a+y log 6 - s log c. 6. Log Vah = ^ log a+-| log 6. 7. Log (a^- a;") = log (a + a;) + log (a -a;). ' • 8. Log Va? - x') = ^ log (a+a;) + |^ log (a -a;). 9. Log a^Va^'"^ log a. Va" - x' > 10- Log (-^_^^N2 =i {log (a- a;) -3 log (a+a;)}. COMMON LOGAEITHMS. 498. The Base of tlie common system of logarithms is 10. This base is most couvenient for numerical calculations, because our numerical system is decimal. 439. In this system every number is conceived to be some power of 10, and by the use of fractional exponents may be thus, approximately, expressed. s ' 430. Eaising 10 to different powers, we have 10° = 1; hence O = logl. 10' =10; hence 1= log 10. 10^ = 100; hence 2= log 100. 10' = 1000 ; hence 8 = log 1000. etc. etc. Also, 10~' = .l; hence -l=log.l. 10^' = .01; hence -2 = log.01. 10-= = .001; hence -3 = log.001. LOGARITHMS. 317 43i. Hence the logarithms of all numbers between 1 and 10 will be + a fraction; between 10 and 100 will be 1 + a fraction; between 100 and 1000 will be 2 + a fraction ; between 1 and .1 will be -1+a fraction ; between .1 and .01 will be —2 + a fraction ; between .01 and .001 will be - 3 + a fraction. 432. Thus it has been found that the log. of 76 is 1.8808, and the log. of 458 is 2.6608. This means that 10'-™ = 76, and 10''-«™ = 458. 433. When the logarithm consists of an integer and a deci- mal, the integer is called the charaeteristie, and the decimal part the mantissa. Thus, in 2.660865, 2 is the charaeteristie, and .660865 is the mantissa. PRINCIPLES OF COMMON LOGARITHMS. Pein. 1. The charaeteristie of a logarithm of a number is one less than the number of integral places in the number. For, from Art. 430, log 1 = and log 10-1 ; hence the logarithm of numbers from 1 to 10 (which consist of one integral place) will have for the characteristic. Since log 10 = 1 and log 100 = 2, the logarithm of numbers from 10 to 100 (which consist of two integral places) will have one for the characteristic, and so on ; hence the characteristic is always one less than the number of integral places. Pein. 2. The characteristic of the logarithm of a decimal is neg- ative, and is equal to the number of the place occupied by the first significant figure of the decimal. For, from Art. 430, log .1= -1, log .01= -2, log .001 = -3; hence the logarithms of numbers from .1 to 1 will have —1 for a characteristic; the logarithms of numbers between .01 and .1 will have — 2 for a charac- teristic, and so on ; hence the characteristic of a decimal is always negative, and equal to the number of the place of the first significant figure of the decimal. Pein. 3. The logarithm of the product of any number multi- plied by 10 is equal to the logarithm of the number increased by 1. 318 EXPONENTS AND LOGAEITHMS. For, suppose log M= m ; then, by Prin. 3, Art. 426, log {Mx. 10) = log M+\os 10 ; but log 10"= 1 ; Hence log (JHxlO) = »i + l. Thus, log (76 X 10) = 1.880814 + 1 ; or log 760 = 2.880814. . Pein. 4. The logarithm of the quotient of any number divided by 10 is equal to the- logarithm of the number diminished by 1. For, suppose log M= m ; then, by Prin. 4, Art. 426, " log (il/^10)=?log j¥-log 10; Hence, log {M-i-10) = m — l. Thus, log (458 ^ 10) = 2.660865 - 1 ; or log 45.8 = 1.660865. Pbin. 5. In changing the decimal point of a number we change the characteristic, but do not change the mantissa of Us logarithm. This follows from Principles 3 and 4. To illustrate : log 234 = 2.369216. log .234=1.369216. log 23.4 = 1.369216. log .0234=2.369216. log 2.34 = 0.369216. Thus we see that the characteristic becomes negative, but not the mantissa. The minus sign is written over the mantissa to show that it only is negative. EXEECTSES ON LOGABITHMS. 434. Common Logarithms are used to facilitate the ope- rations of multiplying, dividing, etc. Tables of logarithms are constructed and used for this purpose. 433. We shall give the logarithms of a few prime numbers to four decimal places, and show how they are used. log 2 = 0.3010 log 7 = 0.8451 log 17 = 1.2304 log 3 = 0.4771 log 11 = 1.0414 log 19 = 1.2787 log 5 = 0.6990 log 13 = 1.1139 log 23 = 1.3617 MULTIPLICATION WITH LOGAEITHMS. 436. Numbers are multiplied by means of logarithms by taking the sum of their logarithms. (See Art. 426.) LOGARITHMS. 319 1. Find the logarithm of 2x5. OPERATION. Solution. From Prin. 3, Art. 426, the log. of log 2 = 0.3010 2x5 equals the log. of 2 plus the log. of 5; log log 5- 0.6990 2 = 0.3010, log 5 = 0.6990; their sum is 1.0000. log 10 = 1.0000 Find by the use of the logarithms given in Art. 435 the fol- lowing : 2. log (3x7). 5.log 5x10. 8. log 2 X 7 X 1,3. 3. log (5 X 7). 6. log 7 X 10. 9. log (5 x 17 x 23). 4. log (7 X 11). 7. log 13 X 10. 10. log (7 X 19 K 23). NoTE.-r^In actual practice with a table we find the number correspond- ing to the logarithm of the product, and thus obtain the product of the numbers. 437. The logarithms above given will enable us to find the logarithms of many numbers of which ^he prime numbers are factors. Find the following: 1. log 4. 5. log 20. 9. log 56. 13. log 1.15. 2. log 6. 6. log 26. 10. log 85. 14. log .230. 3. log IQ. • 7. log 30. 11. log 8.5. 15. log .380. 4. log 15. 8. log 42. 12. log 115. 16. log .0035, DIVISION WITH LOGARITHMS. 438. Numbers are divided by means of logarithms by sub- tracting the logarithm of the divisor from the logarithm of the dividend. 1. Findthelog. of 5^2. OPERATION. Solution. From Prin. 4, Art. 426, the log. log 5 = 0.6990 of the quotient of 5 divided by 2 equals log 5 log 2 = 0.301 minus log 2; log 5 = 0.6990, etc. log (5 ^2) = 0.3980 Find by the logs, given in Art. 435' the logs, of the following : 320 EXPONENTS AND LOGAEITHMS. 2. 3 6. 1- ]0. -ff. 14. If- 3. f 7. ^3^-- 11. 03 • 15. -.Vf. 4. i- 8. ¥• 12. If- 16. IJ:f 5. -¥-• 9. ¥• 13. m- 17. ax.3X.05 .7X. 11 + 22 Note. — For an explanation of the nature and use of the anthmetical comphntent see Trigonometry. POWEES AND BOOTS WITH LOGARITHMS. 439. The powers or roots of numbers are readily obtained by logarithms, according to Prin. 5, Art. 422. 1. Find the log. of 7'- SoLTTTiON. By Prin. 5, Art. 426, log 7' equals opebatioit. log 7 multiplied by 3 ; log 7 = 0.8451 ; multiplying log 7 = 0.8451 by 3, we have 2.5353; hence log 7', or log 343 3 = 2.5353. log 7' qr 343 = 2.5353 « Find the logarithms of the following : 2. 3^ 6. 13^ 10. 3^. 14. 2»x3l 3. 5'. 7. 15*. 11. 7l 15. 3=^-^.05'. 4. 7K 8. 17'. 12. .07^ 16. .07' X. 0141 5. IV. 9. 191 13. .01*. 17. .09^^.021^ Note. — Teachers who wish to give their pupils a knowledge of the use of the tables and numerical computation with logarithms will find the subject presented in my Geometry and Trigonometry. EXPONENTIAL EQUATIONS. 440. An Exponential Equation is an equation in which the unknown quantity is an exponent ; as, a'° = b, af = a, J"" = c, etc. 441. Such equations are most readily solved by means of logarithms. LOGARITHMS. 321 1. Given a' = h, to find x. Solution. Given, a? = h. Taking log. of members, a; log a = log 6. log b log a' Whence, 21 = 1581 2. Given 5"^ = 10, to find K. Solution. Given, 5"^ = 10. Taking log., a; log 5 = log 10. wi, log 10 1.000 Whence, x = -r^ — = — ^^^" = i 4Qnfi ' log .5 0.6990 •*'*"''• 3. Given 5^=1, to find a;. Solution. Given, 5^ =|. Raising to x power, 5' = i-. Taking log., 2 log 5 = a; log 7 - a; log 3. Whence, x=\ ~ 'log 7- log 3' ^ 2 X. 6990 „„„„„ ^'' * = .845l3l771=3-^^«»- EXAMPIiES. 4. Given 5* = 8, to find x. Am. x = 1.2918. 5. Given 4^ = 8*, to find a;. Ans. x = 0.6. 6. Given a" = he, to find a;. • Ans. x = — ^^ ^—. log a 7. Given a" = &^c', to find x. Ans. x = ^^ ^; log a 8. Given 5* = 30, to find x. Ans. x = 0.9464. 322 EXPONENTS AND LOGARITHMS. „ „. ab'-e , , £ J i log (wt£+c) -lo g a 9. Given = d, tofindx. Ans.x^-° j— ^ . 1 i log h 10. Given ab' = c, to find x. Ana. x = i^J-j— - , log 6 - n log a 11. Given a""+" = 6, to find x. Ans. x = " ,„ j^g „ — ■ log ^ 12. Given m^n^ =p, to find x. Ans. x = ;^^+ j^^- 2 log 3 13. Given a'"^ - 2ar = 63, to find x. Ans. x = -^~ . 14. Given S'"' + 3"^ = 6, to find x. Ans. x = 0.6308. 1 n ^ A logi(m±l/m'-4) 15. Given Ji"' + - = m, to find a;. Ans. x = .^ . 16. Given a^+ft" = 2m and a'' - 6" = 2re, to find a; and y. ^'**-''" log a ' ^ log 6 • 17. Given a? = y", and a;' = 2/', to find x and 2/. ^M. a; = 3f , 2/ = 2|^. 18! In a geometrical progression, given a, r and s, to find w. Ans. See page 268. 19. In a geometrical progression, given I, r and s, to find n. Ans. See page 268. 20. In compound interest, if P represents the principal, R=l+r, the rate, J. the amount, and t the time, show that ^ = Pxi?' = P(l+r)'. 21. From the above formula derive the following formulas : 1. log A = log P+t log (1 +r) ; 3. log (1 +r) = ^°^^~^° ^; 2.1ogP=log^-nog(l+rO; 4.< = ^^-^5^. Note. — Exponential equations of the form x'' = a cannot be solved by elementary algebra. Numerical forms like a;* = 10 may be solved by Pouble Position. SEOTIOI?^ XII. PERMUTATIONS, COMBINATIONS, BINO- MIAL THEOREM. PERMUTATIONS. 442. Permutations are the different orders in which a number of things can be arranged. Thus, the permutations of a and b are ab and ba ; the permu- tations of a, b, and c, taken two at a time, are ab, ba, ac, ca, be, cb. 443. Things may be arranged in sets of one, of two, of three, etc. Thus, the three letters a, b, and c may be arranged in sets as follows : Of one, a, b, c. Of two, ab, ac ; ba, be ; ca, cb. Of three, abe, acb ; bac, baa ; cab, eba. Notes. — 1. It is convenient to let P^ represent tlie number of permu- tations when taken two together ; Pj, the number when taken three to- gether, etc. ; Pr, the number when taken r together. 2. The. term permutations is sometimes restricted to the case where the quantities are taken all together, while the term arrangememis oi variations is given to the grouping by twos, threes, etc., the nurdber in the group being less than the whole number of things. PROBLEMS. 444. To find the number- of permutations or arrangements that can be formed of n things taken two at a time, three at a time, etc. 323 324 PERMUTATIONS, COMBINATIONS, BINOMIAL THEOREM. Let a, b, 0, d k represent n things. First, if we reserve one of the n letters, as a, to place before each of the others, there will remain n — 1 letters; and placing a before each of the n — 1 letters, there will be w— 1, in which a stands first; as ab, ac, ad, .... ak. Similarly, if we put b before each of the other letters, there will be n— 1 arrangements in which 6 stands first. Similarly, there will oe n — 1 arrangements in which c stands first. Hence, since each of the n letters may be first, in all there will be n(n — 1) arrangements of n things taken two at a time. Second, let us now find the number of arrangements of there letters- taken three at a time. If we reserve one letter, as a, there remain n — 1 letters; the number of permutations of ra — 1 letters taken two together, from what has been shown, is (w— l)(n— 2). Putting a before each of these, there will be (n — l)(n — 2) perm utations of n letters taken three to- gether, in which a stands first. Similarly, there are (n— l)(n — 2) per- mutations, in which 6 stands first, and so for each of the n letters ; hence the whole number of permutations of n letters taken three together is n(n-l)(?i-2). Similarly, we find that the number of permutations of n letters taken four at a time is n(n — l)(m — 2)(re— 3). From these cases it may be in- ferred, by analogy, that the number of permutations of n things taken r at a time is n(«-l)(n-2)(n-3) .... (n-r'^). 445. We shall now prove that the formula for the permuta- tion of n things taken r at a time, derived above by analogy, is true. Suppose it to be true that the number of permutations of n letters taken r — 1 at a time is n{n-l) .... {n-{f^l-l)}, or n(n — l) .... (w— r-H- 1), then we can shrfw that a similar formula will give the number of permu- tations of the letters taken r at a time. For, out of the w— 1 letters b, 0, d we can form (Art. 444), (n-l)(ra-2) .... (n-l-r-l-l), ^, (n-l)(re-2) .... (re-j^^=I), d'mutations each of r - 1 letters, iu which a stands first. Similarly, we PERMUTATIONS. 325 have the same number when 6 stands first, and as many when stands first, and so on. Hence, on the whole, there are n(n-l)(n-2) .... (n-f^) permutations of n letters taken /• at a time. This proves that if the formula holds when the letters are taken r~l at a time, it will- hold when they are taken r at a time. But it has been shown to hold when they are taken three at a time ; hence it holds when they are taken /our at a time ; hence also it must hold when they are taken five at a time, and so on ; therefore the formula holds universally. Note. — The method of reasoning employed in Art. 445 is called Mathe- matical Induction, it is based on the principle that a truth is universal if when it is true in n cases it is true in n + 1 cases. It is regarded as a valid method of demonstration, while the method by analogy or pure induction is not. 446. To find the number of permutations of n things taken all together. In the formula Just proved put r = n, and we have P„=n(n-l)(n-2) .... 1. 447. For the sake of brevity n(w-l)(»i-2) . . . '. 1 ia often written \n, wliich is read "factorial n." Thus, \n_ denotes Ix2x3x . . . . xn; that is, the product of the natural numbers from 1 to w inclusive. 448. Any number of r things, or combination of r things, will produce \ T permutations. For by Article 446, the r things which m8,ke the given combination can be arranged in \r different ways. 449. To find the number of permutations of n things taken all together when some occur more than once. Let there be n letters, and suppose a occurs p times, 6 occurs q times, and c occurs r times, the rest, d, e, f, etc., occurring but once. Let N represent the required number of permutations. Now if in any of the permutations we suppose the p letters a to be changed intoi> new letters, different from the rest, then without changing 9R 326 PERMUTATIONS, COMBINATIOITS, BINOMIAL THEOREM. the situation of any of the remaining letters we could, from their inter- change with one another, produce \p permutations ; hence if the p letters a were changed into p different letters, the whole number of permutations would be iVx I p. Similarly, if the q letters b w6re also changed into q new letters, different from any of the rest, the whole number of permuta- tions would he Nx \px \q. So also if the r letters c were also changed, the whole number of permutations would be Nx Ipx \qx \r. But this number would be equal to the number of permutations of n different things taken all together; that is, to \n. Thus, ^x Ip X Ig X |r = In. in Hence N= \px \q> 4:^0. To find the number of permutations of n things when each may occur once, twice, thrice, . ... up to r times. Let there be n letters, a, b, c, . . . . Taking them one at a time, we shall have w arrangements. Taking them two at a, time, a may stand before a, or before any one of tlie remaining letters ; similarly, b may stand before 6, or before any one of the remaining letters, and so on; thus there are rax re, or n^, different arrangements. Taking them three at a time, each one of the n letters may be combined with the n' arrangements, making n x n\ or n' in all. Similarly, when the letters are taken r at a time, the whole number of permutations will be n'. 4:51. If they are taken n at a time, or all together, r becomes n, and the number of permutations becomes Ji". EXAUIPIES. 1. How many permutations can be formed of the letters in the word chair, taken three together? Solution. Here n = 5, and r = 3; hence substituting in the formula P3 = ra(n-1) (re-f^)), we have 5x4x3 = 60. 2. In how many ways may the letters of the word home be written ? Ans. 24. 3. In how many ways can 5 persons arrange themselves at table so as not to sit twice in the same order? Ans. 120. 4. In how many different ways, taken all together, can the 7 prismatic colors be arranged ? Ans. 5040. COMBINATIONS. 327 5. The number of permutations of a set of things taken four together is twice as great as the number taken three togetlier ; how many things in the set ? Am. 5. 6. In how many ways can 8 persons form a circle by joining hands ? • Ans. 5040. 7. How many permutations can be made of the letters of the word Caracsas, taken all together? Aiis. 1120. 8. How many permutations can be made of the letters of the word Mississippi, taken all together ? Ans. 34650. 9. The number of permutations of n things taken four together is six times the number taken three together; find the value of n. Ans. w = 9. to. The number of arrangements of 15 things, taken r to- gether, is tea times the. number taken (r—1) together; find the value of r. Ans. r = 6. 11. In how many ways can 2 sixes, 3 fives, and 5 twos be thrown with 10 dice?. [JJO 12. In how many difierent ways can six letters be arranged when taken singl)"^, two at a time, three at a time, and so on, until they are taken all at a time ? Aiis. 1956. Note. — Find the sum of the different permutations. COMBINATIONS. 433. The Combinations of a set of things are the difierent collections that can be formed out of them without regarding the order in which they are placed. ■ Thus, the combinations of the letters a, b, c, taken, ttvo together, are ab, ac, be; ah and ba, though difierent permutations, form the same combination. 453(. Each combination of things when taken two together, as ab, gives two permutations; and when taken three together, as abe, gives 3x2x1 permutations. 328 PERMUTATIONS, COMBINATIONS, BINOMIAL THEOREM. 4S4;. In general, each combination or collection of r things gives \V_ permutations. ^ Thus the combination dbc gives the permutations -abo, acb, etc. ; that is, the permutations of three things taken all together, which by Art. 446 is 1 3, or 3 X 2 X 1. Similarly, it is seen that the combination of r things gives I r permutations. A:55. To find the number of combinations that can be formed out of n things taken r at a time. ■ The number of permutations of n things taken ?• at a time is n{n—l) . . . ." n — r — l) (Art. 444), and each collection or combination of /■ things produces i r permutations (Art. 454) ; hence the number of combinations of n things taken r at a time equals the number of permutations divided by j r ; or, letting Cr represent the number of combinations, we have ^ n(n-l)(n-2) . . . . (n -r-1) Or — ; . [r 4:56. If we, multiply both numerator and deDominator of the previous expression by \n-r, we have a= r \n- 4:57. The number of combinations of n things taken Y at a time is the same as the number of them taken U — r at a time. For the number of combinations of n things taken n — rata, time is that is, n{n- -l)(n- -ii) (n- -n- -r- -1) 1 n- 'JL n(n- -l)(n- -2) {r- -1) Multiply both numerator and denominator by \r, and we obtain In . Cn-r = - 'jz . which, by Art. 456, is the number of combinations of n \r \n — r things taken r at a time. COMBINATIONS. 329 458. Hence in finding the number of combinations taken r together, when r > \n, the shortest way is to find the number taken (;ii — r) together. 459. To find the value of r from which the number of eom- binations of n things taken r at a time is greatest. mi, i- 1 /^ n(n — l)(n — 2) .... ln-r-1) , The formula 0, = -^^ -^ — r—'- ^^ ' may be written Ix2x3x . . . . r „ n 71—1 n-2 n — r — 1 G. = -x--x--....— — . Now, it is seen that the numerators of this formula decrease from right to left by unity, and 'the denominators increase by unity ; hence at some point in this series the factors become less than 1 ; therefore the value of Or is greatest when the product includes all the factors greater than 1. Now, when n is an odd number, the numerator and denominator of each factor will be alternately both odd aind both even ; so that the factor greater than 1, but nearest to 1, will be that factor whose numerator ex- ceeds the denominator by 2. Hence, in this case the value of r must be such that n — 1 n — r—l = r + 2, or r — . 2 When n is even, the numerator of the first factor will be even and the denominator odd ; the numerator of the second factor will be odd and the denominator even, and so on alternately ; hence the factor greater than 1, but nearest to 1, will be the factor whose numerator exceeds the denom- inator by 1. Hence, in this case the value of r must be such that y^ n — 1 — l=r+l, orr=-. Note. — For other principles of Permutations and Combinations see works on Higher Algebra. EXAMPLES. 1. How many combinations can be formed from the letters of the word Prague, taken three together? SoLlTTiON. Here n = 6 and r = 3; hence substituting in the formula, _ wfw-l) . ■ • • n(n-f^i ) 6x5x4 we have 0,= ^ lx'2x3 330 PEEMUTATIONS, COMBINATIONS, BINOMIAL THEOREM. 2. How many combinations' may be made of the letters a, b, e, d, e, /, taken three together ? four together ? Ans 20 ; 15. 3. How many spans of horses can be selected from 20 horses ? How many double spans ? Ans. 190 ; 4845. 4. How many combinations of 3 or of 5 letters can be made out of 8 letters ? Ans. 56, 5. How many combinations can be made of the letters of the word longitude, taken four at a time? Ans. 126. 6. How many combinations may be formed of 16 things taken 5 at a time ? Ans. 4368. 7. How many diiferent parties of 6 men can be formed out of a company of 20 men ? ' Ans. 38760. 8. A guard of 5 soldiers is to be formed by lot out of 20 soldiers ; in how many ways can this be done ? how often will any one soldier be on guard? Ans. 15504; 3876. 9. In how many different ways can a class of 6 boys be placed in a line, one being denied the privilege of the head of "the class? Ans. 600. 10. The number of the combinations of n things taken four together is to the number taken two together as 15:2; find the value of n. Ans. n = 12. .11. The number of permutations of n things taken 5 at a time is equal to 120 times the number of combinations taken 3 at a time ; find n. Ans. n = 8. 12. A and B have each the same number of horses, and find that A can make twice as many different teams by taking 3 horses together as B can by taking 2 horses together ; liow many horses has each? Ans. 8. THE BINOMIAL THEOREM. 400. The Binomial Theorem i^ derived on page 162 by induction ;• we shall now demonstrate it by a more rigid method of reasoning called mathematical induction. 461. By actual multiplication we obtain THE BINOMIAL THEOEEM. 331 (1) (x + aXx + b)=x' + (d + b)x + ab. (2) (x + aXx + b)(^x + c)=xP + (a + b + c)x'' + (a6 + ac + hd)x + abe. (3) {x + a)Q£ + b)(x + c){x + d)=x* + (a+b + c-^d)x\ + {ab + ac + ad + be + bd + cd^x'. + (abc + bed + eda + dab)x + abed- 463. Examining these results, we observe certain laws in the development : 1. The number of terms is one more than the number of bino- mial factors involved. 2. The exponent of Jc in the first term is the same as the number of binomial factors, and decreases by unity in each succeeding term. 3. The coefficient of x in the first term is unity. The coefficient of x in the second term is the sum of the second letters, a, b, c, of the binomial factors. The coefficient of x in the third term is the sum of the products of the second letters taken two at a lime. The coefficient of the fourth term is the sum of the products of the second letters taken three at a time, and so on. 4. The last term is the product of all the second letters of the binomial factors. 463. We shall now show that these laws always hold, what- ever be the number of the binomial factors. Suppose the laws to hold for n — 1 factors, so that (x+n){x+b) .... {x + k) = X^'-+pa:?'~^+qx''-' + raf'^+ . . u, where JO = the sum of the letters a, b, o, ■ . . . A. 5= the sum of the products of these letters taken two at a time. ?- = the sum of the products of these letters taken three at a time. M = the product of all these letters. Then multiply by another factor, a; + Z, and arrange the product accord- ing to the powers of x ; thus, {x+a){x+b)(,x+c) .... {x+k){x+l) = x^+(p-H)oi^' + (q+pl)X^'+{r+ql)3i^-'+ .... +ul. 532 PERMUTATIONS, COMBINATIONS, BINOMIAL THEOREM. The laws (1) and (2) evidently hold in this expression. Now, p+l =a+6 + c+ .... h+l = the sum of the letters a, b, c, . . . . k, I. Also, q+pl = q+l{a + b + o+. . . . +k) = the sum of the products of all the letters a, b, c, . . . . k, I, taken two at a time. Also, r+ql = r-^l{ab + ao+b* = a''(l +2/)", if 2/ = -. We may then ex- pand (1+2/)", and multiply each term by a", and thus obtain the expan- sion of (fls+a;)". THE BINOMIAL THEOEEM. 335 471. The sum of the coefficients of the terms in the expansion of (l+a;)"is 2". For {\ + xY = \ + nx+'^^^^~^^^+ na?^^+a?'. Kow, since this is true for all values of x, it is true when x = l ; whence (l + l)n^2'' = l + w + ^'*— y+ n+1; that is, the sum of the coefficients is 2". 472. The sum of the coefficients of the odd terms in the ex- pansion (1+a;)" is equal to the mm of the coefficients of the even terms. If we let x= —1, the expansion of (l+a;)" becomes 1.2 1.2.3 ^^■' in which we have the sum of the odd coefficients, minus the sum of the even coefficients equal to zero ; hence the two sums are equal. 473. Since the two sums are equal, each is one-half of 2" (Art. 471), or 2" H- 2 = 2'^'. Notes. — 1. It may also be shown that in the expansion of (a + x)'^ the middle term will have the greatest coefficient when n is even ; and the two middle terms will have equal coefficients when n is odd, and be the greatest terms. 2. This demonstration of the binomial theorem is restricted to n being a positive integer. The theorem is also true when n is negative ot fractional ; but the demonstration is too difficult for a work on elementary algebra. We shall assume that n is general, and give examples showing its appli- cation. EXAMFIiES. 1. Expand (h+yY'. Solution. In the general formula (Art. 465) substitute 6 for a, y for X, and — 2 for n, and we have Eeducing, we have (6 ^ryy = j^ - "j? + f I - y, + etc 336 PERMUTATIONS, COMBINATIONS, BINOMIAL THEOREM, 2. Expand (1+2/)*. Solution. In the general formula put 1 for a, y for x, and \ioxn; substitute and reduce, and we have (1 +2/)i = 1 + 1^/ - i2/H t1j3/» — rf ^2/« + etc. 3. Expand (l + 2a;-a;')i Solution. Put^/ for 2K-a;^• then (H-2a;-a;''')i--=(l+2/)i. (1 +2/)i = 1 + i3/ - i2/H tV2/' - jf j2/1 + etc. -l + J(2a;-a:2)-J-(22;-a;'^)HxV(2a;-a;')'-Tf^(22;-a;=i)Hetc. Eeducing, (1 + 2« — a;')^ = I + a; — a;H a;' — fa;* + etc. 4. Expand (1 - xY^. Ans. l+a;+a;^+a^+a;*+a^+ etc. 5. Expand (l+a)-*. Ans. 1 -|a + fa''-,^a'+-j^a*-etc. 6. Expand (l-a;)i . 1 ^_ 2x' _ 2. 5a;° _ 2.5.8a:' _ ^"^•■^"3 3.6 3.6.9 3.6.9.12 7. Expand 7" -^ Jws. a + 2aa; + 3aa;' + 4a.T^ + 5aa;*+etc. (1 - a;)^ 8. Expand lA^^p. ^ns. a - - - -^^ - ^^^ - ^— , - etc. 9. Write the coefficient of of in (1 -x)~'', and coefficient of r- /i w ^1 (r'+l) (r+2)(r- + 3 ) a;' in (1 - a:)-'. Ans. r+1; 1 23 ' 10. Write 4th term of (a; + 21/)", and 6th term of (Sx-y)-'*. Ans. M^)(^.»-y ; _3_L^^l^J_\3.)-i-y. 11. Write the (r-+l)th term of (1 - x)-^ and the 5th term of ro i a\\^ a (r+l)(r+2) , 9:8 . 7. 6„, 5,, , (3x^-4?/^)'. Ans. ^ '-^ -7f; rg 3^a;^4Y. 12. Write the middle term of (a+a;)'", and two middle terms 110 [9 of (a + a;)". Ans. , ■ ,^ aV; .^ (aV-KnV). 13. (l + x-a;'')* = l'+4a; + 2a;''-8a;'-5a;* + 8a;' + 2x«-4x' + a;l 14. (l+a;+xO^ = l + -+-^-— , etc. 15. If the 6th, 7th, and 8th terms of (x+iffaxQ respectively 112, 7, and \, find x, y, and n. An$. x = 4,y = ^, n = 8. PUBLICATIONS OF SOWER, POTTS & CO., PHILADELPHIA. Brooks's Normal Geometry and Trigonometry. By the aid of Brooks's Geometry the principles of this beautiful science can be easily acquiredin one term. 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The articles on Notes and Cards, Titles and Forms of Address and Salutation, are invaluable to every lady and gentleman. Westlake's Common School Literature. A scholarly epitome of English and American Literature, containing a vast fund of informa- tion. More cuUttre can be derived from it than from many much larger works. Lloyd's Literature for Little Folks. The gems of child-literature, arranged to furnish easy lessons in Words, Sentences, Language, . Literature and Composition, united with Object-Lessons. For children in Second Reader. Handsomely illustrated. The boolc is the delight'of all children. > , PRICE PER »ET. Pelton's Outline Maps* $25.00 1. Physical and Political Map of the 'Western Hemisphere 7 ft, by? ft 2. Physical and Political Map of the Xlastern Hemisphere 7 " 7* 3. Map of the United States, British Provinces, Mexico, Central America and the West India Islands 7 " 7 " 4. Map of Surope.... 6 " 7 " 5. Map of Asia 6 " 7 " 6. Map of South America and Afvica 6 " 7 " Pelton's Key to full Series of Outline Maps. Pelton's Key to Hemisphere Maps. This beautiful series of Maps is so well known that a lengthy description seems to be hardly necessary. It is the only set on a large scale exhibiting the main features of Physical in connection with those of Political and Local Geography. Notwithstanding the many outline maps that have been published since Pelton's series originated this method of teaching Geography, the popularity of these elegant maps is undiminished. .^^^ Sample copies ?ent to TeacherE and School Officers for examination upon receipt of two-thirds of retail prices, except those marked (*). Introduction Supplies furnished upon most liberal terms. . Catalogues and Circulars sent free upon applica tion. Correspondence and School Reports solicited. Address SOWER, POTTS & CO., PUBLISHERS AND BOOKSELLERS, B30 Mwrlait St., Philadelphia , 9?