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To accompany Apgar's Drawing Book, consisting of a Triangulation of each of the five Continents, of the great Lakes, of the British Isles, and of France ; measurements of the New England and Middle States, and geometrical figure of the United States. Price per Set, 25 Cts. The Geographical Question Book. Prepared for Warren's Common School Geography, but adapted to all accurate maps. Retail Price, 3a Cts. arV19289 Cornell University Library 3 1924 031 252 988 olin.anx The original of this 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/cu31924031252988 AN ELEMENTARY ALGEBRA BY D. B. JiAGAR, Ph - d -> PRINCIPAL OF STATE NORMAL SCHOOL, SALEM, MASS. PHILADELPHIA : COWPERTHWAIT & COMPANY. HAGAR'S Mathematical Series. Retail Price. I. Primary Lessons in Numbers . . $ .30 II. Elementary Arithmetic 50 III. Common School Arithmetic ... 1.00 IV. Dictation Problems and Key to Common School Arithmetic . i.co V. Elementary Algebra 1.25 VI. Algebraic Exercises and Key to Elementary Algebra. (i N press.) 1.25 VII. Ft tmi/mtattap^ r.vny-n^Tjy q „ T. rrnr , TTnTT ) CORNE LL university!) In continuation of the AdO'De itfiti, a complete series of Higher Mathematical Text-books, designed to meet the wants of College and University Classes, is in preparation by Eminent Professors of Mathematics. Entered, according to Act of Congress, in the year 187S, by DANIEL B. HAGAR and HENRY B. MAGLATHLIN, In the Office of the Librarian of Congress, at Washington. Westcott & Thomson, Deacon & Peterson, Sttreotypcrs and Electrotypers, Philada, Printers, Philada. INTRODUCTION. In this manual the author has endeavored to apply to an elementary course of Algebra the same general plan of treat- ment which was adopted in his Series of Arithmetics. Every important subject is approached by means of simple suggestive questions, which lead the pupil directly to the ap- propriate definitions of terms, and to the principles involved in the subject. This is believed to be a great improvement upon the common plan of presenting definitions dogmatically, without any preliminary consideration of the things to be defined. The plan of combining mental exercises with written work, which has proved valuable in the Arithmetics, is also an im- portant feature of this book. Among other characteristics of the work are the early intro- duction of equations, and the frequent use of practical prob- lems which are designed to interest the learner at every stage of his progress. Great care has been taken to combine clearness and concise- ness of treatment with fullness of topics, so that, in a small compass, there is furnished a course of Algebra sufliciently comprehensive for Grammar-schools, High-schools and Acad- emies. „ INTRODUCTION. Bearing in mind that the work is elementary in its scope, the author has left to the higher Algebra -which is to form a part of this Series of Mathematical Text-books, the full discussion of many subjects that are too abstruse for the easy comprehension of beginners, and has purposely avoided the introduction of nice distinctions, which, though proper enough at a later stage in the course of study, would be likely to perplex the young stu- dent. He has preferred to use the mathematical terms positive and negative in the sense which general usage has established, rather than to attempt a limitation of their signification by applying them only to quantities to denote their essential relations, regardless of the algebraic signs which stand before them. For the convenience of such learners as may desire to study only the most essential portions of Algebra, some topics have been excluded from the body of the work and placed in an Appendix. The author has been led to the preparation of this Course of Mathematics by the popular and just demand for text-books which are brief and yet complete, and which are so graded and mutually consistent as to secure for the learner the greatest economy of time and labor. He ventures to hope that the kind approval so widely be- stowed upon his Series of Arithmetics may be extended to the work now presented to the public. CONTENTS. Section Paoe I. Introduction 7 II. Algebraic Expres- sions 14 III. Algebraic Processes 17 IV. Addition 21 V. Subtraction 28 VI. Multiplication ... 34 VII. Division 42 VIII. Review Problems . . 50 IX. Equations 51 X. Problems 55 XI. Factors 59 XII. Divisors 67 XIII. Multiples 72 XIV. Review Problems . . 75 XV. Fractions 77 XVI. Reductions 80 XVII. Addition 87 XVIII. Subtraction 89 XIX. Multiplication ... 92 XX. Division 96 XXI. Review Problems . . 101 XXII. Simple Equations . . 102 XXIII. One Unknown Quan- tity 105 XXIV. Problems 108 Section Page XXV. Two Unknown Quantities 114 XXVI. Problems 122 XXVII. More than two Un- known Quantities 128 XXVIII. Problems 130 XXIX. Review Problems . 131 XXX. Involution 134 XXXI. Evolution 138 XXXII. Radicals 147 XXXIII. Rationalization . 160 XXXIV. Radical Equations 102 XXXV. Review Problems . 164 XXXVI. Pure Quadratics . 166 XXXVII. Problems 169 XXXVIII. Affected Quad- ratics 171 XXXIX. Problems 179 XL. SimultaneousQuad- ratic Equations . 183 XLI. Problems 188 XLII. Ratio and Propor- tion 191 XLIII. Arithmetical Pro- gression 199 XLIV. Geometrical Pro- gression 204 XLV. Problems 209 XI/VT. General Review . 213 APPENDIX. XLVII. Generalization ... 225 XLiVIII. Negative Solutions 229 XLIX. Indeterminate and Impossible Problems 231 1* L. Imaginary Quanti- ties 235 LI. Binomial Theorem . 238 LII. Logarithms 243 5 Elementary Algebra. SECTION I. INTRODUCTION. ARTICLE 1. — 1. Name some thing that can be measured. -£\. With what can it be measured ? 2. What measures may be used in expressing the distance, or length, between two places ? The area or surface of a farm ? The capacity or volume of a room ? The size of a block of stone ? The time between two dates ? The magnitude of a number ? 3. What do you call every thing that can be measured or computed ? 4. Is 6 a quantity ? 5 miles ? 4 bushels ? Pride ? Space ? Redness ? Length ? 5. The quantity 6 apples contains how many single things ? What is each single thing ? The quantity 10 pounds contains how many single things ? What is each single thing ? 6. What is a single thing in 12 dollars ? In 8 days ? In 4 a's ? In 3 x's 1 In 25 ? 7. What name is given to one of the things of which any quantity is composed ? 8. What is the unit in 3 acres? In 5 dollars? In 11 s's? In 9 tens? 7 INTRODUCTION. 9. Express the number five by single marks ; by a figure ; by a Roman letter. * 10. Express the number or quantity six by marks; by a figure ; by Roman letters. 11. You have now expressed number or quantity in what ways? 12. In the Roman notation, what letter represents one? Five? Ten? Fifty? One hundred? One thousand? 13. Letters, then, may be used to express what? Five is frequently expressed by the letter V, but suppose we agree to express it by a, or b, or c, or any letter whatever; what then will be the value of the letter used ? 14. If the value of a is five, what is the value of 2 a's ? Of 3 a's? 15. If we agree to denote five apples by a and six apples by b, what will a plus b denote ? 16. If we let a stand for seven, 2a for twice seven, 3a for three times seven, and so on, 10a will stand for what? 20a? 4a plus 5a ? 8a plus 2a ? 12a less la ? 9a less 6a ? 17. John has 4 equal piles of apples, each pile containing 15 apples. How many times 15 apples has he ? 18. If we denote 15 apples by b, how many times b will de- note the entire quantity of John's apples ? What, then, will denote the entire quantity ? 19. What will denote the entire quantity if we denote the quantity in one pile by c? By dl Bye? 20. From these examples what may you infer in regard to using the same letter to represent different quantities ? In re- gard to using different letters to represent the same quantity ? 21. George tells me that he has a certain number of cents, and that his sister has twice as many, but he does not tell me the number. INTRODUCTION. 22. If I denote the unknown number of cents which George has by x, what will denote the number which his sister has ? What, then, will denote how many they together have ? 23. If Zx stand for the number they both have, which George at last tells me is 15, what does lx stand for? 2a;? How many cents has George ? How many has his sister ? .24. I walked a certain distance one day, twice as far the second day, and three times as far the third day, and then found I had walked 60 miles. Tell me how far I walked the first day ? 25. What quantity in this case is unknown to you ? If you represent that unknown quantity by y, by what will you repre- sent the distance I walked the second day ? The third day ? What will represent the entire distance ? 26. If 6y stands for 60 miles, for what does y stand ? 2y ? 32/? 27. What letters are commonly used to express known quan- tities ? To express unknown quantities ? Definitions. 2. Quantity is anything that admits of measurement or com- putation. 3. The Unit of a quantity is one of that quantity. 4. A Symbol of a quantity is a character used to express a quantity. 5. Known Quantities are expressed by figures or by the first letters of the alphabet. Thus, quantities expressed by 1, 2, etc., or by a, b, etc., are under- stood to be those whose values are known or determined. 6. Unknown Quantities are expressed by the last letters of the alphabet. 10 INTRODUCTION. Thus, quantities expressed by x, y, z, etc., are understood to be those whose values are unknown or undetermined. 7. Numerical Quantities are those expressed by figures. 8. Literal Quantities are those expressed by letters. 9. A Sign is a character used to express a process or to abbreviate an expression. 10. Algebra is the method of reasoning about quantities by means of symbols employed to express the quantities, and of signs employed to express their relations. Signs. 11. The Sign of Addition is -(-, and is called plus. When placed between two quantities, it means that they are to be added. Thus, a+b is read "a plus b," or 6 added to a. 12. The Sign of Subtraction is — , and is called minus. When placed between two quantities, it means that the one on the right is to be taken from that on the left. Thus, a — b is read " a minus b," or b subtracted from a. 13. The Sign of Multiplication is X, and is read times, or mul- tiplied by. When placed between two quantities, it means that they are to be multiplied together. Thus, a x 6 is read " 6 times a," or a multiplied by b. The sign X is often omitted, and the multiplication is ex- pressed by writing the letters together, the one after the other. Thus, ab expresses the multiplication of a by 6. 14. The Sign of Division is -h, and is read divided by. The dividend is placed at the left of the sign and the divisor at the right of it. Thus, «-?- b is read " a divided by b." INTRODUCTION, 11 Division may also be denoted by writing the dividend above and the divisor below a short horizontal line. Thus, — denotes the division of a by b. b J 15. The Sign of Equality is — -, and is read equals or equal. When placed between expressions, it denotes that they are equal. Thus, a+b = x — y, is read " aplus b equals x minus y." 16. The Signs of Aggregation, the Parentheses, ( ), the Brackets, [ J, and the Vinculum, , denote that the quanti- ties included or connected by them are to be treated as a single quantity, subject to the operation indicated. Thus, (a+c)b, [a+c]6, or a+cxb, each denotes that a+c is to be multiplied by 6. Coefficients and Exponents. 17. A Factor of a quantity is any one of the multipliers used in producing that quantity. Thus, a and 6 are factors of the quantity ab. 18. A Coefficient of a quantity is a number prefixed to the quantity to show how many times the quantity is taken. Thus, in 5a the 5 is the coefficient of the quantity a, and shows that a is taken 5 times. Numerical coefficients are expressed by figures, literal coeffi- cients by letters, and mixed coefficients by both letters and figures. Thus, in 5ab the 5 is the numerical coefficient of ab, and 5a is the mixed coefficient of 6. When no coefficient of a quantity is expressed, the coefficient 1 is understood. Thus, a is the same as la, and bx as Xbx. 19. An Exponent of a quantity is a number written at the right and above the quantity, to denote how many times it is taken as a factor. 1 2 JNTR OD VCTION. Thus, in a 3 the s is the exponent of the quantity a, and shows that a is taken 3 times as a factor ; and in a n , which is read a to the nth, or a, nth, the ™ shows that a is taken n times as a factor. When no exponent of a quantity is expressed, the exponent 1 is understood. Powers and Hoots. 20. A Power of a quantity is the result obtained by taking that quantity one or more times as a factor. Thus, a 2 , which' equals ax a, is the square, or second power, of a ; and a", which equals ax ax a, is the cube, or third power, of a. When a quantity has no exponent expressed, the first power of the quantity is understood. Thus, a is the first power of a. 21. A Root of a quantity is a quantity which, taken as a factor one or more times, will produce the quantity. Thus, a is the second, or square, root of a 1 , which equals ax a, and the third, or cube, root of a s , which equals ax ax a. 22. A Root of a quantity is denoted by the Radical Sign, -j/, or by a fractional exponent. 23. The number written in the opening of the radical sign is the Index of the root. When the radical has no index expressed, the index 2 is understood. Thus, -/a, ■$> a, or a*, denotes the second, or square, root of a, and fa, or afr, denotes the third, or cube, root of a. Axioms. 24. Axioms are self-evident truths. Algebraic reasoning is based upon definitions and the following axioms : 1- If equals be added to equals, the sums will be equal. 2. If equals be subtracted from, equals, the remain- ders will be equal. INTRODUCTION. 13 3. If equals be multiplied by equals, the products ivill be equal. 4. If equals be divided by equals, the quotients will be equal. 5. If a quantity be both increased and diminished by another quantity, the value of the former will not be changed. 6. // a quantity be both multiplied and divided by another quantity, the value of the former will not be changed. 7. Quantities which are equal to the same quan- tity are equal to each other. 8. The whole of a quantity is equal to the sum of all its parts. 9. Like powers and like roots of equal quantities are equal. MXEJtCISES. 25.— Ex. 1. 13 + 16= how many? Arts. 29. 2. 11 + 10 + 9= how many? 3. 17 + 13 - 8 = how many ? Arts. 22. 4. 19 + 6 + 11 = how many ? 5. (18x11)-!- 2= how many? Ans. 99. 6. What is the value of 38 - (10 + 15) ? 7. What is the value of ( 14+14 ) X JJ ? Ans. 112. 4 8. What is the value of 97 - 31 - 22 ? 9. What is the value of 1 l±g + 24 ~ 10 ? Ans. 4. 11 7 10. 5x(8 + 3-6)=howmany? 11. (7 - 2) x (10 - 3) = how many ? Ans. 35. 12. What is the value of 8 + 21-2x8-5? 13. What is the value of (14 + 6 - 8) + (8 - 4) ? Am. S. 14 ALGEBRAIC EXPRESSIONS. SECTION II. ALGEBRAIC EXPRESSIONS. 26. — Ex. 1. If a express a quantity, what will express five times that quantity ? 2. If a express one quantity and b another quantity, by what sign will you connect them to express their sum ? 3. In the expression a+b, what are the parts connected? In a — b, what are the parts connected ? 4. If a = 8 and 6 = 3, what is the value expressed by a+b? Bya-6? Definitions. 27. An Algebraic Expression is a quantity expressed in alge- braic language. Thus, 5a is the algebraic expression of 5 times the quantity denoted by a. 28. The Terms of an algebraic expression are the parts con- nected by the sign + or — . Thus, ab and cb are the terms of ab + cb, and x and y the terms of x-y. Terms are positive or negative according as they have the sign + or — . Thus, in ab — cb, the term ab is positive, and the term — cb is neg- ative. 29. Similar Terms are those which contain the same letters having the same exponents. Thus, 2xy 2 and — 1xy L are similar terms. 30. Dissimilar Terms are those which contain different letters or different exponents of the same letter. Thus, 4a 2 6 and 3cec' are dissimilar terms ; also 4a 2 6 and 4a6 2 . ALGEBRAIC EXPRESSIONS. 15 31. A Monomial is an algebraic expression consisting of one term. Thus, aft 2 , — bx, etc., are monomials. 32. A Polynomial is an algebraic expression consisting of more than one term. Thus, a — c and a + ab'-b are polynomials. 33. A Binomial is a polynomial consisting of two terms. Thus, aV + cd and xy-3x 2 y are binomials. 34. A Trinomial is a polynomial consisting of three terms. Thus, ax+xy — ab and cd+bx+b are trinomials. 35. The Degree of a term is the number of its literal factors. Thus, 2x, which contains only one literal factor, is of the first degree, and 6a6 2 , which contains three literal factors, is of the third degree. 36. The Numerical Value of an algebraic expression is the re- sult obtained by substituting for its letters definite numerical values, and performing the processes denoted by the signs. Thus, the numerical value of 6a+6' 2 — c, when a = 2, 6 = 5 and c = ll, is 6x2+5x5 — 11, which equals 26. EXERCISES. 37. Express algebraically — 1. The sum of x and y. Ans. x+y. 2. The value of 2 times a diminished by b. 3. Five times x, added to three times y, diminished by c times d. Am. 5x+8y-cd. 4. The diiference of a and c multiplied by the sum of a and b. Arts, (a-c) (a+6). 5. Two times a multiplied by 6 square, plus four times b multiplied by c cube. Arts. 2ati i +4be'. 6. a third power into 6 fourth power, plus two times a second power into c, minus three a second power into b third power. 16 ALGEBRAIC EXPRESSIONS. 7. Three times x plus y, divided by seven times the product of a multiplied by b. . Sx+y 8. a minus b, divided by a plus b. rf 9. a minus b, multiplied by c into x. Ans. (a - b)cx. 10. The square root of (a — c). Ans. -y/a — e. 11. The cube root of a, plus the fourth root of x plus y. Ans. -^'a + f^x+y. 12. Write a polynomial of three terms. 38. What are the numerical values of the following expres- sions, when a = 3, b = 5, c = 2, d = 7, m = 2, and n = 3 ? 1. a+b — 2c. Ans. 4- 2. 2a+6 2 +36. 3. 4a*-6 + 5e. -4ns. ^i. 4. (a+6 3 )c. -4ms. 256. 5. (a+d)(b-c). 6. cd(6 + c) + a6. -4ns. Ii5. _ , a+b 7. ad+ . c 2 d + 2c ' 9. c 2 +4a 3 -c 2 a+6 2 . 10. If x = 6 and y = 8, what is the value of 2x+(x+y)x + 151 Ans. 111. 11. If a = 15, 6 = 10 and c = 25, what is the value of 1 /F+ v / 4c-2a? ^Ins. -10. 12. If # = 11, y = 9 and a = 20, what is the value of (»+2/) (#-y)+l/5a? 13. If a = 5, 6 = 6, m = 4 and n = 10, what is the value of 10a+6^l00W- 1 /w i ? ^Ins. 61$. 14. If a = 4, 6 = 10, x = 12 and y = 16, what is the value of («+6)(a6 + 8)+/^-22/? ALGEBRAIC PROCESSES. 17 SECTION III. ALGEBRAIC PROCESSES. 39. A Problem is something to be done or a question to be solved. 40. Algebraic Processes are the means employed in the solu- tion of problems whose conditions can be expressed in algebraic language. PMOBLEMS. 41. — Ex. 1. Henry has twice as many books as Arthur, and both together have 24. How many books has each ? Solution. Let x equal Arthur's x = Arthur's number ; number, then 2 times a;, or 2a;, will Sx — Henry's number. equal Henry's number, and 3x will x + Sx = SA equal the number both together have, g x _ g » which is 24. x= 8, Arthur's number. If 3x equal 24, x, or the number Sx = 16, Henry's number. Arthur has, must equal one-third of 24, or 8, and 2x, the number Henry has, must equal 2 times 8, or 16. 2. A man is four times as old as his son, and the sum of their ages is 55 years. How old is each of them ? Ans. The man, 44 years ; the son, 11 years. 3. A farmer gave for a farm and its' stock S8000, and the farm cost 5 times as much as its stock. What was the cost of each? Ans. The farm, $2500 ; -Its stork, $500. 4. The sum of two numbers is 200, and the larger is 3 times the smaller. What is the larger number ? 5. My salary is $3000 a year, and the portion of it I spend is 4 times the portion I save. What portion of it do I spend ? Ans. 6. My horse and carriage are together worth $750, and the carriage is worth twice as much as the horse. What is the value of each ? 2* 18 ALGEBRAIC PROCESSES. 42. — Ex. 1. A plow, a harrow and a cart cost together $96. The harrow cost twice as much as the plow, and the cart 5 times as much as the plow. What did each cost ? Solution. Let x equal the x = the cost of the plow ; cost of the plow ; then 1x will Sx = the cost of the harrow; equal the cost of the harrow, 5x = the cost of the cart. 5x the cost of the cart, and 8x x + 2x + 5x = $96 the cost of all together, which g x = $ s nwm b a ., the number Willie has, must be one- fifth of 45, or 9, and 6x, or the number John has, must be 6 times 9, or 54. 2. The difference of two numbers is 40, and the larger is 5 times the smaller. What are the numbers ? Ans. 10 and 50. 3. The difference between the ages of a mother and her daughter is 24 years, and the mother's age is 3 times that of the daughter. What is the age of each? 4. Five times the amount of my money diminished by three times that amount is equal to $62. How much money have I? Am. $81. 5. Alice has 36 more books than Susan, and her number is 4 times Susan's number. How many books has each ? 44. — Ex. 1. Alfred and Daniel have together a cents, and Daniel has 4 times as many as Alfred. How many cents has each? Solution. Let x equal Alfred's number; then 4a; must equal Daniel's x^ Alfred! » number; number, and 5a; must equal the num- p:^ D aniel's number. ber both have together, which is a. x+J/X = a If 5a; equals a, x, the number Alfred 5x = a has, will equal one-fifth of a, or |; * = f» Al f red ' s nun^er. and 4x, the number Daniel has, must ^.ta ^.^ ^ equal 4 times — , or — . 5 5 o 2. My horse and carriage cost me a, and the cost of the car- riage was twice that of the horse. What was the cost of each ? 20 ALGEBRAIC PROCESSES. 3. The sum of three numbers is a, the second is 3 times the first, and the third is 4 times the first. What are the num- bers? 4. The difference between two numbers is a, and the larger equals 7 times the smaller. What are the numbers ? „ ^ . a 7a Ans. -, — . 6 6 5. Divide the number b into such parts that the larger part shall be 4 times the smaller. 6. Albert has a more cents than his brother, and his number is 3 times that of his brother. How many cents has each ? Ans. Albert, — : Ms brother, -. 2 '2 7. If a in the last problem equals 12, what is the value of the results? 8. Divide the number m into three parts that shall be to one another as 1, 2 and 3, and find the value of each, if m equals 60. . 1st part, - = 10; 2d part,— = 20; 3d part, — =30. Test Questions. 45. — 1. What is Quantity t The unit of a quantity ? A symbol of a quantity ? What are known quantities ? Unknown quantities ? Nu- merical quantities ? Literal quantities ? What is a sign ? Algebra ? 2. What is the Sign of Addition ? Of subtraction ? Of multiplica- tion ? Of division ? Of equality ? Of aggregation ? 3. What is a Factor of a quantity ? A coefficient ? What coefficient is understood when no coefficient is expressed ? What is an exponent ? What exponent is understood when no exponent is expressed ? 4. What is a Power of a quantity ? A root of a quantity ? How is a root of a quantity denoted ? 5. What are Axioms f Upon what is Algebraic reasoning based ? 6. What is an Algebraic Expression? What are the terms of an algebraic expression? What are similar terms? Dissimilar terms? What is a monomial? A polynomial? A binomial? A trinomial? The degree of a term ? The numerical value of an algebraic expres- sion? 7. What are Algebraic Processes ? What is a problem ? ADDITION. 21 SECTION IV. ADDITION. 46. Addition is the process of uniting two or more quantities to find their sum. CASE I. Terms Similar with like Signs. 47. — Ex. 1. A man earned once two dollars on one day, and twice two dollars on another day. How many times two dol- lars did he earn in the two days ? 2. In one week John saves a certain sum of money, Harry saves twice that sum, and Robert saves three times that sum. How many times the sum do they together save ? If a; stands for the sum which John saves, what should stand for what they all save ? 3. A merchant gained in January a certain amount, in Feb- ruary three times as much, and in March four times as much. How many times the gain in January was the entire gain? Let x represent the gain in January, what will represent the entire gain ? 4. Once any quantity, plus three times that quantity, plus four times that quantity, is how many times that quantity ? 5. Edward lost a number of cents, George lost four times as many, and Richard five times as many. How many times Edward's loss was the loss of the three boys ? If - x represent Edward's loss, what will represent George's loss? Richard's loss ? The entire loss ? 6. A grocer loses by one customer a sum represented by y, by another a sum five times as large, by another a sum seven times as large, and by another a. sum twice as large. "What will represent the whole loss? If y stands for ten dollars, what is the grocer's whole loss ? 22 ADDITION. WRITTEN EXERCISES. 48. — Ex. 1. Henry has 2 apples, Alfred 3 apples and Arthur 9 apples. How many apples have they all ? Solution. 2 apples, 3 apples and 9 apples are 14 apples. 2a But let a denote 1 apple, then la will denote 2 apples, 3a 3a three apples and 9a nine apples. 9a 2a, 3a and 9a are 14a, the apples they all have. l^a 2. Henry has lost two apples, Afred 3 apples and Arthur 9 apples. How many apples have they all lost ? Solution. Let —a denote 1 apple lost; then, —2a will — 2a denote 2 apples lost ; — 3a, 3 apples lost, and — 9a, 9 apples — 3a lost. — 9a — 2a, — 3a and - 9a are — 14a, the apples they all lost. — ij^a 3. What is the sum of la, 6a and 11a? 4. What is the sum of - Gbx, - lObx and - 136* ? 49. Rule for Adding Similar Terms having like Signs.— .Add the coefficients, and to their sum prefix the common, sign, and annex the common literal part. PROBLEMS. (1.) (2.) (3.) (4.) (5.) 3a Ibx % -4s' 2 -26e s 5a bbx by - X* -36c 3 6a 4bx 3y -3s 2 -26c s la bx y -lx* - be" a 2bx J! - a? - 6c 3 22^ - 16.-B 2 (6.) (7.) (8.) (9.) 2a6 2 z — 15ac 4x+ 3y 10a6- 2cd laVx -lOae lx+ fy 5ab- 4cd 13ab 2 x — 9ac 5x+ 2y 8ab- led ab 2 x — ac 3k + ly Sab - lOcd 23ab 2 x 19x + 16y ADDITION. 23 10. What is the sum of bz, bbz and Ibz ? 11. What is the sum of 5ao—d, 3ae — 7d, Bae — 6d and 8ac-12d? 12. Add — 3c(_a — x), -A.c(a-x), -7c(a — x), -8c(a-x) and -10c(a-.r). Ans. -8%c{a-x). CASE II. Terms Similar with unlike Signs. 50. — Ex. 1. A man gained $100 one day and lost $50 the next day. What was the net result of the two days' transac- tions ? If we give to gains the sign + and to losses the sign - , what will be the answer ? 2. If a man gained $50 one day and lost $100 the next day, what was the result ? If we give to gains the sign + and to losses the sign — , what will be the answer ? 3. If John has 20 cents and owes 15 cents, what is his finan- cial condition ? What is it if he has 15 cents and owes 20 cents ? If what one has in possession is regarded as positive and what one owes is made negative, what expression will show John's condition in the first case ? In the second case ? 4. In January a merchant gains a certain sum, in February he loses three times as much, and in March he gains six times as much. Let x stand for the gain in January ; what will stand for the result of the three months' transactions ? 5. Twice any quantity, plus four times the quantity, minus three times the quantity, minus five times the quantity, plus seven times the quantity, is how many times the quantity ? WRITTEN EXERCISES. 51. — Ex. 1. John earned in one week 8 dollars and spent 5 dollars, and the next week he earned 6 dollars and spent 7 dollars. How many dollars had he at the end of the second week? 24 ADDITION. Solution. Let d denote a dollar earned and — da, dollar 8d spent; then, 8d and Qd denote the dollars earned, and — 5(2 —5d and — Id the dollars spent. 6d 8(2 and 6d are 14<2, -Id and — bd are -ltd, and —lid —7d united with 14(2 cancels 12(2 of that quantity. Hence, their 2d sum is 2d, which shows that John had at the end of the week 2 dollars. 2. A speculator gained at one time 200 dollars, at another lost 150 dollars, at another gained 300 dollars, and at another lost 400 dollars. "What was the result of the transactions ? Solution. Let d denote a dollar gained and —da dol- 200d lar lost; then, 300(2 and 200(2 denote the dollars gained, —ISOd and -400d and - 150(2 the dollars lost. SOOd 300(2 and 200(2 are 500c2, and -400(2 and - 150(2 are -JftOd -550(2. 500(2 united with - 550(2 cancels -500(2 of that - 50d quantity. Hence, their sum is — 50(2, which shows that the result of the transactions was a loss of 50 dollars. 3. What is the sum of 3a, la, - 4a, - 3a and 5a? 4. What is the sum of 4ac, lac, - 6ac and - 2ac? 52. Rule for Adding Similar Terms having unlike Signs.— Add the coefficients of the positive and the negative terms separately. To the difference of these sums prefix the sign of the greater, and annex the com- mon literal part. Pit O BLUMS. :i.) (2.) (3.) (4.) 5a labx -6ax 2 13xyz la 3abx -5ax? — llxyz 8a — 5abx 8ax 2 4xyz 3a — 6abx ax 2 3xyz a — labx - 30X 2 — Vlxyz 2a — bax* 5. What is the sum of 3am 2 , 7am 2 , -3am 2 , am' and -9am 2 ? ADDITION. 25 6. What is the sum of Mb, - 4a 3 b, a% 9a 3 b and Ua 3 b ? 7. Add Gac — d, 4ac+3d, 5ac—7d, 4ac+2d and — 5ae+d. Am. Hac — 2d. 8 Add 10o6 - 4cd, Bab + 2cd, - Gab + led and 12o& - 9cd. 9. Add 3x(a — c), 5x(a — c), — 3a;(a — c), — l(te(a — c) and 4x(a — c). -4«s. — x(a — c). 10. Add 4o? - 3, Sx^-ftf+lOi? and a; 2 +6?/ 4 - 62'. ^.«s. 6x , -y i +10*. 12. Find the sum of a; 3 + (a - 6)s» + hx, 14a? - 3(a - 6)a; 2 - 2a; and -5a 3 + 7(a-6> 2 -a;. Ans. 10x 3 +5(a-byjt?+2x. CASE III. Terms SimOar or Dissimilar, with like or unlike Signs. 53. — Ex. 1. A earns one day « dollars, and another day b dollars. How many dollars does he earn in both days ? 2. B gains a dollars on Monday and loses b dollars on Tues- day. What will represent the result of his two days' labor ? 3. Peter has in the morning a cents. During the day he pays out and receives money as follows : Receives 2a cents, pays out b cents, receives a cents, pays out 3a cents, receives 4b cents, receives 5a cents, and pays out lb cents. What will be his financial condition at night ? 4. Sx+4y-2x — 7y, equals what? WKITTJEjy EXMB.CISES. 54. — Ex. 1. What is the sum of 3a, lb and -erf? Solittion. The sum 3a and 76 is 3a +76, and the 3a sum of 3a +76 and — cd is 3a + 76 — cd. Since the 76 terms are dissimilar, the sum admits of no simpler —cd form - Sa+7b-cd 26 ADDITION. 2. What is the sum of 6a +86, 7a-4b + 2ab and 2b-9ab? Solution. For convenience in adding, similar 6a + 8b terms are written in the same column. la — Jfi + Sab Beginning with the column at the right to add, we %b — 9ab find that — 9a6 and +2ab are —Tab; 2b, —46 and 13a + 6b-7ab 86 are 66 ; and la and 6a are 13a. Hence, the sum isl3a+66-7a6. 3. What is the sum of 4a6-a; 2 , 3x 2 — 2ab and 2ax+2bx1 4. What is the sum of 3a+a, 4a+36 + 2c and 2a-46-c? This case evidently includes the two preceding cases, hence the following — 55. Rule for the Addition of Algebraic Quantities.— Write similar terms in the same column, add each column, and connect the results with their proper signs. (1-) (2.) 3ab + 2ba? 2x z + tf+2z* -5ba?-&ay> x*-2y i -3bx 2 8a? +z* 3. Add x 3 - 2ax? + a?x + a 3 ,a?+ 3ax 2 and 2a 3 - ax 1 - 2a?. 4. Add Ba 2 + bae - 2¥, - 2a s - 4ab - b 3 and 5a 8 - 4ac + lab. 5. What is the sum of 3a 2 +4j/a, 5a -2\fx, 5a 2 +12 and b + Zx/x+Tl Ans. 8a?+5a+5-^/x + 19+b. 6. What is the sum of 3a+26-5, a+56-c and 3a-2c + 3? 7. Add cf+ab + V-c, Ztf-W, 4a?+bab and -3a6-36 2 +c. 8. Add 7x + By+8z-4, 5s-7 + 3x-8y, 3y-5x+6-2x and 2-4x+3y-2z. 9. Add a?+ax+x\ 3a 1 -4ax+2x* and a 2 +x 2 +a + x. Ans. 5a?-Sax+4a?+a+x. ADDITION. 27 10. Add ia + b-Z(a-x), 3a+7i-3c, 3&+7(a--z) + 9c-4, 4(a-a) — 2a+7 and a+2(a-x) + b+c-3. 11. Add 13a\x+5y) + 21, - 10a\x+5y) + 16a, - Ba\x+5y) - 8 and —13 — 8a. 12. What is the sum of 7^-16 + 5y, 24:- l /xy-15x ,i +3xy and3z 2 -14 + 28t/? Ans. -Sx^+Sxy-yxy+SSy-e. 13. Add Qa{4cX + ly) + Uab\ Vatf-cd+by, 17a(4x + 7y)-5by, lOby -23ab 2 + 2cd and - a(4x + ly) - ed. ' Ans. 22a(4x+7y) + 6by. 56. Dissimilar Terms having a common factor may be added by enclosing in a parenthesis the sum of the other factors, and annexing the common factor. 1. What is the sum of 2ax-bx+5cx? 2dX Solution, lax is equal to (2a)x, — bx is equal to _ . ( — b)x, and Hex is equal to (5c)». Hence, their sum _ i S (2a-b + 5c)x. ( 2a-b + 6c)x 2. What is the sum of ay 1 , by 1 , - acy 1 and 2xy l ? 3. What is the sum of (a+b)x and as? 4. What is the sum of xy i +ay i , — 3y 2 and - 5e«/ 2 ? Ans. (x — 4a — 3)y\ 5. What is the sum of 4mx-C and b + dxl 6. Add (a-b)+x, c(a-b)-2y and 2x+y. Ans. (l+c^a-fy+Sx-y. 7 Find the sum of (a- Bb+o)x and iax+Sbx. 8. Find the sum of 3(a -b),ax- bx and - xy. 9. Find the sum of dbx+nx, —da? and cx'+dm. Ans. (a&+n)a;+(c-<2)a; 2 +dm, 28 SUBTRACTION. SECTION V. SUBTRACTION. 57. Subtraction is the process of taking one quantity from another, or of finding the difference between two quantities. The quantity from which the subtraction is made is called the Minuend, and the quantity which is subtracted is called the Subtrahend. CASE I. All the Terms Positive. 58. — Ex. 1. A gains 6 dollars and B gains 4 dollars ; how much better off is A than B ? 2. If John pays 3 cents for an orange and sells it for 5 cents, and we give gains the sign + , what will represent his gain ? 3. A gains 6 dollars and B gains 4 dollars ; how much worse off is B than A? 4. If a boy earns 3a dollars and spends 5a dollars, and we give losses the sign -, what will represent his financial con- dition ? WRIXTMN MXMRCI8E8. 59. — Ex. 1. From 11 apples take 7 apples. Solution. — Let a denote 1 apple ; then 11a will 11a denote 11 apples, and la will denote 7 apples. The 7a difference between 11a and 7a is 4a, or 11a — 7a = 4a. fa This is the same as changing the sign of the sub- or, trahend 7a, making it —7a, and then adding it 11a — 7a — %a to the 11a. 2. From 7a take 11a. Solution. Taking from 7a all we can of 11a, there remains 4a to be subtracted, or — 4a. This is the same as changing the sign of the subtra- hend 11a, making it —11a, and then adding it ~^ a to the 7a. 0T,7a-lla=-4a 7a 11a SUBTRACTION. 29 3. From a take b+c. Solution. Taking b from a, we have a— b, and taking also c, we have a — b — e; or a — (b + c) = a — b — c, in which, on remov- ing the parenthesis, the signs of the subtra- hend are changed from + to — 4. From 9a; take lOz. 5. From 136c take 116c. 6. From 14m 2 re 2 take 16m ! »*. 7. From 4a take 3a +26. 8. From a?c+b 2 c take 56 2 c. 9. From 7j' + 3s take 5y 2 + 5z. 10. From ax 2 +bx 2 take aa; 2 +36c+6a: 2 . 11. From 7a+2ac take 6a+9ac+6. 12. From 3a6 + 2a6 2 +ac take 2a6 + 3a6 2 + 9. Ans. ab — atf + ac — 9. a b + c a—b — e or, a-(b + e) = a -b-c Ans. — x. Ans. — Smn 2 . Ans. a- ■2b. Ans. 2y 2 - ■2z. Ans. a — lac -b. CASTS II. One or more of the Terms Negative. 60. — Ex. 1. A loses 6 dollars and B loses 4 dollars ; tow much worse off is B than A? 2. John loses 6 dollars and Bay loses 4 dollars ; if we give losses the sign -, how can we represent how much John is worse off than Bay ? 3. A loses 6 dollars and B loses 4 dollars ; how much better off is B than A? 4. A gains 6 dollars and B loses 4 dollars ; how much better off is A than B ? 5. What is the difference between 6 dollars loss and 4 dollars loss? Between 6 dollars gain and 4 dollars loss? 3* 30 SUBTSA CTION. 6. A gains 6 dollars and B loses 4 dollars ; how much -(-2y* -Sax+lC). 34 MULTIPLICATION. Test Questions. 66. — 1. What is Addition t What is the rule for adding similar terms with like signs ? For adding similar terms with unlike signs ? For addition of algebraic quantities 1 How may dissimilar terms hav- ing a common factor be added ? 2. What is Subtraction f What is the subtrahend ? The minuend ? The difference ? What is the rule for subtraction of algebraic quanti- ties ? How may dissimilar terms having a common factor have their difference expressed ? 3. What does a, Parenthesis indicate? How may the subtraction of any quantity be expressed? When a parenthesis is preceded by minus, how may it be removed from the expression ? What change of signs must be made when any number of terms of an expression are in- closed in a parenthesis preceded by the minus sign ? SECTION VI. MULTIPLICATION. 67. Multiplication is the process of taking one of two quanti- ties as many times as there are units in the other. The quantity taken is called the Multiplicand, the quantity ■which shows how many times the multiplicand is taken is called the Multiplier, and the result is called the Product. The multiplier and multiplicand are called Factors of the product. EXERCISES. 68. — Ex. 1. If a man earns 5 dollars a day, what will lie earn in 3 days ? 2. If a man spends 4 dollars a day, what will he spend in 3 days? 3. If a man earns a dollars a day, and the sign + is given to that which he earns, what will express his earnings for 3 days ? MULTIPLICATION. 35 4. If a man spends a dollars a day, and the sign — is given to that which he spends, what will express his spendings for 4 5? 5. Let x represent a man's gains ; what will represent six times his gains ? 6. Six times the quantity +x is how much? 7. Let — x represent a man's losses ; what will represent six times his losses ? 8. Six times the quantity —a is how much? 9. If y stand for what a man earns in one day, what will stand for his earnings in a days ? In bo days ? 10. If — y represent a person's daily expenses, what will re- present his expenses for 4 days ? For a days ? 11. The result of taking a plus quantity +a times has what sign? The result of taking a minus quantity +a times has what sign ? 12. How much is 3 times +5? 4 times —7? a times plus xl b times minus y 1 13. A plus quantity multiplied by a plus quantity gives what kind of a quantity? A minus quantity multiplied by a plus quantity gives what kind of a quantity ? 14. If you take 3 times 4 dollars and subtract the product, how will you express the result ? 15. What result is obtained by subtracting the product of 5 times 7 apples ? 6 times a dollars ? a times z units ? 16. What is the result of + 6 x ( — 4), if it means that plus 6 is to be taken 4 times subtractively ? 17. A plus quantity multiplied by a minus quantity gives what kind of a quantity ? 18. If a person's income is expressed with the sign +, what sign shall be given to his outgoes ? If gains are plus, what are losses ? If earnings are plus, what are expenses ? 36 MUL TIPLICA TION. 19. A's income is $1000; his outgoes, $400. How will you express his net income ? If from this expression you take away — $400, the outgoes, what will be the net income ? 20. A boy earns 25 cents and spends 15 cents. Express the amount saved. If he does away with his spending, the expres- sion 25 cents - 15 cents will then become what? Taking away — 15 cents from the expression 25 cents - 15 cents, is equiva- lent to adding how much to that expression ? 21. I agreed to pay a boy 12 cents an hour for 7 hours' work, but I find that he played 3 hours out of the 7. How much ought I to pay him ? 22. If — 12 stands for the sum paid for 1 hour's work, what will stand for the sum payable for 7 hours' work ? If the boy worked not 7 hours, but 7-3 hours, - 12 has been taken how many times too many ? Then from — 84 we must take what ? Taking - 36 from - 84 is equivalent to adding what to - 84? 23. What is the result of - 6 x ( - 4), if it means that - 6 is to be taken 4 times subtractively ? 24. What is the result of - 7 x ( - 5), if it means that - 7 is to be taken 5 times subtractively ? 25. If the - sign of a quantity becomes + when the quan- tity is subtracted, a minus quantity multiplied by a minus quantity gives what ? 69. Principles. — 1. The coefficient of the product is equal to the product of the coefficients of the factors. Thus, since the factors of a product may be taken in_any order, the product of 3a by 26 is the same as 3 x 2 x a x b, or 6 x ab, which is 6a6. 2. The exponent of a letter in the product is equal to the sum of its exponents in the factors. Thus, since the exponent of a letter denotes the number of times the letter is taken as a factor, the product of a 3 by a 2 is the same as aaa x aa, or aaaaa, which is a 3 . M VL TIPLICA TION. 37 3. Two factors having like signs give a plus product, and two factors having unlike signs give a minus product. Thus, since a positive multiplier denotes that the multiplicand is to be taken additively, the product of +a by + 6 is found by taking +a as many times as there are units in 6, and adding the result, which gives + ab ; and since a negative multiplier denotes that the multiplicand is to be taken subtractively, the product of — a by — 6 is found by taking — a as many times as there are units in 6, and subtracting the result, which gives + ab. A positive multiplicand taken subtractively gives a negative product, as +ax ( — 6) gives — ab; and a negative multiplicand taken additively gives a negative product, as — a x ( + 6) = — ab. CASE 1. A Monomial by a Monomial. 70. — Ex. 1. What is the cost of 26 tons of coal at 3a dollars per ton? • 3d Solution. The cost must be 26 times 3a dollars. 26 times , 3a is the same as 3 x 2 x a x 6, or 6 x a6, which is 6a6. — — 6ab 2. Multiply 3a 3 by 2a 2 . Solution. The product of 3a 3 by 2a 2 is the same as „ 3 x 2 x a 3 x a 2 , which, by principles (Art, 69), is 6 x a 5 , or 6a 5 . — 3. What is the product of -babe by 36c? 4. What is the product of -7aa; 2 by -36a;? 71. Rule for Multiplying a Monomial by a Monomial.— Find the product of the numerical coefficients of the two factors; annex to this product the different let- ters of both factors, giving to each letter an expo- nent equal to the sum of its exponents in the two factors, and make the result positive when the fac- tors have like signs, and negative when they have unlike signs. 4 38 MVLTIPLICA TION. PROBLEMS. (1.) Sab (2.) -6a 3 (3.) ,9a 2 6 m (4.) ab 1 3x> -l&B* — ab -2afy 3a 2 6" -9a 3 6 m+J 6xy (5.) — 2mn (6.) 86dc 2 (7.) - 6aV (8.) ll&cfc 3 mn 2 2abV - 2eV -36dc 2 9. Multiply -8bd*by -h -8d 1 b-2a+b 18. Multiply a 3 + arty + ay 2 +y s by a;-y. 19. Multiply 2c +£ by 2c+£- 20. Multiply a n + J" by a - b. Ans. a n+1 + ab n - a n b - b n+ '. 76. The multiplication of polynomials may be indicated by enclosing each of the factors in a parenthesis and writing them one after the other ; and the algebraic expression is said to be expanded when the multiplication thus indicated is performed. 1. Indicate the multiplication of 3a6 2 — cd 3 by a^/ 2 +9. Ans. (Sa¥-cd 3 )(xf+9). 2. Expand (2a; 3 -19a; 2 +26a;-16)(a;-8). Ans. 2x i -85x> + 178x i -224x+128. 3. Expand (y*+\)(y+l). 4. Expand (x'+xy+y^x-y). Ans. a?—y*. 5. Expand (3ac-6)(6a;-l). 6. Expand (c™+d")(C"+d"). Ans. tP+tordr+d*. 7. Expand (a+b+c)(a+b + e). 8. Expand (a; 2 -a;+l)(a; 2 +a;+l)(a! 4 -a; 2 +l). Ans. a?+x l +l. 4* 42 DIVISION. SECTION VII. DIVISION. 77. Division is the process of finding how many times one quantity is contained in another ; or, Division is the process of separating a product into two fac- tors, one of which is given. The quantity to be divided or separated into two factors is called the Dividend, the quantity or factor given to divide by is called the Divisor, and the result of the division is called the Quotient. The part of the dividend, if any, remaining undivided is called the Remainder. EXEUCISES. 78. — Ex. 1. How many times is 2a contained in 6a? Solution. — la is contained in 6a 3 times, since the product of 3 by 2a is 6a. 2. How many times is 5x contained in 15a;? In 25z? 3. How many times is a contained in aby ? In lab ? 4. What is the coefficient of the quotient of 6a divided by 2 ? Pf 25a; divided by 5»? 5. If we divide ab by a, what is the quotient ? If we divide aj by b, what is the quotient ? 6. If we omit from the term xy the factor x, by what will that term have been divided ? 7. If we omit from the term a 2 the factor a, what will be the result ? If we omit from a 3 the factor a, what will be the re- sult? 8. If we divide a' by a, what is the quotient ? If we divide a 3 by a 2 , what is the quotient ? 9. If we divide a 5 by a 2 , the quotient is a 3 ; what is the expo- nent of the factor in the quotient ? What is the exponent of DIVISION. 43 the factor a in the dividend diminished by the exponent of the factor a in the divisor ? 10. If we multiply +3 by +2, the product is +6 ; what then is the quotient of +6 divided by +3? by +2? 11. If we multiply -3 by —2, the product is +6; what then is the quotient of + 6 divided by - 3 ? by - 2 ? 12. What is the quotient of + 10a divided by —5a? 13. If we multiply + 3 by - 2, the product is - 6 ; what then is the quotient of - 6 divided by + 3 ? 14. What is the quotient of - 6a divided by + 3a? 15. If we multiply —3 by +2, the product is —6; what is the quotient of — 6 divided by — 3 ? 16. What is the quotient of — 6a divided by — 3a ? 79. Principles. — 1. The coefficient of the quotient is equal to the coefficient of the dividend divided by the coefficient of the divisor. Thus, 15ax -s- 5a = Zx, for Zx x 5a = 15ax. 2. To omit a factor from a term is to divide by that factor. Thus, axy with a omitted, or xy, is the same as axy -s- a, which is equal to xy ; for xy x a = axy. 3. The exponent of a factor in the quotient is equal to its expo- nent in the dividend diminished by its exponent in the divisor. Thus, a 5 -s- a 2 , or a 5 with the factor a? omitted, is equal to a?, for 4. The quotient is positive when the dividend and divisor have like signs, and negative when the dividend and divisor have un- like signs. Thus, + ah divided by & = + a, for + a x ( + 6) = + ab ; — ab divided by — 6 = + a, for + a x ( — 6) = - ab ; + ab divided by — b = — a, for — a x ( — 6) = + ab ; — ab divided by + 6= —a, for — ax( + &) = — ab. 44 DIVISION. CASE I. A Monomial by a Monomial. 80.— Ex. 1. Divide 18a6 by 6a. Solution. Dividing the dividend 18a6 by 6 and a, the factors of the divisor, by rejecting those factors 18ab-*-6a = 8b from the dividend, we have the quotient 36 ; for 36 is such a quantity as, multiplied by the divisor 6a, will give the dividend 18a6. 2. Divide 8a 5 6 by 4a 2 . Solution. Dividing the dividend 8a B 6 by 4 and a?, factors of the divisor, by rejecting those Scfib *- JfO? = %a?b factors from the dividend, we have the quotient 2a»6. 3. Divide -6a 2 6by 2a. Ans. -Sab. 4. Divide 12a 5 6 s by -3a a 6. Ans. -^a s 6 2 . 81. Rule for Dividing one Monomial by another.— Divide the numerical coefficient of the dividend by the co- efficient of the divisor; annex to this quotient the letters of the dividend, giving each an exponent equal to its exponent in the dividend less the expo- nent in the divisor, and make the result positive when the dividend and divisor have like signs, and negative when they have unlike signs. When the dividend and divisor contain an equal literal factor, it is cancelled, and does not appear in the quotient. PROBLEMS. 1. Divide 12a6c by 3c. Ans. J^ab. 2. Divide -12a6cby -4a6, DIVISION. 45 3. Divide 24x 2 y by 3xy. Ans. 8x. 4. Divide — 16x 2 yV by — 4xz. Ans. Jpatfz. 5. Divide 26ax>y 2 by - 2xy. 6. Divide 24aW by - 3aW. Ans. -8a*bV. 7. Divide 20aW«/ 3 by 5&V*/. 8. Divide a n 6 3 by ab. Ans. a" Ll b 2 . 9. Divide -24a 2 6nVby -3a»V. 10. Divide -Fbyb\ Ans. -b. 11. Divide a m+ " by a 2 "- -4ms. a™ - ". 12. Divide -12a 2 6y z 3 by -abtfz. 13. Divide a n+1 6 B by a& 2 . .4ns. a n b n ~\ 14. Divide -9(a + 6) 2 by -3(a+6). 15. Divide 12(a"-6 m ) 2 by A{a n -b m ). 16. Divide (x-yf by (x-yf. 17. Divide 8a m -W by - 2a m+I 6 2 c". 4ns. - 4a- 2 &"- 2 c 3 -". CASE II. A Polynomial by a Monomial. 82.— Ex. 1. Divide 6a 2 6 2 -4a 2 6+2a 2 c by 2a. Solution. Dividing the first term of the 2a )6ab 2 — 4a 2 b +@a?a dividend by 2a, we have 36 2 ; dividing the 3b' -Sab +ac second term by 2a, we have — lab; and dividing the third term by 2a, we have ac. Uniting these results by their proper signs, we have, as the entire quotient, 36 2 — 2a6 + ac. 2. Divide a?+ac by a. Ans. a + c. 3. Divide 6a* - 3aV by 3a". Ans. 2a? - ax\ 46 DIVISION. 83. Rule for Dividing a Polynomial by a Monomial.— Di- vide each term of the dividend by the divisor, and connect the several results. PROBLEMS. 1. Divide Ax* - 8a; 2 + 16a; by 4a;. Ans. x 2 -2x+ 4. 2. Divide 3a 4 - 12a' + 15a 2 by -3a 2 . Arts. -a?+4a- 5. 3 Divide 16a 2 6-30a 3 6 by 4a6. 4. Divide 4aa;y — 4a+12a6 by 4a. Ans. xy — l+Sb. 5. Divide -14aa; 3 +7a^ by -Ix. 6. Divide 10a 2 a;-15a: 2 by 5x. Ans. 2a? -8x. 7. Divide 15a 2 6c-10a s 6 4 cy+5aW by -babe. 8. Divide 4a 2 6-6ao 2 by -2ao. 9. Divide ab 2 +ac — a by a. 10. Divide 2a{x+yf+Qa\x+y) by 2a. Ans. (x+yy+Sa(x+y). 11. Divide aa; 2 (c - d) — a?x(e - a" 2 ) by ax. 12. Divide x n+1 -x n+2 +x n+s -x n+4 by x\ Ans. x-x 2 +x 3 — x*. CASE III. A Polynomial by a Polynomial. 84.— Ex. 1. Divide x 3 + 2x> -3a; by a; 2 +3a;. Solution. The divisor and dividend x i +3x)x s +2x' i -8xi < x-l for convenience are arranged according to x^+Sx 1 the powers of x, beginning with the high- _ ^a _ ## est power. Since X s , the first term of the —x^ — Sx dividend, must equal tile product of a; 2 , the first term of the divisor, by the first term of the quotient, we divide x 3 by a; 2 , and have x for the first term of the quotient, x times the whole divisor is a; 8 + 3a; 2 , which subtracted from the whole dividend leaves —x 2 — 3x. DIVISION. 47 Since —a; 2 , the first term of this new dividend, must equal the pro- duct of x 2 , the first term of the divisor, by the second term of the quotient, we divide — x 2 by a: 2 , and have — 1 for the second term of the quotient. — 1 time the whole divisor is —a: 2 — 3a;, which subtracted from the last dividend leaves no remainder. Hence, the quotient is a;— 1. 2. Divide a 2 + '2ab + ft 2 by a + b. 3. Divide a 2 - lab + b 2 by a - b. Ans. a+b. Ans. a — b. 85. Rule for Dividing a Polynomial by a Polynomial. — Ar- range the divisor and dividend according to the powers of one of their letters. Divide the first term of the dividend by the first term of the divisor, and write the result for the first term of the quotient; -multiply the whole divisor by it, and subtract the product from the dividend. Consider the remainder as a new dividend, find the second term of the quotient in like manner as before, and thus continue till the first term of the divisor is not contained in the dividend. If there be at last a remainder, write it, with the divisor under it, as a fractional part of the quotient. PROBLEMS . Divide a? - a 3 by x - a. 2. Divide a 2 +x 2 by a — x. Solution. x - a)x 3 — a 3 [a; 2 + ax + a' x s — ax 2 a Solution. -xjcf + x 2 {a+x+ ■ a — x ax 2 — a 3 Oi -ax ax 2 —a 2 x ax + x 2 a 2 x—a? ax — x 2 a 2 x—a* 2x 2 3. Divide a 2 -x % by a -x. Ans. n + r. 48 DIVISION. 4. Divide a; 2 +6a;+5 byz+1. 5. Divide m 3 - 10m 2 + 27m, - 18 by m - 6. -4«s. m 2 — J/,m + 8. 6. Divide 6a; 2 +13a^+62/ a by 2»+3y. .Atis. 8x+Sy. 7. Divide a 2 +2a& + 6 2 by a + b. 8. Divide a 3 — y 3 by a; - y. Ans. x'+xy+y 2 . 9. Divide a; 2 — a; -6 by a;- 3. 10. Divide b - 36 2 + 36 s - b* by b - 1. ^n*. -b+2b 2 - b\ 11. Divide 3a 2 - 13aa; + 14a; 2 by 3a - 7a;. J.ns. a - £c. 1 2. Divide a; 2 - 7a; + 12 by x - 3. 13. Divide a; 2 +a;-72 by x+9. Am. x-8. 14. Divide 7a? - 24a; 2 + 58a; -21 by 7a; -3. Ans. x 2 -8x + 7. 15. Divide m*+4m+3 by m 2 +2m+l. 16. Divide a; 4 -13a; 2 +36 by a; 2 +5a;+6. 17. Divide a 2 +2a6 + 6 2 -e 2 by a + b-c. 18. Divide x l +y i by x+v. Ans. x 3 — x 2 y+xy* — y 3 + — 2— x + y Zero and Negative Exponents, and Reciprocals. 86. The Reciprocal of a quantity is the quotient arising from the division of 1 by that quantity. Thus, the reciprocal of a is -, and the reciprocal of a+b is . a a + b 87. A Zero Exponent, or °, is used to preserve the trace of a letter, which might otherwise disappear in the process of division. Thus, the quotient of x s y 2 divided by x*y may be written xfiy. DIVISION. 49 88. A Negative Exponent of a letter in a quotient indicates that the exponent of that letter in the divisor is greater than it is in the dividend. Thus, the quotient of a? divided by a 5 is a* -5 , or a~ z 89. Principles. — 1. Any quantity whose exponent is zero is equal to 1. For, a n -f- a" is a"-" = a" ; also, a™ •*■ a" = 1. Hence, cp = l. 2. Any quantity with a negative exponent is equal to the recip- rocal of that quantity with an equal positive exponent. a 3 1 1 For, a? -*-(£ = a-' 1 ; also, a 3 ^-a 5 = -— = — . Hence, a -2 — — -, and a? a 1 a' 3. Any factor may be transferred from the divisor to the divi- dend, or from the dividend to the divisor, if the sign of its expo- nent be changed from + to — , or from — to +. PROBLEMS. 1. Divide -a&byaft. Ans. — a°6°, or — /. 2. Divide Sab 2 x by 2aW. 3. Express as a fraction the value of x~ m . Ans. — . 4. Express as a fraction the value of y~ 3 . 5. Express without a fraction the. value of ■ — - . Ans. x"' m . 6. Express without a negative exponent — - a~ s b'\ o 7. Divide -SSa-'b-'xby lla 3 b^ Ans. -Sa^Vx. 8. Divide 4a*6 2 by 2aW. 9. What is the reciprocal of 5a 2 ? Ans. , or 5a~'. oar 10. Divide 25aty by 5*y. 5 50 REVIEW PROBLEMS. SECTION VIM. REVIEW PROBLEMS. 90.— Ex. 1. What is the sum of 2a +36 -4c, -3a + 46-c, 4a+76 + 7e,a-6-4cand -5a + 26-6c? Ans. -a+15b-8c. 2. From 5x* - 2xy + By 1 subtract -4x?-2xy+7y 1 3. Prom 2a; 2 - Bxy + y take 4a; 2 + 4xy - 2y*. 4. What is the sum of 7 - xy, ab + 2 and Zxy — 8c - 6 ? 5. From ax — be+3ax+7be take 46c — 2aa:+6c + 4aa;. 6. Multiply c+1 by c+1. -4ras. c 2 +^c + i. 7. Expand (m — n)(m — w). 8. Divide 12a; 4 -192 by 3a; -6. 9. Multiply x*+4xy+y 2 by a; +2/. -4ns. x'+Sx'y+Szyt+i/ 3 . 10. Multiply a 3 + a 2 6 + a6 2 +6 s by a -b. 11. Divide 20a 4 -12a s a; by 4a" 3 . 12. Divide a' - 6 2 - 26c - c 2 by a - b - c. 13. Simplify the expression 15a 2 6° + c"d 3 -5a 2 +a;. 14. What is the reciprocal of x^zl 15. What is the product of 7x+3y~4z by 7a; + 4a? 16. What is the value of (1 + 2a;+a; 2 )(a; - 1) ? 1 7. What is the value of (a? + 6 8 ) + (a 2 - ab + 6 2 ) ? Ans. a+b. 18. What is the value of (9aa^+7a 3 6eW)(4aW) divided by 2a 3 6? Ans. lSab-Wxy + UaVtfx. 19. Divide a; 4 +(26 2 -a 2 )a; + & 4 by x 2 +ax+b\ EQUATIONS. 51 20. What is the value of (a-2)(a-l)(a+l)(a + 2) divided by a 2 -a-2? Test Questions. 91. — 1. What is Multiplication f The multiplicand? The mul- tiplier ? The product ? What are the factors of the product ? What principles of multiplication are given ? What is the rule for multiply- ing a monomial by a monomial ? A polynomial by a monomial ? A polynomial by a polynomial? How may the multiplication of poly- nomials be indicated ? 2. What is Division ? The dividend 1 The divisor ? The quotient ? A remainder ? What principles of division are given ? What is the rule for dividing a monomial by a monomial? A polynomial by a monomial ? A polynomial by a polynomial ? 3. What is the Reciprocal of a quantity? What is a zero expo- nent? A negative exponent? To what is any quantity equal whose exponent is zero ? To what is any quantity with a negative exponent equal? How may any factor be transferred from the divisor to the dividend ? SECTION IX. EQUATIONS. 92. — Ex. 1. By the use of what sign can you express the equality between two quantities ? 2. What does Sx + 6x = 50 -5 express? What terms in the expression are on the left of the sign of equality ? What terms are on the right of the sign of equality ? 3. If the terms on each side of the sign of equality in the expression Bx + 6x = 50 -5 be united, what will the expression become ? 4. If 9x be equal to 45, what is the value of a? If Bx+Qx be equal to 50 - 5, what is the value of xl 52 EQUATIONS. Definitions. 93. An Equation is an expression of the equality of two quantities. The first member of an equation is the quantity on the left, and the second member is the quantity on the right, of the sign of equality. Thus, 4x + 10a; = 70 is an equation of which 4a; + 10a; is the first mem- ber, and 70 is the second member. 94. The Transformation of an equation is the process of changing its form without affecting the equality of the two members. 95. The Solution of an equation is the process of finding the value of the unknown quantity or quantities contained in the equation. Transposition of Terms. 96.— Ex. 1. What is the value of a;+2-2? What is the value of 5-2? 2. If x + 2 = 5, what will express the taking of 2 from each member of the equation ? What will the equation become on uniting the terms of each member ? 3. If 7 be taken from each member of the equation x+ 7 = 15, what will the equation become ? 4. The sum of two numbers is 20, and the larger number is 4 more than the smaller. What is the smaller number ? 5. What is the value of x — 5 + 5 ? What is the value of 7 + 5? 6. If x - 5 - 7, what will express the adding of 5 to each member of the equation ? What will the equation become on uniting the terms of each member ? 7. John gave away 5 books, and had 13 left. How many had he at first ? EQUATIONS. 53 8. Andrew has 2 less doves than Charles, and both together have 12 doves. How many doves has each? WRITTEN EXERCISES. 97. — Ex. 1. Transpose - c in the equation x+a — c = b to the second member of the equation. Solution. Adding e to both members, which ° 7 SB + Gt C — D will not affect their equality (Ax. 1, Art. 24), we have x + a = 6 + e, in which — c has been transposed with the sign changed. x + a ~ b + c 2. Transpose c in the equation x+a=b+c to the first mem- ber of the equation. Solution. Subtracting c from both members, which will not affect their equality (Ax. 2, Art. 24), _ we have x + a — c = b, in which e has been trans- posed with the sign changed. a — c — o 3. Transpose cy in the equation a+by = x — cy to the first member of the equation. Ans. a+by+q/ = x. 4. Transpose ab in the equation ab + cx = 2ab to the second member of the equation. Ans. ex = 2ah — ab = ab. 98. Rule for the Transposition of the Terms of an Equation. —Any term may be transposed from one member of an equation to the other, if its sign be changed. PBOBZEMS. Transpose the terms of the following equations, so that all the terms containing the unknown quantity x shall be in the first member, and all the other terms in the second member. 1. 3d—2be = ac-6x. Am. 6x = ae+2bc — 8d. 2. 5x +50 = 4^+56. Ans. 5x~4x = 56-50. 3. 4z + 12 = 2z-:k+21. 4. fix - 7a = 2ab - ax. Ans. 5x+ax = 2ab + 7a. 5* 54 EQUATIONS. 5. 48 -2* = 60 -4a;. 6. 2a;-46 = a;+2. Ans. 2x-x = 4b+2. 7. What is the value of a; in the equation 2x + 42 = 66 — Ax ? Solution. Transposing, we have 2a; + 4a; = 66 2x+42=66—4x — 42; uniting terms, we have 6a; = 24, and divid- 2x+4x = 66— Jfl ing by the coefficient of the unknown term, which 6x = SJf does not affect the equality of the members (Ax. 4, x = 4 Art. 24), we have a; = 4. This value of x may be verified by substituting it in the given equa- tion. Thus, 2x4+42 = 66-4x4, or 8 + 42 = 66-16. 8. Given 5a; - 4 = 2a; + 14, to find x. Ans. x = 6. 9. Given 2a; - 1 = 5a; - 7, to find x. 10. Given 11a; + 16 = 9x+ 26, to find x. Ans. x = 5. 11. Given 5a; + 50 = 4a; + 56, to find x. 12. Given 4a; - 8 = - 3a; + 13, to find x. Ans. x = 3. 13. Given Sax— Sab = 12c, to find x. Ans. x = 6 + — . a 14. Given 4a;- 8 = 3a; +20, to find x. 15. Given 5a; - 15 = 2a; + 6, to find x. Ans. x = 7. 16. Given 40 - 6a; - 16 = 120 - 14a;, to find x. 1 7. Given ax + a = aft, to find a;. Ans. x = b — l. 18. Given a;+55-2a; = 5a;-ll, to find x., 19. Given (x + 2)3 = (a; - 2)7, to find x. Ans. x = 5. 20. Given nx — ex = nx — 2x, to find x. 21. Given 2a; - 2(30 - x) = 60 - 2a;, to find x. Ans. x = £0. 22. Given (2a;+8)5 = (32+a;)3, to find x. 23. Given (6a; - 7)(2a; - 3) = (4a; - 5)(3a; - 4), to find x. 24. Given 12a; - 8 - (8a; - 6) - (12 - 3a;) = 0, to find x. 25. Given (x - 7)(x - 3) - 1 = (x - 9) (x - 2), to find x. EQUATIONS RE QUIRING TRANSPOSITION. 55 SECTION X. PROBLEMS PRODUCING EQUATIONS RE- QUIRING TRANSPOSITION. 99. The Statement of a problem consists in expressing in algebraic language the conditions of the problem. An equation is thus obtained whose solution is the solution of the problem. 100. — Ex. 1. What are the two numbers whose sum is 42, and whose difference is 18 ? X = the smaller number ; x+ 18 = the larger number. Solution. Let x denote the smaller number; then a; + 18 will denote the larger. By the con- ditions, x+x+ 18 = 42. Transpos- x + x + 18 = ^M 2x = 2h ing and uniting terms, we have 2x = 24, and dividing, x - 12 and * = f ' ^ e f^ nun f er ' ,o_oa x+18 = 30, the larger number. 2. Divide 68 into two such parts that 84 diminished by the greater shall be equal to 3 times 40 diminished by the less. Solution. Let x denote the less x = the less part; part, then 68 — x will denote the 68 — x = ihe greater part. greater part, and by the conditions of 81),— (68 — x)=3{40—x) the question 84- (68- a) =3(40 -a), 16+x = 120-3x or 1 6 + x = 1 20 — 3x. Transposing and fa = 104 uniting terms, we have ix = 104; x = 2Q whence, a; = 26 and 68-2 = 42. 68-x = J$ 3. Divide 100 into two such parts that 120 diminished by the greater is two times the less. 4. Three men engaged in a speculation requiring an outlay of $8500. Of that amount A contributed a certain sum, B con- tributed as much as A and $1000 more, and C contributed as much as both A and B, lacking $1500. How much did each contribute? Arts. A, $2000 ; B,$S000; C, , 56 EQUATIONS REQUIRING TRANSPOSITION. 5. Divide 179 into two such parts that one part shall exceed twice the other by 17. 101. General Rule for Solving Problems Producing Equations. —Denote one of the unknown quantities by x; and from the given conditions find an expression for each of the other unknown quantities, if any. Express in algebraic language the processes that would be necessary to verify the value of the un- known quantity or quantities, if they were already known. Determine the value of the unknown quantity or quantities in the equations thus formed. PROBLEMS. 1. A farmer has 400 sheep so distributed into two flocks, that the greater number increased by 25 is equal to twice the less diminished by 4 times 25. What is the number in each flock ? Ans. 225 in the greater ; 175 in the less. 2. A has $800, and B $500 ; what sum must B give to A in order that A may have twice as much as B ? 3. Two persons, A and B, speculate with equal sums of money; A gained $126 and B lost $87, and then A had twice as much money as B. What sum had each at first ? 4. B is twice as .old as A, and 22 years ago he was 3 times as old. What are their ages ? Ans. A's, 44- yenrs ; B's, 88 years. 5. A father is 3 times as old as his son ; 4 years ago the father was 4 times as old as his son was. What is the age of each? 6. A is twice as old as B, and 7 years ago their united ages amounted to as many years as now represent the age of A. What are their ages ? EQUATIONS REQUIRING TRANSPOSITION. 57 7. A laborer engaged to work for 40 days, upon the condi- tions that for every day he worked he should receive $2 and board, but for every day he was idle he was to pay $.80. At the end of the time he received $38. How many days did he work, and how many was he idle ? Solution. x= number of days he worked, 40— x = number of days he was idle; $%x=sum earned, $£0{lfi— x) = sum paid. $2x - $.80{Jft -x) = $38 * f2.80x = $70 Ve=S5, number of days he worked; JfO—x=lS, number of days he was idle. 8. A person hired a laborer to do a certain work on agree- ment that for every day he worked he should receive $2, but for every day he was idle he should forfeit $.50. He worked twice as many days as he was idle, and received $42. How many days did he work ? 9. A crew which can pull at the rate of 9 miles an hour on still water finds that it takes twice as long to go up a river as to go down. At what rate does the river flow ? Ans. S miles an hour. 10. From two towns, which are 187 miles apart, two travelers set out at the same time with the intention of meeting. One of them goes 8 miles and the other 9 miles a day. In how many days will they meet ? 11. If $56 should be added to the money I have, its amount would be trebled ; how much money have I ? Ans. $28. 12. The sum of $310 was raised by A, B and C together ; B contributed $30 more than A, and C $40 more than B. How much did each contribute? 13. In a mixture of wine and water there are two gallons of wine for every three gallons of water. If a gallon of wine be 58 EQUATIONS REQUIRING TRANSPOSITION. added, the mixture becomes half wine and half water. Of how many gallons did the mixture consist ? Ans. 5 ; 2 of wine and 3 of water. Note. Let 1x = number of gallons of wine, Zx = the number of gal- lons of water, and 2x + 1 = the number of gallons of wine after 1 gallon had been added. 14. After 34 gallons had been drawn out of one of two equal casks, and 80 gallons out of the other, there remained just 3 times as much in one cask as in the other. How much did each cask contain when full ? Ans. 103 gallons. 15. After A had received $10 from B,,he had as much money as B and $6 more, and together they have $40. How much had each at first ? 16. A courier who travels 50 miles a day had been gone 5 days when a second was sent to overtake him, in order to do which he travels 75 miles a day. In what time will the second overtake the first ? 17. A garrison had such a quantity of brea"d as would last 6 weeks, if distributed to each man at the rate of 10 ounces a day. Having immediately discharged 1200 men, the com- mander found that he had bread enough to last 8 weeks, allow- ing each man 12 ounces per day. What was the number of men at first in the garrison ? 18. A has $600 and B $460 ; if A increases his capital by $4 per month, and B increases his by $1 per month, in how many months will A have twice as much money as B ? Ans. 160. Test Questions. 102. — 1. What is an Equation f Which is the first member of an equation ? The second member ? 2. What is the Transformation of an equation ? The solution of an equation ? The rule for the transposition of the terms of an equation ? 3. Of what does the Statement of a problem consist ? What is the general rule for solving problems producing equations? FACTORS. 59 SECTION XI. FACTORS. 103. — 1. Of what two integers is ab the product? 2. What two integers multiplied together produce xy ? What three produce abc ? 3. What integers are factors of bed ? Of 2a; ? 4. Of what integers other than 1 is a'x the product ? 5. What common factors other than 1 have Bac 2 d and a?cx ? 6. Give the sets of integral factors greater than 1 which when multiplied together will produce 4a 2 6 4 . 7. Of what quantities are a 2 and d the factors ? 2, x and y, the factors ? 8. In a 2 how many times is a a factor? In 16a; 3 how many times is 2 used as a factor ? How many times is x used as a factor ? Definitions. 104. The Factors of a quantity are the integers which, being multiplied together, will produce that quantity. ■105. A Prime Quantity is an integer that has no integral factors except itself and 1. Thus, 17, 23 and a are prime quantities. 106. Quantities are prime to each other when they have no common factors except 1. Thus, 17 and 20, and a and 6 are prime to each other. 107. A Composite Quantity is an integer that has other fac- tors besides itself and 1. Thus, 12 and ab are composite quantities. 60 FACTORS. 108. The Square of a quantity is the product obtained by taking the quantity twice as a factor. Thus, 4 is the square of 2 ; a 2 , the square of a. 109. The Square Root of a quantity is one of the two equal factors of that quantity. Thus, a is the square root of a 2 , and Sab is the square root of 9a 2 6 2 . 110. One quantity is divisible by another when the latter is a factor of the former. Thus, a 2 b is divisible by a, a 2 or 6. COMPOSITION. 111. A Theorem is something to be proved or demonstrated. 112. Composition in algebra is the process of producing com- posite quantities. Composite quantities may be produced by actual multiplica- tion, and, in some instances, by abridged methods derived from the following theorems : Theorem I. 113. The square of the sum of two quantities is equal to the square of the first, plus twice the product of the first by the second, plus the square of the second. For, let a and b represent two quantities, then a+b will denote their sum, and (a+b) 2 = {a+b){a + b) = a?+2ab + b\ PROBLEMS. 1. Find the square of m+n. Ans. m 2 +2mn+n i . 2. Find the square of a+e. 3. What is the square of x+2y? Ans. x'+4xy+4y 2 - 4. What is the square of c + 3 ? 5. What is the square of 3 + a*? Am. 9+6a 3 +a*. 6. What is the square of 3a +26? FACTORS. 61 Theorem II. 114. The square of the difference of two quantities is equal to the square of the first, minus twice the product of the first by the second, plus the square of the second. For, let a and b represent the two quantities, then a — b will denote their difference, and (a-b)' = {a-b)(a-b) =a?-2ab + b' 1 . PMOSZEMS. 1. Find the square of x- y. Ans. x*-2xy+y*. 2. Find the square of m - n. 3. What is the square of a 2 - 1 ? Ans. a 1 - 2a 2 + 1. 4. What is the square of 3a 4 - 8a 2 ? 5. Expand (5^ -2cd)(5x i -2cd). Theorem HI. 115. The product of the sum and difference of two quantities is equal to the difference of their squares. For, let a and 6 represent the two quantities, then a+b will denote their sum, and a—b their difference, and (a + 6) (a— 6)= a 2 — 6 2 . PSOBZEMS. 1. Find the product of (x+y) by (x-y). Ans. x 1 -y 1 . 2. Find the product of (m+w) by (m-m). 3. What is the product of (2a +6) by (2a- 6) ? Ans. fap-b*. 4. What is the product of (w*+l) by (n 4 -l)? 5. Expand (2a -46)(2a+46). 6. Expand (7x 1 +4y)(7x 2 -4y). 62 FACTORS. FACTORING-. 116. Factoring is the process of finding the factors of a com- posite quantity. Composite quantities may be factored by actual division, and, in some instances, by abridged methods derived from the pre- ceding theorems (Art. 113-115), and, also, by the following theorems. Theorem I. 117. The difference of any two equal even powers of two quan- tities is divisible by the sum of the quantities. For, let a and b represent two quantities, a being greater than b ; then, by actual division, {a--b i )^-(a+b) = a-b, (a i -b i ) + {a+b) = a?-a;'b+aV i -b 3 , (a«-6 6 )H-(a+6) = a 6 -a*6 + a s 6 2 -a 2 6 3 +a6 4 -6 5 , and so on. Theorem, II. 118. The difference of any two equal, powers of two quantities is divisible by the difference of the quantities. For, let a and b represent two quantities, a being greater than b ; then, by actual division, (a 2 -6 2 )4-(a-6) = a+6, (a 3 -& 3 )-(a-6)=a 2 +a&+& 2 , (a 4 - &') -*- (a - b) = a? + a 2 & + a& 2 + b\ and so on. Theorem III. 119. The mm of any two equal odd powers of two quantities is divisible by the sum of the quantities. For, let a and 6 represent two quantities, a being greater than b ; then, by actual division, (a 3 +6 s )-(a+6) = a 2 -a&+6 2 , (a 6 +6 s )H-(a+6) = a*-a 3 6+a 2 6 2 -a6 3 +6 4 , (a' + V) + (a + b) = a 6 - a 5 6 + a 4 & 2 - a s 6 3 + a?b l - a& 5 + 6 6 , and so on. FACTORS. 63 CASE I. A Monomial into its Prime Factors. 120.— Ex. 1. Find the prime factors of lOaV. Solution. The factors of 10 are 10 = 2x r > 2 and 5: of a 2 , a and a; and of b 3 , „i — „„„ b, b and 6. Hence, the prime fac- b 3 = bxbxb tors of Wd'b 3 are 2x5xaxax& 10a?b 3 = 2 x 5 x a x a x b x 6 x b xbxb. 2. Find the prime factors of 6 — 48m'nx": Ans. 8m 2 nx*(J}i/ + i/ t + 6). CASE III. A Trinomial into two Equal Binomial Factors. 124. A Trinomial can be resolved into two equal binomial factors when two of the terms are squares and positive, and the other term is twice the product of their square roots. 125.— Ex. 1. Find the factors of a*+2ab + b\ Solution, a* is the square of a, b 2 is the square of b, and 2ab is twice the product of a and 6. Hence, a* + Sab + V- = the trinomial is, by Art. 113, the square of (a + 6), and (a+6)(a+6) has for its two equal factors (a+6) and (a+6). FACTORS. 65 2. Find the factors of a 2 -2a&+6 2 . Solution, a 2 is the square of a, & 2 is the square a 2 — Sab + 6 2 = of 6, and — 2a6 is minus twice the product of a and (a — b){a — b) b. Hence, the trinomial is, by Art. 114, the square of (a — b), and has for its two equal factors (a — b) and (a — 6). 3. Find the factors of a? — 2xy+y i . 4. Find the factors of e*+2ed+d\ 126. Rule for Resolving a Trinomial into two equal Binomial Factors. — Find the square roots of the square terms, and if twice the product of these roots equals the other term, these roots connected by the sign of the other term will be one of the two equal factors. PROBLEMS. 1. Find the factors of a 2 + iax+4x^. Ans. (a+2x) and (a + 2x). 2. Find the factors of 25a 2 - 10* + 1. 3. Find the factors of c 2 -2c+l. Ans. (c — 1) and (c- 1), 4. Factor l+A-xy+Ax , y'. 5. Factor 1 - 2a 2 + a\ Ans. (1 - a 1 ) and (1 - a 2 ). 6. Kesolve9a 2 o 2 +24a 2 M + 16a 2 factors. Hence, by Prin. 1, Art. 136, the greatest common divisor is the product of 3, a 2 and b 2 , or 3a 2 6 2 . DIVISORS. 69 2. Find the greatest common divisor of 6a 3 xy and 18a 2 by 2 . Am. 6a 2 y. 3. Find the greatest common divisor of 15x 5 y 2 , 18ax*y and 12&zy. Am. Sx'y. 138. Rule for Finding the Greatest Common Divisor of Mono- mials.— Resolve the quantities into their prime fac- tors, and find the product of all the common fac- tors. FKOBLEMS. 1. Find the greatest common divisor of I2a 2 xy and 16acx 2 y. Ann. Jf.axy. 2. Find the greatest common divisor of 10a 2 bc, QabVd and 4a 3 b 3 c. 3. Find the greatest common divisor of — 24ac 2 cferty and 28abex 2 y 2 . Am. ^acx 2 y. 4. Find the greatest common divisor of 4a 2n 6V, 6a n b 3 x 2 y and 9>a 3n bm 2 x. 5. Find the greatest common divisor of lla 3 bcV, 7b 3 e'm 2 x" and 19aWjj. 2 . CASE II. Greatest Common Divisor of Polynomials. 139. — Ex. 1. "What is the greatest common divisor of 3 a +8a;+15 and z z +9z+20? Solution. Since any quan- , , „ „-. . „ „„, w ... . ., „ t j. . - x 2 +8x+lS)x 2 + 9x+30U tity is the greatest divisor ot 2 v itself, if a; 2 + 8a; + 15 be a di vi- sor of x* + 9x+W, it must be z + 5)a?+8z+15{x+S the greatest common divisor x required. We find it is not a ° x + 1& divisor of x 2 + 9x + 20, since, Sx+15 on trial, x + 5 remains. If a; +5 be a divisor of x 2 + 8x +15, it must also be a divisor of x 2 + 9x + 20, which is once x 2 + 8a; + 15, plus a; +5. On trial it is found to be a divisor of x 1 +82; + 15. Hence, a; + 5 is the greatest common divisor of the given quantities. 70 DIVISORS. 2. What is the greatest common divisor of 2a; 2 -7a; + 5 and SV-7a; + 4? Sx 1 - 7x+4 .2x 1 -7x + 5)6x -*)6x* -lJfX+8^3 6x 2 -Six +15 7)7x- -7 X- -l)2x i - 2x 2 - -7x + 5{2x- -2x -5 -Bx+5 -Sx + S Solution. If we divide 3a; 2 — 7a; + 4 by 2a; 2 — 7x + 5, the quotient of 3x 2 divided by 2a; 2 is a fraction. To avoid this, we multiply the dividend by the factor 2, which will not change the common divisor (Prin. 4, Art. 136), and then divide. If we now divide 2a; 2 — 7a; +5 by 7a; — 7, the first term of the quotient will be fractional. We therefore cancel the factor 7 in the new divisor, which does not change the common divisor (Prin. 4, Art.- 136), and dividing by the remaining factor, x — 1, find it to be the greatest com- mon divisor required. 3. What is the greatest common divisor of a; 3 +a; 2 -2a; and a; 4 +2a; 3 +a; 2 +2a;? Arts. x 2 +2x. 4. What is the greatest common divisor of x 2 — 2x — 80 and a; 2 + 2a; -120? Am. x-10. 140. Rule for Finding the Greatest Common Divisor of Poly- nomials.— Divide the greater quantity by the less, and if there be a remainder, divide the divisor by it, and so continue to divide till there is no remainder. ■The last divisor will be the greatest common divisor. When there are more than two quantities, find the greatest com- mon divisor of two of them, and then of that divisor and a third quan- tity, and so on for all the quantities. In the process, a factor may be introduced into the dividend to make it divisible, or a factor may be cancelled from either divisor or dividend, which is not contained in the other. The signs of either divisor or dividend, or both, may be changed without changing the common divisor. DIVISORS. 71 PMOBLEMS. 1. Find the greatest common divisor of a 2 + ax and a 2 + 2ax + x 2 . Ans. a + x. 2. Find the greatest common divisor of a 2 — ax and a? — lax + x\ 3. Find the greatest common divisor of Sx 2 +x — 2 and 3a; 2 + 4k -4. ^ns. &;-£. 4. Find the greatest common divisor of a; 2 — 7a; +10 and 4a; 3 - 25a; 2 + 20a; +25. 5. Find the greatest common divisor of a; 2 — 9a; — 36 and x 2 -15* +36. SOLTTTIOH. a; 2 - 9a; - S6,)a: 2 -I5x+36{1 x l - 9x-86 Cancelling the factor 6 and , changing the signs, j x — lzjx — 9x—36{x+8 x 2 -lzx 3x~36 3x—36 6. Find the greatest common divisor of a? — 9a; 2 + 23a; — 12 and a; 3 -10a; 2 + 28x-15. Ans. x 2 -5x + 3. 7. Find the greatest common divisor of 3a; 3 — 24a; — 9 and 2a; 3 -16a; -6. 8. Find the greatest common divisor of x'+x — 10 and x* — 16. 9. Find the greatest common divisor of a 2 - 46 2 , a 2 + ab — IV and2a 2 + 3a&-26 2 . 10. Find the greatest common divisor of x 2 - 6 and x 3 — 8. 11. Find the greatest common divisor of a; 2 +a;-6 and 3a; 2 + 6a; -24. 72 MULTIPLES. SECTION XIII. MULTIPLES. 141. — Ex. 1. What quantity is a times 36? c times xyt 2. What quantity contains an integral number of times 2, 3, x and y? 3. What is the least quantity that will contain 4a and lb an integral number of times ? 4. What are the prime factors of 4a and 76 ? What is the product of all the prime factors of 4a and lb 1 Definitions. 142. A Multiple of a quantity is any integral number of times that quantity. Thus, 3a 2 6 2 is a multiple of ab. 143. A Common Multiple of two or more quantities is any quantity which is an integral number of times each of them. Thus, 24a z 6 2 is a common multiple of 3ab and iab. 144. The Least Common Multiple of two or more quantities is the least quantity which is an integral number of times each of them. Thus, 12ab is the least common multiple of flab and 4ab. 145. Principles. — 1. A multiple of a quantity contains all the prime factors of that quantity. Thus, 3os 2 6 2 , a multiple of ab, must contain the prime factors of ab. 2. A common multiple of two or more quantities contains all the prime factors of those quantities. Thus, 24a 2 6 2 , a common multiple of Zab and 4 V or x+-. 2 2. Eeduce to an integer. a Solution. Performing the division denoted, we a? + ae . have a + c, the integer required. a „ , 10a + 3-2a; . 3. Eeduce ■ to a mixed quantity. 5 A n 3 -® X Ans. 2a + ■ . 167. Rule for Reducing a Fraction to an Integer or a Mixed Quantity.— Divide the numerator by the denominator. PROBLEMS. 1. Reduce — to a mixed quantity. ~2 , ^2 (t + it- 2. Reduce to a mixed quantity. Ans. a+x+ ■. a-x a -x o t> j 3a 2 + 5a# 6. Keauce — to a mixed quantity. 4. Reduce to a mixed quantity. Ans. b+- c c af + l 5. Reduce to a mixed quantity. x—l 84 REDUCTION OF FRACTIONS. vy2 2ff2* + X 6. Reduce to an integral quantity. a — x a 3 -6 3 7. Reduce — to an integral quantity. Arts. a 2 +ab+b 2 , a — b a s b 8. Reduce — - to the form of an integral quantity. C Ob Solution. Since the fraction may be regarded a?b s , _ 2 ,_ as an expression of division, we transfer the fac- o 2 d? tors of the denominator to the numerator, which may be done by changing the signs of their exponents (Prin. 3, Art. 89 and obtain a s bc-'*d- s , the form required. 9. Reduce — — to the form of an integral quantity. a -y 2 10. Reduce ■ to the form of an integral quantity. a 2 - b 2 Ans. (o — 6)(a+6) _1 . Dissimilar Fractions to Similar Fractions. 168. Similar Fractions are such as have the same denomi- nator. Thus, — and — are similar fractions. a a 169. Dissimilar Fractions are such as have different denomi- nators. Thus — and ^tE are dissimilar fractions, c a 170. Fractions are said to be reduced to a Common Denomi- nator when they are changed to equivalent fractions with de- nominators alike. c c x n era d dxn dn REDUCTION OF FRACTIONS. 85 171. — Ex. 1. Reduce - and — to similar fractions. d n Solution. Since multiplying both terms of a fraction by the same quantity does not change its value, we multiply both terms of — by n, the d m rnxd dm denominator of — , and have — ; and multiply n nxd dn n dn both terms of — by d, the denominator of — , and have . Hence, n * ' d' dn c . m .en , dm , . . . ., . ,. — and — equal — and , which are similar tractions. d n dn dn 2. Eeduce — , — and to their least common denominator. 2b 3a 46 2 Solution. The least common multiple a ax g a £ g^j, of the denominators is 12a& 2 , and as this ~^r = _, „ , = ^ , 2 contains only such factors as are required to compose the denominator, it must be the £2 _ ^c x 4b 2 _ ffOS'c least common denominator of the fractions & a Sa x Jib' Wab' equivalent to the given fractions. Multi- >y a y a x g a Zla? plying both terms of -f- by 6a&, of ^ by W W x 3a 12ab % am a * 7a u o v, 6ct 2 5 20S 2 c 21a 2 ... 4b', and of — - by 3a, we have ——, — — - and — — - which are the 46 2 VLab 1 12a& 2 12a& 2 fractions reduced so as to have the denominator required. 3. Eeduce — , and to their least common denomi- 4» 6a; 2 12^ 9x? 2abx . 5 nator. Ans. -. , and 12x 3 12a? 12x 3 172. Rule for Reducing Dissimilar Fractions to Similar Frac- -tions.— Multiply both terms of one or more of the fractions by any quantity that will make the de- nominators alike. Or, — Multiply both terms of one or more of the frac- tions by any quantity that will give each a denomi- nator equal to the least common multiple of the given denominators. 8 86 REDUCTION OF FRACTIONS. PROBLEMS. 1. Reduce — , and to similar fractions. 3a 12c 86 56Vc 22a*V , lSdV Ans. , and 24abc' 24abc %4abc 2. Reduce — , — and — to similar fractions. 36 4c 5z 3. Reduce — and to similar fractions. 56 3c 6ac , Sab + 5b 2 Ans. and . 15bc 15bc 36 2a; 4a; 4. Reduce — , — and a+ — to similar fractions. 4 3 5 5. Reduce and to a common denominator. 3 a ax + a? , 6x + 8 Ans. ■ • and . 8a 8a 6. Reduce — — , — - — ■ and • to their least common 26a; dabxy 3aex denominator. 7. Reduce a?+- and to their least common de- V ay-\ ax 2 y 2 H a 2 y — x 2 y — a , cy nominator. Ans. — - and ay'-y ay'-y _ , a? ab , 3a 2 -2a6 ± ^ . . o. Keduce -, and to their least common a + b a—b a 2 — V , . . . a\a-b) ab(a+b) , 8a 2 denominator. Ans. — -, — '- and — 2ab tf-W ' a ! -6 2 a 2 -6 2 9. Reduce and to a common denominator. a + x a?+£ax + x* . a 2 -2ax+x* and , ADDITION OF FRACTIONS. 87 SECTION XVII. ADDITION OF FRACTIONS, 173. Addition of Fractions is the process of uniting two or more fractional quantities, to find their sum. 174. Similar Fractions may be added by finding the sum of their numerators, and placing it over their common denomina- tor. Hence, fractions may be added by means of a common denominator. __ a , c a + c 175. — Ex. 1. "What is the sum of and — - — b b Solution. The sum of (a + c) a + c a — c a + c + a — c 2a divided by 6, and (a — c) divided 6 6 6 6 by b,is the same as (a + e) + (a — c) i. -ii, i ffl + e+a-c ,.,. i,.2a divided by o, or , which is equal to — . 2. What is the sum of — and -? L o Solution. The sum of 3a divided by 2, ^ and a divided by 3, is the same as the sum of a_ 9a , 2a 11a 8 — and — , or — — . 6 6 6 9a 6 3x x — 2 3. What is the sum of -=- and — — ? Am. 7 5 oo 176. Rule for Adding Fractional Quantities. — Reduce the fractions, if necessary, to similar fractions; add their numerators, and write under the sum the common denominator. Sax 3 9a 2x3 6 ax 2 2a 3x2 6 2a 11a 6 6\ 22x- U 88 ADDITION OF FRACTIONS PMOBZEMS. . „ ,,, „ * 2x 5x , 2 . 8x + 2 1. Find the sum of -, — , — and -. Ans. . 3 3 3 3 S „-,.,., . 3a 2 2a , 36 2. Find the sum of . — and — . 26 5 la . 105a? +28a?b+ SOW Ans. — 70ab 3. Find the sum of -, - and -. a b d ,„.,,, „ x 3x Ax , x - 1 . 9x-l 4. h uid the sum of — , — , — and . Ans. . 5a 5a 5a 5a 5a , „ ,,, e 5a; -9 , 2a; + 11 5. Fuid the sum of and . a; + 3 a;-2 6. Find the sum of and -. Ans. a — 2 a-3 a? — 5a + 6 „ . . . x + y . x — y — 2z 7. Add *- and ^- — . 2 2 q nj a a a a 8. Add — — -, and a 2 - 1' a - 1 a + 1 9. Add ^ and ^. Ans. ^^ 1-a; 2 1+a; 2 l-x i 177. When there are Mixed Quantities or Integers, the frac- tions and integers may be added separately and the results united. 1. Find the sum of a+ and b . c a Solution. Sbx , . 3x* , . , 2bx Sx* a+ 1-6 =a+b+ o a o a , Sabx-Scx 1 = a+b-\ SUBTRACTION OF FRACTIONS. 89 2. Find the sum of 2x, Zx+ — and x . Ans. 6x — . 5 9 43 -b* 6 2 3. Find the sum of 11a, and 3a .. ' 3 5 4. Find the sum of a+ and 2a-b+- a 2 -b 2 (fl-Vf i — b + - a 3 -a?b-aV + tf 5. Find the sum of lb, — — and 3z 2 Ax 8a + Sax + 6x 2 Ans. 7b +- 12x* SECTION XVIII. SUBTRACTION OF FRACTIONS. 178. Subtraction of Fractions is the process of taking one frac- tional quantity from another, or of finding the difference be- tween two fractional quantities. 179. Similar Fractions may be subtracted, one from another, by finding the difference of their numerators, and placing it over their common denominator. Thus, - - - = — - — . Hence, bob fractions may be subtracted by means of a common denomi- nator. 180— Ex. 1. Subtract - from -. Solution, a divided by e, minus b divided by e, a b _a — b is the same as (a — b) divided by c, or - a ~ b c c o c 8* 90 SUBTRACTION OF FRACTIONS. 2. Subtract - from -. a e Solution. Beducing the given frac- a b ad bo ad— be ..,„,'. , a ad c d cd cd od tions to similar fractions, we nave — = — i e cd , b bo , ad be . . , ad— be and — = — . and — - - — ; is equal to ; — . d cd cd cd cd 3. Subtract — from . Ans. — . 4 8 8 4. Subtract ^ from /. Ans. * * . ly 6c Jficy 181. Rule for Subtracting one Fraction from another. — Reduce the fractions, if necessary, to similar frac- tions, and write the difference of the numerators over the common denominator. PROBLEMS. « ^ 2a . 3c A 8x-15c 1. From — take — . Ans. — — . 5a 4a Wa, n _ s+J,, a-b 2. From take — — . Ji z c , c . 2bc 3. From take -. Am. a- a + b a 2 - b 2 . _, 9x + 7 . , 6a + 4 4. From — - — take — - — . 4 5 5. From take -. Ans. x-1 x+1 x*-l 6. From — take -. n SUBTRACTION OF FRACTIONS. 91 1 * 1 %u 7. From take . Ans. ——- . x + y x-y x—y' 8 . From ^±1 take tJ.. Ans . **+*V-? m y x xy _ „ x-+y A . 1x 9. From take . V x + y 7 7 566 10. From take . Ans. 5a -4b 5a + 46 25a* -16V 182. When there are Mixed Numbers, they may be expressed as fractions ; or, if convenient, the fractional part of the subtra- hend may be first subtracted, then the integral part, and the results united. « _, _ 2a , , ,, a — 3e 1. From 7a + — take Sa . o b Solution. /„ , 2a\ /. a—3e\ „ „ , Sa , a {7a + -y(3a —)=7a-Sa + - + - =4a+ -So b 3a- So b „ _- „ 3«-2a i , „ Aa-Bx 2. From 6a + take 2a + . a x 3. From 5a-^ take 2a+^-. Ans. 8a+2- y ~. 4 12 <2» 4. What is the value of 8y'- ^-? 6 5. What is the value of ( lx+ f\-( Zx -TZ~\ ? ab - x> Ans. 4x+-~ -• 6(6 - x) 92 MULTIPLICATION OF FRACTIONS. SECTION XIX. MULTIPLICATION OF FRACTIONS. 183. Multiplication of Fractions is the process of finding a product, when one or both of the factors are fractions. CASE I. A Fraction by an Integer. 184 —Ex. 1. Multiply j by c. b Solution, e times a divided by 6 is ae divided by a ac , ac b b b,or-. 2. Multiply ^ by b. Solution, b times a divided by o 2 is — ; or, T7 x & = it = t 6 2 6 2 b since a fraction is multiplied by either multiplying its numerator or dividing its denominator, 6 times — is — . 3. Multiply - by d. Ans. — . e c 4. Multiply — by y. 185. Rule for Multiplying a Fraction by an Integer. — Mul- tiply the numerator, or divide the denominator, by the integer. When there are factors common to the numerator and denominator, they can be cancelled. MULTIPLICATION OF FRACTIONS. 93 PROBLEMS. 1. Multiply — by e. be 2. Multiply —— by m 2 . ^ws. . on dn 3. Multiply — ^— by a&V 1 g 4. Multiply by a 2 -x 2 . Ans. 5(a+x). a — x rr H u . , 2a 2 m . , ,, 5. Multiply ■ by a 2 -n\ a — n 186. When the multiplicand is a Mixed Number, it may be expressed as a fraction, or the integral and the fractional parts may be multiplied separately, and the results united. 26 1. Multiply x+— by 2a. a Solution. (x+— ) xSa=Sax+^=Sax+4b. 2. Multiply 6a - e + — by 5m. Ans. 80am - 5cm + . * J bm J b 3. Multiply a+l+- by a- y. Ans. a 2 +a-ay-y+ — — . x x .„,,., . ie , i» t a 2 m+bcm 4. Multiply a + — by — . Ans. an an 4b 2 Ab 2 16b* 5. Multiply 3a + by 3a . Ans. 9a 2 — 5e J 5c 85c 2 94 MULTIPLICATION OF FRACTIONS. CASE II. An Integer or a Fraction, by a Fraction. 187— Ex. 1. Multiply c by £ b Solution, c multiplied by a is ac, and c multi- « _ 22. plied by a divided by b must be ac divided by b, or b b ac b' 2. Multiply - by p a b Solution. — multiplied by a is — , and — multi- — x — = — d d d d b bd plied by a divided by b, must be — divided by b, or — . d bd 3. Multiply - by -. Ans. —. ej bd bd x 4ic M- x ^ 4 Multiply - by — . Ans. — . 188. Rule for Multiplying an Integer by a Fraction.— JlfztZ- &yoZi/ the integer by the numerator of the fraction, and write the result over the denominator. 189. Rule for Multiplying a Fraction by a Fraction. — Mul- tiply the numerators together for the numerator, and the denominators for the denominator, of the product. If there- are mixed numbers, reduce them to fractions before multi- plying. Factors common to numerator and denominator can be cancelled. PROBLEMS. 12 9 1. Multiply — by — . Ans. Za J la 21a* MULTIPLICATION OF FRACTIONS. 95 2. Multiply— by— . 3. Multiply |- by - % . Ans. - ^ . 7a; 5 35 4. Multiply --by-—. 5. Multiply by . Am. . * J 206 J 34a 2 8ab 6. Multiply by . ax 1 cm 7. Multiply ^ by ^ . Ans. H* c+d a+a; c+d 8. Multiply ^^ by 6 ab a+B x x 9. Multiply — by — . Ans. 10. Multiply 2a 2 -3a; by — . 11. What is the value of 5y 2 x (y 2 j? J.ws. 5y*— ^-. x+a ya x] 12. What is the value < 13. What is the value of 3a; x °^- x — ? la a + b Ans — 3 -~ Sx _ Sx ( x *-l) ' 2a? +2ab~ 2a (a + b)' 14. What is the value of (a + — \(b - — V \ a -b/\ a + b) 96 DIVISION OF FRACTIONS. SECTION XX. DIVISION OF FRACTIONS. 190. Division of Fractions is the process of finding a quotient, when the divisor or dividend, or both, are fractions. CASE I. A Fraction by an Integer. 191— Ex. 1. Divide —by c. Solution. Dividing the numerator of a fraction ac a divides the fraction (Prin. 3, Art. 159) ; hence, to divide 6 b etc — by c, we divide the numerator, ac, by c, and, writing the result a over the denominator, 6, we have t- 2. Divide - by c. b J Solution. Multiplying the denominator of a fraction a _ a divides the fraction (Prin. 3, Art. 159) ; hence, to divide 6 be — by c, we multiply the denominator, 6, by c, and, writing the result a under the numerator, we have — • bo 4a 2 2 3. Divide by 2a 2 . Arts. . 5xy 5xy 4. Divide — - by 6y. Ans. — . 26c 2 J " Ab- -j ae+6«^ , 0/ ,. . ae+bd 12. Divide— — — by2(a-6). Ans. 2(a+6) J 4(a?-b 2 )' 98 DIVISION OF FRACTIONS. CASE II. An Integer or a Fraction, by a Fraction. 193. A fraction is inverted when the denominator is taken for the numerator, and the numerator for the denominator. Thus, — inverted is — ; — inverted is _. b ay x 194. A Complex Fraction is a fraction which has a fraction in one or both of its terms. Thus, — is a complex fraction, and is, also, an expression of the m division of one fraction by another. 195.— Ex. 1. Divide a by-, a r, j- -j ji • a j j. -j j c axd ad Solution, a divided by e is -, and a divided a -s- — = = — e ace by c divided by d must be d times as great, or — . 2. Divide - by - . b d Solution. Eeduced to similar fractions, a e _ad be ad , a ad , c bo „, ,. b d bd bd be we have - = -—,, and - = — ■ The quotient, i><* <™. 6 bd d bd or, then, of the fractions is the same as that of a c _ axd _ ad .. . .... ad b d bxo be their numerators, which is — . be The same result is obtained by inverting the given divisor, and mul- tiplying numerator by numerator and denominator by denominator. 3. Divide a'b by — ■. Ans. . n m . „. . n x . 2d xy 4. Divide -by — . Ans. — —. ■ 6 J y 1M DIVISION OF FRACTIONS. 99 196. Rule for Dividing an Integer or a Fraction by a Fraction. -Multiply the dividend by the divisor inverted. If there be mixed numbers, reduce them to fractions before dividing. Shorten the process of division as much as possible by cancelling. tttOBIMMS. 1. Divide 16a 2 6 3 by Am. . J 3 c o tv- -j l. 13&B . aey 2. Divide xy by . Am. • etc 18b 3. Divide — by — . b 3 36 . -. . , 3ab , 4e . 8a?b 4. Divide — by — . Am. . c a Jf55x. x-5a x — 3a_ a ~4a 9~ ~18* 11. Clear of fractions, — - — + 6 12 Ans. 8x-3m?+8x-2n i = 7mn. SECTION XXIII. SOLUTION OF SIMPLE EQUATIONS WITH ONE UNKNOWN QUANTITY. 205. Simple Equations with one unknown quantity may be solved by so transposing the terms of the equation that the unknown quantity shall stand alone as one member; and the other member will then denote the value of the unknown quantity. 206. The Root of an equation is the value of the unknown quantity. The root is verified, or the equation satisfied, when, this value being substituted in the given equation, the two members equal each other. 207. — Ex. 1. What is the value of x in the equation +1=»-14? 3 Solution. Clearing of the fractions, x+15 - ? _ ? we have x + 15+3 = 3x -42; transpos- 3 ing, we have x—3x = -42-15-3; x+15+3 = 3x-42 uniting the terms, we have — 2x= x — 3x=—J l 2—lB—3 — 60; and dividing by —2, the eoef- —2x=-60 fieient of x, we have x = 30. x = 30 106 SIMPLE EQUATIONS : ONE UNKNOWN QUANTITY. 2x — 8 x 2. What is the value of a; in the equation = 38 +- ? 4 6 Am. x = 120. 208. Rule for solving Simple Equations with one unknown quantity.— Clear the equation of fractions, if it have any. Transfer the terms, so that all those containing the unknown quantity shall be in the first member, and all the others shall be in the second member; unite the terms in each member, and divide both mem- bers by the coefficient of the unknown quantity. PROBLEMS. 1. Given x + — i — = 11, to find x. Ans. x = 6. 2 3 2. Given 36 - — = 8, to find x. 3. Given — v- = x — 7, to find x. Ans. x = 15. 5 3 4. Given — v =x — 2, to find x. 2 7 5. Given — + = 29, to find x. Ans. x = 12. 4 6 6. Given — - — = 8-2, to find x. 7. Given -+- = c, to find x. a b Solution. Clearing of fractions, we £:_i_:E_ have bx+ax = abc; factoring the first a b member, we have (a+b)x = abc; and bx+ax = abc dividing by a + b, the coefficient of x, we (a + b)x = abc . a6e abc have x= — x = a+b a+ b SIMPLE EQUATIONS: ONE UNKNOWN QUANTITY. 107 8. Given - = — t- — , to find x. x c m 9. Given ax+ 6 = cx+d, to find x. Ans. x = . 10. Given 3a; + = x+a, to find x. n „. 9a; 3*-10 nn a;+15 OK . ~ ■, 11. Given +29= +65, to find x. 2 5 5 Solution. Transposing and unit- 9x Zx — 10 x+15 .? . ., . , 9x 4x+5 2 5 ' " 5 ing similar terms, we nave ■ 2 5 9x Jjx+5 _ = 36 ; clearing of fractions, we have 2 5 4te-ta-10-860 ; transposing and ^Sx-10^360 uniting, we have 37a; = 370; and di- viding by the coefficient of a;, we have = a; = 10. x = 10 12. Given 56 - ^ - 48 - — , to find x. 1Q _,. a; -16 3a;+4 14a: -6 A „ . 13. Given — - — — , to find x. Ans. x = 2. 4 5 4 14. Given^+90=a;+— +53. iK _,. a;-3 x „. a;+19 „ , 15. Given ~^-+g = 20 — - to find x. Ans. x = 9. 16. Given 3a; 2 - 10a; = 8a;+a; a , to find x. Solution. Dividing by x, we have the sim- Sx 1 -10x = 8x+x 2 pie equation, 3a;— 10 = 8 +x; transposing and Sx —10 = 8 +x uniting, we have 2a; = 18; and dividing by the Sx = 18 coefficient of a;, we have a; = 9. x= 9 17, Given ax 2 +bx = ex i +dx, to find x. Ans. x = ~ . 108 SIMPLE EQUATIONS: ONE UNKNOWN QUANTITY. 18. Given 3a* 3 - &abx i = ax i +Zax l . 19. Given x 2 — Wx + 30 = » 2 - 5x + 6, to find x. Ans. % = £. 20. Given 12*' 2 +25a;+7 = 12x s +24a;+12, to find x. 21. Given — = bm+n+-, to find x. Ans. x = . x x bm+n 22. Given x + ax - be = ab + ex, to find *. 23. Given — +- — = , to find x. ab — axbe — bxac — cx ab + bc — ac Ans. x = . _. „. ax+bx+ex , „ , a+ -c 24. Given = x + a, to find *. a+b SECTION XXIV. PROBLEMS PRODUCING SIMPLE EQUATIONS WITH ONE UNKNOWN QUANTITY. 209. — Ex. 1. What number is that, the sum of whose fourth part and fifth part is 9 ? Solution. Let x denote the number; x = the number then - will denote the fourth part, and - A 4 5 the fifth part. Hence, by the conditions - = the fifth part; x x of the problem, we have — I — = 9. Clear- ~ ~ 4 5 5 + ^ =9 ing of fractions, we have hx + 4x = 180; ■£ ^ . . . , Sx+Ax=180 uniting terms, we have 9a; = 180 ; and divid- - = ,-. ing, we have x = 20. x = S0 2. A certain sum of money increased by f of it, and dimin- ished by $75, is $175. What is the sum ? SIMPLE EQUATIONS : ONE UNKNOWN QUANTITY. 109 Solution. Let x equal the sum; 2a; 2 then — will denote - of it, and by the 3 3 ' J conditions of the problem cc + $75 is $175. Transposing —$75, and unit- ing it with $250, we have x+— = $250. Clearing of fractions, we have 3x+2x =$750; uniting terms, we have 5x = $750; hence, a; = $150. x = the sum ; 2x. 2 of the mm. x+ — -$75=$175 x+- --$250 o 3x+2x = $7S0 Sx = p50 x = $150 3. A line is 31 inches long ; it is required to cut it into two parts, such that one part shall be 3 inches more than -| of the other part. What is the length of each part? Am. 16 inches, one part ; 15 inches, the other. 4. A man bequeathed $161 to two sons ; to one he left $17 more than half as much as to the other. What did he leave to each? 5. Out of a cask of molasses, -J- of which had leaked away, 21 gallons were drawn, and then the cask was found to be half full. What was the capacity of the cask ? Soltjtion. Let x denote the ca- pacity of the cask in gallons ; then - o will denote what leaked away, and by the conditions of the problem 21+— will be one-half of the capacity of the cask, or — . Clearing of frac- 2i x = capacity of the cask; — = what leaked away. 21^A 126+2x = 8x -x=-126 x= 126 tions, we have 126 + 2a; = 3a;; transposing and uniting terms, we have -x= -126; whence, a; = 126. 6. When I had spent from my purse $70 more than f of the money there had been in it, I found only \ of the money left. How much money had I at first ? 10 110 SIMPLE EQUATIONS: ONE UNKNOWN QUANTITY. 7. What is a man's age, if f of his age less 10 years is just f of his age? Ans. 80 years. 8. A cistern was f full of water. After 17 hogsheads had run in, it was found to be f full. What number of hogsheads is its capacity? 9. A alone can do a piece of work in 9 days, and B alone can do it in 12 days. In what time will they do it if they work together ? Soi/utios'. Let x denote the time in which they can do the work by working together; as A does \ of the work in 1 day, in x days he will do — of the work, and in like manner B will do — of the work. Denoting the entire work by 1, we have, by the conditions of the problem, — + — V \.Ji = 1. Clearing of fractions, we have 4x+3x = 36; whence, x = 5$-. 10. A man can do a piece of work in 8 days, and his son can do the same in 3 times as many days. In what time can they do it by working together? Arts. 6 days. 11. A certain canister of tea will last a woman 12 days, and will last her husband -| of that time. How long would it last them, using it together ? 12. A cistern can be filled with water by means of one pipe in a hours, and by means of another in b hours. In how many hours could the cistern be filled by both pipes running to s ether? i M .Aw». a + b 13. A cistern can be filled by means of one pipe in 6 hours, and by means of another pipe in 8 hours; and it can be emptied by a tap in 12 hours, if the two pipes are closed. In x= = the time ; X 9~ =what A can do; X - what B com do. X X 9 IS' = 1 4x+3x = = 36 7x = = 36 x= -4 SIMPLE EQUATIONS: ONE UNKNOWN QUANTITY. Ill what time will the cistern be filled, if the pipes and the tap are all open ? 14. A gentleman gave in charity $46, a part of which he distributed in equal portions to 5 poor men, and the rest in equal portions to 7 poor women. It was found that a man and a woman had together $8. How much was given to each man, and how much to each woman ? 15. A has $900 and B $700. What sum must A give to B in order that B may have ^ as much as A ? Ans. XX 16. A man can row 6 miles Solution. an hour with the current of a x=number of miles out; river, and 4 miles an hour x . .i , i_. — =time with the current: agamst the current, his rowing 6 in each direction being uni- x n xt j, , . — = time against the current. form. How tar can he go in 4 order that the time between leaving and returning to the g H place from which he started x = 6 , number of miles out. shall be 2^ hours ? 17. A person walked to the top of a mountain at the uniform rate of 2\ miles an hour, and without stopping walked down again by the same way at the uniform rate of 3| miles an hour. He walked 5 hours. How far did he walk ? Ans. 7 miles up, and 7 miles down. 18. If a person have only a hours at his disposal, how far can he ride in a coach which travels b miles an hour, and return home in the time, walking back at the rate of c miles an hour ? , abc „ Ans. miles. b+c 19. A starts from a certain place, and travels at the rate of 7 miles in 5 hours ; B starts from the same place 8 hours after A, and travels in the same direction at the rate of 5 miles 5 = = the distance A travels. -(x- -S) = = the distance B travels. -{x 3 K -*)• _7x ~ 5 25x- 200 = =21x 4z = =300 a; = = 50 7* 5 = = 70, the distance A travels. . 1x '~ 5 7x ; whence, a; = 50, and — = 70. 5 112 SIMPLE EQUATIONS: ONE UNKNOWN QUANTITY. in 3 hours. How far will A travel before he is overtaken byB? Solution. Let a; denote x = the number of hours A travels; the number of hours which x— 8= the number of hours B travels; A travels before he is over- 2? taken; then B travels x — 8 hours. A travels 7 miles in 5 hours ; therefore, he travels £ of a mile in one hour, and |a; miles in x hours. B travels f of a mile in one hour, and §(x — 8) miles in x — 8 hours. But, when B overtakes A, they have trav- elled the same number of miles; hence, we have $(* — 8) = 20. A privateer, running at the rate of 10 miles an hour, discovers a ship 18 miles off, running at the rate of 8 miles an hour. How many miles can the ship run before it is overtaken ? Arts. 72. 21. A clock has two hands turning on the same center. The swifter makes a revolution every 12 hours, and the slower every 16 hours. In what time will the swifter gain one com- plete revolution on the slower ? 22. A man loaned $1000, a part at 4 per cent., and the rest at 5 per cent. The whole yearly interest received was $44. What sum was lent at 4 per cent. ? Solution. x = the sum lent at fy per cent. $1000— x = the sum lent at 5 per cent. Lx — — = yearly interest on sum lent at Jf per cent. — = yearly interest on sum lent at 5 per cent. 100 " " * J^_ 5($1000-x) =f or4x+ ^ BO oo-Sx = $UOO 100 100 x = $600, the sum lent at Jj. per cent. SIMPLE EQUATIONS ONE UNKNOWN QUANTITY. 113 23. A person lent $900, a part at the rate of 4 per cent., and a part at the rate of 5 per cent., and he received equal sums as interest from the two parts. How much did he lend at each rate? 24. A manufacturer adds to the cost price of goods 20 per cent., to give the selling price. Afterward, to effect a rapid sale, he deducts from the selling price of each article a discount of 10 per cent., and then obtains on each article a profit of What was the cost price of each article ? Am. 25. A and B join capital for speculation, B contributing $250 more than A. If their profits amount to 10 per cent, on their joint capital, B's share of them is equal to 12 per cent, of A's capital. How much does each contribute ? 26. A man has a number of coins which he tries to arrange in the form of a square. At the first attempt he has 130 over. When he increases the side of the square by 3 coins, he has only 31 over. How many coins has he ? Solution. x = the number of coins in each row at the first attempt; X 2 = the number of coins in the square at first attempt ; x' + 130 = the whole number of coins. (x+3) 2 = the number of coins in the square at second attempt; (x+3) 2 +31 = the whole number of coins. x 2 +130=(x+3) 2 +31 x 2 +130 = x 2 +6x + 9 + 31 6x = 90 x=15 x 2 + 130 = 355, the number of coins he has. 27. A colonel wishes to arrange his men in a solid square. In the first formation he has 39 men over. When he increases the side of the square by 1 man, he wants 50 men to complete the square. How many men has he ? 28. A regiment was drawn up in a solid square : when some time afterward it was again drawn up in a solid square, it was 10* 114 SIMPLE EQUATIONS: TWO UNKNOWN QUANTITIES. found that there were 5 men fewer on a side. In the interval 295 men had been removed from the field. What was the original number of men in the regiment ? Am. 102 Jf. men. 29. At a public meeting a resolution was carried by a ma- jority of 9 ; but if \ of those who voted for it had voted against it, it would have been lost by 3 votes. How many voted ? SECTION XXV. SIMPLE EQUATIONS, WITH TWO UNKNOWN QUANTITIES. 210. Independent Equations are such as express essentially different conditions, or cannot be reduced to the same form. Thus, — = 6 and 6x — 4y = 4 are independent equations; but x +y = 7 and 2a; + 2^ = 14 are not independent equations, because one can be directly obtained from the other. 211. Simultaneous Equations are any two or more equations in which the same unknown quantities express the same values. Thus, in each of the equations x +y = 10 and 5x — 2y = 8 the value pf x is 4, and of y is 6. When two or more unknown quantities in an equation are to be de- termined, there must be as many independent and simultaneous equa- tions as there are unknown quantities, and from these must be deduced a single equation containing only one unknown quantity. ELIMINATION. 212. — Ex. 1. If x+y be added to x — y, what expression will represent the result ? 2. If 2x+y be added to x-y, what will represent the sum? 3. If x+2y be added to x — 2y, what will represent the sum? 4. If the equation x-y = 2 be added to the equation x+y = 16, what expression will represent the sum ? SIMPLE EQUATIONS: TWO UNKNOWN QUANTITIES. 115 5. If the value of x in the equation x+y = 18 is 9, what is the value of y ? 6. If the equation Zx — 2y = l be added to the equation x+2y = 13, what will represent the sum? What is the value of x ? What is the value of y ? 7. If x-2y be taken from 2x — 2y, what will represent the result ? 8. If the equation x - 2y = 1 be taken from 2x - 2y = 8, what will represent the result ? What is the value of x ? Of y ? Definition. 213. Elimination is the process of deducing from given simul- taneous equations one or more equations, containing together fewer unknown quantities than the given equations. CASE I. By Comparison. 214. Elimination by Comparison consists in finding an ex- pression of the value of the same unknown quantity in each of the equations, and in forming a new equation from those values. 215.— Ex. 1. Given z + 3y = 9 and 3s+2y = 13, to find x and y. Solution. Transposing 3y in x+3y= 9 (1) equation (1), we obtain (3). Trans- 3x+2y = 13 (2) posing 2y in (2) and dividing by x = 9—Sy (3) ' 3, we obtain (4). Placing these 13— Sy ... two values of a; equal to each other, " g ' we obtain (5). Clearing of frac- lS—Sy ... tions, we obtain (6). Uniting and 3 transposing, we obtain (7). Divid- 13—2y = 27—9y (6) ing by 7, we obtain (8), or y = 2. 7y = H (7) Substituting 2 for y in 3, we ob- y=% (8) tain (9). Uniting, we obtain (10), x = 9-6 (9) ora; = 3. x = 3 (10) 2. Given 5x + 2y = 45 and 4x+y' = S3, to find the values of x and y. Am. x -= 7 ; y = 5. 116 SIMPLE EQUATIONS : TWO UNKNOWN QUANTITIES. 216. Rule for Elimination by Comparison. — Find an ex- pression for the value of the same unknown quan- tity in each of the equations. Form a new equation by placing these values equal to each other, and solve the equation. PROBZJEMS. 1. Given x+y = 10 and 2x — 3y = 5, to find x and y. Am. x = 7; y = 8. 2. Given 9a; -4$/ = 8 and 5x+Zy = k\, to find x and y. 3. Given Sx+7y = 99 and llx-5y = 87, to find x and y. Ans. x = 12 ; y =9. 4. Given 9x - Ay = 8 and 13»+ 7y = 101, to find x and y. 5. Given -+- = 18 and — - = 21, to find x and y. 5 6 2 4 * Ans. x = 60; y = S6. 6. Given Zx + 1 = 36 and 6y - 2x = 32, to find x and y. o v — 3 a;— 2 7. Given 2«-^—— = 4 and 3y+— — = 9, to find x and y. o o ^Ins. x = 2; y = 3. 8. Given = - and =-, to find x and y. y 8 3,-2 6' * 9. Given — —+x = 15 and " +y = 6, to find a; and $/. -4ns. x = 10 ; y=5. x 4x -2 10. Given -+3y = 7 and — = — = 3^-4, to find * and y. x+3y=9 (1) 3x+2y = 13 (2) x = 9-3y (3) 27-9y+2y=13 (4) -7y=-U (5) y=2 (6) x=9-6=3 (7) SIMPLE EQUATIONS: TWO UNKNOWN QUANTITIES. 117 CASE II. By Substitution. 217. Elimination by Substitution consists in finding an expres- sion of the value of one of the unknown quantities in one of the equations, and substituting this value in another equation. 218.— Ex. 1. Given »+3y = 9 and 3a+2y = 13, to find x and y. Solution. From equation (1) by transposing Zy, we obtain (3). Substi- tuting this value of x in equation (2), we obtain (4). Uniting, we obtain (5). Dividing by —7, we have (6), or y = 2. Substituting the value of y in equation (3), we obtain (7), or a; = 3. 2. Given 8x + 1y = 100 and 12* - 5y = 88, to find x and y. Ans. x = 9 ; y=4- 219. Rule for Elimination by Substitution.— Find an ex- pression for the value of one of the unknown quan- tities in one of the equations. Substitute this value for the same unknown quantity in the other equa- tion, and solve the equation. PROBLEMS. 1. Given 4x+ 9y = 46 and 8* - lSy = 30, to find x and y. Ans. x = 7; y = £. 2. Given 7x+5y = 33 and 13a — lly = 41, to find x and y. 3. Given x+ 2y = 15 and 5x — 19y = 17, to find x and y. Ans. x = ll; y = 2. 4. Given 3«+- = 36 and Gy — 2a = 32, to find x and y. 5. Given 7#+2y = 30 and 5x+3y = 34, to find x and y. Ans. x = %; y=8. 118 SIMPLE EQUATIONS: TWO UNKNOWN QUANTITIES. op 1] or 11 6. Given -+J = 8 and --^ = 1, to find x and y. 2 3 3 2 * 7. Given- — £ = 11 and — +-f = 2»+7, to find x and «. 4 3 5 4 9 c „. 9a; 4 _ , 18 20 .„ 8. Given = 7 and — h — = 16, to find * and y. x y x y Ans. x = S; y = 2. 9. Given 4#-^ = ll and 2x-4y = 0, to find x and y. 10. Given — —*- = 5 and 2a = ", to find a; and v. 4 3 * CASK III. By Addition or Subtraction. 220. Elimination by Addition or Subtraction consists in adding two equations, or subtracting one equation from another, when the coefficient of the same unknown quantity in each is the same, or has beenlnade the same. 221.— Ex. 1. Given 8a;+72/ = 100 and 12a;-5y = 88, to find x and y. Solution. Multiplying equation (1) by 5, and equation (2) by 7, we obtain equations (3) and (4), in which the coef- ficients of y are the same. Adding equa- tions (3) and (4), we obtain (5). Unit- ing, we obtain (6). Dividing by 124, we obtain (7), or a; = 9. Substituting 9 for x in (1), we obtain (8). Transpos- ing and uniting, we obtain (9). Divid- ing by 7, we obtain (10), or y = 4. 8x+7y = 100 (1) 12x-6y = 88 (2) 40x+85y = 500 (3) 8JfX-85y = 616 (4) 40x+84x = 500+616 (5) 124a; =1116 (6) x = 9 (7) 73+7y = 100 (8) 7y = 28 (9). y=4 (10) x+8y=9 (1) 3x+2y=13 (2) 8x+9y = 27 (3) 7y=U (4) y = 2 (5) x+6 = 9 (6) x = 8 (7) SIMPLE EQUATIONS: TWO UNKNOWN QUANTITIES. 119 2. Given x+3y = 9 and 3»+2y = 13, to find x and y. Solution. Multiplying equation (1) by 3, we obtain (3). Subtracting equation (2) from equation (3), we obtain (4). Di- viding by 7, we obtain (5), or # = 2. Sub- stituting 2 for y in (1), we obtain (6). Transposing and uniting, we obtain (7), or x = 3. 3. Given 7% + 3y = 42 and 8y — 2x = 50, to find x and y. Ans. x = 3 ; y = 7. 222. Rule for Elimination by Addition and Subtraction. — Mul- tiply or divide one or both of the equations, if necessary, so that one of the unknown quantities shall have the same coefficient in both equations. Then, if the signs of the terms containing this quantity are alike, subtract one equation from the other ; if unlike, add the two equations. Of the different methods of elimination, each has its advantages in application to particular problems. When the coefficient of one of the unknown quantities is 1, the method by substitution is often the most convenient. In general, however, elimination by addition or sub- traction, as it does not give rise to fractions, is the most simple, and is therefore to be preferred. PROBLEMS. 1. Given Gx-7y = 42 and 7x - Qy = 75, to find x and y. Ans. x = %l; y = l%- 2. Given Bx — Ay = 18 and Sx+2y = 0, to find x and y. 3. Given - - ^ = 20 and °^- - ^^ + - = 35, to find x andy. 2 4 5 4 3 Ans. x = 60; y = Jfi. 4. Given 5x + fy = 22 and 2a; -f 3y = 13, to find x and y. 120 SIMPLE EQUATIONS: TWO UNKNOWN QUANTITIES. 5. Given 3x+7y = 79 and 2y-\x = 9, to find x and y. Ans. x = 10 ; y = 7. 6. Given Bx + _ = 36 and Gx-2y = 48, to find x and y. o 7 . Given ^±% . 10 _ V and 4*^3* _ 3s find sandy. 5 3 6 * J.ns. x=4-j y = 9. 8 . GiY e n l^^ + fci = 2 and ^±K + 2/ = 9, to find a; and y. 9. Given -+- = 2 and bx-ay = 0, to find a; and y. a b Ans. x = a; y = b. y O n. A 10. Given 2a; +- = 21 and %+— — = 29, to find x and y. 5 6 223. In finding the values of unknown quantities in the fol- lowing equations, the learner may use any of the methods of elimination which may be deemed most convenient. 1. Given 10a; = 69 +^ and 10v = 49 +-, to find x and y. 5 " 7 Ans. x = 7; y = 5. 2. Given 6a;+5« = 70 and — + y - = 9^ to find x and y. a 2 4 2 " 3. Given 6x+8y = 52 and 3x+7y = 32, to find x and y. x = 6; y = 2. 4. Given —+^ = 11 and —+f = 16, tofinda;andy. 7x 9 9x 2 SIMPLE EQUATIONS: TWO UNKNOWN QUANTITIES. 121 5. Given = - and = -, to find x and y. V 3 y + 1 4 Ans. x = 4; y-lS. 6. Given 3x+8y = 60 and 3y+6x = 42, to find a; and y. 7. Given "= -2 and - + " = 8, to find a; and y. 35 2 3 .An*. x = 6; y = 25. 8. Given — + — = 16+ and - = 16A, to find x and y. 3 2*23 9. Given — + -^ = 20 and — +— = 2a;-7, to find x and y. 4 8 5 4 * 10. Given bx+lly = 146 and llx+5y = 110, to find x and j/. J.»is. x = 5; y = ll. 11. Given ^ zjL +2y = 6 and 4v - ^±-^ - 3. 4 3 " 3 12. Given — h - = - and = -, to find a; and y. x y a x y b 2ub 2ab Ans. x = ; # = — - a+b a—b 13. Given ax = by and x+y = c, to find a; and y. 14. Given 4x+y = 11 and — = - — , to find a; and «/. 5a; 3a; i& 15. Given 12x-7y = 3 and 9a; + 5?/ = 33, to find a; and y. Ans. x=2; y = 8. 16. Given IHzHS. = g and ^Lt^S. = 16, to find * and y. 3 4 17. Given 7^+13^ = 660 and 8|y+5|z = 398, to find y and z. , _ , „,„ * J.ns. y = Hi z = 86. 11 122 SIMPLE EQUATIONS: TWO UNKNOWN QUANTITIES. SECTION XXVI. PROBLEMS PRODUCING SIMPLE EQUATIONS WITH TWO UNKNOWN QUANTITIES. 224. — Ex. 1. My herd of cattle is such that if the number of cows be added to four times the number of the oxen, the sum is 29 ; and if the number of oxen be added to four times the number of cows, the sum is 26. What is the number of each ? x = the number of cows ; y = the number of oxen. x+4y = 29 (1) y+4x = 26 (2) 4y + 16x = 104 (3) = (2)x4 lBx = 75 (4)-(8)-(l) x — B (5) = (4) -s-15 y+S0=S6 (6) = (2) after substitution. y = 6 (7) = (6) after transposition and reduction. Solution. Let x denote the number of cows, and y the number of oxen. Then, by the conditions of the problem, we have x + 4^ = 29 and 3/+4a; = 26. Multiplying equation (2) by 4, we obtain (3). Subtract- ing equation (1) from (3), we obtain (4). Dividing by 15, we obtain (5), or £ = 5. Substituting the value of x in equation (2), we obtain (6). Transposing and uniting, we obtain (7), or y = Q. 2. A's and B's ages are such that if twice A's age be added to three times B's, the sum will be 140 ; and if B's be added to four times A's, the sum will be 130. What are their ages ? 3. A farmer sold to one person 9 horses and 7 cows for and to another, at the same prices, 6 horses and 13 cows for the same sum. What were the prices ? Am. Ahorse,$24; a cow, $12. 4. There are two numbers such that if twice the less be added to the greater, the sum will be 18; and if three times the greater be diminished by the less, the remainder will be 19. Find the numbers. Ans. The greater, 8 ; the less,5. SIMPLE EQUATIONS: TWO UNKNOWN QUANTITIES. 123 5. Two pieces of cloth, measuring together 57 yards, were sold for $36. The first was valued at $.75 per yard, and the second at $.50. What was the number of yards in each piece? Ans. In one piece, SO ; in the other, 27. 6. A purse contains dimes and dollars. Add a dime, and then there will be twice as many dimes as dollars. Add a dol- lar to the original contents of the purse, and then there will be more dimes than dollars by 2. What number of dimes and dollars does the purse contain ? Solution. Let x denote the num- ber of dollars, and y the number of dimes. Then, by the conditions of the problem, we have y+l = 2x and 2/=a; + l + 2. Transposing equation (1), we obtain (3). Comparing equa- tions (2) and (3), we obtain (4). Transposing and uniting, we obtain (5), or a; = 4. Substituting the value of x in equation (3), we obtain (6), ory = 7. 7. The ages of two persons are such that if to the sum of their ages 22 be added, the sum will be double the age of the elder, and if 1 be taken from the difference of their ages, the remainder will be the age of the younger. How old is each? 8. A says to B, "If you will give me half of your money, I shall have $100." B replies, "I shall have $100 if you will give me a third of your money." How much had each ? 9. A liberty pole consists of two parts ; one-third of the lower part, added to one-sixth of the upper part, is equal to 28 feet ; and five times the lower part, diminished by six times the upper part, is equal to 12 feet. What is the height of the pole? Ans - 108 feet. 10. The numbers in two opposing armies are such that the sum of both is 21110; and twice the number in the greater army, added to three times the number in the less, is 52219. What is the number in the greater army ? Ans. 11111. x — the number of dollars ; y = the number of dimes. y + l = 2x (1) y=x+l+2 (2) y = 2x-l (3) Sx-l=x+l+2 (4) x = 4 (5) • y=8-l=7 (6) 124 SIMPLE EQUATIONS: TWO UNKNOWN QUANTITIES. 11. There are two numbers such that the first added to half the second gives 35; the second added to half the first gives 40. What are the numbers ? 12. A certain number expressed by two figures is equal to six times the sum of the digits, and if 117 be subtracted from three times the number, the order of the figures is reversed. What is the number ? Solution. x = the number expressed by the tens' figure. y = the number expressed by the units' figure. 10x+y = the number expressed by the figures. 10y+x = the number expressed by the figures reversed. 10x+y = 6(x+y) (1) 8(10x+y)-117=10y+x (2) 4x = 5y (3) = (1) simplified. 29x-7y = 117 (4) = (2) simplified.. Lx —y= — , value of -y from (3) o 117-29X , , ... , value of —y from (4) 7 Jjx _ 117- '-, two values of —y are equal; whence, 5 7 117x = S85 ; whence, x = B. /Substituting value of x in (3), we have y = Jf. Hence, 5Jf is the number. 13. There is a number expressed by two figures, the sum of whose digits is 13, and if 27 be subtracted from the number, the digits will be inverted. What is that number ? 14. A says to B, " Give me $1 of your money, and I shall have twice as much as you will have left." " Yes," says B, " but give me $1 of your money, and I shall have three times as much as you will have left." How much money has each ? Ans. A,$2J.; B, $2~. 15. Two numbers are such that if one were increased by 18, it would be double the other, and if the second were diminished by 11, it would be one-third of the former. What are the numbers ? SIMPLE EQUATIONS: TWO UNKNOWN QUANTITIES. 125 16. A fraction becomes equal to 2, when 7 is added to its numerator, and equal to 1, when 1 is subtracted from its de- nominator. What is the fraction ? Solution. x = the numerator. y = the denominator — = the fraction. y Then, by the conditions, \ X + 7 % (1) y -2—-1 (2) I y-i Reducing (1), x=2y-7 (3) Reducing (2), x= y—1 ' (4) From (3) and (4), 2y-7=y-l (5) whence, y = 6 (6) and, x=6 — l = 5 (7) From (6) and (7), x 5 y~ 6 17. What fraction is that whose value will be \, if its nume- rator be increased by 3 ; and whose value will be \, if its denomi- nator be diminished by 3 ? 18. A person has two horses, and a saddle worth $50. If the saddle be sold with the first horse, it will make his value double that of the second ; but if it be sold with the second, it will make his value three times that of the first. What is the value of each horse? Arts. One, $30 ; the other, $Jfi. 19. If B were to give $25 to A, they would have equal sums of money. If A were to give $22 to B, he would then have twice as much as A. How much money has each ? 20. A says to B, " Give me $30 of your money, and my money will equal 4 times yours ; but should I give you $25 of my money, yours will equal 3 times mine." How much money has each? 126 SIMPLE EQUATIONS: TWO UNKNOWN QUANTITIES. 21. A and B together perform a certain work in 30 days. At the end of 18 days B is called off, and A finishes it alone in 20 days. In what time could each perform the work alone ? - Solution. Let x = the number of days in which A could perform it alone, and y = the number of days in which B could perform it alone. nn. 1 , 1 1 , 18 t 18 , 20 . , .„ , - K Then, — H — = — ,and h V — = 1; whence, £ = 50 andw = 75. x y 30 x y x Here the process is shortened by multiplying the first equation by 18, and, without clearing of fractions, subtracting the result from the second equation. 22. A pound of tea and 5 pounds of sugar cost $1.30, but if tea were to rise 50% and sugar 20%, the same quantities would cost $1.77. Find the price of tea and sugar. Ans. Tea, $.70 per lb.; sugar, $.12. 23. A workman having worked 12 days, and been idle 5, received $25. When he worked 16 days, and was idle 7, he got $33. What were his daily wages and his maintenance? 24. If A and B each gain $24, A will have twice as much money as B, but if each lose $24, A will have three times as much money as B. How much has each ? Ans. A, $168; B, $72. 25. There are two kinds of tea — one worth $1 per pound, the other worth $1.25 per pound ; how many pounds of each must be taken so that 112 pounds of the mixture may be worth $130^ 26. What are the two numbers whose sum is 133, and whose difference is 19 ? 27. On making up the roll of an army after a battle, it was found that the number of effective men was only 714 more than half the number before the battle. Of the remainder, the wounded were twice as many as the slain, the prisoners one- third as many as the slain, and the prisoners also one-third as many as were left for immediate service, while the number of wounded exceeded the number of prisoners by 677. What was the original strength of the army ? SIMPLE EQUATIONS : TWO UNKNOWN QUANTITIES. 127 28. A sum of money was divided equally among a certain number of persons. If there had been six persons more, each would have received $2 less than he did ; and if there had been three persons less, each would have received $2 more. Find the number of persons and how much each received. Solution. x = the number of "persons. y = the number of dollars each received. xy = the number of dollars divided. By the conditions, i^+8)(y-»)~xy (1) \(x-8)(y+2)=xy (2) Expanding (1), xy + 6y—2x — 12 = xy or, 6y-2x = 13 (3) Expanding (2), xy + 2x—3y- 6 =xy or, Sx-3y = 6 (4) Adding (3) and (4), 3y = 18 whence, y = 6 Substituting value of y in (4), 2x — 18 = 6 whence, x = 12 29. There is a certain rectangular floor, such that if it had been two feet broader and three feet longer, it would have been sixty-four square feet larger. But, if it had been three feet broader and two feet longer, it would have been sixty-eight square feet larger. Find the length and breadth of the floor. 30. A number of posts are placed at equal distances in a straight line. If to twice that number we add the number of the feet between two consecutive posts, the sum is 68. If •from four times the number of the feet between two consecutive posts we subtract half the number of posts, the remainder is 68. Find the distance between the extreme posts. Ans. (24-1)20, or 460, feet. 31. A certain company in a hotel found, when they came to pay their bill, that if there had been three more persons to pay the same bill, they would have paid one dollar each less than they did ; and if there had been two fewer persons, they would have paid one dollar each more than they did. Find the num- ber of persons and the number of dollars each paid. 128 SIMPLE EQUATIONS : SEVERAL UNKNOWN QUANTITIES. SECTION XXVII. SIMPLE EQUATIONS WITH MORE THAN TWO UNKNOWN QUANTITIES. 225. Equations with more than two unknown quantities can be solved by the methods of elimination given for solving equa- tions with two unknown quantities. C3x+2y-4z = 8") 226. — Ex. 1. Given 1 5x-3y +3z=33 V , to find x, y and z. (_7x+ 2/+5z = 65J Solution. From the ^ ^.^ ^ r 3x+Sy- Ip=8 (1) three given equations three unknown quan- < 5x—3y + 3z=S3 (2) obtain two equations, **• C ?» + » + 5z=65 (3) which shall not con- te+ty-ifc-** (4) „ , 10x-6y + 6z=66 (5) tarn some one of the 1 & +2y +10z = lS0 (6) ' Two equations wilh two J -^a; — 6z = 90 (7) say J/. Multiplying unknown quantities, | llx+lJ,z=122 (8) equation (1) by 3 and 133x-J$z = 6S0 (9) equation (2) by 2, we 83x+42z = 366 (10) obtain (4) and (5), in f 166x = 996 (11) i ■ i ii. ie • 0«6 emwrfion wtfft one I X = 6 (12) which the coefficients ,,-,-(. tmhnown quantity, | z=A (13) of y are the same. j w = S (14) Multiply (3) by 2, making the coefficient of y the same as in (1). Adding (4) and (5), we have (7), and subtracting (1) from (6), we have (8). We now have two equations with but two unknown quantities. We are now to obtain one equation with but one unknown quantity. Let us eliminate z. Multiplying equation (7) by 7, and (8) by 3, we obtain (9) and (10). Adding equations (9) and (10), we obtain (11). Dividing (11) by 166, we obtain (12), or x = 6. Substituting this value of x in (8), and reducing, we get (13), or z = 4. Substituting the value of x and z in (3), we obtain (14), or y = 3. I SIMPLE EQUATIONS : SEVERAL UNKNOWN QUANTITIES. 129 Cx+y-z=Z\ 2. Given ■< x+z -y = 5 > , to find x, y and z. (y+z-x=7) x+y-z=8 (1) x+z-y=5 (2) Solution. Adding equations (1) y+z— x = 7 (3) and (2), we obtain (4) ; whence, x = 4. Sx = 8 (4) Adding (1) and (3), we obtain (5); %-<$■ hence, y = 5. Adding (2) and (3), we Sy = 10 (5) obtain (6); whence, z = 6. y = 5 ■s=12 (6) z=6 ( x+2y-Sz = - 4") 3. Given < x+Sz — y= 9 j- , to find », y and z. (,#2/+2z- a;= 15 ) _4ws. x=S; y = 4> z = 5. Cy+z = 1-\ 4. Given -< x+z = 9 > , to find a, 2/ and z. r &e + 7y-llz= 10 1 5. Given ■< 5x-10y + 3z=-— 15 > , to find a, y and z. (-ta+122/- z = 31 ) Ans. x = 8 ; y = 7; z = 5. ■x+ y+ z=29"\ 6. Given -i > , to find a;, y and z. -+-^+--=10 2 3 1 1+7+1=15) , to find a, y and 3. >> 4: O I 4 5 6 / F+! = 3\ 8. Given <| +-=5) , to find », y and z. 10 c '* + *=4l .4k.s. x = a; y = 2h ; z = Sc. 130 SIMPLE EQUATIONS: SEVERAL UNKNOWN QUANTITIES. 9. Given ■) Z-ll=26 \ > to find x > & z and M * SECTION XXVIII. PROBLEMS PRODUCING. SIMPLE EQUATIONS WITH MORE THAN TWO UNKNOWN QUANTITIES. 227. — Ex. 1. Find three numbers, A, B and C, such that A added to half of B, B added to a third of C, and C added to a fourth of A may each be 1000. Ans. A, 6Jfi; B, 720; G, 8Jfi. 2. Divide $200 among three persons, A, B and C, so that twice A's share +$80, three times B's share +$30, and four times C's share +$40, may each equal the same sum. 3. A man, talking with his wife and son of their ages, said that his age added to that of his son was 16 years more than that of his wife; the wife said that her age added to that of her son made 8 years more than that of her husband, and that all their ages together amounted to 88 years. What was the age of each ? Ans. Husband, Jfi ; wife, 86 ; son, 12. 4. A and B working together can earn $40 in 6 days, A and C together can earn $54 in 9 days, and B and C together can earn $80 in 15 days. Find what each man alone can earn per day. 5. At an election there were two members of the legislature to be chosen, and there were three candidates, Lawton, Bowen and Stone. Lawton obtained 1056 votes, Bowen 987, and Stone 933. Now, 85 voted for Bowen and Stone, 744 for Bowen only, and -98 for Stone only. How many voted for Lawton and Stone? How many for Lawton and Bowen? How many for Lawton only? Ans. For Lawton only, 1J/B ; for Lawton and Stone, 750; for Lawton and Bowen, 158. REVIEW PROBLEMS. 131 6. A's money, together with twice that of B and C, amounts to $1050 ; B's, together with thrice that of A and C, amounts to $1400 ; and C's, together with four times that of A and B, amounts to $1650. How much money has each ? 7. A man bought 10 horses, 120 cows and 46 colts. The price of 3 cows is equal to that of 5 colts. A horse, a cow and a colt together cost a number of dollars greater by 300 than the whole number of animals bought ; and the whole sum ex- pended was $9366. Find the price of a horse, a cow and a colt, respectively. Am. A horse, $420; a cow, $35; a colt, $21. 8. A, B and C can do a piece of work in 15 days ; A and B together do four-thirds of what C does ; and C does twice as much as A. Find the time in which each alone could do the work. SECTION XXIX. REVIEW PROBLEMS. 228 —Ex. 1. Given { ^-S^ } ' t0 find x and y ' Ans. x = 6; y = 5. 2 - Given {2-S = 39}' t ° finda;alld2 '- (f + By = 23] 3. Given < " ~„, > , to find x and y. 4. A gentleman divided $1.20 among three poor persons ; to the second he gave twice, and to the third three times, as much as to the first. How much did he give to each ? 132 REVIEW PROBLEMS. 5. If a certain number be multiplied by m and by n, the sum of the products will be a. What is the number ? Solution. x = the number. mx+nx = a Factoring, {m-Yn)x = a a Dividing by (m+ri), x = m+n 6. A man bought an equal number of tons of coal for $5 and for $7 a ton, and the cost of the whole was $492. How many tons of each did he buy ? Ans. J/.1. 7. A owes $1200 and B $2500, but neither has money enough to pay his debts. " Lend me," said A to B, " the eighth part of your money, and I shall be enabled to pay my debts." B answered, " I can discharge my debts if' you will lend me the ninth part of yours." How much money had each ? 8. What fraction is that to the numerator of which if 1 be added, the value will be \ ; but if 1 be added to the denomi- nator, its value will be \ ? Ans. ^. 9. Two persons, A and B, have the same income. A saves one-fifth of his yearly, but B, by spending $50 per annum more than A, at the end of four years finds himself $100 in debt. What is their income ? 10. The grading of a square piece of land at $2 a square yard cost as much as the inclosing of it at $5 a linear yard. Required the side of the square. Solution. X = the number of yards in the side of the square. Tft = the number of yards around the square. x 2 = the number of square yards of grading. $5*. 4% = $20x = price of inclosing. $2xx 2 = $2x* = price of grading. $2x i = $20x Dividing by $2x, x = 10 REVIEW PROBLEMS. 133 11. A general, ranging his army in the form of a solid square, finds he has 284 men to spare ; but increasing the side by one man, he wants 25 to fill up the square. How many soldiers has he ? Ans. £4000. 12. A company of 180 persons consists of men, women and children. The men are 8 more in number "than the women, and the children 20 more than the men and women together. How many of each are in the company ? Ans. 44 men ; 36 women ; 100 children. 13. Given \z + Sy- 2 = 31, 3x-y-2z = and 6x+2y+3z = 45, to find x, y and z. 14. It is required to divide 90 into four such parts that if the first part be increased by 2, the second diminished by 2, the third multiplied by 2, and the fourth divided by 2, the sum, difference, product and quotient shall all be equal. 15. There are three numbers such that their sums, taken two by two, are 11,12 and 13, respectively. What are the numbers ? Ans. 5, 6 and 7. lest Questions. 229. — 1. What is the degree of an Equation f What is a simple equation ? What is the rule for clearing a simple equation of fractions? 2. How may a simple equation be solved ? What is the root of an equation? How is the root verified? What is the rule for solving a simple equation with one unknown quantity ? 3. What are Simultaneous Equations ? What are independent equa- tions ? 4. What is Elimination f Elimination by comparison ? Elimination by substitution ? Elimination by addition or subtraction ? 5. What is the Rule for elimination by comparison ? For elimination by substitution? For elimination by addition or subtraction? How may equations with more than two unknown quantities be solved ? 12 134 INVOLUTION. SECTION XXX. INVOLUTION. 230. — Ex. 1. What is the product of a by a? What power of a does a 2 represent ? 2. What is the product of a 2 by a ? What power of o does a 3 represent ? 3. How many times is a taken as a factor in -producing a 2 ? In producing a 3 ? 4. Of what factors is a 2 the product ? Of what factors is ab the product ? 5. Is a 3 the product of equal or unequal factors ? Is 6 the product of equal or unequal factors ? 6. How will you find the third power of 2 ? Of a;? 7. What is the fourth power of a ? The second power of — a ? The* third power of — a ? Definitions. 231. A Perfect Power is the product of equal factors. Thus, a 2 , 6 3 , 4, 9, etc. are perfect powers. 232. An Imperfect Power is the product of unequal factors. Thus, ab, cd+xn, 6, 15, etc. are imperfect powers. 233. Inyolution is the process of finding the powers of a quantity. It is performed by successive multiplications, the quantity being taken as many times as there are ones in the index of the power. Thus, a is involved to the third power by taking a three times as a factor. involution: 135 234. Principles. — 1. All powers of a positive quantity are positive. For, taking a positive quantity any number of times as a factor must give a positive result. Thus, axa = d i ; d'xa = a 3 . 2. All the even powers of a negative quantity are positive, and all the odd powers are negative. For, taking a negative quantity twice as a factor, or multiplying it by itself, gives a positive result ; multiplying that result by the quan- tity which is negative, gives for the third power a negative result. Thus, ( - a) x ( - a) = a 2 ; d'x( — a) = - a 3 ; ( — a 3 ) x ( - a) = a 4 , etc. CASE I. Involution of Monomials. 235. — Ex. 1. Find the cube or third power of 5aV Solution. 5a?x taken three times (5a?x) 3 = 5a?xx5a i xx5a'x as a factor is equal to £>a?x x 5a?x x oa?x, — g x S x 5a?a?d*xxx or to 5 x 5 x haPaPcPxxx, whose product _ ? gr„6 r 3 is 125a e x s . 2. Find the square of - Sax. Ans. S«V. 236. Rule for Involution of Monomials.— Involve the nu- merical coefficient to the required power, multiply the exponent of each letter by the number denoting the exponent of the required power, and give the proper sign to the result. A fraction is involved by involving both numerator and denominator. PROBLEMS. 1. Find the cube of lab. Ans. S4Sa 3 ¥. 2. Find the square of 6a 2 . 3. Find the third power of - aW. Ans. - a*b 9 . 4. Find the fourth power of abV. 5. Find the fifth power of - aW. Ans. - a I0 6 15 c M . 6. Find the mth power of a&e 2 . 7. Find the third power of x-*f~. Ans. x~ 3n i/^ m . 136 INVOLUTION. 8. Find the fifth power of - Sax*. 9. What is the wth power of x 2 f ? 10. What is the nth power of -3»y ? Ans. i^asV- Here, n may be any number whatever ; hence, the nth. power of the given quantity may be even or odd, and consequently may be either positive or negative, as is indicated by the sign ± . When a certain number is given as the value of n, the power is positive if the number is even, and negative if odd. 6a 2 11. What is the second power of ? c Solution. /6a?V- 6a? y 6a? V c / c c 12. What is the cube of — ? Ans. - — - . 4a 64a? 4a 2 13. What is the fifth power of — ? ox a~ n c 2 a~~ mn c 2m 14. What is the with power of — — — ? Ans. — . fl~ n h rt-nmASin— mn 15. What is the value of ( - 4a6Vm, 2 ) 4 ? 16. What is the value of (6a 2 6 VV) 3 ? Ans. 0iga , &-W. CASE II. Involution of Polynomials. 237. — Ex. 1. Find the square and the cube of (a +b). Solution. The square a+b of a+b, or [a+b){a+b), is, a +b by actual multiplication, , ,„ J + ab+b 2 a 2 + 2ab + b\ and the cube — — _ , , ,„ ,, , . . a?+2ab+b 2 , square of [a+b). of a+b, or (a+b)(a+b) a+b (a + b), is, by actual mul- a 3 +8a*b + ab* tiplication, a s +3a 2 6 + 3ee& 2 +d'b+Sab 2 + b 3 + b 3 . a s +Sa 2 b + 3ab 2 + b s , cube of (a+b). 2. Find the cube of (a - b). Ans. a 3 - Sa?b + Sab 2 - b 3 . INVOLUTION. 137 238. Rule for Involution of Polynomials.— Use the given quantity as a factor as many times, as are indicated by the exponent of the required power. PROBLEMS. 1. Find the square of (a+e). Am. a 2 +2ac+c 2 . 2. Find the square of (a+2x). 3. Find the square of (o-c). Am. a? — %ac + c*. 4. Find the square of (2a +c). 5. Find the square of (ac — b). Am. a 2 c 2 — 2abc + b 2 . 6. What is the cube of (m+n) ? 7. What is the cube of (2a +36) ? Am. 8a 3 +36a?b+54ab 2 +27b\ 8. What is the cube of (2+z) ? 9. What is the value of (3 - 2z) s ? Am. 27-5J/X+ S6x 2 - 8x\ 10. Expand (x -2) 4 . 11. Expand (a + b + cf. Ans. a 2 +2ab+2ac + b 2 +2bc+c 2 . 12. Involve 1+x-x 2 to the third power. 239. Any Polynomial may be squared without recourse to multiplication by observing the following case and the princi- ples drawn therefrom. Let a+b + c+dbe any polynomial ; then by actual multiplication, (a+b+c+dy=a 2 +2ab+b 2 +2ac+2bc+c 2 +2ad+2bd+2cd+d 2 . Changing the order of terms, we have , (a+b+c+d) 2 =a 2 +b 2 +c 2 +d 2 +2ab+2ac+2bc+2ad+2bd+2cd. Factoring and changing the order, we have {a+b + c+d) 2 = a?+2ab+b 2 +2{a+b)c+c 2 +2{a+b+c)d+d' 1 . 12* 138 EVOLUTION. 240. Principles. — 1. The square of any polynomial consists of the square of each term, together with twice the product of every pair of terms; or, 2. The square of any polynomial consists of the square of the first term, plus twice the product of the first term into the second, plus the square of the second, plus twice the sum of the first two terms into the third, plus the square of the third, etc. PXOBZEMS. 1. Find the square of a — b+c. Ans. a?+V + c i -2ab + 2ac-2bc. 2. Find the square of x+y+z. 3. What is the second power of a+m — nl Ans. a 2 +m 2 +n 2 + 2am-2an — 2mn. 4. Expand (2+Zx+4x 2 y. 5. Expand (2a - b + c - d) 2 . Ans. 4a 2 + b 2 + c 2 + d 2 - 4ab + 4ac- 4ad - She + 2bd - 2cd. SECTION XXXI. EVOLUTION. 241.— Ex. 1. What is the product of&xj? Of xxxxx? What is one of the equal factors of 6 2 ? Of x 3 ? 2. What is the second or square root of 9 ? The third or cube root of 27 ? 3. What is the square root of a 2 ? What is the cube root of a 8 ? Of -a 3 ? EVOLUTION. 139 Definitions. 242. Evolution is the process of finding a root of a quantity. It consists in determining the quantity which, when involved to the power corresponding to the required root, will equal the given quantity. This process is sometimes called extraction of the root. 243. In Fractional Exponents the numerator denotes a power, and the denominator a root. Thus, a s , or ffaF, denotes the cube root of a square. 244. Principles. — 1. An odd root of a quantity has the same sign as the quantity. Thus, the cube root of a 3 is a, and the cube root — a 3 is - a. 2. An even root of a positive quantity is either positive or negative. Thus, axa = a 2 , and — ax—a = a 2 ; hence, the square root of a 2 is either a or — a ; that is, it is either positive or negative. 3. An even root of a negative quantity is impossible. Thus, there is no square root of — a 2 , for if any quantity be multi- plied by itself, the result is a positive quantity. Hence, an indicated even root of a negative quantity is called an impossible or an imaginary quantity. CASE I. Evolution of Monomials. 245.— Ex. 1. Find the cube root of 125aV. Solution. To cube a monomial yffia&—fffi xa txi-6atoi we cube the numerical coefficient, and multiply the exponents of the letters by 3 ; hence, to find the cube root of the given monomial we reverse the process ; that is, we extract the cube root of its coefficient, and divide the exponents of the letters by 3. The cube root thus found is 5a 2 x. 140 EVOLUTION. 2. Find the square root of 36a 2 6*. Solution. Since finding the rfg&&= /SSxoM- ±6aV square root of a quantity is the reverse of squaring the root, we extract the square root of the co- efficient of the given quantity, and divide the exponents of its letters by 2. The square root thus found, which may be either positive or negative (Art. 244, 2), is ± 6ab\ 3. Find the fourth root of 81x?y l . Ans.±3x 2 y. 246. Rule for Evolution of Monomials.— Extract the re- quired root of the numerical coefficient, divide the exponent of each letter by the number denoting the index of the required root, and give the proper sign to the result. The root of a fraction is found by taking the root of the numerator and of the denominator. PROBLEMS. 1. What is the cube root of- 8x 3 y e ? Ans. - 2xy 2 . 2. What is the fourth root of 625<»y ? 3. What is the cube root of -2Wc la ? Ans. -£«W. 4. Find the cube root of — 64a 6 2/ 3 . Ans. - 4a 2 y. 5. Find the mth root of 7°aV. 6. Find the fifth root of aV°. 7. What is the value of ^625xYz w ? 8. What is the value of tf - 8a-"6V s ? Ans. -2a~ l ¥x~i. 9. What is the square root of ? Solution. 16a? i/lfia? ha \ 9b 2 " V~9W~ J*' EVOLUTION. 141 10. What is the fourth root of ^_^! ? 16ay 11. What is the third root of- ^- ? Am. - ^. 27a; 9 Sx 3 OO-5A10 12. What is the fifth root of ^=± ? CASE II. Square Root of Polynomials. 247.— Ex. 1. Find the square root of a?+2ab + b 2 . Solution. From the nature of the a 2 +2ab + b\a + b square of a polynomial (Art. 239, Prin. 2), a 2 the first term of the given square will be 2a ) 2ab + b 2 the square ofthe first term of the required (2a+b)b = 2ab + b 2 root. The first term of the square is a 2 , and its square root is a, which we write as the first term of the root. Subtracting a 2 from the whole expression, we have left 2ab + b 2 , which must be two times the product of the two terms of the root plus the square of the last term of the root (Art. 239, Prin. 2). Dividing 2ab, the first term of the dividend, by la, which is double the first term of the root, we obtain b, the other, or second, term of the root. Adding the b to the trial divisor la, we obtain the complete divisor 2a + b. Multi- plying this by 6, the second term, and subtracting, we have no remain- der. Hence, a+b is the root required. 2. Find the square root of a 2 +2ab + b 2 — 2ae — 2ba+c 2 . a?+2ab+b 2 -2ac-2bc+c\a+b-c a 2 2a ) 2ab+b 2 {2a + 6)5 = 2ab+b 2 2a+2b ) -2ae-2bc+c 2 (2a+2b- c)c=-2ac-2bc+e 2 Solution. We find the first two terms of the root, a+b, as before. Dividing the remainder by twice this root, we obtain —c, the third term of the root. Adding the — e to the trial divisor, we obtain the complete divisor, 2a+ 2& — e. Multiplying this by — c, the third term of the root, and subtracting, we have no remainder. Hence, a+b — c is the root required. 142 EVOLUTION. 3. Find the square root of 4x 2 + 12xy+9y 2 . Ans. 2x+Sy. 248. Rule for Extracting the Square Root of Polynomials. — Arrange the terms according to the powers of some letter. Find the greatest square root of the first term, and place it at the right for the first term of the re- quired root, and subtract its square from the given polynomial. Divide the first term of the remainder by double the root already found, and annex the result to the root, and also to the divisor, for a complete divisor. Multiply the complete divisor by the second term of the root, and subtract the product from the divi- dend. Continue the process, if there are other terms, as before. PROBZMMS. 1. Find the square root of 9z 4 - 24z 2 + 16. Ans. 8x?~4. 2. Find the square root of 16x i -8x i cy+e 2 y i . 3. Find the square root of 36« 6 + 12a 3 + 1. Ans. 6a? +1. 4. Find the square root of 1 -2x+5x 1 -4x 3 + ix i . 5. Find the square root of x l — 4x? + 8a; + 4. Ans. x i — 2x — 2. 6. What is the value of i/W+iax+x 2 - 4ay - 2xy+y'l 7. What is the value of (9x 2 +S0xy+25y*)* ? 8. What is the value of v /a?-2ab+b*+2ae-2bc+c"! 249. Any Trinomial that is a perfect square has two of its terms squares, and the other term is double the product of their square roots. Hence, to obtain the square root of a tri- nomial which is a perfect square, — EVOLUTION. 143 Arrange the trinomial according to the powers of either of its letters, if necessary ; extract the square roots of the two extreme terms, and unite their roots by the sign of the middle term. PROBLEMS. 1. What is the square root of 1 6a 2 -56a&+496 2 ? Ans.J,a-7b. 2. What is the square root of 64a 2 +48a6e+96V? 3. What is the square root of 4w 2 +12m»i+9re 2 ? CASE III. Square Root of Numbers. 250. A Method of extracting the square root of numbers • may be derived from the nature of algebraic polynomial 251. Principles. — 1. If a square expressed by more than two orders of figures be separated into periods of two figures each, commencing with the units, the number of figures in the root is indicated, and also the orders of the squares in which are to be found the squares of the numbers in the orders of the root. For the square of 1 is 1, the square of 10 is 100, the square of 100 is iOOOO, etc. 2. The square of any number expressed by more than one order of figures consists of the square of the tens, plus twice the product of the tens by the units, plus the square of the units. For any number expressed by more than one order of figures may be regarded as an algebraic polynomial, and as composed of two parts, units and tens. Thus, if t denote tens and u denote units, any number will be denoted by t+u, and (j!+m) 2 = « 2 +^m+m 2 = < 2 +(^+m)m. Then, let t = 3 and M = 7, and t+u = S7 ; hence, 37* = 30H 2x30x7 +7*= 1869. 144 EVOLUTION. 252.— Ex. 1. What is the square root of 2209? Solution. Since , . t u the given number is 2209 = f+2tu+u\47 expressed by 4 orders If = IB =t 2 ' of figures, it may be 2t = 2x4 tens = 80)309= 2tu + u 2 = {2t+u)u separated into two . u = 7 periods, and its square 2t+u = 87 root will consist of 2 87x7 = 609= (2t +u)u figures, and will ex- q o press tens and units. Eepresenting the tens by t and the units by u, we have 2209 = i 2 +2to + u 2 . Then t 2 , or the greatest square of tens which is less than 2200, is 16 hundreds, whose root is 4 tens. Subtracting 16 hundreds from the given number, and its equal t" from the expression above, 609 remains, which is equal to 2tu+v?, or (2t+u)u. Dividing this remainder by 2t, that is, by 80, we obtain 7, the value of u. Then (2t+u)u, that is, 87x7, is equal to 609. Subtracting this product and its equal (2t+u)u, from the quantities above them, there is no remainder. Hence, the square root of 2209 is 47. 2. What is the square root of 55225 ? Solution. ^-, .... t u u 5Bs\25 = e+gtu+u\23 5 & = 4 =f 2t = 2x2tens= 40)152 = u = 3 2t+u =43 43x8= 129 = 2tu+v? = 2{t+u)u (2t+u)n 21 = 2x23 tens = 460) 2325 = u =5 2l+u =465 465x5= 2325 = 2tu+u i = (2t+u)u (2t +u)u In the same manner as in the preceding solution, we find 23, the tens and units of the tens of the root of the greatest square of tens which is contained in 55200, and have the remainder 2325. We now consider 55225 as the square of 23 tens and some number of units of units. Eepresenting the 23 tens of the root, which have been EVOLUTION. 145 found, by t, and the units of the root to be found, by u, we may repre- sent the square of the root by t' i + 2tu+'j?. The equal of* 2 , or 230 2 , has been subtracted; hence, the remainder, 2325, is equal to 2tu+v?, or (2*+u)«. Dividing the remainder by 21, that is, by 460, we obtain 5, the value of«. Then (2t+u)u, that is, 465x5, is equal to 2325. Subtracting this product and its equal, (2t+u)u, from the quantities above them, we have no remainder. Hence, the square root of 55225 is 235. If the given number had b3en a decimal fraction, its periods would have been pointed off to the right, beginning with units ; for the square of .1 is .01, the square of .09 is .0081, etc. 3. Find the square root of 77841. Ans. 279. 253. Rule for the Extraction of the Square Root of Numbers. — Point off the given number into periods of two orders each, beginning with the units and proceed- ing toward the left and right. Find the greatest square in the highest period, considered as units, and place its root at the right for the first figure of the required root. Subtract this square from the highest period, and to the re- mainder bring down the next period for a divi- dend. Divide the dividend, omitting its right-hand order, by twice the root already found, and write the quotient for the second figure of the required root. To the divisor add the part of the root found by it, multiply the result by that part of the root, and subtract the product from the dividend. Continue the process, if there are other periods, as before. When occurs in the root, instead of indicating the multiplication by and subtracting, it is simpler to annex a cipher to the divisor, and to the dividend bring down another period. 13 146 EVOLUTION. If there be a remainder after all the periods have been used, periods of decimals may be formed by annexing ciphers, and the work continued. When the given quantity is a fraction whose numerator and denomi- nator are not both squares, reduce the fraction to a decimal before ex- tracting the root. PROBLEMS. 1. Find the square root of 177241. Am. 421. 2. Find the square root of 3249. 3. Find the square root of 165649. Am. 407. 4. Find the square root of 41.2164. 5. Find the square root of 2.5. Ans. 1.5811. 6. Find the square root of 17.3056. 7. What is the square root of .001849 ? Ans. .048. 8. What is the square root of 2 to two decimal orders? 9. What is the square root of ? Ans. — H 3481 59 10. What is the square root of — to three decimal orders ? H 26 11. What is the value of -J2— ? Ans. 1—, \ 121 U 12. What is the value of \l— to three decimal orders? 13. What is the square root of .008723 to three orders ? 4 14. What is the square root of 6- to three decimal orders? RADICALS. 147 SECTION XXXII. RADICALS. 254.— Ex. 1. What is expressed by j/li?? By 9o^? 2. In what two ways may the root of a quantity be indi- cated ? 3. In 3|/a, what shows how many times ya is taken ? 4. Name two different expressions having the same quantity under the same radical sign. 5. Can the root indicated by i/4a 2 be exactly obtained? Can the root indicated by y / 2a be exactly obtained? Definitions. 255. A Radical is an indicated root of a quantity. _ i Thus, j/a, 64 3 , i/2ax i , etc., are radicals. 256. The Coefficient of a radical is the quantity prefixed to the radical to show how many times it is taken. Thus, in l-^ast?, 2 is the coefficient of j/aje 5 . 257. The Degree of a radical is indicated by the index of its root, or by the denominator of its fractional exponent. i i Thus, i/ab, x* , (ax) , are radicals .of the second degree, and i fxy, tym, (ZaV) T , are radicals of the third degree. 258. Similar Radicals are those which have the same quan- tity under the same radical sign. Thus, ^~aR>, 2 v /~aW, (a 3 b) 3 , are similar radicals. 148 RADICALS. 259. A Rational Quantity is a quantity whose indicated root can be exactly obtained, and an Irrational Quantity, or Surd, is a quantity whose indicated root cannot be exactly obtained. Thus, i/9a? and f27a s are rational quantities, and j/36 and -tyWx* are irrational quantities. 260. Principles. — 1. The root of any quantity is equal to the product of the roots of its factors. Thus, j/M= j/4 x /16 = 2x4 = 8. 2. The product of the same roots of two quantities is equal to the same root of their product. Thus, 1 /4x 1 /T6= 1 /64. 3. The quotient of the same roots of two quantities is equal to the same root of their quotient Thus, 1 /64-V'4= 1 /16. CA.SB3 T. A Radical to its Simplest Form. 261. A Radical is in its Simplest Form when it contains no factor whose indicated root can be obtained. - 262.— Ex. 1. Reduce ^SOa^ to its simplest form. Solution. We separate the quantity under the radical sign y«taW-|/ifldWxffa^ into two factors, 16a 2 6 s , which = \/16a'b 2 x y/ '5a= J/ab]/ '5a is a perfect square, and 5a, which is a surd. Then, by Prin. 1, Art. 260, 1 /80a 8 5 s = 1 /16a 5 5 5 x\/5a; hence, y/SOa'b 2 is equal to the product of the square root of 16a 2 o 2 by ]/5a, or 4ao/5a. 2. Reduce 5^''24i? to its simplest form. Solution. We separate the quantity under the radical sign into two factors, 8x*, 5fy21t& = 5^8^ x Sx which is a perfect cube, and 3x, which is _ * >/-op> x ys~ a surd. Multiplying the coefficient 5 by 2x, the cube root of 8a; 3 , and that product = 5xSxx {/Sx by the surd, {/3x, we have lOx^/Sx, the =10x1 Sx simplest form of the given quantity. RADICALS. 149 3. Reduce y / 27a 5 & 3 to its simplest form. Ans. 8a?b-\/8ab. 263. Rule for Reducing a Radical to its Simplest Form.— Separate the given radical into two radical factors, one of which is rational and the other a surd. Find the root of the rational factor, multiply it by the coefficient of the given radical, and place the prod- uct as the coefficient of the surd. PROBLEMS. Reduce the following radicals to their simplest form. 1. t/ISS. Ans. 8i/2x. 2. S-i/leys. 3. |/150a 2 6. Ans. 5 aA /6b. 4. ^55W. 5. 2j^54aW. Ans. 6ab-tfJa?. 6. 20V75^- 7. ■fiax'+bx 6 . Ans. x-fta+bx 3 . 8. 5i/25a-25. J 9. a6(a 3 6 2 -a 2 6 2 ) . Ans - dV^/a-l. 10. 7i/{.d'-f%a+b). 264. When the Radical is fractional, to have only an integral quantity under the radical sign, Multiply both terms of the fraction by that quan- tity which will make its denominator a perfect power of the same degree as the root indicated. 13* 150 RADICALS. Reduce to the simplest form — Solution, 8 3. +- (to) 6 - Ww' -4ns. -y^oT .A»*. ^il. Ans j/6\ 8. °4- 9. 4h 10. WI 11. «J£- Ans. -yio. Au- IT-yi^- CASE II. A Rational Quantity to the Form of a Radical. 265. — Ex. 1. Reduce baW to the form of a cube root. Solution. Since any quantity is equal to the 5ab 2 = ■&(6ab l y cube root of its cube, 5ab' is equal to the cube root of (5ao a ) s , or to fVZEcm; = V^BaW 2. Reduce 5xy to the form of the square root. Ans. -y/25xy. 266. Rule for Reducing a Rational Quantity to the Form of a Radical. — Involve the quantity to the power denoted by the given root, and place the result under the corresponding radical sign. RADICALS. 151 PROBLEMS. 1. Reduce 2oc to the form of the square root. Ans. j/^aV. 2. Eeduce Bx'y~ l to the form of the cube root. Ans. $"27x*y-*- 3. Eeduce — 3x to a radical of the third degree. 4. Reduce •*— - to a radical of the second degree. Ans. « / — 5. Reduce x* +y to the form of the cube root. 267. The Coefficient of a radical can be placed under the radical sign by involving the coefficient to the power denoted by the radical, and multiplying the quantity already under the radi- cal sign by the result. PROBLEMS. 1. Place the coefficient of 1a\/bx under the radical sign. Solution. 2ai/5x = ■/ {2af x5x = i/20a?x 2. Place the coefficient of 4-j/cd 2 under the radical sign. o 3. Reduce --ftmn to a radical without a coefficient , 27mn Ans. 64 4. Reduce Zay'Vc' to a radical without a coefficient. 5. Place the factor 2 of the coefficient of 2ab\/ Bc'x under the radical sign. Ans. ab-tyfU&x. 6. Place the coefficient of 5 = 12a'b 3 reduced, is 12a z 6 3 . 14 158 BADICALS. 2. Involve bay'x to the third power. Ans. 125a*x. 279. Rule for Involution of Radicals.— Involve the ra- tional and radical parts to the required power, and reduce the result to its simplest form. When the quantity to be involved is affected by a fractional exponent, the involution may be performed by multiplying the exponent of each letter by the exponent of the required power. Dividing the index of the root produces the same effect as multiply- ing the fractional exponent. PROBLEMS. 1. What is the second power of ay's? Ans. 8a?. 2. What is the third power of 5^0^? o ■ _ g 3. Involve -j/3 to the third power. Ans. -g\/3- 4. Involve 3y / 2a 2 6 to the fourth power. Ans. 162a?b-$ / 2a7b. 5. Involve -j/'oVto the fourth power. 1 6. What is the fourth power of ^VK"? ~ «') ? Ans. at — SaV+x*. i 7. What is the sixth power of (a+bfl CASK IX. Evolution of Radicals. 280.— Ex. 1. Find the square root of 4ay9F. Solution. Since the root of a quantity /TV, /ora_ ± o a /~&fi3 is equal to the product of the roots of its factors (Art. 260), we take the square root of the coefficient, which is ±2a, and the square root of the quantity under the sign, which is j/36 3 ; uniting these, we have, for the entire root, ± 2a J /36 3 . RADICALS. 159 2. Find the cube root of 54|/3a! Solumon. The coefficient, ^'^S^^xV^ which is not a perfect cube, is composed of the factors 27 and = fWx $ S-/Sa = S\/S-^Sa 2, the former of which, 27, is a perfect cube. We take the cube =3y / - i /TIa=3fl2a root of 27, which is 3. We square the 2, since it is not a perfect cube, and introduce it as factor under the sign. The 12a under the sign is not a perfect square ; hence, we denote its root by multiplying the index of the sign by the index of the required root. We have then for the entire root 3^ 12a. 281. Rule for the Evolution of Radicals. — Extract the re- quired root of the rational and radical parts, if possible, and reduce the result to its simplest form. If the rational part is not a perfect power, intro- duce it under the radical sign; and if the radical part is not a perfect power, multiply its index by the index of the required root. When the quantity to be evolved is affected by a fractional exponent, the evolution may be performed by dividing the exponent of each letter by the index of the required root. PROBLEMS. 1. Extract the square root of 9 i^'S". Am. ±8-yJ. 2. Extract the square root of 25a l b 1 c. 3. Extract the cube root of -~\/3a. Arts. r^Sa. / Q 4. What is the cube root of —a 4 ? 5. What is the square root of a\l~? Am. =•= — i^ax 3 . \ x x 6. What is the cube root of --»/-? 3\3 160 RATIONALIZATION. 7. What is the cube root of 27a V? Ans. SaHK 8. What is the fourth root of a*fi- 5 c*? cr xor = a SECTION XXXIII. RATIONALIZATION. 282. Rationalization is the process of removing the radical sign from a quantity. This transformation is often of great utility, especially in finding the numerical values of fractional radicals. CASE I. Rationalization of any Monomial Surd. 283. — Ex. 1. Rationalize y a and o?. Solution. Multiplying ]/a by -\/a, we have i/ax 1 /a = « a, which is a rational quantity. 1 2 Multiplying a? by a? (which is the same quan- tity, with such a fractional exponent as will make the sum of the frac- tional exponents equal to 1), we have a, which is a rational quantity. 4 3 2. What factor will rationalize a/t Ans. a 7 . 284. Rule for the Rationalization of any Monomial Surd.— Multiply the surd by the same quantity with such a fractional exponent as, when added to the given exponent, shall be equal to one. PROBLEMS. 1. What factor will rationalize y^ 2 ? Ans. ^a. 2. What factor will rationalize 5j/c? ? 3. What factor will rationalize B-^aFb ? Ans. ^aF. 4. What factor will rationalize ^(a-ft) 2 ? RATIONALIZATION. 161 CASE II. Rationalization of Binomial Surds of the Second Degree. 285. — Ex. 1. Rationalize -[/a+i/b. Solution. The product of the sum and the differ- _ _ ence of two quantities is equal to the difference of their . — ,-r r- - - - Va-Vb squares; hence, i/a+\/b multiplied by j/ a— \/ b is « — 6 equal to a — b, which is rational. 2. What factor will rationalize j/a — -\/b ? Am. -y/a + j/6. 286. Rule for Rationalization of Binomial Surds of the Second Degree.— Multiply the binomial by the same expres- sion with the sign of one of the terms changed. PROBLEMS. 1. What factor will rationalize a +3j/8? Ans. a — S-y/8. 2. Rationalize -j/7 - y/5. 3. What factor will rationalize j/5-j/a? Ans. -\/T> + \/^i, 4. Rationalize 5 + j/3. CASK III. nationalization of either Term of a Fraction. 287. — Ex. 1. Reduce — - *° a fraction whose denominator is rational. V G Solution. Multiplying both terms of the fraction by -j/e, which does not change the a x V c _ a V° value of the fraction, we obtain — 5—, in ycyo c which the denominator is rational. 2. Rationalize the denominator of J_ ns v 3 l/3" 8 ' 288. Rule for the Rationalization of either Term of a Frac- tion. — Multiply both terms of the fraction by such a factor as will render either term rational, as may be required. 14* 162 RADICAL EQUATIONS. FBOBI.EMS. 1. Rationalize the numerator of *—^. l/c V 2. Rationalize the denominator of — -. 3. Rationalize the denominator of *^—z • Am. = y #■ a ac 4. Rationalize the denominator of ^3 a 3 + yV 5. What factor will rationalize the denominator of — ^ ? ya 6. What factor will rationalize the denominator of 3-y'2 SECTION XXXIV. RADICAL EQUATIONS SOLVED LIKE SIMPLE EQUATIONS. 289. Radical Equations are those which contain the un- known quantity under the radical sign. In the process of solving such equations, it will be found necessary first to rationalize the surd expression of the unknown quantity, and then to find its value. \ 290.— Ex. 1. Find the value of a; in 1 /»+4 = 9. Solution. Transposing and uniting (1), ^/x+4 = 9 (1) we obtain (2). Squaring both members of (2), V x = s (2) we obtain (3), or x = 25. * = SS (3) RADICAL EQUATIONS. 163 2. Find the value of a; in \/x + 9 = \/x + 1. Given, yx+9 =-\/x+l Squaring, x+9 = x+2\/x+l. Transposing, — %-\/~x =—8 Dividing by — #, y/x — lf Squaring, x = 16 291. Rule for Solving Radical Equations. — Transpose the terms, so that the radical part shall stand as one side of the equation; then involve each side to a power of the same degree as the radical. If there be still a radical part, transpose and involve as be- fore ; and, finally, fund the value of the unknown quantity as in ordinary simple equations. PROBLEMS. 1. Given x/x+i = 8, to find x. Ans. x = 60. 2. Given i/3x+i = 5, to find x. [2x 3. Given •»/ 1- 5 = 7, to find x. Ans. x = 6. \ 3 4. Given 5+y7x+8 = 13, to find x. 5. Given -y/12+x = 2 + -\/x, to find x. Ans. x = 4- 6. Given 1 /a;-32 = 16 - -\/x, to find x. 7. Given \/ax - b~x = e, to find x. Ans. x = c a—b 8. Given j/z-16 = 8 - -\/x, to find x. 9. Given 2 + j/3a; = |/4 + 5x, to find x. Ans. x = 12. 10. Given j/z+ll - 5 = \/x - 4, to find x. 164 REVIEW PROBLEMS. -- ^- x+ax ,/^T / 11. Given — —=11^, to find *. Ans. x=—. yx x 1+a 12. Given . 7 J -x - 6 = 14, to find *. 13. Given 3 + 5j/«+4 = 28, to find a. .Aws. a; = 21. 14. Given (10 8. Given h = -, to find x. Ans. x = ± 1. 4+x 4—x 3 9. Given ^(3* 2 + 5) - -(x> -r 21 ) = 39 - 5a; 2 , to find x. SECTION XXXVII. PROBLEMS PRODUCING PURE QUADRATIC EQUATIONS. 301. — Ex. 1. If a certain number increased by 6 is multi- plied by the same number diminished by 6, the product is 64. What is the number ? Solution. Let x = the number ; then, x+6=one factor ; and x—6 = the other. By conditions, (x+6)(x-6) = 6j. or, x*-86 = 64 Transposing and uniting, x 2 = 100 Extracting the square root, x = =t 10 Using the positive value, we have the answer 10. 2. What number is that the fourth part of whose square being subtracted from 8, leaves a remainder equaJ to 4 ? 15 170 PURE QUADRATIC EQUATIONS. 3. The distance to a certain village is such that if 96 be subtracted from the square of the number of miles, the remain- der will be 48. What is the distance ? Ans. 12 miles. 4. Two men were talking of their ages ; one said that he was 94 years old. " Then," said the younger, " if the number of years in your age, plus the number in mine, be multiplied by the difference of those numbers, the product will be 8512." What is the age of the younger ? 5. An army was drawn up with 5 more men in depth than in front ; but when the front was increased by 845 men, the army was arranged in 5 lines. What was the number of men in the army ? Ans. 4550. 6. What two numbers are such that f of the greater is equal to their difference, and the difference of their squares is 128 ? 7. There is a rectangular field whose length is § of its breadth. A part of this field equal to \ of the. whole having been measured off, there remain 625 square rods. What are the dimensions of the field ? Ans. Length, SO rods ; width, 25 rods. 8. A man being asked how much money he had, replied, that if the square of the number of dollars he had were mul- tiplied by 7, the product would be 1575. How many dollars had he? 9. William bought two pieces of cloth, which together mea- sured 36 yards. Each piece cost as many shillings a yard as there were yards in it, and the entire cost of one piece was 4 times that of the other. How many yards were there in each piece ? Ans. 24 in one ; 12 in the oilier. 10. There are in a certain house two square parlors. The side of the larger is \\ times that of the smaller, and it takes 180 square feet more of carpeting to cover the floor of the one than of the other. What is the length of one side of each room? Ant. 18 and 12 feet, respectively. AFFECTED QUADRATIC EQUATIONS. 171 SECTION XXXVIII. AFFECTED QUADRATIC EQUATIONS. 302. — Ex. 1. What powers of the unknown quantity are contained in the equation a; 2 — 8a; +16 = 33? 2. In the equation x* + 8x = 20, what is the highest power of the unknown quantity ? Of what degree, then, is that equa- tion? 3. If x 2 + 10a; +25 is the square of x+5, what is the square root of a? + 10a; +25? 4. In a; 2 +10a; what term is wanting to make it the square of z+5? 5. In £ 2 +6a; what term is wanting to make it the square of a;+3? 6. In a; 2 +10a; what term is wanting to make it a perfect square ? 7. In x*+6x what term is wanting to make it a perfect square ? Definitions. 303. An Affected Quadratic- Equation is an equation which contains both the square and the first power of the unknown quantity. 18a; Thus, x 2 -\ =27 is an affected quadratic equation. o An affected quadratic equation is sometimes called a Com- plete Equation of the second degree. 304. Principles. — 1. Every affected quadratic equation can be reduced to the general form ax'+bx = c. 172 AFFECTED QUADRATIC EQUATIONS. \%x Thus, the equation x 2 -\ = 27, when cleared of fractions, becomes o 3a; 2 +18a; = 81. Denoting 3, the coefficient of a 2 , by a; 18, the coefficient of x, by 6; and 81, the value of 3a; z +18a;, by c, we obtain ax 2 +bx = c. Now, as a may stand for any coefficient of x 2 , b for any coefficient of x, and e for any value of the terms containing the unknown quantity, the equation ax 2 +bx = c is a general form for affected quadratic equations. 2. Affected quadratic equations depend for their solution on the form of a squared binomial. Thus, (x+a) 2 = x 2 +2ax+a 2 . Here a 2 is evidently the square of half of 2a, the coefficient of x. If, then, we have the expression x 2 + lax, which is not a perfect square, we can make it a perfect square by adding to it the square of half the coefficient of x. Hence, if we have the equa- tion x 2 +2ax = b, we add a 2 to each member, thus making the first mem- ber a perfect square, and preserving the equality of the two members, and have the equation x 2 + 2ax + a 2 = b + a 2 . This process of adding the square of half the coefficient of x to both members of the equation is called completing the square. 305. — Ex. 1. What is the value of x in the equation a?+8x = 9? Solution. If the square of half the x 2 +8x = 9 (1) coefficient of x, or 4 2 , be added to the first x 2 +8x+4 2 =S5 (2) member, it will be in the form of a squared x+Jf = ± 5 (3) binomial. Adding, then, 4 2 , or 16, to x= ±5 — 4 (4) the first member to complete the square, x = + 1 (5) and 16 to the second member to preserve x= — 9 (6) the equality, we obtain (2). Extracting the square root of both members, we obtain (3). Transposing, we ob- tain (4) ; whence (5), or x= +1, and (6), or x= — 9. Two values, therefore, of the unknown quantity are obtained, and it will be found, on substitution, that the value of x 2 +8x is 9, whether 1 or — 9 be put in the place of x. 2. What is the value of * in the equation » 2 — 8x = — 12 ? Solution. Completing the square x 2 — 8x =—12 (1) by adding 4 2 , the square of half the co- x 2 — 8x+4 2 = 4 (2) efficient of a;, to each member, we obtain x — 4 =±2 (3) (2). Extracting the square root of both x = ±2+4 (4) members, we obtain (3). Transposing, x = 6 (5) we obtain (4); whence (5), or a; = 6, and x=2 (6) (6), or x = 2. X 2 - Id-' til 4 (2) 7 / 7 V 81 x'- --X + 1 4 \ sr 64 (3) X- 7 _ 8~ 8 (4) x= ± 9 - + 8 7 8 (5) x= 2 (6) x= 1 (7) AFFECTED QUADRATIC EQUATIONS. 173 3. What is the value of a; in the equation - 4a; 2 +7a; = - 2 ? -J l x i +7x=-2 (1) Solution. Dividing both mem- bers by — 4, we obtain (2). Complet- ing the square, we obtain (3). Ex- tracting the square root, we obtain (4). Transposing, we obtain (5), whence (6), or x=2, and (7), or 1 4. What is the value of a; in the equation a?+Qx = — 8 ? Solution. Completing the square, we obtain (2). Extracting the square root, we obtain (3). Transposing, we obtain (4). Whence (5), or x= -2, and (6), or x= —4. From these solutions it appears that the two roots of an affected quadratic equation may have different signs or may both have the same sign. 5. What is the value of a; in the equation 5a; 2 — 20a; = — 15 ? Ans. x = S, or 1. 6. What is the value of a; in the equation 11a; 2 — 88a; = 99 ? Ans. x = 9, or — 1. 306. Rule for Solving Affected Quadratic Equations.— Re- duce the given equation, if necessary, to the general form, ax 2 +bx=c. Divide each member of the equation by the co- efficient of x 1 , and complete the square by adding to each member the square of one-half the coefficient of x in the resulting equation. Extract the square root of both members, and solve the resulting equation. 15* x*+6x =-8 (1) x 2 +6x+3*= 1 (2) x+ 3 = ±l (3) x = ±l-3 (4) x = -2 (5) x= —4 (6) 174 AFFECTED QUADRATIC EQUATIONS. In the general form, bx and c may be either positive or negative, in- tegral or fractional (Art. 304). The square root of a negative quantity is impossible ; therefore, when the sign of the term containing x % is negative, change the signs of all the terms in the equation, which is equivalent to dividing by — 1. PROBLEMS. 1. Given %*+6x = 40, to find x. Ans. x = 4,or- 10. 2. Given 3a; 2 - 12a; = 96, to find x. Ans. x = 8,or -4. 3. Given x % — Zx = — 2, to find x. Am. x = l,or 2. 4. Given a; 2 +6 = 5x, to find x. 5. Given 7a; 2 + 28a; - 224 = 0, to find x. Ans. x=-8,or 4. 6. Given 3a; - 54 = - a; 2 , to find x. 5 11 7. Given x = 2 + — .to find x. Ans. x = 2-,or --. Ax 2 2 9 x 8. Given - - - = 2, to find x. x 3 9. Given 4a; +22 = — —, to find x. Ans. x=-6,or S-. x-3 4 10. Given x+ = 5, to find x. Ans. x = 4,or 4- x-3 Here, the two roots of the equation are alike, both in signs and nu- merical values. Hence, the equation is said to have equal roots. 11. Given a; 2 +10a; = 24, to find x. Ans. x = 2,or -12. 10 „. 2a;+ll _ x-5 , „ . 12. Given = 5 , to find x. x 3 13. Given a;(a;+3) = 180, to find x. Ans. x = 12, or - 15. AFFECTED QUADRATIC EQUATIONS. 175 14. Given x 2 + 3x - 54 = 0, to find x. 15. Given 2x 2 + Ylx = — 16, to find x. Ans. x= -2, or - 4- 307. Another method of completing the square, which has the advantage of avoiding fractions, is often to be preferred to the preceding method. Let the equation be brought to the form ax' i +bx = e, in which a and b are whole numbers and prime to each other, and c is either a whole number or a fraction. Multiply each member of this equation by a, the coefficient of x 1 , and by 4, the smallest even square number, and it becomes jci¥ + JfCtbx = 4ac, in which the first term is a perfect square, and the second term is divis- ible by 2. Regard iaW+iabx as the first two terms of the square of a binomial. The first term of the binomial must be the square root of Aa'x 1 , or lax. The iabx must be twice the product of the first term of the binomial by the second ; hence, the quotient of Aabx divided by 4ax, which is b, must be the other term of the binomial. Adding the square of b, or 6 2 , to the first member of the above equa- tion to make it a perfect square, and to the other member to preserve the equality, we have the equation, 4a*x' +4abx+b*=4ac+b\ The first member of this equation is a complete square, and its several terms are whole numbers. Extracting the square root, we have 2ax+b= ± i/.£ae+& 2 , whence, by transposing 6 and dividing by 2a, we have x = -b±-[/Jiac+b' When b is an even number, — will be a whole number, and it will fbY be sufficient to multiply each member by a, and then add / - I to each member. Hence, the following rule: — 1 76 AFFECTED Q UADBA TIC EQ UA TIONS. 308. Rule for Solving Affected Quadratics.— Reduce the given equation to the form ax l +bx = c, in which the coefficients a and b are integral and prime to each other. If the coefficient of x be odd, multiply the equa- tion by four times the coefficient of x % , and add the square of the coefficient of x to both members; or, If the coefficient of x be even, multiply the equa- tion by the coefficient of x*, and add the square of one-half of the coefficient of x to both members. Extract the square root of both members, and solve, the resulting equation. rieoBzxurs. 1. Given x'+Gx+i = 22 - x, to find x. Solution. Given, x*+6x+4=22-x By transposition, x*+7x=18 Multiplying by 4, 4z 2 +28x=72 OompUting the square, 4x'+ 28x +7* =121 Extracting the square root, whence, 2x+7=±U ±11-7 x= 2 or, x=2 and, x=—9 Given x 2 — 5x = — 6, to find x. Ans. x = 3, or Given x* -13a; = 68, to find X. 4. Given a? + 25a; = - 100, to find x. Ans. x=-5,or - 20. 5. Given x % - ^ = 27, to find x. 6. Given 2x 2 - 7x + 3 = 0, to find x. Am. x = S,or 1. AFFECTED QUADRATIC EQUATIONS. 177 7. Given 3a; 2 + 10a; = 57, to find x. 8. Given 2a; 2 - 12a; + 3 = 83, to find x. 9. Given 4a; 2 - 12a;+9 = 0, to find x. 10. Given x'-6x = 2, to find x. Ans. x = 8± ^JT. 11. Given — = 1, to find x. x—1 x+1 a; 2 12. Given — - - 4x = 7, to find the approximate values of x to Ji three orders of decimals. Ans. x = 9.^77, or — 1.4.77. < j j 13. Given 2-|/a; 2 — 4a;+4a; = l, tofinda;. Ans. x = — -, or — . 14. Given x+5 = y/x 1-5 + 6, to find x. 15. Given 2ar , +a; = ll, to find the approximate values of x to three orders of decimals. Ans. x = 2.108, or —2.608. 16. Given 7a; 2 - 63a;+ 56 = 0, to find x. 17. Given 2 2 -8z+20 = 0, to finds. Ans. z=4±y^J. 18. Given y* — 2my = b 2 , to find y. Ans. y = m± -^/WTrr?. 309. Any equation in the Quadratic Form — that is, any equation which contains but two powers of the unknown quan- tity, the exponent of the one power being twice that of the other — can be solved by the application of the rules given for the solution of affected quadratics. • For, all such equations take, or can be made to take, the form ax M +bx n =c, in which " may have any value whatever, and the two terms of the first member are so related that we can supply the third term, in order to complete the square of a binomial. 178 AFFECTED QUADRATIC EQUATIONS. Ex. 1. Given a; 6 - la? = 8, to find x. Solution. x s -7x s =8 the square, re 6 — 7a 3 +(- I = — 7 9 Extracting the square root, x 3 = ± — a; s = -±- or, x 3 = 8, or —1 Extracting the cube root, x = 2 or, —1 2. Given X s - 10s 3 = 459, to find x. Am. x = S,or tf~^W. 3. Given x i - 9a; 2 + 20 = 0, to find x. Solution. Given, x i -9x 2 +20 = Transposing, x* — 9x* = —20 (9 \ 2 1 — I = — 9 1 Extracting the square root, x 2 = ± — whence, x % = — ± — 2 2 or, x 2 =B, or If Extracting the square root, x = ± y/7>, or ± 4 4. Givenl6x 4 -12a; !! +2=0,tofinda;. jlws.a;= ±-,°» - ^a/-- 5. Given 4a; 6 - 3240 = 12a; s , to find x. 6. Given 2a; 4 - 16a; 3 = 18, to find x. Ans. x =±8,or± \/~l. AFFECTED QUADRATIC EQUATIONS. 179 SECTION XXXIX. PROBLEMS PRODUCING AFFECTED QUAD* RATIO EQUATIONS. 310. — Ex. 1. What two numbers are such that their differ- ence is 12, and their product 64 ? Solution. X = one number ; x+lS=the other number. By conditions, x[x\12) = 61^ or, x 2 + lSx = 64 Completing the square, X 1 + 12x + 6 2 = &J+ 86 = 100 Extracting the square root, x+6=±10 whence, x= — 6±10 = 4, or —16 and, x+12 = 16,or—4 Hence, the two numbers are 4 and 16, or —16 and —4, as either pair of values satisfies the problem. 2. What two numbers are such that their sum is 15, and their product 54 ? Ans. 9 and 6. 3. What is that number from the square of which if we take 7 times the number, the remainder will be 44 ? Ans. 11, or — J.- 4. The ages of a man and his wife amount to 42 years, and the product of the numbers expressing their ages is 432. What is the age of each? Ans. The man, 21/. years; the wife, 18. 5. A wholesale shoemaker received an unexpected order for 990 pairs of shoes, to be finished by the end of the month. He found that if these were divided equally among his men, each would have allotted to him 12 pairs more than he could get finished by the given time at his ordinary rate of working. He therefore engaged 54 more men, and got the work executed by the time specified. How many men had he at first ? 180 AFFECTED QUADRATIC EQUATIONS. 6. A merchant sold a quantity of flour for $39, and gained as many per cent, as equalled the number of dollars in the price of the flour. What was the price of the flour? Solution. {the number of dollars in the price of the flour ; x = the rate of gain per cent. ; x x 2 x x = = the qain. 100 100 * x* By conditions, = 89— x * ' 100 whence, x 2 +100x=8900 Completing the square, x*+100x+ (50)* =8900 +2500=6400 Extracting the square root, x+50= =80 whence, x=80, or —180 The cost of the flour was $30. The other value of x, not satisfying the conditions of the question, is not admissible. 7. A person invested a certain sum of money for goods, which he sold again for $24, and thereby lost as many per cent, as equalled the number of dollars invested. How much did he invest? Ans. $40, or $60. 8. There is a rectangular field whose length exceeds its breadth by 20 rods, and its area is 6300 square rods. What are its length and breadth ? 9. A man planted a rectangular field with 8400 trees at equal distances, having 50 trees more in the longer rows than in the shorter ones. What was the number of trees in each of the longer rows ? Ans. 120. 10. A trader computes that, during the time he has been in business, he has made $6300 clear profit. His neighbor, however, who has been three years less time in business, has made the same sum, owing to his clearing $27 per annum more. How long is it since the first commenced business? 11. A farmer sold a number of tons of hay for $112, and observed that if he had sold one ton more for the same money, each ton would have brought him $2 less. Eequired the num- ber of tons sold and the price per ton. AFFECTED QUADRATIC EQUATIONS. 181 Solution. x = the number of tons sold ; IIS _ f the number of dollars in x I the prise per ton. Us Us m By conditions, * ' x+1 x Clearing of fractions, USx = USx +112— Sx 2 — Sx whence, x 1 +x=56 Completing the square, 4x*+Jx+l = SS4 + 1 = SS5 Extracting the square root, Sx+1 = ± IS whence, x=7,or—8 and, = 16, or —Ik x The number of tons sold was 7, and the price per ton was $16. The negative value of x is not admissible. 12. The sum of $144 was divided equally among a certain number of persons. If there had been two persons less, each would have received f 1 more. How many persons were there? Ans. 18. 13. A man bought a certain number of sheep for $80. If he had bought 4 more sheep for the same money, they would each have cost him $1 less. How many sheep did he buy ? 14. Among a certain number of poor persons 110 bushels of coals were equally divided. If each person had received 1 bushel more, he would have received as many bushels as there were persons. Eequired the number of persons. Ans. 11. 15. A cistern is supplied with water by two pipes ; by one of them it can be filled in 6 hours sooner than by the other, and by both together in 4 hours. Find the time in which each pipe will fill it. Solution. x — the number of hours in which one will fill it; x+6 = the numbe r of hours in which the other will fill it. By conditions, — I =— " ' x x+6 4 whence, x=6, or —% and, X+6 = 1S, or —2 The first will fill it in 6 hours, and the second in 12 hours. The neg- ative value of x is not admissible. 16 182 AFFECTED QUADRATIC EQUATIONS. 16. A and B can perform a certain piece of work in 14f days, and A alone can perform it in 12 days less than B alone. Find the time in which A alone can perform it. 17. Two travellers, wishing to meet, set out from two towns, A and B, which are 120 miles distant from each other ; the first goes 6 miles a day, and the other 1 mile a day more than the number of days in which they meet. In how many days will they meet ? Ans. 8. 18. The length of a rectangle exceeds its breadth by 12, and the sum of the squares of the length and breadth is 20880. What are the sides of the rectangle and the area ? Ans. Breadth, 96; length, 108 ; area, 10368. 19. In a concert-room 800 persons are seated on benches of equal length. If there were 20 fewer benches, it would be necessary that two more persons should sit on each bench. Find the number of benches. Solution. x = the number of benches ; SOO f the number of persons on one x I bench in first case ; 800 f the number of persons on one X -20 ~ i bench in or -80 second case. 800 x-%0 800 X X =3 = 100, By conditions, whence, Rejecting the negative value, we have 100 as the numher of benches. 20. Divide 10 into two parts, such that the product of the parts shall be 24. 21. Two detachments of infantry are ordered to a station which is 39 miles distant. They begin their march at the same time ; but one party, by travelling one-fourth of a mile an hour more than the other, arrives one hour sooner. Required the rates of marching per hour. Ans. 3 and 8- miles. SIMULTANEOUS QUADRATIC EQUATIONS. 183 22. A had 40 yards of cloth, and B 90 yards, which they sold together for $42. Now, A sold for $1, a third of a yard more than B sold for the same money. How many yards did each sell for $1 ? 23. The difference of two numbers is 2, and the difference of their cubes is 152. What are the numbers? 24. Two boys, John and William, start at the same time to walk a distance of 75 miles ; but John walks 1£ miles per hour faster than William, and finishes his journey 8^- hours before him. How many miles per hour did each walk ? Ans. John, J/.-; William, S. SECTION XL. SIMULTANEOUS QUADRATIC EQUATIONS. 311. A Homogeneous Equation is one in which the sum of the exponents of the unknown quantities in each term contain- ing such quantities is the same. Thua, x 1 — 2/ 2 = 12 and x 2 — xy-Yy i = VL are each homogeneous. 312. A Symmetrical Equation is one in which the unknown quantities are similarly involved. Thua, aj 2 +3/ 2 = 25 and x % y — xy i = § are each symmetrical. 313. Simultaneous Quadratic Equations (Art. 211) contain- ing two unknown quantities cannot all be solved by the rules for quadratics. The cases which can be treated in an elementary work are necessarily only such as can be solved by comparatively simple processes. 4x + 2y = 18 (1) Sxy= SO (2) 9-y x = - 2 (3) 4Sy- Sy 2 so 2 (4) J l 5y-5y' i = 100 (5) Sy*-45y=-100 (6) yi- 9y=-20 (7) -36y + 81 = 1 (8) 2y-9 = ±i (9) 9± 1 y ~ 2 (10) ofy, y=S,or4, (ID x=2, or 2^ (12) 184 SIMULTANEOUS QUADRATIC EQUATIONS. 314. — Ex. 1. Given 4x+2y = 18 and 5xy = 50, to find x and y. Solution. From (1), Substituting in (2), Clearing of fractions, or, Dividing by 5, Completing the square, Extracting the square root, .whence, Substituting in (2) these values of y, we have, Here one of the equations is simple and the other quadratic. 2. Given 2x — 2y = ll and xy = 20, to find x and y. Ans. x = 8, or— 8^ > y = 2j, or —8. 3. Given 2x+3y = 8 and x 2 +xy+y 2 = 7, to find x and y. Solution. 2x+Sy =8 (1) x 2 + xy+ y* = 7 (2) Assuming lhaty = vx, and substituting this value of y in (1) and(2), we have, (2+8v)x =8 (3) (i + « + « J )a s = 7 (4) Dividing (4) 6y to *""* * an< * 2A 5. Given ] „ . , x ~ , v ~ . \ , to find x and y. I 2z*+xy-5y*=4 j » f a;= » ,„. _ 5 -Arcs- ■! ' „ i# = 2, or-2. "■«"<°{;-;: 2 :;:"}.«»«'»' d »' ?, or -46, or IB. 7. Given } _ ^ 2 _ 1 , to find x and y. 8. Given 4 „ J „ ^ .? , , to find a; and y. I 3xy+2x+y =485 J 9. Given {* ^jj^}, to find* and y j/= ± 3, or=f=5i/5. 316. Some Quadratics and Higher Equations, which are sym- metrical, may be readily solved by artifices, especially such as are suggested by the relations existing between the sum, differ- ence and product of the unknown quantities. SIMULTANEOUS QUADRATIC EQUATIONS. 187 1. Given x]+y 1 = 25 and ay = 12, to find £ and y. Solution. X i + y2= 26 (1) xy = 12 (2) Multiplying (2) by 2, 2xy = 24 (3) Adding (1) and (3), x 2 + 2xy+y* = 49 (4) or, {x+ y y= 49 (5) whence, x+y = ± 7 (6) Subtracting (3) from (1), x 2 -2xy+y 2 = 1 or, (x~y) 2 = 1 (7) M/Aemce, x—y = ± i (8) .From (6) amd (8), £ = ± 3, or ±4 (9) and, y = ± 4, or ±3 (10) Here, the equation, which can be solved by substitution, is more readily solved by an artifice which avoids the process of completing the square in the solution. 2. Given \ . , ._ [ , to find * and y. (y=±^or ±7. 3. Given x 2 —y 2 = 24 and a+j/ = 6, to find x and y. Solution. 4. Given x 2 -y 2 =M (1) x +y = 6 (2) Dinding (1) % (2), X -v = 4 (3) Adding (3) to (2), 2x =10 (4) whence, x = 5 (5) Subtracting {3) from (2), 2y = 2 whence, y- 1 3 n j^ 2 : 2 n, to find x and y. 188 SIMULTANEOUS QUADRATIC EQUATIONS. 5. Given x+y = 5 and rf+y 3 = 65, to find x and y. Solution. a; +y = 5 (1) x 3 +y 3 = 65 (2) Dividing (2) ii/ (1), x 3 +y 3 65 x + y 5 (3) or, x 2 - xy + y 2 =lS (4) Squaring (1), x 2 +2xy + y 2 = 25 (5) Subtracting (4) /ran (5), Sxy =12 (6) Dividing by 3, xy = 4 (7) Multiplying by 4, 4xy =16 (8) Subtracting (8) /rom (5), x 2 -2xy+y 2 = 9 (9) Extracting the square root, x-y = ±# (10) From (1) and (10), v x = 1, or Jf (11) and, y =4, or 1 (12) Here, after (4) was found, the solution could have been made also by combining (1) and (4), and applying rule 1. 6. Given or or \ , , _ [ , to find x and y. Arts. \ ' (x 3 ~y 3 = 8)' " \y=0, SECTION XLI. PROBLEMS PRODUCING SIMULTANEOUS QUADRATIC EQUATIONS. 317. — Ex. 1. Into what two parts can the number 40 be separated, so that the sum of their squares shall be 818 ? Ans. 17 and 23. 2. The difference of two numbers is f of the greater, and the sum of their squares is 356. What are the numbers ? SIMULTANEOUS QUADRATIC EQUATIONS. 189 3. The product of two numbers is 192, and the sum of their squares is 640. What are the numbers ? 4. The sum of the squares of two numbers is 170, and the difference of their squares is 72. What are the numbers ? 5. A farmer having sold 7 lambs and 12 sheep for $50, found that the number of lambs he had sold for $10 was 3 more than the number of sheep he had sold for $6. What was the price of each? Solution. x = the number of dollars each lamb sold for, y = the number of dollars each sheep soldfor ; ■ — = the number of lambs sold for $10. — = the number of sheep sold for $6. y By conditions, — = — \rS (1) x y and, 7x+12y = 50 (2) From (2), z = y-* (3) Substituting this value of x in (I), 70 = (50 - 12y)l -+s\ whence, 9y l -2y = 75 2 /i\ 2 Completing the square, y* — y + 1 - 1 = whence, y = 3, or 1Y 676 81 _25_ 9 250 and, x=2, or ' 21 From the nature of the question the negative result is not admissible. The price of the lambs is $2 each, and of the sheep, $3. 190 SIMULTANEOUS QUADRATIC EQUATIONS. 6. What are the two numbers whose sum multiplied by the greater is 60, and whose difference multiplied by the less is 8 ? 7. There is a certain number expressed by two digits. The sum of the squares of the digits is equal to the number increased by the product of its digits, and if 36 be added to the number, the digits will be reversed. What is the number ? . Ans. 87 or 48. 8. The difference of two numbers is 3, and the difference of their cubes is 279. What are the numbers ? 9. The sum of two numbers is 20, and the sum of their cubes is 2240. What are the numbers ? Ans. 12 and 8. 10. A and B invest in a certain partnership, having a joint capital of $100. A leaves his money in the partnership for 3 months, and B leaves his for 2 months, and each at last realizes $99 of capital and profit, the monthly rate of profit being uniform. What sum did each originally contribute ? Solution. x — the number of dollars in A's capital, 100— x= the number of dollars in B's capital; y = the monthly rate of gain. By conditions, x+3xy = 100-x+Sy(100-x) (1) and, x+8xy = 99 (2) it ,■.^ 100- Sx ,„. Fromd), y^^Z^o (3) Substituting this value in (2), x 2 + 395x = 19800 whence, x=Jfi and, 100-x = 55 Hence, A's capital was $b&, and B's was $55. 11. A grazier bought as many sheep as cost him $300, and after reserving 15 out of the number, sold the remainder for $270, and gained $.50 a head. How many sheep did he buy ? RATIO AND PROPORTION. 191 12. Two traders jointly invest $500 in business. One of them let his money remain 5 months, the other only 2 months, and each received $450 capital and profit. How much did each invest? Ans. One, $200; the other, $300. Test Questions. 318. — 1. What is a, Quadratic Equation? A pure equation? A pure quadratic equation ? "What are the given principles of quadratic equations ? What is the rule for solving pure quadratic equations ? 2. What is an Affected Quadratic Equation? What are the given principles of affected quadratic equations ? What is the rule for solving an affected quadratic equation ? Give the method which avoids frac- tions in the process of solution. •3. What is a Homogeneous Equation? A symmetrical equation? What cases of simultaneous quadratics can be treated in an elementary work? What are the rules for solving some simultaneous quadratic equations ? SECTION XLII. RATIO AMD PROPORTION. 319. Eatio is the relation which one of two similar quantities bears to the other, with respect to magnitude. It is ascertained by the division of the first of the given quantities by the second. Thus, the ratio of a to 6 is — . b 320. The Sign of ratio is the colon ( : ). Thus, 5 : 6 denotes the ratio of 5 to 6, or -. ' ' 6 192 RATIO AND PROPORTION. 321. The Terms of a ratio are the two quantities whose mag- nitudes are compared. The first term is called the Antecedent, and the second term is called the Consequent. 322. A Simple Ratio is a ratio each term of which is a single quantity. A Compound Ratio is a ratio formed by multiplying together the corresponding terms of two or more simple ratios. Thus, a : b is a simple ratio; and (a : b)(c : d), or — x — , is a com- b a pound ratio. A Ratio of Equality exists when the antecedent and consequent are equal. A ratio of greater inequality exists when the antecedent is greater than the consequent ; and of less inequality, when the antecedent is less than the consequent. The ratio is said to be direct when the antecedent is divided by the consequent ; and inverse or reciprocal when the conse- quent is divided by the antecedent. 324. Principles. — 1. Both terms of a ratio may be multiplied or divided by the same quantity without affecting its value. For, a : & = — ; multiplying both terms of — by n, we have — , and ob bn dividing both terms of — by n, we have — , the value in each case bn ' b being unchanged. 2. A compound ratio is the ratio of the product of its antece- dents to the product of its consequents. For, (6 : 2) (5 : 3), or |i| is equal to |±|. RATIO AND PROPORTION. 193 PROBLEMS. 1. What is the ratio of 15 dimes to 3 dollars? 2. Which is the greater ratio, 3 : 4, or 2 : 3 ? 3. Find the ratio compounded of 4 : 15 and 5 : 6. 4. Express as a fraction the ratio of 5 to 17. 5. Which is the greater ratio, 5 : 6, or 123 : 148 ? Ans. 5 : 6. 6. Which is the greater ratio, 5 : 6, or - : - ? Ans. - : -. 5 6 5 6 7. Two equal glasses are both full of mixtures of wine and water. In the one the ratio of the wine to the water is 1 : 7 ; in the other, the ratio is 1 : 9. When the two are poured into the same vessel, is the ratio of the wine to the water greater or less than 1:8? Ans. Greater. PROPORTION. 325. Proportion is an equality of ratios. Thus, a : b = c : d expresses a proportion. 326. The Sign of proportion is a double colon ( : : ), which, instead of the sign of equality, may be placed between two equal ratios. Thus, a : b :: c : d denotes a proportion. Each ratio is called a couplet, and each term a proportional. 327. The Antecedents of a proportion are the antecedents of its ratios. The Extremes of a proportion are its first and fourth terms, and the means are its second and third terms. 328. A Mean Proportional of two quantities is a quantity that serves as both the means of a proportion in which the two quantities are the extremes. Thus, 6 is a mean proportional in a : b : : b : c. 17 194 RATIO AND PROPORTION. 329. A Continued Proportion is a proportion in which each antecedent, except the first, is the same as the preceding con- sequent. Thus, a : 6 : : 6 : c : : c : d is a continued proportion. 330. Quantities are said to be in proportion by alternation when antecedent is compared with antecedent and consequent with consequent ; in proportion by inversion when the antece- dents are made consequents and the consequents antecedents ; in proportion by composition when the sum of antecedent and consequent is compared with either antecedent or consequent ; and in proportion by division when the difference of antecedent and consequent is compared with antecedent or consequent. Theorem I. 331. In any proportion the product of the extremes is equal to the product of the means. La a:b : : c : d; then, a b ~ c " d ! Clearing of fractions, ad= = 6c. Theorem II. 332. If the product of two quantities be equal to the product of two other quantities, either two may be made the extremes of a proportion, and the other two the means. Let ad = bc; Dividing by bd, — = — / o d or, a:b :: c: d. RATIO AND PROPORTION. 195 Theorem III. 333. In a proportion, either extreme is equal to the product of the means divided by the other extreme, and either mean is equal to the product of the extremes divided by the other mean. Let a : b : : c : d. By Theorem I, ad— be; . ..... be , be . ad , ad whence, by aivmon, a= — ,■ a= — / o= — , ana c= — . d a o b Theorem IV. 334. The mean proportional between two quantities is equal to the square root of their product. Let a :b: :b:c. By Theorem I, 6 2 = = ac; Extracting square root, 6 = = l/ac. Theorem V. 335. If four quantities be in proportion, they will be in pro- portion by alternation. Let a: b : : c : d. By Theorem I, ad^ = 6c. Dividing by do, a c 6 = d t or, a: c : :b: d. Theorem VI. 336. If four quantities be in proportion, they will be in pro- portion by inversion. Let a : b : : e : d. By Theorem I, bc = ad; whence ( Theorem II), b : a : : d : c. 196 RATIO AND PROPORTION. Theorem VII. 337. If four quantities be in proportion, the first together with the second will be to the second as the third together with the fourth is to the fourth. Let a: b :: c : d; 41 a c then, — = — . b a 1 to each member, — + 1=— + 1; b a a + b c + d _ b ~ d '' a+b : b :: c+d : d. Theorem, VIII. 338. If four quantities be in proportion, the excess of the first above the second will be to the second as the excess of the third above the fourth is to the fourth. Lei a : b : : e : d; then, — = —. o a a c Subtracting 1 from each member, T~l—^—l: a — b c — d S dT' whence, a—b :b :: c—d : C?., Theorem IX. 339. If the corresponding terms of two or more proportions be multiplied together, the products will be proportionals. RATIO AND PROPORTION. 197 Let and Then, and Multiplying, a:b : : C : d, e:f: ■9- h. a b~ c V e f" 9 h' ae Off dh t ae-.bf: ■■eg :dh. Theorem X. 340. If four quantities be in proportion, like powers or roots of those quantities will be 'proportionals. Let a:b::c:d; then, Raising to the nth power, Extracting the nth root, whence, and, PROBLEMS. 1. In 4 : 7 : : 8 : x, find x. Ans. x = 11/.. 2. In 5 : x : : x : 45, find x. 3. In x+4 : x+2 : : x+8 : x+5, to find x. Ans. x = 4. 4. Find a fourth proportional to ab, cd and ax. 5. Find the number to which, if 2 and 5 be successively added, the resulting sums will be the ratio of 5 : 11. Ans. \. 17* a b~ c ~'d' a n b n ~ e" d a ' i a" * b n i -£T. d n «" : : b n : : c» : d n ; a n :6" < ::c" : d 198 RATIO AND PROPORTION. 6. Find a fourth proportional to \, \ and \. 7. Find a mean proportional to 2 and 8. 8. Divide $1000 between two persons so that their shares shall be in the ratio of 7 to 9. Solution. x = number of dollars in first person's share; 1000 —x = number of dollars in second person's share. By conditions, X : 1000— x ::7:9 By Theorem I, 9x = 7000-7x whence, x= 437. SO and, 1000- x = 562.50 Hence, the first person's share is $437.50, and the second person's share, $562.50. 9. Divide the number 56 into two parts, such that one shall be to the other as 3 to 4. Ans. %b; 32. 10. A and B are at present of the same age. If A's age be increased by 36 years, and B's by 52 years, their ages will be as 3 to 4. What is the present age of each ? 11. A, B and C make a joint stock. A puts in $60 less than B, and $68 more than C ; and the sum of the shares of A and B is to the sum of the shares of B and C as 5 to 4. What did each put in? Ans. A,$lJfi; B,$200; G,$72. 12. Find two numbers whose sum is to their difference as 3 to 2, and whose difference is to their product as 1 to 5. 13. A person in a railway-car observes that another train running on a parallel line in the opposite direction occupies 2 seconds in passing his train. But if the two trains had been proceeding in the same direction, one would have taken 30 ARITHMETICAL PROGRESSION. 199 seconds to pass the other. Compare the speed of the two trains. Solution. x : y = the ratio of the rates. Sy conditions, x—y : x+y :: % : 30 whence, x : y :: 8 : 7 That is, the speed of the two trains is as 8 to 7. 14. There is a rectangular field which contains 360 square rods, and whose length is to its breadth as 8 to 5. What are the length and breadth ? Ans. Length, &£ rods ; breadth, 15 rods. 15. In a court there are two square grass-plots, a side of one of which is 10 yards longer than a side of the other, and their areas are as 25 to 9. What are the lengths of their sides ? Ans. 15 yards ; 25 yards. SECTION XLIII. ARITHMETICAL PROGRESSION. 341. A Progression is a succession or series of quantities increasing or decreasing according to some fixed law. 342. The terms of a progression, or series, are the quantities of which the series is formed. 343. An Arithmetical Progression is a series formed by add- ing a constant quantity. The constant quantity added is called the Common Difference, 200 ARITHMETICAL PROGRESSION. and the progression is ascending when the common difference is positive, and descending: when it is negative. Thus, a, a+d, a+2d, a+Sd, . . . is an ascending arithmetical progression in which the common differ- ence is + d, and a, a — d, a — Sd, a — 3d, . . . is a descending arithmetical progression in which the common differ- ence is — d. 344. In an Arithmetical Progression having a limited num- ber of terms there are fire elements to be considered : 1. The first term, a; 3. The number of terms, n; 2. The last term, If 4. The common, difference, 3 rtiTl ipTlfip J\ hence, pr*(r-l) fSOOO xl.OT'x (1.07-1) ir—i 1.07* -l or, x = $731.67. 14. A farmer owes on his farm $1000 ; what equal annual payments will in 10 years discharge the debt, at 6 per cent, compound interest ? Ans. $185.87. 15. Suppose the debt of a certain city to be $18,000,000 ; what equal annual payments will in 25 years discharge it, at 6 per cent, compound interest ? 16. If a man travel 50 miles the first day, and 45 miles the second, and so continue to travel 5 miles less each day, how far will he have gone on his journey at the end of the eighth day? GENERAL REVIEW. 213 Test Questions. 362. — 1. What is Ratio f The sign of ratio ? What are the terms of a ratio ? What is a simple ratio ? A compound ratio ? A ratio of equality ? Of greater inequality ? Of less inequality ? What are the given principles of ratio ? 2. What is Proportion? The sign of proportion? What are the antecedents of a proportion ? The consequents ? The extremes ? The means ? What is a mean proportional of two quantities ? What is a continued proportion ? What are the given theorems relating to pro- portion ? 3. What is a Series or Progression ? An arithmetical progression ? The common difference? What are the given theorems relating to arithmetical progression ? 4. What is a Geometrical Progression ? The rate or ratio of a geo- metrical progression? An infinite series? The sum of an infinite series ? What are the given theorems relating to geometrical progres- sion? SECTION XLVI. GENERAL REVIEW. 363. — Ex. 1. Express in algebraic form the sum of x, y and s, divided by the product of x and the square of y. 2. Combine in one sum 3afy s — 10y*, — x'y'+by*, Sx^ — Qy* and x 1 y*+2y i . 3. Reduce to its simplest form the expression 5a — 46 + 3e+ (-3a+26-c). Ans. 2a-2b+2c. 4. Multiply 2a; — y by 2x — y. 5. Prove that the continued product of x — 3, x + 3, x — 4 and a: + 4isa; 4 -25r 2 +144. 214 GENERAL REVIEW. 6. Divide a'-b* by a+b. 7. Divide 5a>y-40 2 a;y + 25^ by -5xy. 8. Divide as 4 — y l by a; — y. .4ns. a?+x > y+xy 2 +i/ s . 9. Expand ( - 5aJVy)( - batftfy'). 10. Find the square of 6a a +2a 2 6. 11. Factor x*— Ixy+y*. Arts. (x-y)(x—y). 12. What is the greatest common divisor of 8a 2 xy, — 12bxy 2 and 20cafy? 13. Show that the greatest common divisor of 10a* and 15a; 2 is 5x. 14. Show that the least common multiple of a?bc and 2a6 2 d is 2a?¥cd. 15. State in algebraic form that the product of a, b and c, divided by the difference of c and d, is equal to the cube of a divided by the sum of b and the square of c. 16. What is the least common multiple of a, c+b and e — bt 17. Express algebraically a quantity with a numerical co- efficient and a literal exponent. 18. What is the value of 4a 8 +(6 2 -c)(a 2 -c 2 ), a being equal to 3, b to 4 and c to 2? 19. What is the greatest common factor of a 2 +2a6 + 6 2 and a+b? Ans - a+b - x 1 - 1 20. Eeduce to its lowest terms. xy + y GENERAL REVIEW. 215 21. What is the least common denominator of — , — , ? 26 a c Ans. 2abc. 22. What is the sum of and ? a+x a — x 23. What single fraction expresses the sum of -, - and — ? b d y 24. Divide by — . Ans. . 5x 5 x 25. What is the product of by — - ? a zax — 2a; 2 26. Multiply l by . Am. . x-y x+y x-y At . 2 27. Divide ~p~ by . x s — 1 x + y 28. What is the value of a; in 2a; = —+4? 4 4 3a; — 1 11— 4a; 12 29. Show that the value of a; in + — - — = — -, is 2. 7 3 7 30. Divide 21 into two parts, so that ten times one of them shall exceed nine times the other by 1. 31. A says to B, "You are 4 years more than twice as old as I am ; 20 years ago you were 3 times as old as I was." What is now the age of each ? 4a; -3 32. Solve the equation, 7a; — +5£ = 8a; + 6£. , a d -~ a 33. Solve the equation, mx+a = nx + d. Ans. x = . 34. Solve the equation (m+n)(m-x) =m(n-x). 216 GENERAL REVIEW. 35. Solve the equation — - — = x + 36. 36. Solve the equation az+b 2 = a? + bz. 37. What number is that to which if its third and its fourth he added, the sum will exceed its sixth by 17 ? 38. A man being asked the price of his watch, replied : " If you multiply the price by 4, and to the product add $70, and from this sum subtract $50, the remainder will be equal to $220." How much did he give for his watch ? 39. Two persons received equal sums of money ; the first paid away $25, and the second $60. It then appeared that the former had twice as much as the latter. What sum did each receive ? Ans. 40. A certain sum of money at interest amounts to $297.80 in .8 months, and in 15 months to $306. What is the sum, and what is the rate of interest ? 41. A person bought two casks of vinegar, one of which held just 3 times as much as the other. From each of them he drew 4 gallons, and - then found that there were 4 times as many gallons remaining in the larger as in the other. How many gallons were there in each ? Ans. 12 in one ; 36 in the other. 42. Find the value of x and y in the equations a?+y = 36 and x -y = 12. 43. Given 4»+5t/ = 32 and x+Sy =■ 15, to find x and y. 44. Given = 8 and — y = 11, to find x and y. 6 2 45. Given x = }(y-2) + 5 and 4y-\(x + 10) = 3, to find x and y. Ans. x = 5 ; y = 2. GENERAL REVIEW. 217 46. If a privateer discovers a ship, 18 miles off, sailing away at the rate of 8 miles an hour, and pursues her at the rate of 10 miles an hour, how long will the chase last ? 47. Coffee and sugar were formerly cheaper than now, coffee by 2 cents and sugar by 4 cents a pound ; and the price of the former was double that of the latter. The prices now are as 3 to 2 ; at what rate are they sold ? 48. A steamboat whose speed in still water is 10 miles an hour, descends a river whose velocity is 4 miles an hour, and returns in 10 hours. How far did she proceed? Ans. Jfi miles. 49. A farm, consisting of pasture and arable land, is hired at an annual rent of $390, the pasture being valued at $1.50 per acre, and the arable land at $3. Now the number of acres of arable land is to half the excess of the arable above the pas- ture as 5 to 1. Required the quantity of each. 50. A farmer, disposing of his stock, sells to one person 9 horses and 7 cows for $1200 ; and to another, 6 horses and 13 cows for the same sum. What was the price of each ? Ans. Cows, $48; horses, $96. 51. What fraction is that which, by the subtraction of 4 from both of its terms, becomes \, and by the addition of 9 to both, becomes f ? 52. A rectangular lot of land was sold at the rate of $2.25 per foot, for $5400. Had its length been 10 feet less, its value would have been $4725. What were its length and breadth? Ans. Length, 80 fee} ; breadth, SO feet. 53. Three persons, A, B and C, have certain sums of money, such that A and B together have $120; A and C together have $140; and B and C together have $150. What sum has each? 54. Given 5« + 3y = 65, 2y-z = 11 and 3a;+4z = 57, to find x, y and z. Ans. x = 7; i/ = 10; z = 9. 19 218 GENERAL REVIEW. 55. Given 3+21/+ 3a = 17, 2x+3y+z = 12 and 3x+y + 2z = 13, to find x, y and z. 56. Three persons, A, B and C, were talking of their ages. A said the sum of their ag& was 90 years. B replied, that if his age were taken from the sum of the other two, the remainder would be 30 years. C said, if his age were taken from the sum of the other two, the remainder would be \ of his age. Required their ages. 57. What is the mth power of xfz" ? Ans. x^z"" 1 . 58. What is the cube of a?+2a - 4 ? Ans. a e +6a*-4.0a ! '+96a-64. 59. What is the square root of x^tfz* ? 60. What is the cube root of -8aW? Ans. -SabV. 61. Wh^t is the square root of a 2 x i -2ahxy i +Vy i ' > - 62. What is the square root of x l - 2x* + 3 x* - 2x + 1 ? 63. Reduce y'SOaPx* to its simplest form. Ans. ^ax 2 ^/5a. 64. What is the sum of y/iax 2 and Sxi/9a ? 65. Express - 7a 2 6 in the form of the square root. Ans. l /49a l b\ 66. What is the sjim of Bx^/^aaad -7zj/a? 67. What factor will rationalize i/a+i/'bl Ans. -|/o--|/6 5x* 68. Solve the equation 3ar + 7 = +35. Ans. a;= ±4- x 2 69. Solve the equation 3a; 2 - 29 = — + 510. GENERAL REVIEW. 219 70. Find two numbers whose sum is to the less as 10 to 3, and the difference of whose squares is 640. 71. A number of boys set out to get some apples, each car- rying as many bags as there were boys, and each bag being capable of holding 6 times as many apples as there were bags. After filling their bags, they found the whole number of apples was 1536. How many boys were there ? 72. If A's money were increased by half of B's, it would amount to $54 ; and if B's remaining sum were trebled, it would exceed three times the difference of their original sums by $6. How many dollars had each at first ? Arts. A, $10; B, 73. The fore wheel of a carriage makes 6 revolutions more than the hind wheel in going 120 yards, and the circumference of one is a yard less than that of the other. Find the circum- ference of each. 74. A person changed a. half-eagle for 25 pieces of foreign coins, some of which were reckoned as 30 to the half-eagle, and the others 15. How many did he get of each ? Arts. 20 of one kind ; 5 of the other. 75. A market-woman having bought some eggs at 2 for 1 cent, and as many more at 3 for 1 cent, sold them at 5 for 2 cents and lost 4 cents by the transaction. How many cents did she pay out ? Ans. 100. 76. A farmer has two cubical stacks of hay : the side of one is 3 yards longer than the side of the other, and the difference of their contents is 117 solid yards. Required the side of each. Ans. 5 yards ; 2 yards. 77. Divide 100 into two parts, so that one shall be a mul- tiple of 7, and the other a multiple of 11. 78. Two-thirds of a certain number of poor persons received 220 GENERAL REVIEW. $3 each, and the rest $5 each, the whole sum spent being $110. How many poor persons were there ? Arts. SO. 79. Two girls together carried 25 eggs to market. They sold them at different prices ; but each received the same amount. The one would have sold them all for 12 cents, and the other for 13 cents. How many did each sell ? 80. A and B together carried 100 eggs to market, and each received the same sum. If A had carried as many as B, he would have received 18 cents more for them, and if B had car- ried as many as A, he would have received only 8 cents. How many had each ? 81. Solve the equation x* — Ylx +20 = 0. Ans. x = 10, and 2. 82. Solve the equation 2x = 4+-. x 83. Solve the equation %x — %x = 9. 24 84. Solve the equation x + = Sx — 4. x-1 85. The expenses of a party are $10, and if each pays 30 cents more than there are persons, the bill will be settled. How many persons are there in the party ? 86. Find two numbers whose product is 100, and the differ- ence of whose square roots is -3. Ans. 25 ; 4- 87. I bought a number of* pieces of cloth for $33.75, which were sold again at $2.40 a piece, and as much was gained by the transaction as one piece cost. Find the number of pieces. 88. A person bought cloth for $60. If he had bought 1 yard less for the same money, each yard would have cost $.25 more. How many yards did he buy ? 89. A grazier bought a certain number of oxen for $240, and after losing 3 he sold the remainder for $8 a head more than they cost him, thus gaining $59 by his bargain. What number did he buy ? Ans. 16. GENERAL REVIEW. 221 90. A and B begin trade, A -with 3 times as much stock as B. They each gain $50, and then 3 times A's stock is exactly equal to 7 times B's. What were their original stocks ? 91. A man travelled 105 miles, and then found that if he had not travelled so fast by 2 miles an hour, he would have been 6 hours longer in performing the journey. Find his rate of travelling. Ans. 7 miles an hour. 92. A party at a tavern had a bill of $20 to pay, but two of the party having gone off, those who remained had each $.50 additional to pay. How many were there at first ? 93. Insert two arithmetical means between 1 and 3. 94. A and B could do a piece of work in 4 days. A works alone for 2 days, and then they finish it in 2-| days more. In what time could they have done it separately ? Ans. A, in 5- days ; B, in 16 days. 95. A and B engaged in trade, A with $275 and B with $300. A lost half as much again as B, and B had then re- maining half as much again as A. How much did each lose ? Ans. A, $135 ; B, 96. A regiment of 1000 men is drawn up in an oblong form, so that the difference of the numbers in the two unequal sides is 117. Kequired the numbers in rank and file . 97. A gentleman pays $180 more for his carriage than for his horse ; and the price of the carriage is to that of the horse as the latter is to 15. What is the price of each ? Ans. Carriage, $21fi ; horse, 98. A father left $210 to three sons, to be divided in sums that form a geometrical progression, so that the first should have $90 more than the last. What were their legacies ? 99. Find two numbers, whose sum multiplied by the greater produces 130, and whose difference multiplied by the less gives 21. 19* 222 GENERAL REVIEW. 100. A person took $6 with Mm to distribute equally among some poor persons ; but, as 5 of them were absent, the remainder received 10 cents apiece more than they otherwise would have received. How many poor persons were there? 101. A merchant bought a piece of cloth for $40, and after cutting off 4 yards, sold the remainder at $2.50 per yard, and received what the whole cost him. How many yards were there, and what did they cost ? Ans. 20 yards, at $2. 102. Find the greatest common divisor of a 6 — 5a*«+10aV - lOaV + 5ax* - x? and cf+af-ax 2 - a 2 x. Ans. a 2 - 2ax + x 2 . 103. A broker bought two cabinets at a sale, having been told that a $10 gold piece was hidden in one of the two. If in the one, he thought it should be worth twice the other ; but if in the other, it ought to be worth three times the first. What value did he put upon the cabinets without the gold piece ? _< t_ 104. Multiply a* +6" +c * by a'^+e". 105. Two boats were sent out from a ship on a whaling ex- cursion, with crews in the ratio of 3 to 5. Meeting some time after, they considered it would be better to divide the hands equally between the two boats, and did so by removing 6 men from one boat to the other. What was the whole number of men sent out ? Ans. Jfi. 106. Find the value of a; in the equation ■ 107. A party of laborers were sent to remove a bank of earth containing 350 cubic yards ; but just as they were com- mencing, four of them were disabled by an accident, in conse- quence of which each of the rest had ten additional yards of earth to remove. What was the number of the party ? Ans. H. 108. A traveller, having proceeded 175 miles by railway, complained of the slowness of the train, and showed that, if it GENERAL REVIEW. 223 had only gone five miles per hour faster, he would have ac- complished his journey in less time by an hour and three-quar- ters. What was the rate at which the train went ? Am. '20 miles per hour. 109. Find the least common multiple of (x + 2af, (x -2a) 3 and z?+Aa 2 . 110. Find two numbers such that, if 6 be added to each, they shall be to each other as 4 to 5, and if 4 be taken from each, they shall be to each other as 2 to 3. Ans. 14 ; 19. 111. Given -h — = a, — \-- = b, and — h - = c, to find x, y and z. x y x z y z 112. Given x — j/aT= 1, to find x. Ans. x = £(3 ± y^). 113. A gentleman bought separately two contiguous fields containing together 40 acres. Each of them cost as many dol- lars per acre as there were acres in the field, and the prices of the two were in the proportion of 4 to 9. "What were the areas of the two fields ? 114. A person rents a certain number of acres of pasture land for $70. He keeps 8 acres in his own possession, sub- lets the remainder at 25 cents per acre more than he gave, and thus covers his rent and $2 over. What was the number of acres? 115. A and B are two towns situated 18 miles apart on the same bank of a river. A man goes from A to B in 4 hours, by rowing the first half of the distance and walking the second half. In returning, he walks the first half at the same rate as before, but the stream being with him, he rows 1\ miles per hour more than in going, and accomplishes the whole distance in 3|- hours. Find the rates of walking and rowing. Ans. 4— miles per hour walking ; ^.-, rowing at first. 224 GENERAL REVIEW. 116. Reduce the surds j/300 and \/lW to the same radical, and find their sum. 117. Find two numbers in the ratio of 5 to 6, such that their sum has to the difference of their squares the ratio of 1 to 7. Am. 35; Jfi. H-n r> ■, , J ■ * 21 23 118. Solve the quadratic — h = — . 7 x+5 7 119. An orange peddler bought a number of oranges at the rate of 5 for 2 cents. He then arranged the good and bad in two separate baskets, containing equal numbers, and sold the one basketful at 3 for 1 cent, and the other at 3 for 2 cents. In selling them, he was told he would make no profit upon them ; but when he had sold the whole he found he had gained 6 cents. How many oranges did he buy and sell ? Am. 360. 120. Divide 4a; - 12 by \^x - 3. Am. fyx - 3. 121. A and B engage in partnership with a capital of $5000 ; A has his money in for 3 months, and B for 2 months, and each at last realizes $4950 of capital and profit. What was the original contribution of each? Am. A, $2250; B, $2750. 122. The seventh term of a geometrical progression is 1000, and the common rate is \$%. What is the first term ? Arts. 790.31 + 123. A certain rectangle contains 300 square feet ; a second rectangle is 8 feet shorter and 10 feet broader, and also contains 300 square feet ; what is the length and breadth of the first rectangle? Am. 20 feet; 15 feet. 124. What two numbers are such that their product is 45, and the difference of their squares is to the square of their difference as 7 is to 2 ? Am. 9 and 5. 125. Two detachments of infantry being ordered to a station 39 miles from their present quarters, begin their march at the same time ; but one party, by marching \ of a mile per hour faster than the other, arrive there an hour sooner. Required their rates of marching. Am. 3± and 3 miles per hour. APPENDIX. SECTION XLVII. * GENERALIZATION. 364. A General Problem is one in which all the quantities are represented by letters. The result obtained by the solution of a general problem ex- presses the value of the unknown in the terms of the known quantities. 365. Generalization is the. process of solving a general prob- lem, and interpreting the resulting expression. The formulas, or general expressions, derived from the solu- tion of general problems, when interpreted, form rules for the solution of all similar problems. 366. A problem is generalized when letters are made to rep- resent its known quantities. EXEMCISJES. 367. — Ex. 1. The sum of two numbers is s, and their differ- ence is d. What are the two numbers ? ^ Solution. Let £ = the greater num- x = the greater; ber, and y = the less ; then, by the eondi- H = t x+y = s (1) (2) tions, x+y=s, and x— y = d. —/1 Adding equations (2) and (1), and Sx — s+d (3) subtracting (2) from (1), we obtain equa- Sy = s — d (4) tions (3) and (4). Whence, z = ^±^ x = ^~ (5) and 3/ = — ; — -. But s and d may be any y = (6) quantities whatever; hence, the values of x and y are general, and 225 226 GENERALIZATION. equations (5) and (6) are formulas for finding two numbers when their sum and difference are given. Hence, the following, — 36S. Rule for finding two numbers from their sum and differ- ence. — Add the difference to the sum, and divide by 2, for the greater of the two numbers; subtract the difference from the sum, and divide by 2, for the less of the two numbers. 1. The sum of two numbers is 391, and their difference is 53. What is the greater number ? 2. A and B hire a pasture together for $65, and A is to pay $13 more than B. How much is each to pay ? Ans. A, $89; B, 3. In an election, the aggregate of votes for A and B was 9637, and B's majority over A was 593. How many votes did each receive ? 369. — Ex. 1. Divide the sum S among A, B and C, in the proportion of the numbers m, n and p. Solution. Let x = A's share ; then x = ^> s s ^j re . nx , px — , and m : p : : x : ■*—. m m nx , px „_ m:n::x: — , and m :p :: x : — . m-.m: x:— = B's share; m .Hence, ^ represents B's share, and m:p :: x:^-=C's share. m — represents Cs share. By the nx px , m x+ ■" " m m conditions, x + H— = £; whence, mS m m m+n+p mS nS , pS an( i — tL nx nS m + n+p m + n+p m + n+p are expressions for their several px pS shares, and, being general, can be m m+n+p applied in the solution of all similar problems. Hence, the following, — GENERALIZATION. 227 370. Rule for dividing a sum into parts proportional to given numbers. — Multiply the sum by each of the propor- tional numbers, and divide the several products by the sum of the proportional numbers. 1. Divide $5400 among A, B and C in proportion to the numbers 7, 9 and 11. 2. Three men are in company. A puts in $8 as often as B puts in 85, and as often as C puts in $3. They gain $1200 ; what is each man's share of it ? Am. A's, $600; B's, $375; Cs, $225. 3. Four men, A, B, C and D, hire a pasture in common. A puts in 6 oxen for 5 months, B puts in 5 oxen for 3 months, C puts in 5 oxen for 5 months, and D puts in 5 oxen for 4 months. They are to pay $90 ; how much is each man's share of it ? 371. — Ex. 1. A cistern can be filled by three pipes ; by the first in 2 hours, by the second in 5 hours, and by the third in 10 hours. In what time can it be filled by all the pipes run- ning together ? Generalize the problem. x = the number of hours in the lime required; — = the part filled in 1 hour, x a b e x bcx+ acx +abx=abc (ab+ac+bc)x= aba abe x=- ab+ao+bc Solution. Represent in the problem, 2, 5 and 10 by a,b and e respectively, and it becomes general. Let cc = the time required for all to fill the cistern ; then — = the part of the cistern filled in one hour by all running together, and, by the conditions, -+— H — = — . a b c x 228 GENERALIZATION. Clearing of fractions, factoring and dividing, we have the general answer, — hours. That is, ab+ac+bc The required time for three agencies operating together to accom- plish a certain result, is the product of the given times, divided by the sum of the products of the times taken two and two. Substituting for the letters, in the general answer, their respective values, we have the particular answer, 1\ hours. 2. Two men are employed to do a piece of work ; the first can do it in 10 days, and the second in 15 days. How many days would it take both, working together? Generalize the problem. 3. What principal at interest at 6 per cent, will, in 3 years, amount to $472 ? If a represent the amount, r the rate per cent, and t the time, what will be the general answer ? Particular Ans. $JfiO ; general Ans. . 100 + r t 4. A gentleman distributing some money among some beg- gars, found that, in order to give them 8 cents apiece, he should want 5 cents more ; he therefore gave them 7 cents each, and had 4 cents left. How many beggars were there ? Generalize " ' Particular Ans. 9 ; general Ans. . a — c 5. The fore wheel of a carriage is 12 feet in circumference, and the hind wheel 18 feet; what will be the number of feet passed over, when the fore wheel has made 40 revolutions more than the hind wheel ? If n denote the number of revolutions, a the circumference of the fore wheel, and b the circumference of the hind wheel, what will be the general answer ? Particular Ans. 1440 ', general Ans. . b — a 6. A workman agreed to work 40 days. For each day he worked he was to receive $2, and for each day he was idle he was to forfeit $.50. He received $50. How many days did NEGATIVE SOLUTIONS. 229 lie work ? If n denote the given number of days, a the number of dollars for each day he worked, b the number of dollars for- feited for each day he was idle, and c the amount received at the end of the time, what will be the general answer ? Particular Ans. 28 ; general Ans. . a + b 7. A merchant has two kinds of tea ; the one cost 90 cents per pound, the other 50 cents. He wishes to mix both kinds together in such quantities that he may have 50 pounds, and each pound, without profit or loss, may be sold for 80 cents. How much must he take of each to make up the mixture? Generalize the problem. Particular Ans. 87 j lbs. of the best; 12j lbs. of the other. General Ans. - — lbs. of the best ; - — of the other. a—b a—b SECTION XLVIII. NEGATIVE SOLUTIONS. 372. Negative Solutions are such as produce results which have the minus sign. EXERCISES. 373. — Ex. 1. The length of a garden is 8 rods and its width is 5 rods. How much must be added to the length, that the garden may contain 30 square rods ? Solution. Let 2 = the quantity which x = the quantity ; must be added. Then, (8 + x)5 = area (8 +x)5= area of garden. of garden; hence, 40 + 5x = 30; whence, 4O+Sx = 30 5a; =—10, or x= -2, which satisfies the 5x=—10 conditions of the problem algebraically, but x = — 2 not arithmetically. But, since adding a negative quantity is equivalent to subtracting an 2:i 230 NEGATIVE SOLUTIONS. equal positive quantity, the negative result indicates that the problem, to be consistent arithmetically, should read : The length of a garden is 8 rods, and its width is 5 rods ; how much must be subtracted from its length that the garden may contain 30 square rods ? 2. A father was 50 years old when his son was 20 ; in how many years afterwards was the father four times as old as the son? Solution. Let x = the number of years. x = the number. Then, 20+K = 5°jtf . w hence, cc=-10. But 20+ x = ^^ the problem, in the exact sense of its enuncia- SO +JfX=50+x tion, is impossible, for when the father was 50 3x= —30 years old and the son 20, the father was less x = — 10 than four times as old as the son ; hence, the time when he was just four times as old as the son must have been before the given time. The — 10 must then be understood as indicating that the time was before and not after, and the enunciation of the problem should be modi- fied accordingly. From these illustrations may be drawn the following in- ferences. 374. — 1. The negative solution of a problem by an equation of the first degree indicates some inconsistency or absurdity in the conditions of the problem as enunciated. 2. That in such cases an analogous problem consistent in its conditions can generally be formed by changing the terms of the absurd condition to those entirely opposite. Solve, interpret and properly modify the enunciation of the following problems : 1. A man when he was married was 45 years old, and his wife 20. How many years before was he twice as old as she ? Ans. — 5. The — 5 indicates a wrong condition ; that is, their ages bore the given relation 5 years after, not before, their marriage. INDETERMINATE AND IMPOSSIBLE PROBLEMS. 231 2. A had $150 and B $120. They each gained a certain sum, when it was found that A's money was to B's in the ratio of 3 to 2. What did each gain ? 3. What number is that whose fourth part exceeds its third part by 16 ? Ans. - 192. That is, the enunciation contains a contradiction ; it should read : What number is that whose third part exceeds its fourth part by 16 ? 4. What fraction is such that if 2 be added to its numerator its value is \, or if 2 be added to its denominator its value is \ ? Ans. 5j. 5. A man worked 12 weeks, having his wife with him 7 weeks, and received $46. He afterwards worked 8 weeks, having his wife with him 5 weeks, and received $30. How much did he earn per week for himself, and how much did his wife earn? Ans. Man, $5; wife, —$2. That is, the man earned $5 per week for himself, and was charged $2 per week for expense of his wife, when she was with him. SECTION XLIX. INDETERMINATE AND IMPOSSIBLE PROBLEMS. 375. Zero, or the symbol 0, may be used to denote the ab- sence of value, or to represent a quantity less than any assign- able value. 376. Infinity, or the symbol oo, is used to denote a quantity greater than any assignable value. A quantity which is less than any assignable value is said to be infinitely small. 232 INDETERMINATE AND IMPOSSIBLE PROBLEMS. 377. An equation of the first degree with one unknown quantity may be reduced to the general form, ax = b; whence, x = —\ a in which we know that if a and b are both finite, the value of the unknown quantity will be a determinate finite quantity. But, if a -or b, or both, be equal to 0, the values of the un- known quantity assume peculiar forms. 1. Let 6 = 0, then the value of z becomes °-o, a where a denotes some finite quantity. For, with a given denominator, the less the numerator of a fraction, the less is the value of the fraction ; and when the numerator becomes 0, the value of the fraction becomes infinitely small, or 0. Hence the solution is not possible in finite numbers. 2. Let a = 0, then the value of x becomes a — = oo. For, with a given numerator, the less the denominator of a fraction, the greater is the value of the fraction ; and when the denominator be- comes 0, the value of the fraction becomes infinitely large, or oo. Hence, it is evident that in this case no finite value satisfies the equa- tion. 3. Let a = and 6 = 0, then the equation becomes x = -. This is equivalent to x x = 0, an equation which any finite value whatever of x will satisfy. The solution is then indeter- minate. When, however, - is derived from a fraction whose terms contain a common factor, the fraction may have a determinate value. Thus, let x = — — '■ , and a = l ; the value of x is then — . Dividing the terms of the fraction by a — 1 , we have x = = — . In this case the value a + 2 3 of x is determinate. INDETERMINATE AND IMPOSSIBLE PROBLEMS. 233 378. An Indeterminate Problem is one whose conditions are satisfied by different values of the same unknown quantity. Such a problem will be found to produce either an equation whose members are merely the repetition the one of the other, or a less number of independent equations than there are un- known quantities to be determined. 379. An Impossible Problem is one whose conditions are absurd or contradictory. EXERCISES. 380. — Ex. 1. What number is that the sum of whose £, £ and \ is equal to the number increased by its ^-? Solution. Let x = the num- x = the number. rv* /yt FY* ry* ryi syt ber; then, by the conditions, — I — — I — I — = aH ' ' ' '23 2 3 4 12 +* = X + JL. Clearing of frac- 6x + 4x*3x-lSx+x (1) 4 12 5 13x = 13x (2) tions, we have equation (1), and, lSx- 13x = (3) by uniting, we have the identical \ ~ 1 — \ J 6 _ . 0X3 = (5) equation (2). Transposing and g x = - (6) factoring, we have (3) and (4) ; whence (5) and (6), or £ = -. The problem is therefore indeterminate. 2. What number is such that the sum of its \ and its •§, diminished by 1, is equal to its \\ increased by 2 ? Solution. Let £ = the num- ber; then, by the conditions, — + 1 = h2. Clearing of frac- tions, we have equation (1), and by uniting, transposing and fac- toring, we have equations (2) and (3). Whence (4) and (5), orz = co, which indicates that no finite number will Batisfy the conditions. The problem is, therefore, impossible. 20* x=the number. 4 3 12 x + 8x-12=llx+24 (1) llx-llx = 36 m (11- ll)x = 36 (3) 0xx = 36 (4) 36 x= — = oo (5) 234 INDETERMINATE AND IMPOSSIBLE PROBLEMS. 3. Given y+z = 21 and y-x = l%, to find x and y. Solution. Subtracting the second equation from the first, we have x+z = &, an equation which admits of any number of values of x and z. The problem is therefore indeterminate. Zx 4. Given x+y = 14, x — y = 2 and — = 3. y Solution. From the first and second equations, we find x = 8 and y = 6. But the third equation requires 3 times x divided by y to be 3, which cannot be fulfilled for those values of x and y. The problem therefore is impossible. 5. A piratical vessel is 10 miles ahead of a sloop of war, and the two are sailing at equal rates per hour. In how many hours will the sloop overtake the pirate ? Ans. x = — =oo. Impossible. 6. What number is such that if 3 be subtracted from twice the number, the remainder will be equal to one-third of the excess of 6 times the number over 9 ? Ans. *-?. JM*^. 7. A man being asked how many fish he had caught, replied, " If 5 be added to one-third of the number that I caught yester- day, it will make half the number I have caught to-day ; or if 5 be subtracted from three times this half, it will leave the number I caught yesterday." How many were caught each day? Ans. Impossible. 8. I have money in two purses ; that in the one is to that in the other in the ratio of 5 to 6; if | of the money in the second were to be taken out, the money in the two purses would be equal. How many dollars are there in each ? Ans. Indeterminate IMAGINARY QUANTITIES. 235 Test Questions. 381. — 1- What is a General Problem? When is a problem gene- ralized ? What is generalization ? How may the formulas derived by generalization be interpreted ? 2. What are Negative Solutions? How are negative quantities de- rived ? What does a negative solution of a problem by an equation of the first degree indicate? How can a problem leading to a negative result be generally changed to a problem consistent in its conditions ? 3. What is an Indeterminate Problem? An impossible problem? What is zero used to denote ? Infinity ? SECTION L. IMAGINARY QUANTITIES. 382. An Imaginary Quantity is an indicated even root of a negative quantity. Thus, Y — a and f — 16 are imaginary quantities, and symbolize pro- cesses which it is impossible to perform. They are, however, of considerable use in mathematical analysis, and when subjected to certain rules of calculation, they lead to possible and valuable results. 383. The addition and subtraction of imaginary quantities are performed by the same rules that apply to other radicals ; but with regard to their multiplication or division, the ordinary rules require some modification. 384. Principles. — 1. Every imaginary quantity may he re- solved into two factors, — one a real quantity, and the other the imaginary expression |/ — 1, or V ~ 1- 236 IMAGINARY QUANTITIES. For every negative quantity may be regarded as the product of two factors, one of which is — 1. Thus, since — a = ax( — 1), we have \/ — a=-\/a x ( — l) = -j/a Xj/ — 1; also, since — 4 = 4x( — 1), we have ;/— 4 = -|/4x( — 1) = j/4 x 1 /^T=2 l /'^r The real factor is called the Coefficient of the imaginary- factor \/ -\. 2. The product of two imaginary quantities is real, and the sign before the radical is the reverse of that obtained by the common rule. For the ambiguity in the signs to be prefixed to an even root by the common rule (Prin. 2, Art. 244) is removed when we know the factors which compose the quantity whose root is taken. These factors of an imaginary quantity we may know, as has already been shown. Thus, )/-ax j/ - 6 = (/ax y — l)(|/5x j/ — 1) = i/o5x(-l) = — -y/ab. In like manner it may be shown that - |/ — ax ( — ]/ — 6)= — j/aS"; also, /-ax(-)/-6)= +-\/ab. 3. TAe quotient of one imaginary quantity divided by another is real, and has the sign before the radical the same as that ob- tained by the common rule. • a l S0) zVZ^ -t/«*/_ _ V^h t/5xt/-1 385. By attention to these principles, the algebraic processes with imaginary quantities may be readily performed. y~^5 V~b~xV-l \ 6 ' I fa IMAGINARY QUANTITIES. 237 PS, OBZ EMS. Ex. 1. Multiply 5v/"^3 by ^/~^%. Solution. 5-/=S = B-/S x /=! and 1 /"^l= 1 /lx /^l. (V# x /=T)(i/0x i/ Tr T=5/"6x (^"^7)2= -5 V '6T 2. Multiply 3V ^2 by 2 v '~-i. Ans. -6-/2. 3. Multiply -li/^a by --(A^. -A?is. -7a"|/6. 4. Multiply 1 /^a J by j/^ 1 ^ Ans. ah{ x /~^lf = - ah. 5. Multiply 3-V"^2 by 3 i y~^2. Ans. 11. 6. Multiply 1 - t/^I by 1 - t/^T. ^s. - V" 117 - 7. Divide j/ -a& by -j/ — a. Solution. a 8. Divide (Jt/^T by 6 V / -^ Ans - j 9. Divide 10i/^Y4 by 2 1 / 3 T: • 4ms - V^ 10. Divide 1 + j/=l by 1 - iA^I. ^» s - l/^" 7 11. Divide -S^/^by - 2|/ - 3. Ans. 4, 12. Multiply a+bi/^T by a - &j/ - 1. .4ns. a 2 + Z> 2 . 238 BINOMIAL THEOREM. SECTION LI. BINOMIAL THEOREM. 386. The Binomial Theorem expresses the simplest method known of writing out the different powers of any binomial. This method, discovered by Sir Isaac Newton, may be deduced from the first few powers of any binomials, as a+b and a-b. 1. Let a+b be raised to the 2d, 3d, 4th and 5th powers, by actual multiplication. Solution. 1st power, a+b a + b a 2 + ab ab + b 2 2d power, a*+2ab+b 2 a +b a"+2a 2 b+ab 2 a*b+2ab 2 +b 3 3d power, a 3 +3a 2 b+3ab 2 +b 3 a +b a i +3a 3 b+Sa 2 b i +ab s a*b+3a?V i +8ab*+b i 4th power, a l +4a 3 b+6a 1 b i +4ab 3 +b i a +b a 5 + JfaSb + 6a"b*+ 4a 2 b s +ab l a'b +4a 3 b 2 +6a 2 b 3 +4ab i + 6 5 5th power, a 5 + 5a*6 + 10a?b' 1 + lOa'b 3 + Sab* + 6 5 BINOMIAL THEOREM. 239 2. Let a — b be raised to the 2d, 3d, 4th and 5th powers, by actual multiplication. SOLTJTIOB r. ls< power, a - a - -6 -6 a 2 -a& power, -a&+6 2 2d a 2 - -2ab + b 2 a - -b a 3 - -2a 2 b+ab 2 power, -a 2 b+2ab 2 - & 3 Zd a 3 - -8a?b + 8ab 2 - -6 3 a - -b a 4 - -3a?b+8a 2 b 2 -a& 3 power, -a 3 b+3a 2 b 2 - Sa& 3 +& 4 4th a 4 -4a 3 b + 6a?b 2 -4a& 3 +& 4 a -b a 5 - -4« 4 &+6a 3 & 2 -4a 2 6 s +a6 4 -a%+4a 3 b 2 - ■6a 2 6 3 +^a6 4 -& 5 5th power, a 5 -5a 4 6+i0a s 6 2 -I0a% 3 +5a6 4 -6 5 387. These different powers of a+b and a — b show that cer- tain invariable laws govern the expansion of a binomial. 1. The first, or leading, term of the binomial appears in every term of the power, except the last. Its exponent in the first term is the same as the index of the power, and in the following terms it decreases regularly by one. 2. The second term of the binomial appears in every term of the power, except the first. Its exponent in the second term is 240 BINOMIAL THEOREM. one, and in the following terms it increases regularly by one, be- coming in the last term equal to the index of the power. Thus, in the fifth power of each of the given binomials, The exponents of a are 5, 4, 3, 2, 1. The exponents of b are 1, 2, 3, 4, 5. 3. The coefficient of the first term is one ; that of the second is the same as the index of the power; and, in general, the coefficient of any term, multiplied by the exponent of the leading quantity of that term, and divided by the exponent of the following quantity increased by one, equals the coefficient of the next term. Thus, in the fifth power of a+b, the coefficient of the first term is 1 ; that of the second term is the same as the exponent of the power, or 5 ; the coefficient of the second term, 5, multiplied by 4, the exponent of a, the leading quantity of that term, and divided by 2, the exponent of b (5 x 4 \ = = 1 1, is the coefficient of the third term ; and, in like manner, 10 is the coefficient of the fourth term ; 5 is the coeffi- cient of the fifth term ; and 1 is the coefficient of the last term. 4. When both terms of the binomial are positive, all the terms of the power are positive; but when the second term is negative, all the odd terms, counted from the left, are positive, and all the even terms negative. It will be seen that the coefficients are repeated in the inverse order after passing the middle term or terms, so that most of the coefficients can be written without calculation. When the number of terms is even, the two middle terms have the same coefficient. It may also be observed that the number of terms is always one more than the exponent of the power, and that the sum of the exponents in any term is equal to the exponent of the power. BINOMIAL THEOREM. 241 388. Combining the principles given in the preceding article, we have, when the exponent of the power is n, the general de- velopment, (a + 6)" = „ ii n(n — l) „ ,,, w(w-l)(w-2) „ „, .„ , .„ which is the Binomial Formula, and can be used, in the expan- sion of any binomial. PROBLEMS. 1. Raise a+x to the fourth power. Solution. Letters and exponents, a*, c?x, a?x 2 , ax 3 , x* Coefficients, 1 +4 +6 +4 +1 Combining, a* +4o?x +6a?X 2 +4ax* +a? 2. Raise x — y to the fourth power. Ans. x i — 4x 3 y+6x 2 y 2 — 4xy 3 +y 4 '. 3. Raise x+y to the sixth power. 4. What is the eighth power of x - y ? 5. Find the fourth power of 1+x. Solution. Expanding the terms, l 4 l s x IV lx* x* 1 +4 +6 +4 +1 Rejecting the factor 1, and combining, 1 +4% +6x 2 +4X 3 +X 4, 6. Find the sixth power of 1 — x. Ans. l-6x+ 15a? - 20a? + 15x* - 6a? + x 6 . 7. Develop (a - 1) 5 . Ans. a? - 5o? + 10c? - 10a? +5a-l. 389. The Binomial Formula also applies when either or both terms of a binomial have coefficients or exponents. 21 242 BINOMIAL THEOREM. 1. What is the third power of 2a — 36? Solution. Expanding terms, (Sa) s '- 3{2a) 2 (3b) + 3(Sa)(3b) 2 - (3b) 3 Multiplying factors, 8a? - 36a 2 b + 54ab 2 - 27b 3 2. Develop (a? + b 3 y. Ans. a s +4a e b 3 + 6a 4 b 6 +4aV+V 2 . 3. Develop (l + 2») 3 . Ans. l + 6x + 12x 2 +8a?. 4. Expand (2a- 1) 4 . Ans. 16u* -82a 3 +24a 2 - 8a < + l. 5. What is the square of 9a; + -? Ans. 81x 2 +18+ — . x x 2 6. What is the cube of 2ax-Zbe"! Ans. 8a 3 x s - S6a 2 b\x 2 + S^abVx - 27b 3 c e . 390. A Polynomial may be raised to any power by putting it under the binomial form, and then applying the general for- mula. 1. Find the cube of a+b + e. Solution. Putting a+b+e under the binomial form, we have a+{b+c). Expanding [a+ (6 + e)] 3 , we have a 3 +Sa 2 (b + c)+8a(b+c) 2 +{b+c) 3 . Expanding and multiplying factors, we have a 3 + 3a 2 b + 8a 2 c + Sab 2 + 6abc + Sac 2 +b 3 + 3b 2 e +3be 2 + e 3 . 2. Find the square of a — b+y. Ans. a 2 -2ab+2ay+b i -2by+y 2 . 3. Find the square of a — b — c. Ans. a'-Zab + tf-Sac + Zbc + c 1 . LOGARITHMS. 243 SECTION Lll. LOGARITHMS. 391. The Logarithm of a number is the exponent of the power to which a constant number must be involved to produce the given number. Thus, if 8 is the constant number, 2 is the logarithm of 64, because 8 2 = 64 ; 3 is the logarithm of 512, because 8 3 = 512. 392. The Base of logarithms is the constant number which must be involved to produce the numbers. 393. Common Logarithms are those whose base is 10. Hence, in the common system, is the logarithm of 1, since 10° equals 1 ; 1 is the logarithm of 10, since 10 l equals 10 ; 2 is the logarithm of 100, since 10 2 equals 100 ; 3 is the logarithm of 1000, since 10 3 equals 1000; and, generally, if n be any positive integer, and log. stand for logarithm, we have log. 10 n =n. Again, by means of negative exponents, — 1 is the logarithm of .1, since 10" L equals .1 ; — 2 is the logarithm of .01, since 10~ z equals .01 ; — 3 is the logarithm of .001, since 10 -3 equals .001; and, generally, if n be any negative integer, we have log. =— n. It thus appears that the logarithm of any number between and 10 must be a fraction ; between 10 and 100, 1 plus a fraction ; between 100 and 1000, 2 plus a fraction, and so on. The logarithms of numbers which are not exact powers can only be obtained approximately, and are usually expressed by decimals. 244 LOGARITHMS. 394. The Characteristic of a logarithm is the integral part Of its expression. The decimal part is sometimes called the mantissa. 395. Principles. — 1. The characteristic of the logarithm of any number greater than 1, is one less than the number of inte- gral orders in the given number. For, the logarithm of 1 is 0, of 10 is 1, of 100 is 2, of. 1000 is 3, and so on. 2. The characteristic of the logarithm of any number less than 1, expressed as a decimal, is one more than the number of ciphers between the decimal point and the first numeral figure in the decimal. For, the logarithm of .1 is — 1 ; of .01 is — 2 ; of .001 is — 3, and so on. When the characteristic of the logarithm is a negative number, it is distinguished by being written with a short horizontal line over it. Thus, 1 is written instead of —1, 2 instead of —2, etc. 3. Tlie logarithm of the product of two or more numbers is equal to the sum of the logarithms of those numbers. For, let m and n be any two numbers, x and y their respective loga- rithms, and a the base of the system. Then, by the definition of a logarithm, we have a?=m, (V=n. Multiplying these equations the one by the other, member by member, we have a x +v=mn, in which x+y is the logarithm of the product mn. 4. The logarithm of the quotient of two numbers is equal to the logarithm of the dividend diminished by that of the divisor. For, dividing the equation a x =m by the equation a? = n, member by member, we have - ., m a x ~v= — n in which x—y is the logarithm of the quotient — . LOGARITHMS. 245 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, raising both members of the equation a x = m to any power p, we have a x » = m?, in which xp is the logarithm of m raised to the power p. 6. The logarithm of the root of any number is equal to the logarithm of that number divided by the index of the root. For, extracting the rth root of both members of the equation a x = m we have \ in which r is the logarithm of y'm. 7. Numbers integral, decimal or mixed, having the same suc- cession of numeral figures, have logarithms with the same decimal part. For, the decimal part of a logarithm of a number is the same as that of the product or the quotient of the number, multiplied or divided by 10, 100, 1000, etc. ; and since the logarithms of 10, 100, 1000, etc., are 1, 2, 3, etc., the effect of the multiplication or division must be to change the characteristic of the logarithm of the number, and not to change the decimal part. TABLES OF LOGARITHMS. 396. A Table of logarithms contains the computed logarithms of all integers between 1 and some given number. The computation of the logarithms is generally made by- means of a series developed by the aid of the binomial theo- rem, and is too abstruse for a work of this kind. . 397. The Table given on the next two pages gives the decimal part of the common logarithms of the series of natural numbers from 10 to 999, carried to five decimal orders. The characteristics are not given in the table, but are left to be sup- plied by inspection. (Prin. 1 and 2, Art. 395). 21* 246 LOGARITHMS. Table of Common Logarithms. No. IO 1 2 3 1284 4 1703 5 2119 6 2531 1 2938 8 3342 9 3743 D. 414 00000 0432 0860 II 04139 4532 4922 53o8 5690 6070 6446 6819 7188 7555 378 12 07918 8279 8636 8991 9342 9691 O037 O380 O721 I059 348 13 "394 1727 2057 2385 2710 3033 3354 3672 3988 4301 322 J4 14613 4922 5229 5534 5836 6i37 6435 6732 7026 73'9 300 IS 17609 7898 8184 8469 8752 9033 9312 9590 9866 2140 280 16 20412 0683 0952 1219 1484 1748 201 1 2272 2531 O789 263 17 23 45 33°° 3553 3805 4055 4304 4551 4797 5042 5285 248 18 2SS27 5768 6007 6245 6482 6717 6951 7181 7416 7646 235 19 27875 8103 8330 8556 8780 9003 9226 9447 9667 9885 223 20 30103 0320 0535 0750 0963 "75 1387 1597 1806 2015 212 21 32222 2428 2634 2838 3041 3244 3445 3646 3846 4044 202 22 34242 4439 4635 4830 5025 5218 54" 5603 5793 5984 193 23 36173 6361 6549 6736 6922 7107 7291 7475 7658 7840 185 24 38021 8202 8382 8561 8739 8917 9094 9270 9445 9620 177 25 39794 9967 O140 O312 O483 O654 O824 0993 H63 I330 170 26 4H97 1664 1830 1996 2160 2325 2488 2651 2813 2975 164 27 43136 3297 3457 3616 3775 3933 4091 4248 4404 4560 158 28 44716 4871 5025 5179 5332 5484 5637 5788 5939 6090 "52 29 46240 6389 6538 6687 6835 6982 7129 7276 7422 7567 147 30 47712 7857 8001 8144 8287 8430 8572 8714 8855 8996 142 31 49>36 9276 9415 9554 9693 9831 9969 O106 O243 0379 138 32 50515 0651 0786 0920 1055 1 188 1322 1455 1587 1720 134 33 5i85i 1983 2114 2244 2375 2504 2634 2763 2892 3020 130 34 53H8 3275 3403 3529 3656 3782 3908 4033 4158 4283 126 35 54407 4531 4654 4777 4900 5023 5H5 5267 5388 5509 122 36 55630 5751 5871 5991 61 10 6229 6348 6467 6585 6703 119 37 56820 6937 7054 7177 7287 7403 7519 7634 7749 7864 116 38 57978 8092 8206 8320 8433 8546 8657 8771 8883 8995 "3 39 59106 9218 9329 9439 555o 9660 9770 9879 9988 0097 no 40 60206 0314 0423 0531 0638 0746 0853 0959 1066 1172 107 4 1 61278 1384 1490 1595 1700 1805 1909 2014 2118 2221 105 42 62325 2428 2531 2634 2737 2839 2941 3043 3M4 3246 102 43 63347 3448 3548 3649 3749 3849 3949 4048 4147 4246 100 44 64345 4444 4542 4640 4738 4836 4933 5031 5128 5225 98 45 65321 54i8 55H 5610 5706 5801 5896 5992 6087 6l8l 95 46 66276 6370 6464 6558 6652 6745 6839 6932 7025 7117 93 47' 67210 7302 7394 7486 7578 7669 7761 7852 7943 8034 91 48 68124 8215 8305 8395 8485 8574 8664 8753 8842 8931 90 49 69020 9108 9197 9285 9373 9461 9546 9636 9723 9810 88 50 69897 9984 0070 Oi57 O243 O329 O415 O501 O586 O672 86 51 70757 0842 0927 1012 1096 1181 1265 1349 '433 15^ 84 52 71600 1684 1767 1850 1933 2016 2099 2181 2263 2346 83 53 72428 2509 2591 2673 2754 2835 2916 2997 3078 3159 81 54 73239 3320 3400 3480 356o 3640 37'9 3799 3878 3957 80 LOGARITHMS. 247 Table of Common Logarithms. No. 55 1 4"5 2 4194 3 4273 4 435 1 5 4429 6 4507 •7 4586 8 4663 4741 D. 78 74036 56 74819 4896 4974 5051 5128 5205 5282 5358 5435 55" 77 57 75S87 5664 5740 5815 5891 5967 6042 6118 6i93 6268 76 58 76343 6418 6492 6567 6641 6716 6790 6864 6938 7012 74 59 77085 7159 7232 73°5 7379 7452 7525 7591 7670 7743 73 60 77815 7887 7960 8032 8104 8176 8247 8319 8390 8462 72 61 78533 8604 8675 8746 8817 8888 8958 9029 9099 9169 7i 62 79239 93°9 9379 9449 9518 9588 9657 9727 9796 9865 69 63 79934 O003 O072 O140 O209 O277 O346 O414 O482 055O 68 64 80618 0686 0754 0821 0889 0956 1023 1090 1158 1224 67 65 81291 1358 1425 149 1 1558 1624 1690 1757 1823 1889 66 66 81954 2020 2086 2151 2217 2282 2347 2413 2478 2543 65 67 82607 2672 2737 2802 2866 2930 2995 3059 3 I2 3 3187 64 68 83251 3315 3378 3442 35o6 3569 3632 3696 3759 3822 63 69 83885 3948 401 1 4073 4136 4198 4261 4323 4386 4448 62 70 84510 4572 4634 4696 4757 4819 4880 4942 5°03 5065 62 71 85126 5187 5248 5309 S37o 5431 549> 5552 5612 5673 61 72 85733 5794 5854 59H 5974 6034 6094 6i53 6213 6273 60 73 86332 6392 6451 6510 6570 6629 6688 6747 6806 6864 59 74 86923 6982 7040 7099 7157 7216 7274 7332 7390 7448 58 75 87506 7564 7622 7679 7737 7795 7852 7910 7967 8024 58 76 88081 8138 8i95 8252 8309 8366 8423 8480 8536 8593 57 77 88649 8705 8762 8818 8874 8930 8986 9042 9098 9154 56 78 89209 9265 9321 9376 9432 9487 9542 9597 9653 9708 55 79 89763 9818 9873 9927 9982 O037 O091 O146 O200 O255 55 80 90309 0363 0417 0472 0526 0580 0634 0687 0741 0795 54 81 90849 0902 0956 1009 1062 1116 1 169 1222 1275 1328 53 82 91381 H34 1487 1540 1593 1645 1698 1751 1803 1855 53 83 91908 i960 2012 2065 2117 2169 2221 2273 2324 2376 52 84 92428 2480 2531 2583 2634 2686 2737 2788 2840 2891 5« 85 92942 2993 3044 3095 3H6 3197 3247 3298 3349 3399 5i 86 93450 3500 3551 3601 3651 3702 3752 3802 3852 3902 5° 87 9395 2 4002 4052 4101 4151 4201 4250 4300 4349 4399 50 88 94448 4498 4547 4596 4645 4694 4743 4792 4841 4890 49 89 94939 4988 5036 5085 5'34 5182 5231 5279 5328 5376 49 90 95424 5472 5521 5569 56i7 5665 5713 576i 5809 5856 48 9i 95904 5952 5999 6047 6095 6142 6190 6237 6284 6332 47 92 96379 6426 6473 6520 6567 6614 6661 6708 6755 6802 47 93 96848 6895 6942 6988 7035 7081 7128 7174 7220 7267 46 94 97313 7359 7405 7451 7497 7543 7589 7635 7681 7727 46 95 97772 7818 7864 7909 7955 8000 8046 8091 8i37 8182 45 96 98227 8272 8318 8363 8408 8453 8498 8543 8588 8632 45 97 98677 8722 8767 881 1 8856 8900 8945 8989 9074 9078 45 98 99123 9167 9211 9255 9300 9346 9388 9432 9476 9520 44 99 99564 9607 9651 9695 9799 9782 9826 9870 9913 9957 44 548 LOGARITHMS. 398. The Numbers expressed by not more than two orders of figures are given in the column marked No., and opposite each number in the adjacent column is its corresponding loga- rithm. The first two orders on the left of a number expressed by three orders of figures are given in the column marked No., and the third order is given at the top of another column on the same page. The first order on the left of the mantissa is printed only once, and in the first column of logarithms, for all the ten loga- rithms in the horizontal line ; and where a figure of the man- tissa is printed in light-face type, the first order on the left of the mantissa is to be found in the column in the line next below. 399. The Average Difference of the ten logarithms in the same horizontal line is given in the column marked D. 400. The mantissas of the logarithms of 1, 2, 3, etc., are the same as those of 10, 20, 30, etc. Hence, the preceding table is sufficient for the first 1000 natural numbers. EXERCISES. 401.— Ex. 1. Find the logarithm of 131. Solution. The decimal part of the logarithm, from the table, cor- responding to the given number, is .11727 Prefixing the characteristic (Prin. 1, Art. 395), we have 2.11727 = Log. of 131. 2. Find the logarithm of 144. Ans. 2.15886 3. Find the logarithm of 99. 4. Find the logarithm of .0102 Solution. The decimal part of the logarithm in the table, corre- sponding to the given number, is .00860 Prefixing the characteristic (Prin. 2, Art. 375), we have 2.00860 LOGARITHMS. 249 5. Find the logarithm of .0017 Ans. 81 6. Find the logarithm of 1.29 Ans. 0.11059 7. Find the logarithm of 10.2731 Solution. The logarithm corresponding to 10.2 is 1.00860 The difference in column D, on the same horizontal line as the deci- mal part of the logarithm, is 414, which being multiplied by 731, the remaining orders of the given number, we have 414x781 = 302634. Eejecting from the right of this product as many orders as there are figures in the multiplier 731, we have the part; 303, approximately ob- tained, to he added to the logarithm found. Adding, we have, 1.00860 1.01163=Log. of 10.2781 8. Find the logarithm of 1495. Ans. 8.1746 9. Find the logarithm of 10210. 10. Find the logarithm of .00763 Ans. 8.88275 11. Find the logarithm of .018414 Ans. 2.26515 402. — Ex. 1. Find the number whose logarithm is 2.98182 Solution. The number in the table corresponding to the decimal part of the given logarithm is 959, which according to the characteristic must contain three integral orders. Hence, the required number is the integer 959. 2. Find the number whose logarithm is 1.61066 3. Find the number whose logarithm is 2.15836 Ans. 144- 4. Find the number whose logarithm is 3.23045 Ans. .0017 5. Find the number whose logarithm is 1.52634 6. Find the number whose logarithm is 2.57287 Ans. 374. 250 LOGARITHMS. 7. Find the number whose logarithm is 1.011702 Solution. The number in the table whose logarithm is next less than the given logarithm is 10.2 The given logarithm, 1.01170 The logarithm next less, 1.00860 ; corresponding number, 10.2 , Difference of logarithms, 310 _ „.„ Difference from column D, fylfy Number required, 10.274" 8. Find the number whose logarithm is 1.60552 Ans. 40.82 9. Find the number whose logarithm is 3.88275 Ans. .007634 10. Find the number whose logarithm is 3.17464 11. Find the number whose logarithm is 1.082426 Ans. .1209 Multiplication by Logarithms. 403 —Ex. 1. Multiply 76.4 by 5.4 Solution. Logarithm of 76.4 = 1-88309 Logarithm of 5.4 = 0.73239 Logarithm of 412.56 = 2.61548 404. Rule for Multiplying one Number by another by Log- arithms. — Add the logarithms of the factors, and find the number corresponding to the sum.— (Prin. 3, Art. 395.) PROBLEMS. 1. Multiply 5.3 by 2.8 Ans. U-84 2. Multiply 3.26 by .0025 Ans. .00815 3. Multiply .25 by .003 Ans. .00075 • 4. Multiply 134 by 25.6 Ans. 8 480 4 5. Multiply 1853 by 46. 6. Multiply .0051 by 2.3 Ans. .01178 LOGARITHMS. 251 Division by Logarithms. 405— Ex. 1. Divide 59.45 by .0315 Solution. Logarithm of 59.45 = 1.77415 Logarithm of .0315 = 2.49831 Logarithm of 1887.1 = 3.27584 406. Rule for Dividing one Number by another by Loga- rithms. — Subtract the logarithm of the divisor from the logarithm of the dividend, and fund the num- ber corresponding to the difference.— (Art. 375.) PROBLEMS. 1. Divide 875 by 25. Ans. 85. 2. Divide 410.4 by 5.4 Ans. 76. 3. Divide .008215 by .031 Ans. .265 4. Divide .0023808 by 3.72 Ans. .00064 5. Find the value of ■§ in a decimal expression. Solution. The fraction f is equal to 3 divided by 8. Logarithm of 3 = 0.47712 Logarithm of 8 = 0.90309 Logarithm of j, or .375 = 1.57403 6. Find the value of 3^ or ^ in a decimal expression. Ans. 3.25 7. Find the value of ^f in a decimal expression of two orders. Ans. .44 8. Find the value of -j- 1 ^ in a decimal expression. .Aug. .096 252 LOGARITHMS. Involution by Logarithms. 407. — Ex. 1. Find the second power of 6.4 Solution. Logarithm of 6.4 = 0.80618 Logarithm of Jfi.96 = 1. 408. Rule for Involution by Logarithms. — Multiply the logarithm of the number by the exponent of the power, and find the number corresponding to the product— {Art. 402.) PROBLEMS. 1. Find the fourth power of 1.05 to four orders of decimals. Am. 1.2155 2. Find the fifth power of 1.06 to three orders of decimals. Ans. 1.S88 3. Find the cube of .25 Ans. .015625 4. Find the fifth power of .836 Ans. ,4.083+ 5. Find the cube of 13. Ans. 2197. 6. Find the fifth power of 2.846 Ans. 186.65 Evolution by Logarithms. 409.— Ex. 1. Find the square root of 40.96 Solution. Logarithm of 40.96 = 1.61286 Dividing by 2, we have — 0.80618 whose corresponding number is 6.4 410. Rule for Evolution by Logarithms. — Divide the log- arithm of the number by the integer denoting the degree of the root, and find the number correspond- ing to the quotient.— (An. 402.) LOGARITHMS. 253 PROBLEMS. 1. Find the square root of 1849. Ans. £3. 2. Find the cube root of 2197. Ans. IS. 3. Find the cube root of 10648. Ans. 22. 4. Find the tenth root of 1024. Ans. 2. 5. Find the fifth root .00009 to four orders of decimals. 6. Find the square root of .001849. Solution. Logarithm of .001849 = 81 Adding — 1 to 3, to make it divisible by the index of the root, and adding + 1 to mantissa, to the offset, we have 4+1., and dividing by 2, we obtain whose corresponding number is .043. That is, when the characteristic of the logarithm is.negative, and not divisible by the index of the root, the characteristic may be increased by any number which will make it exactly divisible, provided we pre- fix an equal positive number to the mantissa of the logarithm. 7. Find the cube root of .015625 Ans. .25 8. Find the seventh root of .005846 Ans. 4797 9. Find the sixth root of .0432 to three orders of decimals. Ans. .592 10. What is the value of jX034 to three orders of decimals ? Ans. .508 11. What is the value of T/.0000169 to four orders of decimals? Ans. .0041 12. Find the tenth root of 581.4 to two orders of decimals. Ans. 1.89 22 254 LOGARITHMS. COMPOUND INTEREST. 411. Compound Interest is interest reckoned on interest and principal combined, at specified intervals. The intervals may be years, half years, quarters, etc., accord- ing as the interest is made part of the principal annually, semi- annually, quarterly, etc. 412. Let p denote any principal at compound interest ; r, the rate of interest plus 1, or the ratio of increase by compound interest ; n, the number of years, or other intervals, for which interest is taken ; and M, the amount at the end of n years ; and we have, in general terms, M=pr n . Semee, p = — , and. -iff 413. The compound interest gained in-n intervals is M—p, or pr"— p=p (*•"— 1). 414. The Computations of compound interest are most readily performed by the use of logarithms. Taking the logarithms of both members of the equation M=pi m , we have the useful formulas : — log. M= log. p + n log. r (1) log. p = log. M— n log. r (2) Jog.M-log.p n log. M—loq.p ... log. r LOGARITHMS. 255 EXMJtCISES. 415. To how much will $100 amount in 7 years, at 5% com- pound interest ? Solution. M=pr n = $100xl.05' 1 Log. 105 = 0.02119 7 Log. 105 1 ' = 0. Log. 100 = 2.00000 Log. 140.73=2.14333 2. What will be the amount of $1, at 7% compound interest, in 15 years? Am. $2.75 3. How much will be the amount of $10, at 6% compound interest, in 7 years 6 months, the interest payable semi-annu- ally? 4. How much will be the compound interest of $500 for 5 years, at 6 % ? Ans. $169.10 5. What sum of money at 6% compound interest will amouut to $150 in 10 years? Solution. M $150 P r n 1.06 10 - Log. 150 =2.17609 Log.l.06 w = .25310 Log. 83.75 = 1.92299 Ans. $83.75 6. What sum of money at 6% compound interest will amount to $1000 in 12 years ? 7. What sum of money at 5% compound interest will amount to $5000 in 50 years? Ans. $433.98 256 LOGARITHMS. 8. What sum of money at 7% compound interest will amount to $276 dollars in 15 years? Ans. $100. 9. At what rate must $100 be at compound interest to become $276 in 15 years ? Solution. \p) \ 100 Log. 276 = 844091 Log. 100 = 8.00000 IB) 44091 Log. 1.07 .08939 1.07-1 = .07, or 7%. Ans.7%. 10. At what rate must $10 be at compound interest to amount to $14.80 in 10 years ? Ans. 4