UC-NRLF *B 53E 325 .'■^W' LIBRARY OF THE University of California. OTF'T OK i:u^ P^^.a (^V^-^ Class -1>*--.m!s^:*"-jL .'^■%^^^ ■*U«'^ •♦•^v.:?-i-v.:<- 9 ^, •' m£ ■ ■^>■^^.•'•^'■tt■.• K..• fiOBINSON S MATHEMATICAL SERIES. I NEW ELEMENTARY ALGEBRA ' i OONTAIKINO THB RUDIMENTS OF THE SCIENCE. POE SCHOOLS AND ACADEMIES. | BY HORATIO K ROBINSON, LL. D., AUIHO& OF A FULL COU&SB OF KAIHEUATIC8. NEW YUKK IVISON, PHINNEY, BLAKEMAN & CO^ CHICAGO : S. C. GRIGGS & CO. 1866. RO B IN so N'S the most CoMPLETa,. jfe£a(/¥EAOTiOAL, and most Scientific Series of MATHEiiATicfciK Text-Books ever issued m this eouMn, ^1 ^*6^^£ Robinson's Progressive Table Sook, --•--. Robinson's Progressive Primary Arithmetic,- - - o Robinson's j::'rogressive Intellectual Arithmetic, - « - Robinson's Rudiments of "Written Arithmetic, - - - Robinson's Progressive Practical Arithmetic, - - - Robinson's Key to Practical Arithmetic, - - - - - Robinson's Progressive Higher Arithmetic, - - - - Robinson's Key to Higher Arithmetic, - - - - - Robinson's Arithmetical Examples, Robinson's 'Nqw ^Elementary Algebra, - - . « . Robinson's Key to Elementary Algebra, - - - - - Robinson's University Algebra, Robinson's 'KeY to University Algebra, - - - - - Robinson's I^ew University Algebra, - - - - - Robinson's Key to InTcw University Algebra, - - - - Robinson's IsTew Grcometry and Trigonometry, - - - Robinson's Surveying and K"avigation, Robinson's Analyt. Geometry and Conic Sections, Robinson's Bifferen. and Int. Calculus, (in preparatic i )- Robinson's Elementary Astronomy, Robinson's University Astronomy, Robinson's Mathematical Operations, - - - - „ Robinson's Key to Geometry and Trigonometry, Conio Sections and Analytical Geometry, Entered, according to Act of Congress, in the year 1859. bv HOKATIO N. ROBINSON, LL.D., In the Clerk's OflSce of the District Court of the United States for the Northern District of New York. PREFACE. Within the last twenty years Algebra has been steadily gainiRg ground and favor as an important branch of education, and it is now taught and studied in all the academies, semi- naries, and best public schools in our country. While this fact is indicative of the onward progress of popular education, it also bears testimony to the value of the science as one eminently calculated to discipline the mind and develop the f reasoning powers. Pupils now commence this study at an earlier age than formerly, and hence the necessity of a work elementary in its character, and adapted to the comprehension of the youthful mind. In the preparation of the following treatise the author has constantly kept in mind the existing condition of school and academic education, and has adapted the work to the most approved modern methods of teaching. The author believes this treatise to be superior to other elementary works upon the same subject in the following par- ticulars : beauty of typography, the clear and concise opera^ tions and a^ialyses of the rules and pr^no^p/e5, the great number of examples and their adaptation to the several subjectSy and the progressive character of the work^ so neces- sary to the vigorous development of the intellect. IV PREFACE. Particular attention is invited to tlie articles on Fractions, Simple and Higher Equations, Powers and Roots, and the analyses of the rules and principles, as it is claimed that in them will be found much that is new and valuable. The introductory chapter is adapted to give the pupil a cor- rect comprehension of the utility of symbols, and of the identity and chain of connection between Arithmetic and Algebra, in which the simplicity of Mental Algebra, and the spirit of the author's University Algebra are so blended, that the work can not fail to be a most useful and popular one. While the author takes great pleasure in acknowledging his obligations to several thorough and practical teachers for valuable hints and suggestions, both in theory and application, contributed to this book, he desires to make special mention of the valuable services rendered by J. C. Porter, A.M., in the preparation of this work, whose acknowledged ability as a mathematical scholar, and his long and successful experience as a teacher, ought to afford a sufficient guarantee for its practical character, and adap- tation to the purposes of teaching. August, 1859. CONTENTS. SECTIOISr 1. DEFINITIONS AND NOTATION. Th<* Signs ► page 7 I Definitions of Terms 23 Definitions of Algebraic Quantities.... 9 I Axioms 24 ENTIRE QUANTITIES. Addition 25 Subtraction 33 Multiplication 40 To Square a Binomial 47 To multiply the sum of two quantitiee by their difference 49 Division 63 General Principles of Division 62 Exact Division C4 Reciprocals, Zero Powers, and Nega- tive Exponents 65 Factoring Algebraic Quantities 67 Greatest Common Divisor 72 Least Common Multiple 74 FRACTIONS. Definitions of Terms and Signs 76 General Principles of Fractions 77 Reduction of Fractions 78 Addition 90 Subtraction 93 Multiplication 96 Division 102 Complex Fractions...* 105 SECTION II. SIMPLE EQUATIONS. Definitions 107 Transformations 108 Reduction 113 Proportion, and its relation to Equa- tions ". 122 Problems 124 Two Uriknowu Quantities 139 1* Eliminatioa 140 Problems with two Unknown Quan- " titie.s 148 Three Unknown Quantities 153 Problems with three or more Un- known Quantities 157 Negative Results 161 VI CONTENTS. SECTION III. POWERS AND ROOTS. fnvolution 163 Powers of Monomials 164 Powers of Fractions 166 Powers of a Binomial 168 Newton's Binomial Theorem 171 French's Theorem 174 Evolution 181 Roots of Monomials 181 Square Root of Polynomials 184 Square Root of Numbers ]87 Cube Root of Polynomials 191 Cube Root of Numbers 195 Contracted Method 199 Reduction of Radicals 203 Addition of Radicals 208 Subtraction of Radicals 209 Multiplication of Radicals 210 Division of Radicals 213 Principles relating to the application of Involution and Evolution 214 Simple Equations containing Radical Quantities 21? SECTION lY. QUADRATIC EQUATIONS- Definitions 220 Pure Quadratics 220 problems producing Pure Quadratics. 223 Affected Quadratics, completing the Square 228 Second Method of completing the Square 232 Higher Equations in the Quadratic Form 236 Polynomials under the Quadratic Form 238 Formation of Quadratic Equations, First Method 239 Second Method 241 Factoring Trinomials 243 Rfc)al and Imaginary Roots 244 Problems producing Quadratic Equa- tions 245 Quadratic Equations containing two Unknown Quantities 247 Problems producing Quadratics with two Unknown Quantities 2;'''2 SECTION Y. PROGRESSIONS AND PROPORTIONS. Arithmetical Progression 255 General Applications 268 Problems 260 SpeciUl Applications 261 Geometrical Progression 205 In-lnite Series 268 Geometrical Means 270 Applications < 272 Problems 272 Special Applications .t... 27^ V PROPORTION. Definitions „ 277 General Principlefl, in Twelve Propositions 279 Problems 286 Approximate Roots op ITigher Deqeees 289 Miscellaneous Examples 297, \ ELEMENTAEY ALGEBRA. SECTION 1. DEFINITIONS AND NOTATION. 1. Guantity is anything that can be measured or compared ; as distance, space, motion, time. 3. Mathematics is the science which treats of the relations of quantities. 3. Algebra is that branch of mathematics in which the ope- rations are indicated by signs or symbols, and the quantities are represented by letters. It is universal arithmetic. THE SIGNS. 4. Addition is denoted by the perpendicular cross, +, called plus ; thus, in 4 + 2 -f 9, the sign indicates that 4, 2, and 9 are to be added. ^. Subtraction is denoted by the horizontal line, — , called minus ; thus, in 10 — 7, the sign indicates that 7, the number after it, is to be subtracted from 10, the number before it. 5. Multiplication is denoted by the oblique cross, X ; tbus^ in 5 X 4, the sign indicates that 5 and 4 are to be multiplied together. '^. Division is denoted by a horizontal line, with a point above and one below, —- ; thus, in 18 -f- 6, the sign indicates Pefine quantity. Mathematics. Algebra. Explain the sign of Addi- tion, Of Subtraction. Multiplication. Division (7) 8 ALGEBRAIC QUANTITIES. that 18, the number before it, is to be divided by 6, the number after it. The horizontal line without the points becomes the sign of division when the dividend is written above and the 18 divisor below it; thus, -^ indicates division, the same as 18 -f- 6. 8* Equality is denoted by two horizontal parallels, =, which represent the words, equal to ; thus, in 4 + 8 = 7 + 5; the parallels indicate the equality of the two sets of numbers com- pared ; and the expression is read, 4 plus 8 are equal to 7 plus 5. 0, Inequality is denoted by the angle, ^, the opening always being toward the larger number or quantity ; thus, in 12 -f 7 > 14, the sign, ^, indicates that the sum of 12 and 7 is greater than 14, and the whole expression is read, 12 plus 7 is greater than 14. The expression 6 <^ 4 -f 7 is read, 6 is less than 4 plus 7. 10. A parenthesis, ( ), denotes that the several numbers or quantities included within it are to be considered together, and subjected to the same operation ; thus, (10 -f 4) x 3 indi- cates that both 10 and 4, or their sum, is to be multiplied by 8 ; (10 — 4) X 3 indicates that the difference of 10 and 4 is to be multiplied by 3. 11. A horizontal vinculum, , placed over the num- bers or quantities, is frequently used instead of the parenthe- Bis ; thus, 4+ 2-^3 X 7 is equivalent to (4 + 2 + 3) x 7. 12. The radical sign, V , indicates that the root of the quantity placed under it is to be taken. EXAMPLES. 1. 4-1-7 — 6 =^ how many ? Arts. 2, 36 — 5 — 20 -f 1 = how many ? Ans, 3. — ^p— = how many ? Ans. 3, 5. Ans, 12. Explain the sign of Equality. Inequality. Explain the use of the Parenthesis. Of the Vinculum. The Radical Sign. DEFIi;iTIONS AND NOTATION. Q A^ Q I 2 4. ~ — : X 7 = how many ? Ans, 98, 72 — 2 5 X IS , „ .IP 6. — ^ — I irr — = now many r -4ns. lb. . 86— (8 x2) 7 + 21 , , , ^ I). ^ — — = now many r Ans. 2. 7. (4 + 6) X 12 = how many ? Jws. 120. 8. Show that (75 — 25) -^ 25 = 2. o Qi. .1 100 + 30 — 70 ^ 9. Show that -^ = 1. oO 10. Show that (^?i:^?/— 6) x 5 = 25. 11. Show that (17 + 4 — 11) x 6 > 50. 12. Show that (?^I^-^ ^ 6) + 18 (16 — 12 + 10) X 5 — 67, 1674 — 1569 ^/1765 — 1653^ 7' 14. Show that ;:; «« ^^l 1«'«;,?'. ^"^ ^' ''/irf^f by Sx. By addition, we find that the Jx = lUo. g^j^ q£ ^Yie two numbers is equal to 9aj. X = 12, less ; But by the conditions of the example. Sx = 9C, greater. the sum of the two numbers is equal to 108 ; hence, 9a; is equal to 108 ; and x, the less number, must be J of 108, or 12; and 8a;, the greater number, must be 8 times 12, or 96. 4. The greater of two numbers is 6 times the less, and tbeir sum is 147 ; what are the numbers ? Ans, Less number, 21 ; greater number, 126. 5. A and B together had $100, and B had 3 times as much money as A ; how many dollars had each ? Ans, A, $25 ; B, $75. 6. A man divided 90 cents between two beggars, giving the second four times as much as the first ; what did he give to each ? Ans. First, 18 cents ; second, 72 cents. 7. Two men, A and B, trade in company with a joint capital of $900, of which B put in five times as much as A ; how much did each man put in ? Aiis. A's $150 ; B's $750. 8. In a mixture of 720 bushels of grain, there is three times as much corn as wheat ; how many bushels of each ? ^^^ (Wheat, 180 bu. (Corn, 540 bu. 9. I expended $12570 in the purchase of a house and lot, and the house cost twice as much as the lot ; what did I give forelch? . |Lot, $1190. ' (House, $8380. 10. A and B engage in trade with joint capital, of which B owns four times as much as A. They gain $7500 ; what is each man's share? . . (A's share, $1500. ^^* IB's share, $6000 SO. 1. A man paid 24 dollars for a hat, vest, and coat. The vest cost twice as much as the hat, and the coat cost three times as much as the hat ; what was the pric-e of each ? DEFINITIONS AND NOTATION. Jg OPERATION. Analysis. We represent the Let X = price of the hat ; price of the hat by aj ; as the vest 2.x = price of the vest; ^^^^ ^^^^^ ^^ ^"^^ ^^ *^^^ ^'^^' -^ Sx = price of the coat. '''^^ represent the price of the r; f- vest; and as the coat cost three 6x = 24 dollars. ^-jj^^g ^^ ^^^^^ ^^g ^^^ ^^^^ 3^ ^• X = 4, nat ; represent the price of the coat, 2x = 8, vest ; Since x and 2x and 3a: are equa' Sx = 12, coat, to 6x, the -whole suit cost Gac, But by the conditions of the ex- ample, the whole suit cost 24 dollars ; hence, 6a: are equal to 24 dol- lars ; and a;, the price of the hat, must be | of 24 dollars, or 4 dol- lars ; and 2x, the price of the vest, must be 2 times 4 dollars, or 8 dollars ; and 3x, the price of the coat, must be 3 times 4 dollars, or 12 dollars. 2. A purse of 108 dollars is divided among three men ; A takes a certain sum, B takes three times as much as A, and C takes five times as much as A ; what is each man's share ? {A's, 12 dollars. B's, 36 dollars. C's, 60 dollars. 6. A man, dying, bequeathed $30,000 to his wife, son, and daughter. The will provided that the son should receive twice as much as the daughter, and the wife three times as much as the daughter ; what was the share of each ? r Daughter, $5,000. Ans.J Son, $10,000. (wife, $15,000. 4. Divile the number 91 into three such parts that the second shall be five times the first, and the third seven times ^'^^e first. (First part, 7. AnsJ Second part, 35. (Third part, 49. 5. Divide the number 96 into four such parts that the second 2 Ans. 14 ALGEBRAIC QUANTITIES. shall be three times the first, the third five times the first, and the fourth seven times the first. / First, 6, . ] Second, 18. ^""'■^ Third, 30. Fourth, 42. 6. A farmer purchased a certain number of oxen, three times as many cows, and ten times as many sheep. There were 112 in all; required the number of each kind. TOxen, 8. • Ans.) Cows, 24. [ Sheep, 80. 7. A tax of $936 is assessed upon four persons, according to the relative values of their property. B is worth three times as much as A, C four times as much as A, and D five times as much as A ; what is each man's tax ? / A's, $72. , B's, $216. C's, $288. D's, $360. 8. A man traveled a certain distance on Monday, twice as far on Tuesday, three times as far on Wednesday, and so on till Saturday. The whole week's journey was 504 miles ; what was the distance traveled on Monday ? Ans. 24 miles. S7. 1. Divide the number 72 into three such parts that the second part shall be twice the first, and the third part three times the second. OPERATION. Analysis. We represent the I ot X — first part • ^^^^ P^^*^ ^^ ^ ' ^^ *^^® second part 2x = second part ; jf '^"««, f'^ Srst part, it will be ^ +1 • -I + 2a;; and as the third part is three ba; = thir d part ^j^^^^^ ^j^^ ^^^^^^^ ^^^,^^ j^ ^^,jjj ^^ 9^ = 72. three times 2x, or 6a;. The sum OC = S, first part ; of all the parts, 6a: and 2a; and x, 2x = 16, second part; is 9^, which must be equal to 72, Qx = 48 third Dart ^^^ whole number; hence x, the first part, is I of 72, or 8 ; 2a;, the second part, i? 2 times 8, or 16 ; and 6a;, the third part, i;? 6 times 8, or 48. ' . DEFINITIONS AND NOTATION. 15 2. Three gentlemen contributed $500 for a charitable object. A gave a certain sum, B gave three times as much as A, and C gave twice as much as B ; what did each contribute ? rA, $50. A71S. 3 B, $150. [ C, $:J00. 3. An orchard contains four times as many cherry-trees as pear-trees, and twice as many peach-trees as cherry-trees ; the whole number of trees in the orchard is 156 ; what is the num- ber of each kind ? C Pear, 12. Ans. < Cherry, 48. ^t [Peacli, 96. 4. Pivide the number 147 into three such parts that tlie second part shall be five times the first, and the third part three times the second. ( First part, 7. Ans. < Second part, 35. I Third part, 105. 5. A man performed a journey of 624 miles.. He traveled twice as far by railroad as by stage, and five times as far by steamboat as by railroj^d ; how many miles did he travel by steamboat? ' Ans. 480 miles. 6. A man cancelled a debt of $873 by paying a certain sum on Monday, twice that sum on Tuesday, three times Tuesday's payment on Wednesday, four times Yi^ednesday's payment on Thursday, and so on till Saturday ; what sum was paid on Monday? A^ns. $1. 7. A mere]] ant gnined three times as much by his business the second year as the first, as much the third year as tlie second, and twice as much the fourth year as the third. Ilia entire gain was $9750 ; wliat did he gain the fourth year ? Ans. $15u0. @8. 1 Divide the number 35 into three such parts that the second shall be four times tlic first, and the third one lialf the second. 16 ALGEBRAIC QUANTITIES. Let OPERATION. X = first part ; 4j7 = second part ; 2x = third part. 7a; == 85. X = b, first part ; 4:x = 20, second part ; 2x = 10, third part. Analysis. We represent the first part hy a; ; as the second part is 4 times the first, it will be 4a; ; and as the third part is one half the second, it will be one-half of Ax, or 2x, The sum of all the parts, 2x and Ax and x, is Ix, which must be equal to 35, the whole number ; hence x, the first part, is 4 of 35, or 5 ; Ax, the second part, is 4 times 5, or 20; and 2a;, the third part, is 2 times 5, or 10. 2. A, B, and C, together have 104 dollars, B has nine times as much as A, and C has one third as much as B ; what num- ber of dollars has each I TA, $8. Ans. \ B, 72. [C, 24. 8. There are three trees which together bear 32 bushels of apples. The second bears twelve times as many as the first, and the third one fourth as many as the second ; ho\Y many bushels does the first bear ? Ans. 2 bushels. 4. The sum of four numbers is 510. The second is six times the first, the third is three times the second, and the fourth is one half the third ; what is the fourth number ? Ans. 135. 5. Four men together are taxed 480 dollars. B's tax is four times A's, C's tax is six times B's, and D's tax is one eighth of C's ; what is C's tax ? Ans, 360 dollars. SS, 1. John had a certain number of marbles, James had three times as many as John, and William had as many as both John and James ; they all had 64 ; how many had each ? Analysis. We let x repre- sent the number of marble John had ; as Janies had three times as many, 3a; will represent his number; and as William Let OPERATION. 57 = J ohn's number ; ox = James's number ; 4:x = William's number. Sx = 64. X = S, John's ; i^x = 24, James's ; Aujc «= 32, William's had as many as both John and James, x added to 3a;, or 4a;, will represent William's num ber. Axy and 3x, and a;, are 8a;, which represents the numboi DEFINITIONS AND NOTATION. 17 they all had but by the conditions of the problem, they all had 64 ; hence, 8a; are equal to 64 ; and x is J of 64, or 8 ; 3a; is three times 8, or 24 ; and 4a; is four times 8, or 32. 2. Divide the number 100 into three such parts, that the second part shall be four times the first, and the third part as much as the sum of the first and second. f First, 10. AnsJ Second, 40. (Third, 50. 3. Divide the number 105 into three such parts, that the second part shall be four times the first, and the third part twice the sum of the first and second. ("First, 7. Ans.J Second, 28. (Third, 70. 4. Four men together contribute $5250 to build a parson- age. A gives a certain sum, B gives three times as much as A, C gives three times as much as A and B together, and D gives one third as much as B and C ; how much does A con- tribute ? Ans. $250. 5. A man gave $324 for a horse, carriage, and harness. The horse cost five times as much as the harness, and the car- riage cost one half as much as both horse and harness ; what was the price of each ? C Harness, $36. Ans. } Horse, $180. [ Carriage, $108. 6. Divide the number 1008 into three such parts, that the second part shall be nine times the first, and the third part one fifth of the sum of the other two. C First, 84. Ans. } Second, 756. [ Third, 168. 7. What number is that, which being multiplied by 7, and the product added to the number, the sum, product, and given number will together be equal to 80 ? Ans. 5. 8. A grocer has four kinds of weights. One of the second kind will balance two of the first, one of the third kind will balance two of the second, one of the fourth kird will ba- 2* " B 18 ALGEBRAIC QUANTITIES. lance two of the third, and one of each kind, placed together in the scales, will balance 15 pounds ; what are the denomina- tions of these weights ? Ans. 1 pound, 2 pounds, 4 pounds, and 8 pounds 30. 1. There are three numbers, whose sum is 96. The second is four times the first, and the third is equal to the first subtracted from the second ; what are the numbers ? ^^^^ . r^^rs^T Analysis. We represent the first by OPERATION. . ^, 1 - A ,' .^ n : x; since the second is 4 times the first, XjCj X =^ nrst ; it will be 4x ; and as the third is equal Ax = second ; , to the first subtracted from the second, 8^ = third. it will be 4x minus x, or 3a;. The sum g^ __ 9Q . of all the numbers, 3a;, 4a;, ?nd x, is 8a;, a; = 12 first • which is equal to 96 ; hence x, the first ix - 48,' second ; °""''^f '. '' f. '^^ ^^' °' }^ ' ^' *« f,^ ,, . , second, is 4 times 12, or 48; and 3a;, 3.r = ob, third. xi ..i • ^ • o i.- to o^ ^ the third, is 3 times 12, or 36. 2. The sum of three numbers is 108. The first multiplied by 7 will give the second, and three times the first taken ^rom the second wdll leave the third ; what are the numbers ? Ans. First, 9 ; second, 63 ; third, 86. 3. John has a certain number of marbles, James has five times as many as John, Henry has as many as twice John's subtracted from James's, and Henry's added to John's are equal to 20 ; how many has each ? Ans. John, 5 ; James, 25 ; Henry, 15. 4. Divide the number 8488 into three such parts, that the second part shall be five times the first, and the third part one half the difference of the first and second. .4ns. First, 436 ; second, 2180 ; third, 872 5. The difference of two numbers is 12 times the less num ber, and 5 times the less number subtracted from the greatei is equal tc 120 ; what are the numbers ? Ans {^^^^' ^^' I Greater, 195. DEFINITIONS AND NOTATION. 19 31, 1. A bookseller being asked tlie value of two pencils, answered that the second cost twice as much as the first, and that his price mark, a, would represent the cost of the two ; what was the cost of each ? OPERATION. Analysis. We let x repre- Let w - cost of the first ; ««'^' *''« ''"^' "* '^'^/''^^ ?«"«'•' o J. X? ±\. 1 and, as the second cost twice zx - cost of the second. 1,0 -n v ^i as much, Zx ■will be the cost da; =. a of the second. The cost of a r. ^i , both is 2x added to ar, which 3' ^ ' is 3x; but by the conditions of 2^ the example the cost of both 2x = -^i second -was a ; hence 3a; are equal to a ; x, the cost of the first, is J of a, or -- ; and 2x is two times ^-, or ^-. o ' 00 Note. — The expression — - indicates the division of the number a by o 3, and is read, a divided by 8. 2. If in the example above, the letter a represent 9 cents, how many cents is each pencil worth ? Ans, First, 3 cents ; second, 6 cents. 8. A certain number, represented by the letter c, is divided into two such parts that the greater is four times tlie less ; what are the parts 1 I t^ c * Less, Ans. 5* 4c Greater, - 5 4 Divide the number m into three such parts, that the second part shall be twice the first, and the third part three times the first, / -r^. . '^^^ First, -— D 2m Ans, { Second, -77-. ' 6 ^ Third, -^. 5. Divide the number n into four such pan», that the second part shall be twice the first, the third part as mucl\ as the first Ans. . I Brandy, 20 ALGEBRAIC QUANTITIES. and second; and the fourth part as much as the first, second^ and third. *" Ans. First, ^ ; second, ^^ ; third, ^-^ ; fourth, :r^. 6. B is three times as old as A, and C is four times as old as A ; the sum of their ages is d ; what is the age of C ? Ans. -^-, 7. A cask which held b gallons was filled with a mixture of brandy and water, and there was ten times as much brandy as water; how much was there of each ? j Water, ^ ; 106 11' S2, The following examples do not require for solution the use of equations ; they are given to exercise the pupil in al- gebraic notation, and embrace only given relations of known quantities. 1. If the number of weeks in a year be represented by m, what will express the number of days ? Analysis. There are 7 days in one week, and in m weeks there must be m times 7 days, or 7m days, the answer. 2. A man labored b days at c dollars a day ; how much wages did he earn ? Ans. be dollars. 3. John had m m.arbles, which he sold for m cents apiece ; now many cents did he receive for them ? Ans. vf cents. 4. An orchard contains b rows, in each row are c trees, and each tree bears d bushels of apples ; how many bushels of apples does the orchard produce ? Ans. bed. 5 The height of a rectangular box is h, the breadth is b, and the length is I ; what are the solid contents ? A71S. hbl 6. John has m marbles, James has n times as many as John, and William has n times as many as James ; how many marbles has William ? Ans, mn^. DEFINITIONS AND NOTATION. 21 7. A prize v^^as divided among three men. The first mao received a dollars, the second man h dollars, and the third man c times as many dollars as the other two ; how many dollars did the third man receive ? An^. (a -J- ^)^- 8. A man purchased three books. For the first he gave a dollars, for the second h times as many dollars as for the first, and for the third c times as many dollars as for the other two j vfhat did he give for the third ? An^, {a + ab)G. 9. A man started in business with a capital of c dollars. The first year he doubled his money, the second year he gained b times his first capital, the third year he lost d dollars, and dying, left his property to n children; how much did each receive ? . 2g -{• he — d Ans, . n 10. Four men having joint partnership in a nursery, sold m rows with m trees in a row, at m cents apiece ; the trees were packed in h bunches, and delivered at an expense of h cents a bunch ; what did each man realize by the sale ? m^ — li^ Ans. — J — . 33. To find the numerical value of an algebraic quantity, -vhen the literal factors represent known numbers, we must substitute the given numbers for the letters, and perform upon them the operations indicated. Til^ — — b^ 1. What is the numerical value of — j — , when m = 40, and 6 = 8? OPERATION. Analysis. By m»-- 6' _ 40 X 40 -- 8 X 8 __ the conditions of ~J — J — ^'^'* the example, 7h^ is equal to 40 X 40, or 1600 ; S^ is equal to 8 X 8, or 64 ; and 1600, less 64, is 1536, which, divided by 4, gives 384, the answer. Find the numerical values of the following algebraic expres- sions, in each of which a = 12 5C=«10;m = 8;n = 5 22 ALGEBRAIC QUANTITIES. 2 ao — m\ 3. (a + c)m. 4. {a?—-am)n. 5. (a + G -{■ m + n)am. 6 a^ — mn 4 • 7. am + 'inhi — 2m mn (a^ — cmn)m a Ans. 56. Ans. 176. Ans. 240. .4ns. 3360. Ans. 26. Jns. 10. Ans. 885^. 9. ; ^-. Ans. 32. (a + c + m)7i 10. 7^^ — ^ — , — -^-— . ^ws. 6. (a + c)?i 4- en 11. ('^-^ + ^-"^)«\ ^„s. 27. 7ac~-m'^ 12. -. Ans. di. ba — nv 13. 4 (3a -f 2m). Ans. 208., ^ , 5m(a^ 4- 'rnc — 14m) . ^ 14. — ^^ — :; ^. Ans. 7. 15. — r, -, — 5 ^^^s. 1775. m^ — 2mn + n ^ 16. ac ^^ -^ Ans. 113. a ^^ am 3a >i 10 17. 1 . Ans. 12 a + m 71 18. ?£(^J!^. Jns. 456. Ibn 19. /l!!^_^)^i. ^ns. 16. \ n 0/ 20. (a + c) (m + n). ^Itis. 286. 21. -^^- n. ^ns. 15. m 4- 2n 22. fl2 _|. ^2 — c' — n». ^n8.. 83. TERMS. 23 TERMS. 34« The Terms of an algebraic quantity are the divisions made by the signs, + and — ; thus, in the quantity oa+¥ — mx^ there are three terms, of which 3a is the first, + ¥ is the second^ and — 7nx is the third. 3^. Positive Terms are those which have the plus sign ; as, + a or + b'^d. The first term of an algebraic quantity, if written without any sign, is positive, the sign + being under* stood. 36. Negative Terms are those which have the minus sign ; as, — 2a, or — 3c^d The sign of a negative quantity is never omitted. 37. Similar Terms are terms containing the same letters, affected with the same exponents ; the signs and coefficients may differ, and the terms still be similar. Thus, Ca^ and' 5a^ are similar terms ; Ih'd and — bhi^d are similar terms. 38. Dissimilar Terms are those which have different letters or exponents ; thus, ahc and acd are dissimilar terms ; ax^y^ and d^xy are dissimilar terms. 39. A Monomial is an algebraic quantity consisting of only one term ; as, 4a, ocd, or *lh^x. 4©. A Polynomial is an algebraic quantity consisting of more than one term ; a + 6 ; or odb — 2x -\- c. 41. A Binomial is a polynomial of two terms ; as a + c, or 2.x — y. 4S. A Eesidual is a binomial, the two terms of which are connected by the minus sign ; as, a — b, or ilx — 2y. 43. A Trinomial is a polynomial of three terms ; as x -{- y -{- z, or 2a — 26 -f c\ 44. The Degree of a terra is the number of literal factors it contains, and is found by adding the exponents of the several letters; thus, a and S6 are terms of the first degree; a^ and Define the terms of an algebraic quantity. Positive terms. Negative terms. Similar terms. Dissimilar terms. A Monomial. A Polynomial. A Binomial. A Residual. The Degree of a term. 24 ALGEBUAIC QUANTITIES. 2ab are terms of the second degree ; a^ Sa-b, and bahc ard terms of the third degree. 45. A Homogeneous quantity is one whose terms are all of the same degree ; as, oc^ — 8a;^2/ + xyz, AXIOMS. 46. An Axiom is a self-evident truth. The principles of all algebraic operations are based upon the following axioms : 1. If the same quantity or equal quantities be added t:> equal quantities, their sums will be equal. 2. If the same quantity or equal quantities be subtracted from equal quantities, the remainders will be equal. 3. If equal quantities be multiplied by the same, or equal quantities, the products will be equal. 4. If equal quantities be divided by the same, or by equal quantities, the quotients will be equal. 5. If the same quantity be both added to and subtracted- from another, the value of the latter will not be altered. 6. If a quantity be both multiplied and divided by an- other, the value of the former will not be altered. 7. Quantities which are respectively equal to any other quantity are equal to each other. 8. Like powers of equal quantities are equal. 9. Like roots of equal quantities are equal. 10. The whole of any quantity is greater than any of ite parts. 11. The whole of any quantity is equal to the sum of all its parts. Define a Homogeneous Quantity. An Axiom. Repeat the Axioms giveiL ADDITION. 25 ADDITION. 47. Addition, in Algebra, is the process of uniting iwo or more quantities into one equivalent expression, called their sum. Since, in algebra, the quantities to be added may be either positive or negative, it is necessary to consider here more fully the nature of the signs + and — . Thus far they have been employed to indicate simply the opposite processes of addition and subtraction. They have, however, a wider significance, and indicate not only operations to be performed^ but the quality, or relative character of the quantities to which they are applied. They may denote opposite directions in space, opposite effects in nature, or opposite results in business. Thus, if plus indicate direction north, minus will indicate direction south ; if plus indicate heat, minus will indicate cold ; and if plus indicate gain, minus will indicate loss. CASE I. 48. To add similar terms. 1. A cooper made 7 barrels on Monday, 9 barrels on Tues- day, and 6 barrels on Wednesday ; how many barrels did hi j make in the three days ? AKITHMETICALLY. 7 barrels. . 9 barrels. 6 barrels. Ans, 22 barrels. Define Addition. Explain the nature of the signs -f and — . What is Case I ? Give Analysis. 3 26 ENTIRE QUANTITIES. ALGEBRAICALLY. Analysis. We represent 1 barrel b;^ fche fjy letter b ; then 76 will represent 7 barrels, 96 will represent 9 barrels, and 66 will represent ^ 6 barrels ; and since 7 barrels, 9 barrels, and ^ 6 barrels are 22 barrels, 76, 96, and 66 are 22b 226. 2. A mass of iron and wood is submerged in water by its own gravity. The iron tends to sink with a force of 20 Z, and the wood buoys upward with a force of 16 Z ; what will the whole mass weigh while under water ? ^■.^^,> * m-r^T^r Analysis. Wb Indlcate actual weight by the OPERATION. , . , , . /» , , plus sign, and the opposite force, or buoyancy, by • the minus sign. Since the tendency to sink is 'ZL—L greater by 4Z than the tendency to rise, the mass 4- 4Z has a weight of 4Z ; hence, + 20Z and — 16Z united are + 4Z. Note. — The answer, -}- 4Z, in the above example, is called the alge- braic sum of the two forces, + 20Z and — 16Z, because it shows their united effect, 3. A ship started at the equator and sailed the first day 16 miles north, the second day 20 miles south, the third day 8 miles north, and the fourth day 7 miles south ; how far from the equator, and in what latitude, was the ship at the end of the four days ? FIRST OPERATION ANALYSIS. Weletwrepre- j^ l^T^ — 20m sent 1 mile ; and to distinguish I 3^ 7^ the directions, we indicate dis- -— ^-j jj=— tance north by the plus sign, + 24m-27m = -3m and distance south by the minus sign, writing the positive terms in one column and the nega* tive terms in another column. The whole distance sailed north, is + ICm and + 8m, which is + 24m ; and the whole distance sailed south, is — 20m and — 7m, which is — 27m ; and, as 27 is 3 more than 24, the ship must be three miles south of the equator, ex pressed algebraically thus, — 3m. Explain the diflference between Arithmetical and Algebraic Addition, S8 shown in example 3. ADDITION. 27 SECOND OPERATION. Analysis. In the second operation we + 16m write all the terms in one column, since — 20m they are similar. + 16m and -f 8m are 4- 8m + 24m ; and — 20m and — 7m are — 27m ; __ 7^ and + 24m — 27m are — 3m, the algebraic o — sum of the quantities in the column. Note. — There is a dis%ction between arithmetical and algebraic addi- tion. In the above exanl^^le had the question been, how far the ship sailed, the answer would be 16 -|- 20 + 8 -|- 7 = 51 miles, or 51 w, which is the arithmetical sum of the distances sailed. But the real question is, what distance^ north or south, did tJie ship make from the point of starting ; and the answer is 3 miles south, or — Sw, which is the algebraic sum of the distances sailed. Hence, adding, in algebra, does not always aug- ment. Positive and negative quantities represent things opposite in kind or quality, and, if similar in denomination^ are added or united, by apparent subtraction. From these examples and illustrations we derive the fol- lowing Rule. I. Wlien the signs are alike, add the coefficients, and prefix the sum with its proper sign to the common literal part, II. When the signs are unlike, find the sum of the posi- tive and of the negative coefficients separately, and prefix the difference of the two sums with the sign of the greater, to the common literal part, EXAMPLES FOR PRACTICE. (4.) (5.) (6.) (7.) (8.) 3a 2m' — Sbx — 4a'6o -h 7cd^ 9a 6m' — 56a; — 5a'6c + Scd:" 5a 5m' — 46a; — 12a'6o + 2cd^ 12a 10m' — 26a; — a^6c + cd^ a 5m' — 76a; — 14a'6c + Q^cd:" 2a 7m' — bx — 2a'6c + 4c(? 32a t^^ — 226a; — - ■ " ^ ■ ■ + 23ceP Give the rule for the algebraic addition of similar terms. 28 ENTIRE QUANTITIES. (9.) (10.) (11.) (12.) (13.) — 5a + 3aa?^ + Sar" — 5a' + 36y + 4a + 4aa7^ — 6ar' — Ida? + 96y + 6a — '^ax' — IQaf 4-lOa^ — i06y — 3a — ^ax" + 3x' + Ua? _ i9&y + a 4-5aa72 + 2a? + Qa' _ 26y + oa — 2ax^ — ^a? ^15.' — i96y (14.) (15.) (16.) (17.) bh'd 2mn — 25a'6o 147z» •^Wd 4:mn ZWbo — 25z» Wd — . ' I2mn — na^ho 123» — Wd 16mw 48a'6o — 143» 18. What is the sum of 57?i, 7w, 11m, m, and 12m? "t -4ns. 36m. 19. What is the sum of 4a, 7a, 6a, and 10a ? ^ns. 27a. 20. What is the sum of — 5c^ l^c\ — 28c^ and — c" ? Ans. — 24c\ 21. What is the sum of — 1256e, — 1686c, and 26c? Ans. —2916c. 22. What is the sum of Sxy, — 4^i/, lOxy^ and — 7xy ? Alls, 2xy. 23. What is the sum of 9a'5c, 3a'6c, — 8a'6c, — 2a-6c, 5a*^6c, and — 15a^6c ? Ans. — 8a^6c. 24. Add dni^Xf — 2m^Xf — 7m^a7, and 28m^5?. Ans. 24m^x, 25. Add 7^y, — 15:ry, + 12a;^?/^ — 6xy, and xy, Ans, — xy. In the same manner similar quantities, of whatever kind, may be added by taking the algebraic sum of their coeffi- cients. The pupil will observe that, since 2 times any num- ADDITION. 29 ber whatever, added to three times the same number, are 5 times that number, so 2(a + ^), added to 3(a -f 6), are 5(a + &). (26.) (27.) (28.) (29.) ^c-x) (x+y) _4(2a-6) 4(ar — 2/ + 3) 7(c — x-) — 3(a; + 2/) _7(2a — 6) 7(a;— y+3) lOic — x) 20(x4-y) 8(2a_6) 18(a;#>i/) _3(2a — 6) — 12Cx—y + 3) n(c — x) — ix-y + B) (30.) (31.) (32.) 8a(a + 6) 7(Qx + y — zy 4(62/ + &) 7a(o + b) -SiQx + y — zy — 3(6y + 6) — 5a(a + &) _ 2(6x4- 3/ — 2;)' 7(62/ + 6) Sa(a + 6) 3(6a; + 2/ — z)'' -2(62/ + 6) 83. What is the sum of S(z — m), b(^z — m), — 12(z — m), _- 14(z — m), and 10(2 — m) ? ^ns. — 8(2 — m). 84. What is the sum of — 4(a -r 2&)^ 5(a 4- 26)^ ^ 12(a + 2by, and 20(a + 2by ? Ans. 9(a + 26)1 85. What is the sum of {x + 1), 5(x + 1), 8(0? + 1), and — 8(^ + 1)? Jws. (x + 1). CASE n. 49. To add polynomials. It is evident that dissimilar terms may be added by writing them one after another, connected by their proper signs ; but, if in the same polynomial, or in the different polynomials to be added, there be similar terms, these may be united by Case I, and a reduced expression obtained ; and it is immaterial in wliat order the terms are written in the aggregate sum, since the whole is equal to the sum of all its parts, in whatever order the parts are taken, (Ax. 11). 1. Add Ba + 2hc, 4a — 7bc + m, and x^ — 2a + 36c. What is Case II? 8* 90 ENTIRE QUANTITIES. Analysis. We write 3a, 4a, OPERATION. ^^^ — 2a, in one column, be- 8a + 2bc cause they are Bimilar terms, and 4^ *jIq ^ yfl 2bc, — 7dc, and 36c, in another « 9/7 4_ ^h/» column, for the same reason ; andx* _— and m not being similar to each a;* + 5a — 26c + m other, or to any of the other terms, we write them in separate columns. Commencing at the left, we write x^ in the>«um ; then — 2a, -}- 4a, and -f 3a are + 5a, which we write under the column added ; and •+• 36c, — 76c, and + 26c, are — 26c, which we write under the column added ; finally annexing + m, we obtain the entire sum, a:* + 5a — 26c + m. From this example we deduce the following Rule. I. Write similar terms in the same column, forming as many columns as there are dissimilar terms in the given quantities. 11. Add each column, as in case J, and connect the results by their proper signs EXAMPLES FOR PRACTICE. (2.) (3.) (4.) Sa^ — Acd 4:x'y— a'b' 5ab' — 7 do ^x' — Scd Sx"y+ 9a'b' — m 7ab' + Udc ~ 5a?' -f ^cd z — bx'y — 12a'b' —12ab^— Qdc ^x' — Scd z + 2x'y— 4a*'6' — m + dc 5. What is the sum of 6a6 + 12bc—Scd, Scd — 7ab — 9bc, and 12cd — 2a6 — 56(? ? Ans, 7cd — Sab — 26c. 6. What is the sum of 96* — 3ac + d, W + 7d — Aac, M — 46^ + 6ac, 56' — 2ac — 12d, and W — d? Ans. lSb' — Sac — 2d, 7. What is the sum of 7a6 — m^ + q, — 4a6 — 5m' — 3g, 12a6 + 14m' — z, and — 6m' — 2q ? Ans. 15a6 4- 2m' — 4^ — z. Give analysis. Rule. ADDITION. gj 8. What is the sum of Ga; — 5& + a -f 8, and — 6a -—i 407 + 46 — 3? Ans. 2x — b — 4a-\-5. 9. What is the sum of a + 26 — Sc — lO, 36 — 4a + 5(? + 10, and 56 — c ? Ans. — 3a + 106 + c. 10. What is the sum of 3a -f h — 10, c — d — a, and — 4c 4- 2a — 36 — 7? Ans, 4a — 26 — 3c — (i_17. 11. Add 15a' — 86^c? + 32aV — 126c, 196V — 4a^ + llaV 4- 26c, a'^ — 29aV — 126'^c + 56c, and 9aV — 146c -f- b\\ Ans. 12a'' + 23aV — 196c. 12. Add 5a^6' — 8a26' + x'y + xy\ 4a'6« — 7a«6' — Sxy' + Qx'y, Sa'b' + Ba'b' — Bx'y + dxy\ and 2a26« — a«6^ — 3a;V— Sxy\ Ans. a^b^ + x^y. 13. Add nax' — ^ay", »_38a.77^ — 3ai/* + laf, 8 + 12ai/*, — ^ay^ + 12, and — 34aa;* + bay^ — 9ay^. Ans. — 2a2/» + 20. 14. Add 7x^ — 5c^ + 14m^, — Sx^ -f icx — 17m5r — pq^ 4j?^ -f 12mg + ^pq — z, 2cx — 7mg — 2pq, and 3a;'' — 2ca; — mg — 4pg -f- 32. Ans. llo?'' — ex + mg — 4pq + 2z. 15. Add 7m + 3n — lip, 3a — 9n — 11m, Sn — 4m + bp, and Qn — m + Sp. Ans. 3a — 9m + Sn — Zp, 16. Add 7a — 36 + c -f m, and 36 — 7a — c + m. Ans. 2m, 17. Add X — y — z, and y — x + z. Ans. 0. 18. Add 3(a 4- 6), 4(a + b), and — 2(a + 6). Ans. 5(a + 6). 19. Add 6(m^ — nj + 2c, — 5(m'' — n) + 7c, S(m^ — n) — 4c, and 4(m'' — n) + c. Ans. 8(m^ — n) + 6c. 20. Add 2a(x — y'') — Zmz\ 4a(j7 — 2/') — 5m2;^ and 5a (;r — 2/'-^) 4- 7m2;l Ans. lla(x — 2/0 — mz\ 21. Add 8 aa; + 2(x + a) + 36, 9aa; + 6(x + a) — 96, and 11a; -1- 66 — 7aa; — 8(a; + a). Ans. Wax + 11a?. 82 ENTIRE QUANTITIES. •5©. The Unit of addition is the quantity whose coefficients are added ; thus, in the example, 3^ — 4^ + 7^ = 6^, the unit of addition is x ; for the result, 6^, was obtained by uniting the coefficients of x into one number. Dissimilar terms may often be added, by making some common letter or letters the unit of addition. 1. What is the sum of ax, hx, and ex ? Analysis. We make the conimon let- OPERATION. ter, x, the unit of addition ; thus a times ^/j; X, b times a;, and c times x must be equal T to X multiplied by the sum of a, 6, and c; and since a. b, and c are dissimilar, we ex . indicate their addition, inclose th-o sum Sum. (a + 6 + c)x in a parenthesis, and write it as the co- efficient of X, and thus obtain the sura of the given quantities. EXAMPLES FOR PRACTICE. (2-) (3.) (4.) ax by' lay 2cx Bay' — 2ay Adx If — cy (a + 2c + 4:d)x (b+Sa-h7)f (pa — c)y 5. What is the sum of ex, 2cx and Gar, when x is the unit of addition ? Ans. (So + 6)^7. 6. What is the sum of am^y — 6m^ -{- em^y when m^ is the unit of addition ? Ans. (a — 6 + c)m\ 7. What is the sum of (a H- b)x and (a H- e)Xf when x is the unit of addition ? Ans, (2a + 6 + c)j?. 8. What is the sum of Sx, bx, and (a + b)Xf when x is the unit of addition ? Ans. (a + 2b-{- S)x. 9. What is the sum of 4:axy, — axy, and + cxy, when ary is the unit of addition ? u4ws. (3a -f c)^^/. Define the Unit of Addition. SUBTRACTION. gg SUBTRACTION ^S. Subtraction, in algebra, is the process of finding the difi^3rence between two quantities. CASE I. 53. To find the diiFerence of similar terms. 1. A and B travel north from the same point ; the distance A travels is 7m, and the distance B travels is 47?i ; how much farther north is A than B ? OPERATION. ^ ^ , , „ , Analysis. A must be as much farther Minuend, im j^q^^j^ ^^^j^^ jj^ ^^ g>g distance, 4m, sub- Subtrahend, 4m tracted from A's distance, 7m ; and 7m — Difference, 1^ ^m = 3m, the answer. 2 A and B start from the same point ; A travels north a distance of 7m, and B travels south a distance of 4m ; how much farther north is A than B ? Analysis. To express the distances OPERATION. algebraically, we indicate the different Minuend + 7m directions by opposite signs ; north . by pluSf south by minus. Since A Subtrahend, ^ 4m ^^^^^^^^ ^^ ^^^^^^ ^^.^^^ ^ tv^Yehd Difference, + Hm 4m south, A must be 11m farther north than B, and to indicate his direction from B wo must use the plus sign, thus + 11m. The expression, + 11m, in the last example, is called the algebraic difference of + 7m and — 4m, because it denotes their distance asunder. To subtract, in algebra, is not in all cases to diminish. A positive and a negative quantity are in Define Subtraction. What is Case I ? 84 ENTIRE QUANTITIES. opposite circumstances, or counted in opposite directions; he7ice the difference , or space between them, is their apparent sum. If we demand the difference of latitude between 7 degrees north and 4 degrees south, the answer is 7 -}- 4 = 11 degrees, an operation which appears like addition. From these two examples, we learn that a positive term is subtracted by changing its sign to minus, and a negative term is subtracted by changing its sign to plus. This principle is further illustrated by the example below, in which it is plain that the remainders in the lower line must increase by 2 throughout, since the numbers to be subtracted decrease by 2 throughout ; hence, to obtain the true result, the sign of — 2 and — 4 must be changed to -J-. 3. From 16 16 16 16 16 Take 4 2 — 2 _ 4 Reinainder,'l2" ^ii 16 "~18 "^ 4. From + 7a subtract -f 12a. OPERATION. Minuend, + 7a Subtrahend, -f- 12a Difference, — Ofl Analysis. "VVe change the sign of the subtrahend as in the other examples ; then — 12a and -f 7a, are — 5a, the algebraic difference. 5, From Take Difference, 10a 6a 15a 10a 5a Analysis. Since 5a 5a 5a 5a *he minuends de- — — ■ crease by 5a toward — ^^ the right, and the subtrahends are all equal, the remainders must decrease by 5a, and the last remainder 43 therefore — 5a. We cannot, numerically, take a greater quantity from a less, nor any quantity from zero, for no quantity can be less ^Jian nothing. Hence, in the last two examples, the answer, How is a positive term subtracted ? How a negative term t What is tnderstood by a minus quantity ? SUBTRACTION. 85 — 5a, is not 5a less than nothing, but 5a applied in the oppo- site direction to + 5a. To subtract a quantity algebraically, is to change the direction in which it is reckoned or applied. Thus, we see that by a change of sign, we can find the alge- braic difference between any two quantities whatever. From these examples and illustrations we derive the fol- lowing Rule. Change the sign of the subtrahend, or conceive it to be changed, and unite the terms as in addition. EXAMPLES FOR PRACTICE. Prom Take Rem. (6.) + 4a •^ a + 3a (7.) + 6^ + Sx" (8.) ^lObc — 7bG — 36c (9.) + 4m^z ^12^ Prom Take Rem. (10.) — IQb'G — llb'c + b'o (11.) + lSmd + mnd — 2md (12.) -I- 27h' — h ' + 2Sh^ (13.) _+_2W — 28^» 14. From llx^y subtract — 4:X^y, Ans, 21x^y, 15. From abed subtract — abed, Ans. 2abcd. 16. From 25^m subtract 2Sgm. Ans. — Sgm. 17. From — 166V subtract 46V. Ans, — 206W. 18. From — lls^^ subtract — 12sq^, Ans, sq^ 19. From SOxy subtract 4:0xy. Ans. — lOocy. 20. From 75mn^ subtract — 25mnl Ans, lOOmn^, 21. From — 75mn^ subtract — 25mn^ Ans, — 50mn*. 22. From — ISpqr subtract — llpqr, Ans, — pqr, 23. From Ubx^y subtract llbx^y, Ans, ^- Sboc^y, 24. From 5(a + b) subtract 2(a + 6). Ans, 3(a -i- 6). How may the algebraic difference of any two quantities be found? Give the rule for the subtraction of similar terms 86 ENTIRE QUANTITIES. 25. From la{c — m) subtract — 5a (c — m). Ans. 12a(c — m). 26. From — 11(^2 — %j) subtract — b{x' — y). Ans, — Q(x^ — 2/). 27. From 12(m — n) subtract — 12(m — n). Ans, 24(m — n). 28 Subtract 15^y2; from — oxyz, Ans. — ISxyz. 29. Subtract — Iblm^nq from — l^m^nq. Ans. l^lm^nq, 30. Subtract Vlld'hc from ISOa^Z^c. Ans. — 22a'6c. 31. Subtract lipc" — y"— z^) from 12(a?2 — 2/' — ^'0. ^ns. 5(^2 — 2/2 — z^). 32. Subtract 12a6Q} _ 5) from Ibab^p — q), Ans. Sab(j)< — q). 33. Subtract m\c— 1) from — 2mXc — 1). Ans, — ^m\c — 1). CASE n, 53. To find the difference of polynomials. I. From a subtract h — c. Analysis. We first subtract 6 from a, and obtain for a result, a — 6 ; but OPERATION. ouj. t^uQ subtrahend is not 6. but 6 — c; Minuend, a and, as we have subtracted a quantity Subtrahend, 6 c t^o great by c, our remainder must be 7 too small by c ; we therefore add c to Difference, a ^ b ■{- C ^^^ g^^^. ^^^^^^^ ^^^ ^^^^j^^ ^^^ ^^^^ ^^^ mainder, a — 6 + c. By this example, we have in a more general manner estab- lished the principle, that the sign of a term to be subtracted must be changed. Hence the following Rule. I. Write the subtrahend underneath the minuend, placing similar terms under each other, II. Change the signs of the terms of the subtrahend^ 09 conceive them to be changed. What is Case II? Give Analysis. Rule. SUBTRACTION. 37 III. Unite similar terms as in addition^ and bring down all the remaining terms with their proper signs. From Take Hem EXAMPLES FOR PRACTICE. (2.) (3.) 4a + 2^ — . 3c Sax + 2y '^a'\-4:X — 6c xy — 2y 3a — 2j? -4- 3c oax — xy + 4:y (4.) b a a- 2h \ From iTake •Rem. From Take Rem. From Take Rem. (5.) 2x^ — Sx + y^ — x^ — 4:x ■+ a dx^ + X + y^ — a (6.) 7a + 2 — 5c -a + 2 + c 8a .^G i^ — hy y (8.) %x^ _ Sxy + 2^* + c x^ — %xy + oy'^ — 2c 7a?' + Zxy — 2/' + 3c (10.) Sx^ — Ixy 4- 21a + c — x^ -\r oxy — 4a + 4c (9.) ab + cd — m' ab — cd — 2m^ 2cd+ m' (11.) Sax — *iby + 4:ab — ax — 10 by -f 2a6 12. From Sxy — 20 subtract — xy + 12. Ans, 9xy — 32. 13. From 7a^a; + a subtract 3a^^ — 2a. Ans. ^a^x + 3a. 14. From — 8a; — 2i/ + 3 subtract lOa?— Sy -f 4. Ans. — 18a? •{■ y — 1. 15. From %y^ — 2y — 5 subtract — Sif — 5?/ + 12. Ans. Uy^ + Sy^n. 16. From Tm^ — . 4a6 — c subtract 2m^ + 3c — 8a& —8. Ans, 5m' + 4a6 — 4c + «, 17 From a + 2a; take a — x. Ans. 3^ 4 38 ENTIRE QUANTITIES. 18. From 4a -f 46 take b +a. Ans. 3a + 36. 19. From 4a — 46 take 3a + 56. Ans. a — 96. 20. From 13a26' + 11a — 5a^ + 66, take 7a — ^a" + 66 ^ 10a^6l Ans, 2Za^W + 4a. 21. From 3a + 6 + c — J— 10, take c + 2a — d. u4ns. a + 6 — 10» 22. From3a + 6+c — cZ — 10, take 6 — 19 4- 3a- Ans, c — ' (Z 4- 9. 23. From 2a6 + 6^ — 4c + &c — 6, take 3a^ — c + h\ Ans, 2ab — ^c + hc — 3a' — b. 24. From a' 4- 36'c + a6' ■— a6c, take 6' + a6' — a6c. Ans. a^-\-Wc — h\ 25. From bx^y — 36a? 4- e, take 3a?V + 26^27 -f c^ Ans. 2x^1/ — 56^ — d^ -{-c, 26. From 4m* — m 4- 2ca; — 2/^ ^^^^ V^ — ^^^ — m 4- car. Ans. 7m^ 4- ox — 2y\ Note. — The minus sign before a parenthesis indicated that the whole quantity inclosed is to be subtracted. . 27. What is the value of 3a' — (Sa — x + b)? Ans, 3a' — 3a 4- ^ — 6. 28. What is the value of ^Oxy — (30^?/ — 26' 4- 3c — 4:d) ? An^s, lOxy + 26' — 3c 4- 4 J. 29. What is the value of a' — a — (4a — y — 8a' — 1) f Ans. 4a' — 5a 4- 2/ 4- 1. 80. What is the value of 7m' 4- 26c — (3m' _ 6c — a?) ? -4ns. 4m' 4- 36c 4- x. 81. What is the value a4-^ — ^ — (wi — a — 6)? Ans. 2a 4- 26 — 2m. 54. The difference between two dissimilar quantities may often be conveniently expressed in a single term, by making some common letter or letters the unit of subtraction. Explain the unit of subtraction. SUBTRACTION. g9 1. From ax subtract hx. OPERATION. . Ti. • 'A LL\. L-UL- Analysis. It is evident that o times Minuend, ax x taken from a times x must leave a Subtrahend, bx minus b times X, which is expressed 7 rT~ thus (a — b)x. Remainder, {^a — u)X EXAMPLES FOR PRACTICE. (2.) (3.) (4.) (5.) From 2am my^ axy ex Take cm ny^ — cxy x Rem. (2a — c)m (m — Ti)y^ (a -\- c)xy (c — l)a7 6. From 2abx'^ take bcx^, Ans. (2ab — bc)x\ 7. From 4:xy take mxz. Ans, (iy — mz)x. 8. From ax -i-bx + ex take x-h ax -{■ bx, Ans, (c — l)x. 9. From 3a^ — by take 2a^ — cy, Ans. a"^ + (c — b)y. 10. From bacx^+20aa^y^-'2bm take 3acx^+12ax^y^—207n, Ans. 2aa^(cx + 4y') -— 5m. 11. From (2a + b + c)x take (a + b^x. Ans, (a + c)x, 12. From (3a + c)xy take 2axy + cxy. -4ns. axy, 13. From at/ + 26i/ — c^/ take ai/ + cy. ' Ans. (2b — 2c)y. 14. From mz take n2; — 5z. -4ns. (m — n + 5)z. 15. From 5a^x — 2x take 3a; + 5aj?. ^ns. (5a' — 5a — 5)a:. 16. From — ^G\m^ — 1) take 7c'^(m' — 1). Ans. _10c=^(m' — 1). 17. What is the value of ^cy — x'^y — {my — 2x^y + 2cy) ? Ans. (c + a;' — m)y, 18. What is the value of 3m — z — y — (2z — y — 3m) ? ! Ans, 6m — ^z. 40 ENTIRE QUANTITIES. MULTIPLICATIOK i5fS. Multiplication, in algebra, is the process of taking one quantity as many times as there are units in another. CASE I. 56. When both factors are monomials. 1. Multiply 4a by 3&. OPERATION. Analysis. Since it is immaterial in what A^ order the factors are taken in multiplica- Multiplicand, fttt .. , ,, _ ,. '■. ^y tion, we may proceed thus : 6' times 4 are ipier, -j^2. ^ times a are ah; and 12 times ab Product, 12ab are 12a&, the entire product. 2. Multiplya^ by al OPERATION. Analysis. Since a^ is equal to aaa, and • a^ is equal to aa. their product must bo Multiplicand, a ^ , . , . i , * .i • 2 aaaaa, which is equal to a^; this expo- Multiplier, a_ ^^^^^ 5^ ^^y ^^ ^^^^^ ^^ ^^^jjjg ^j^Q ^^^ Product, a^ given exponents, 3 and 2. • 3. Multiply 3a6^ by 4^>». OPERATION. o^i^2 Analysis. 4 times 3a are 12a ; and 5^ ^ ^^ ^ * A 1.3 ti°^6s ^^ ^re 6^ ; hence the entire product Multiplier, 4 6 jg j2a times 6«, or 12a6«. Product, 12a6'^ Sy. In the examples given above, all the quantities are understood to be positive. It is necessary, however, to investi- gate the law of signs when one or both the factors are negative In arithmetic, multiplication is restricted to the simple idea of Define Multiplication. What i| Case I? MULTIPLICATION. 41 repeating a n amber. In algebra, quantities have two qualities, and are either positive or negative ; and multiplicaticn has the double province of repeating by additions^ and repeating by subtractions, as indicated by the signs of the multiplier. Hence, the full signification of a multiplier, when analyzed, is as follows : I. The plus sign of a multiplier shows that the multiplicand is to be added to zero. II. The minus sign of a multiplier shows that the multi- plicand is to be subtracted from zero ; and III. The value of the multiplier sliows how many times the multiplicand is to he taken by either process. To exhibit the law which governs the sign of a product, according to these principles, we present four examples, as follows : 1. Multiply a by h. OPERATION. , Analysis. The plus sign of the multiplier indi- cates that the multiplicand, + a, is to be added to zero h times, giving + a + a -f a, &c. ; hence h "f ^^ the result will be positive, or + ah. 2. Multiply — a by — 5. OPERATION. Analysis. The minus sign of the multiplier in- — a dicates that the multiplicand, — a, is to be subtracted — h from zero b times, which will change its sign, -\- ah g^'^^^S -{• a -\- a -{- a, &c. ; hence the result will bo positive^ OT -\- ah, 3. Multiply a by — 6. OPERATION. Analysis. The minus sign of the multiplier in- + a dicates that the multiplicand, + a,*is to be subtracted — h from zero h times, which will change its sign, ' "^ giving — a — a — a, &c. ; hence, the result will be negative, or — ah. Explain the difference between Arithmetical and Algebraic multiplica- tion. How do the signs -\. and — before a multiplier affect the product? The value of a multiplier ehows what ? 4* 42 ENTIRE QUANTITIES. 4. Multiply — ahj b. OPERATION. Analysis. The plus sign of the multiplier indi- — a cates that the multiplicand, — a, is to be added to + 5 zero b times, which only repeats the letter, 7T giving — a — a — a, &c. ; hence the result will bo negative^ or — db. Comparing the four examples, we observe that like signs produce plus ; and unlike^ minus. From the foregoing examples and illustrations we derive the following Rule. I. Multiply the coefficients of the two terms together for the coefficient of the product. II. Write all the letters of both terms for the literal part, giving each letter an exponent equal to the sum of its expo- nents in the two terms. III. If the signs of the two terms are alike, make the prO" duct plus ; if unlike, make it minus. Note 1, The value of the product will be the same in whatever order the factors are written. Most algebraists prefer to arrange letters in alphabetical order. EXAMPLES FOR PRACTICE. 5. Multiply 8^ by 7a. Ans, 21ax, 6. Multiply 4y by Sab. Ans. 12ab7j. 7. Multiply 156c by lOa?. Ans. IbObcx. 8. Multiply Qax by 12by. Ans. 72abxy. 9. Multiply 17cd by 3m. Ans. blcdm. 10. Multiply 4pg by 7xy. Ans. 2Spqxy. 11. Multiply 12am by ^bcd. Ans. 60abcdm. 12. Multiply 2bpqr by Sxyz. Ans. Ibpqrxyz. 13. What is the product of a^ by a^ ? Ans. a\ 14. What is the product of x^hj x^? Ans, x^°. 15. What is the product of y^hj y^l Ans. y^^. If the factors have like signs, what must be the sign of the product ? If unlike, what ? Give the rule for multiplication of monomials. MULTIPLICATION. 48 16. What is the product of m^ by m^ ? Ans. m}\ 17. What is the product of b^a^ by 6V ? ^ns. 6V. 18. What is the product of aW by am^ ? J^ns. a^m^ 19. Multiply 4ac by — .3a6. -4ns. —12 a'6c?. 20. Multiply 9a^G by — 4ay. -4ns. — 36a^c2/. 21. Multiply — 2xy by — 2xy. Ans. 4iX^y\ 22. Multiply — lay by 3^?/. .4ns. — 21axy\ 23. Multiply 21^;^ by —Zxy. Ans. — 63^2/'- 24. Multiply — 5a^m by — 4:abm\ Ans. 20a^bm^. 25 Multiply — Tm^ by lOc^m'a:. Ans. — lOc^m^z. 26 Multiply llxy by 2a;'2/'- ^^s. 34a75t/». 27. Multiply Uab'cd^ by — 363c2m. ^ns. — 42a6'^c'<:?^m. 28. What is the value of 3a X 46 x 2c ? Ans. 24:abc. 29. What is the value of 7m^ X 4am x 2my ? Ans. 56am*i/, 30. What is the value of ar* x a;' X ^ ? Ans, of, 31. What is the value of — la'b X 2ab'' X 3a6 ? Ans. — 42a*&*. 32. What is the value of — ba'm X 3a6'c X 2bc^m^ ? Ans. — 30a''6Vm^ 83. Multiply 3(a; + 2/) ^7 2. Ans. 6(a? + y'). 34. Multiply a(^x^ + m) by b, Ans. ab(x^ + m). 35. Multiply (a -f- wi)^ by c. ^ns. c(a + w?,)\ 36. Multiply (a + by by (a + b)\ Ans. (a + 6)\ 37. Multiply 3a(m — n)^ by — a(m — n)l Ans. — 3a*(m — n)*. 38. Multiply 4m(j7^ — y^)'^ by — 2am(^=^ — 2/0- Ans. —SamXx^ — y^y. Note. — "When quantities have literal exponents, powers of the same letter or quantity are multiplied by indicating the addition of the exponents. 39. Multiply a"» by a". Ans. a"' + ". 40 Multiply c"* by c. Ans. c"*+*. How are quantities with literal exponents multiplied? 44 ENTIRE QUANTITIES. 41. Multiply (a — by by (a — by. Ans, (>i — 6)'+^ 42 Multiply a'"(p + qy by a'O + (?)"». CASE II. 58. When one factor is a polynomial. 1. Multiply 46 + 5a' — be by 3a. Analysis. Since the whole multi- OPERATION. plicand is to be taken 3a times, we ., , p^ 2 z, must multiply each of its terms by 4^ -^ Da — ^C 3^ . ^j^^g 3^ ^jj^gg 4^ jg ^2a6 ; 3a times *^^ Sa'^ is 15^3 ; 3a times — &c is — 3a6c ; 12a6 + 15a' — Sabc ai^d "we have for the entire product, 12a6 + I5a^ — ^abc. Hence the Rule. IluUiply each term of the polynomial separately by the multiplier j and write the partial products connected bv their proper signs. (2.) 5a — 8c 2a EXAMPLES FOB. PRACTICE. (3.) (4.) 3ac — 46 2a'— 8c + 5 — 3a he 10a' — 6ac (5.) 12j; — 2ae 4a — Wc + 12a6 (6.) 15c— 76 — 2a 2a'6c — 36c'' + 56c (7.) 4a; — 6 + 3a& 2a6 (8.) 8c' 4- X \xy (9.) — 4x' (10.) 3a' — 2a;' — 66 2aa;' Give Case II. Analysis. Rule. MULTIPLICATION. 45 11. Multiply 36 — 2c by 56(? Ans. l^b'^c — I0bc\ 12. Multiply ^xy — 9 by 6x, Ans. 2Aa;^y — 54^7. 13. Multiply a^ — 2^ + 1 by 4:x\ - Ans, 4:aV---Sx' + 4:x' 14. Multiply lla^6c^ — ISxy by Sax, Ans. SSa'^bc^x — S9ax% 15. Multiply 42c' — 1 by — 4. Ans. — 168c' + 4. 16. Multiply — 30a'6a;V + 13 by — oa\ Ans, l^Oa^bx^y — 65a' 17. Multiply 26 — 7a — 3 by 4a6. Ans, 8a6' — 28a'6 — 12a6. 18. Multiply a + 36 — 2c by — 3a6. Ans. — 3a'6 — 9a6' + 6a6c. 19. Multiply 13a' — 6'c by — 4c. Ans, — 52a'c + 46V. 20. Multiply ISxy — 36 by — 25^. ^ Ans, — S2bx^y + 756a?'. CASE in. 59. Wbieii both factors are polynomials. 1. Multiply 2a + 36 by a + 6. OPERATION. Analysis. To multiply by Multiplicand, 2a -r 36 a+ 6, we must take the mul- Muitipiier, a + 6 tipKcand a times and 6 times, Product by a, 2a'' + 3a6 ^hich is done by multiplying Product by 6, + 2a6 + 36' ^y f ^^^ ^ separately, and „ .. ^ , . o ■> ■ — r 7 . or9 adding the partial products. EnUre Product, 2a^ + 5a6 + 36' -rj th Rule. Multiply all the terms of the multiplicand by each ierm of the multiplier separately ^ and add the partial products. Give Case III. Analysis. Rule. 46 ENTIRE QUANTITIES. EXAMPLES FOR PRACJTICB. Multiply (2.) 2a + 6c (3.) 8^ — 5y a — c X —2y Product, 2a' 4- 5ac _2ac — 5c* 2a' 4- 2>ac — h(^ Bx^ — bxy — Qxy + lOy" Zx' — llxy+lOy' Multiply By Product, (4.) ^x —2z 3a —bd ISax — Qaz — SOdx + lOdz Multiply By (5.) a-h b + c X + y + z Product, ax-i- bx + ex + ay+ by + cy + az + bz + cz. 6. Mult iply Sa^ — 2a6 — b^ by 2a - -4b, Ans, 6a» — IQa'b +'^ab' + W. 7. Multiply x'^ — xy + y^hj X + y. Ans. a^ + y^. 8. Multiply 3a 4- 4c by 2a -— 5c. Ans. 6a'— 7ac — 20c*. 9. Multiply a^ + ay — y^hy a — y. Ans, a* — 2ay^ + y^. 10. Multiply a^ + ay + y'^hj a — y, Ans. a" — y^. p.1. Multiply a' — ay + y^hj a + y. Ans. a' 4- 1/*. 12. Multiply a^ + a^y + ay^ + y^hj a — y. Ans. a* — y*. 13 Multiply y^ — y + lhj y + 1, Ans. t/* 4- 1. 14 Multiply a^ + y'^hj x^ — i/l Ans. a^ — t/*. 15. Multiply a* — 3a 4- 8 by a 4- 3. Ans. a» — a 4- 24. 16. Multiply b" 4- 1 V 4- a;* by 6' — x\ Ans. 5« — 3^. 17. Multiply a* 4- 26 by 2a^ — 46. Ans. 2a* — 86*. MULTIPLICATION. 47 18 Multiply a^ -]' x^ + x^ hj x^ — 1. Ana, x^ — x\ 19. Multiply m + n by 9m — 9n. Ans. 9m^ — dn\ 20. Multiply 2x^ + xy ^ "If by Zx — 3?/. Ans. Ga?' — Zx^y — ^xy^ + 63/*. 21. Multiply m' — 3m — 7 by m — 2. Ans. m^ — 5m^ — m + 14. 22. Multiply a* — 2a«c + 4a'c'— 8ac^ + 16c* by a + 2e. 23. Multiply a;^ — 3^?^ + 3a; — 9 by a? + 3. ^ns. a;* — 6ar^ — 27. 24. Multiply m* — m* + m* — m + 1 by m + 1. Ans. m^ + 1. 25. Multiply m* + w<' + w' + ^ + 1 by m — 1. 26. Multiply 2a3 '+ 5ac' — 2c^ by 2a» — 5ac' + 2c'. ^ns. 4a« — 25aV + 20ac* — 4c^ Note. — The product of two or more polynomials may be indicated by inclosing each in a parenthesis, and writing them in succession ; siich an expression is said to be expanded, when the multiplication has been actually performed. 27. Expand (a + 6) (a + c). Ans. a? + ah + ac + be. 28. Expand (x + By) (x^ — 2/)- Ans. xr^ + Bx^y — xy — By\ 29. Expand (w} + 2c) (m^ — 5c). 80. Expand (a + 6 — c) (a — 6 + c). Ans. a' — l)' + 2hc — &. 31. Expand (a — c — 1) (a + 1). Ans. d^ — ac — c — 1. CASE IV. 60. To square a binomial. If a polynomial be multiplied by itself, the product is the square of the polynomial. A binomial quantity is easily How may multiplication of polynomials be indicated? When are such expressions expanded ? Give Case IV. 48 ENTIRE QUANTITIES. squared without the formal process of multiplying, as will be seen by the analyses of the two following examples. 1, What is the square of a + 6 ? OPERATION. Analysis. Multiplying a -|- 6 by ' a -j- 6 by the common method, wo (i ~\- obtain for the result, a^ which is the ^2 _j_ ^jy square of a; -j- 2a6, which is the -A- ah 4- h^ ^ times product of a and 6 ; and 6^, ■ which is the square of 6. 052 _|. 2ah + 6^ 2. What is the square of fl — 6 ? OPERATION. a — b a — h Analysis. Multi plying in the usual way, we obtain for the result a^ which is the square of a ; — 2a5, which is a^ — ah twice the product of a and — h ; and dj) +6^ ^^ which is the square of 6. Hence o? — 2ah+h' *^® Rule. Write the square of the first term^ twice the pi\ duct of the two terms, and the square of the second ter^m. Note. — The product of the two terms will be minus, when one of them Is minus. EXAMPLES EOR PRACTICE. 3. Square a + c. Ans, d^ + 2ao + &. 4. SquaBe p + ^- Ans, p^ + 2pq + (f 5. Square m — n. Ans, m? — 2mn + ^^ 6. Square x — y, Ans, x^ — 2xy + 2/' 7- Square A + B. An^. A^ + 2AB + B 8. Square A—G. Ans, A' — 2AG+ G\ 9. Square 8a — 2^. Ans, Oa^ — 12aa7 + 4ar*. Give analysis Rule. MULTIPLICATION. 49 10. Expand (m + z) (m + z). Ans. m^ + 2mz + z\ 11. Expand (2a — c) (2a — c). Ans. 4a^ — 4ac + c^. 12. Expand (5^7 — 3) (5a;— 3). ^ns. 25a;2_3o^ + 9 13. Expand (4a + i^) (4a + ^x), Ans, 16a^ + 4:00) + ^a;'. Note. — The square of a binomial may be indicated by an exponent. 14. Expand (m + cy, Ans, m^ + 2cm + c\ 15. Expand (2c — Sdy, Ans, 4c^ _ 12cd + 9cP. 16. Expand (x^ — xy, Ans, a^--2cc^ + x\ 17. Expand (a — 1)1 Ans, a' — 2a + 1. 18. Expand {a^x — ax^y, Ans. aV — 2aV + aV. 19. Expand {y^ — 20) {y'' — 20)> ^ns. 2/* — 402/' + 400. 20. Expand (a:"* — y") {x"^ — 2/"). ^ns. a;''" — 2^?"'?/" + 2/'". - 21. Expand (c^ •— 1) (c'" — 1). ^ns. c'"* — 2c^ + 1. CASE V. 01. To find the product of the sum and difference of two quantities. The sum of two qnantities multiphed by their difference gives a result still more simple than a binomial square. 1. Multiply a + & hy a — h. OPERATION. Analysis. We are required to Sum, a -\- h multiply the sum of a and h, by the , difference of a and b. Multiply in a: Difference, a , •^\, , „ , . ' by the usual process, we find in add- a^ + <^^ ing the partial products, that -f ab . — ab — b^ and — ab reduce' to zero, and the — ^ — product is a^ — 6^ or the difference Product, a of the squares of a VLud b. But since « and b may represent any two quantities whatever, the form of this product embodies a general truth. Hence Give Case V. Analysis. 5 D 50 ENTIRE QUANTITIES. Rule. From the square of the greater quantity ^ subtract the square of the less. NoTB. — The term or quantity having the minus sign in the difference, is supposed to be the less. EXAMPLES FOR PRACTICE. 2. What is the product of m + n by m — n? Ans, w? — n* 8. What is the product of a + c by a — c? Arts, a^ — ^ 5* '!f^ ENTIRE QUANTITIES. 05, In the foregoing examples, no signs being expressed, both dividend and divisor are understood to be positive. To ascertain what sign the quotient should have when one or both the terms of division have the minus sign, we have only to observe what sign must be given to the quotient, in order that the product of the quotient and divisor shall have the same sign as the dividend, according to the law of signs in multi- plication. To exhibit the law which governs the sign cf the quotiei^t, Fe present four examples. 1. +ab-7- + a= + b because -{-ax -{- b = -\- ab. 2. — ab -7--^a= + b '' — a x + b = — ab. *S, + ab-. a = — b " — ax — b = + ab, 4. — a5 _i_ _|_ a = — b " + a X — b = —ab. From these examples, taken in order, we make the following inferences : IsL 4- divided by + e-ives +7 n • ^ ^ i ^ .7 J b •>- or, like signs produce -|- 2d. — divided by — gives + ) 6d. + divided by — gives — 7 n • i ,. ., , ^ . >■ or, unhke signs produce — it/L — divided by + gives — ) G©. These priuciples may be deduced from tlie nature of the signs themselves, by taking another view of division. Division, considered in its most elementary sense, is not merely the converse of multiplication; it is a short process of finding how many times one quantity can be subtracted from ciuothcr of the same kind. When the subtraction is possible, and diminishes the numeval value of the minuend, and brings it nearer to zero, the operation is real and must be marked plus.^ When the subtraction is not possible without going farther from zero, we must take the converse operation, and the converse operation we must mark min^is. Give analyses of the law which governs the signs of the quotient. Give the law- DIVISION. 55 Thus, divide 18a by Qa. ITere, it is proposed to find how many times 6a can be subtracted from 18a ; and as wo can actually subtract it 3 times, the quotient must be -f 3- Divide — 18a by — im. Here, again, the subtraction can actually he performed, and the number of times is o, and, of coarse the quotient is -f- '^• Divide — I8a by Oa, Here, subtraction will not I'educe the dividend to zero ; but addition will, and must be performed 3 times ; but the operation is the converse of the one'proposed, and therefore must be marked by the converse sign to pluSy that is — 3. Again, divide 18a by — Co. Here, if we subtract — 6a it will not reduce 18a ; but the converse opemtion will, and therefore the quotient must be minus, that is, — 3. From all these illustrations we derive the following : Rule. ' I. Divide the coefficient of the dividend by the coefficient of the dimsor^ for a new coefficient. II. Write the lette^^s of the dividend in the quotient ^ giving each an exponent equal to tlie difference of its exponents in the two terniSj and suppressing all letters whose exponents be- come zero. III. If the signs of the terms are alike ^ make the quotifi*'t plus ; if unlike, make it minus. Note. — If the dividend does not exactly contain the divisor, the division may be indicated by writing the dividend above a horizontal line, and the divisor bo1ow, in the form of a fraction ; and the result thus obtained may be simplified by canceling all the facfbrs common to the two terms ; ia^b'^c 2b thus, ^a'^b^c ~ Qii'^b^c"^ := rr:-— . But this process is essentially a case of reduction of fractions; we shall therefore omit all examples of this class till the section on fractions is reached. EXAMPLES fOR PRACTICE. 1 Divide 16a& by 4a. Ans. 4&. 2. Divide 21acd by 7c. Ans. Sad. 3. Divide ab'c by ac. Ans. b\ Kule for division of monomials. 56 ENTIRE QUANTITIES. 4. Divide 6abc by 2c. Ans. SaO. 5. Divide aoc^ bj ax\ Ans. x, 6. Divide Bmx^ by mx. Ans, ^x^, 7. Divide 210c'b by 7c6. u4ns. 30cl 8. Divide 42a;?/ by xy, Ans. 42. 9. Divide — 2 lac by — 7a. Ans. 3c. 10 Divide — 12xy by By. -4ns. — 4^7. 11. Divide 72a&c by — 8c. Ans, — 9ab 12. Divide 2a« by a*. Ans, 2d\ 13. Divide — a^ by a*^. u^ns. — a. 14. Divide 16^ by 4aj. ^ns. 4j;^ 15. Divide Ibaxy^ by — oa^/. u4ns. — bxy'^, 16. Divide *— ISa^x by — 6a^. u4ns. oa\ 17. Divide Qacdxy^ by 2adxy^ Ans, 3c. 18. Divide 12aV by — 3a^^. ^ns. — 4x. 19. Divide 15ai/^ by — Say. Ans, — by. 20. Divide 45(a t- xf by 15(« — ^)^ Ans. 3(a — a;). Note. — In this example, copslder (a — z) as one quantity. 21. Divide 45i/^ by 15y^. ^ns. 3^/. Note. — Examples 20 and 21 are exactly alike, if we conceive (a — x) equal to y. 22. Divide z^ by zK Ans. z\ 23. Divide (x — ^z)^ ^7 (^ — 2/)'- -^^s* (^ — 2/)^- Note. — Observe that examples 22 and 23 are essentially alike. 24. Divide (a + b'^^hj (a + 6). Ans. (a + 6)1 25. Divide x"^ by a;^ -4ns. ^'""■", 26. Divide 6c^ by 3c. Ans. 2c'^-'K 27. Divide (a — c)"' by (a •— c)l ^ns. (a — c)'"--^ 28. Divide 10(a — c) by 5(a — c). Ans. 2. 29. Divide .6a%a + my by ^a + m) -4ns. S^fa + m)^ 30. Divide 52m^c(l — x'y by 13mc(l — x'f, Ans. 4m. 31. Divide 81a*;s'^(4m — qf by 27 zX^m — qy Ans. 3a*(4m — q). division;. 57 CASE II. 67. To divide a polynomial by a single term. I. Divide I2a' — Qa'c + Ba'm by 3al OPERATION. Analysis. The whole divi- Sa^)12a^ — Qa^c + 3a^m dend is divided by 'Sa\ by di- 4^3 2ac 4- m viding each of its terms by Za\ Heuce the Rule. Divide each term of the dividend separately ^ and connect the quotients by their proper signs. EXAMPLES FOR PRACTICE 2. Divide Ibab — 12ax by 3a. Ans. 5b — 4x. 8. Divide — 25a'^x + l^ax^ by — bax. Ans. ba — 2>x. 4. Divide lOab + Ibac by 5a. Ans. 2b -f 3a 5. Divide Wax — Mx by 6x. Ans. ba — 9. 6. Divide Sx^ + 12x by 4:x\ Ans. 2x + Bx-\ 7. Divide Bbcd + 126cx — Wc by 36c?. -4ns. d + 4x — 36 8. Divide lax + lay — lad by — la. Ans. — x — y + d 0. Divide oax^ + 6x^ + 3ax — Ibx by 3a:. A71S. ax^ + 2x + a — 5. 10. Divide 3a6*c + 12ab'x — 3a^6' by 3a6«. Ans. be 4- 46^a; — ab\ II. Divide 25a26x — Iba^cx^ + ba^bcx^ by — ba'^x. Ans. — 56 + 3 ex — abcx. 12. Divide 20a^6^ + 15a'62 + 10a^6 + 5a by ba\ Ans. 4¥ 4- 36^ -f 26 + a-\ 13. Divide 21a + 356 — 14 by — 7. ' Ans. 2 — 3a-— 56. Give Oase II. Analysis. Rule. 58 ENTIRE QUANTITIES. 14. Divide — 12a'bc + Qacx"" — Qab^c by — oac. 4ns, 4ab — ox' + 25^ 15. Divide 6(a + x) + 9(x + ij) by 3. Ans. 2(a + x) + 3(;?; + y). 16. Divide 12(a + ic) — 3c(a + ^) F <^(a + x) by (a + i^). Ans. 12 — 3c -f d, 17. Divide (a + c)' — (a + c)^ by (a -f e). ^ns. (a + c) — (a + c)*. 18. Divide 12(a — 6) + Qc(a — b) + 2(a— 6)by (a— 6). Ans. 12 + 6c + 2 19. Divide (m + n)x^ + (m + ??)a' + (m + n.)c' by (m -f n) Ans. x' -f a'^ 4- g\ 20. Divide (a 1 by + 2(a + &) by (a + ?>}. ^ns. a + Z> + 2. Note. — Wlien a parenthesis has the unit 1 for both coefficient and ex- ponent, and is connected with the other parts of the algebraic expres- sion by -|- or — » ^^ ^''^y be omitted; thus, (a-\- b) -\- 2, is the same as a -\- b -\- 2. But when a parenthesis having the minus sign before it is dropped, the signs of the quantities inclosed must all be changed (63) ; thus, a^ — {a — x), is the same as a^ — a-\- x 21. Divide 2a(a + c) + {a -^^ cy by (a -f c). Ans. 3a + c. 22. Divide 5c(3m — 2c) — (Sm — 2c)' by (S??! _ 2c). Ans. 1g — Sm. 23. Divide (1 — a:) — (1 — x)' by (1 — x). Ans. x CASE ni. 68. To divide one polynomial by another. Since the dividend is always the "product of the divisor hy the quotient souglit, the highest power of any letter of the dividend must be the product of the highest powers of the same letter in tlie divisor and quotient ; and the inferior powers of this letter in the di /idend, must be the products of inferior powers in divisor and quotient. Hence fhe terms of both Give Case TIL DivisiaN. 59 aimsor and dioidend must be arranged irv the order of the powers of one of the letters. 1. Divide 2a' -f 5aV + 2a'b — Qab' + 46* by a' + 2ab \- 46'. OPERATION. a2 _j_ 2ab -(- 4b\ Divisor. 2a' — 2ab -f 6^, Quotient. Dividend, 2a* -f- 2^36 4. Sa^^s _ g^^s ^ 4^>4 2a* 4- 4a36 -f- 8a2i2 1st Rem. — 2a^ — Sa'^b'^ — 6ab^ — 2a4 — ^a'^b'^ — 8a63 2d Rem. a26*^ -f 2ab^ + 46* Analysis. We arrange the terms of both divisor and dividend according to the powers of a, so that in the dividend the exponents of this letter, taken in their order, are 4, 3, 2, 1 ; and in the divisor, 2, 1. Now, according to the principle just stated, the first term of the dividend, thus arranged, must be equal to the first terra of the divisor multiplied by that term of the quotient having the highest power of a; we therefore divide 2a'*, the first term of the dividend, by a^ the first term of the divisor, and ol;)tain 2d^ for the first of the quotient. We next miiiij^ly the whole divisor by this term of the quotient, and subtract the product from the dividend, bringing down as many terms as are necessary for a new dividend. We then divide — 2a^6, the first term of the remainder, by d\ th-e first tetm of the divisor, and obtain — 2ah for the second term of the quotient. We next multiply the lohole divisor by this term of the quotient, and subtract the product from the second dividend, and obtain a second remainder to which we annex another term of the dividend for another dividend. Dividing aW, the first term of this dividend, by a^ the first term of the divisor, we obtain 5^, anothe.r term of the quotient. Lastly, multiplying the whole divisor by this term of the quotient, and subtracting the product from the last dividend, we have no remainder, and the work is finished. From this example we derive the fQllowing E-HLE. I. Arrange both divisor and dividend with refe^ "ence to the powers of one of the letters. II. Divide the first term of the dividend by the first term ^f the divisor^ and write the result in the quotient. Give analj'sis Rule for the division of polyuomiala. go ENTIRE QUANTITIES. III. Multiply the whole divisor^ by the quotitnt thus founds and subt^ct t/ie" product from the dividend. I Y. Arrange the remainder for a new dividend, ivith which proceed as before, till the first term of the divisor is no longer contained in the first term of the remainder. Y. Write the final remainder, if there be any, over the divisor in the form, of a fraction, and the entire result will e the quotient sought. EXAMPLES FOR PRACTICBl 2. Divide a^ + 2ax + x^hj a -\- x. Ans, a -f a>, 8. Divide a^ — 3a^?/ + Say^—^y^ by a — y. Ans, a? — 2«2/ + 2/'« 4. Divide a« + ha^b + hah^ + b'' by a + 6. Ans. a' + 4a6 + 61 5. Divide o:? — ^x^z -f z^\i^ x — z. Ans. x^ — ^xz — 22^ . X — z 6. Divide a^ + la^b + "hab^ -f H' by a^ + ab + b\ Ans. a + 6 7. Divide x"" — ^x' + Tix _ 27 by a? — 3. Ans. 0?' _ 6^ + 9. 8. Divide Ga;* — 96 by ^x — \% Ans. x^ -f 2^2 _j. 4^ ^ g^ 9. Divide ^a^ + 9a^ — 15a by Sa^ — ^a. Ans. 2a^ + 2a + 5. 10. Divide 25x^ — x" — 2x^ — Sx by 5^^ — 4x. Ans. bx" + 4x^ -\- Sx ^ 2. 11 Divide ISa^ — 86^ by 6a + 46. .4ns. Sa — 26. 12. Divide 2^ — 19x' -f 26^ _ 16 by ^ — 8. Ans. 2x^ _ 3 jr + 2 13. Divide 2/^ -f 1 by 2/ + 1. Ans. y^ — V^+y^ — 2/ + 1. 14. Divide y^ — 1 by 2/ — 1. Ans. y^ + y^ + y^-h y^ + y -hi. 15. Divide x"^ — a* by a? — a. Ans, x + a DIVISION. 61 16. Divide 6a' — 3a'b — 2a + 6 by 3a' — 1. Ans. 2a — 6. 17. Divide y^ — 3^/ V + 3i/ V — x^ by y^ — dy^x + ^yx"^ — a^ Ans. y' + oy^'x -f oyx'^ + x^ 18. Divide Ua'b^ — 2ba"b' by Sa'¥ + bah\ Ans. Sa'^b' — ^ab\ 19. Divide 2a* — 2a;* by a — - ^. Ans. 2a' + 2a^^+ 2a^2 _^ 2;2r\ 20. Divide (a — ^)^ by (a — xy, A7is. (a — xy 21. Divide a' — od^x + ^ax^ — a^ by a — x. Ans. c^ — 2ax + x^, 22. Divide a'^ + 1 by a + 1. Ans. a* — 0^ -\- Q^ — a -\- \. 23. Divide 6« — 1 by 6 — 1. Ans. b"^ -\-b' + ¥ ^b^ + b^-X. ^4. Divide 48a^ — 92a'a; — ^^ax" + lOOx^ by 8a — hx. Ans. 16a"^- — 4ax — 20x1 25. Divide 4cZ* — 9(Z^ + 6(Z — 1 by 2cZ^' + 3cZ — 1. .4>26-. 2^'^ — 36Z 4- 1. 26. Divide 6a* + 4a«x — 9a'x^— 3ax^ + 2x* by 2a'^ + 2t^ — x^. Ans. ^a^ — ax — 2x1 27. Divide 3a* — Sa'^^'^ + 3aV + 56* ~ 36V by a'^ — b\ Ans. 3a' — 56' + 3c'. 28. Divide 2x' + "Ixy + 6^/' by x + 2y. Ans. 2x + 3t/. 29. Divide 2mx + Swx + lOmn + ISn' by x + 5n. Ans. 2m + 3/1. 30. Divide cZ* — 3cZ'c — 10c' by d' — 5c. Ans. d' + 2c. 31. Divide m' — c' + 2c2: — 2' by m + c — z. Ans. m — c + z. 32. Divide y^ + B2z^ hjy + 2z. Ans. 2/* — 2^2: H- 42/'z' — 82/2' + I62:*. 33. Divide 12(a + by + 3(a + by by 3(a + b). Ans. 4(a -j- 6)' + a + 6. 84. Divide 3c(m — 5c)^ — (m — 5c)^ by (in — 5c)'. Ans. 8c — w. 6 Q2 ENTIRE QUANTITIES. GENERAL PRINCIPLES OF DIVISION 09, The value of a quotient in division depends apon the relative values of the dividend and divisor; and the sign of the quotient depends upon the relative signs of the dividend and divisor. Hence any change in the value or the sign of either dividend or divisor must produce a change in the value or the sign of the quotient ; though certain changes may be made in both dividend and divisor, at the same time, that will not affect the quotient. The laws that govern these changes are called General Principles of Division. CHANGE OF VALUE. 70. It will be necessary to examine only those changes of value produced by multiplying and dividing the dividend and divisor. Let us take abed for a dividend, and ab for a divisor ; the quotient will be cd, and the operations performed upon divi- dend and divisor will affect this quotient as follows : Dividend. Divisor. Quotient. abed -=r ab = cd ♦ 1. abcde -^ ab == cde ( Multiplying the dividend by e \ multiplies the quotient by e. r) 1^ _^ J _ f Dividing the dividend by d di- ( vides the quotient by d, ( Multiplying the divisor by c di- ( vides the quotient by c viding the divisor by b mul ies the quotient by /;. 5. abcde -^ abe = cd I Multiplying both terms by e I does not alter the quotient. ^ , J , J i Dividing both terras by a does ^, bed -^ b = cd \ :^ \ I not alter the quotient. \^hat determines the value of a quotient in division? 8. abed -f- abe = d 4. abed -r- a = bed ( Dividin \ tiplie DIVISION. (53 In these six operations, the factors employed to operate with are literal quantities, and may represent any numbers whatever ; hence the results are general truths ; they may be stated as follows : Prin. I. Multiplying the dividend multiplies tJte quotient and dividing the dividend divides the quotient. (1 and 2.) Prin. 1 1. Multiplying the divisor' divides tlie quotient^ and dividing the divisor multiplies the quotient. (3 and 4.) Prin. III. Multiplying or dividing both dividend and divisor by the same quantity does not alter the quotient (5 and 6.) Tl. These three principles may be embraced in one GENERAL LAW. A change in the dividend produces a like change in the quotient; but a change in the divisor produces an o1»posite change in the quotient. CHANGE OF SIGN. 72. To investigate the relative changes of signs in division, let it be remembered that when the divisor and dividend have like signs, the quotient is j:>Zms, and when they have unlike signs, the quotient is minus. Then l8^. Suppose the divisor and dividend have like signs ; if either of the signs be changed, they will become unlike, an(a + b), and 12(a'^ — 6"). A^is. 60aV(a^ — b'). 10. What is the least common multiple of m* — 1, m* — 2?/i -h 1, and m' + 2m + 1 ? An,3. The Apparent Sign of a fraction is the sign written before the dividing line, to indicate whether the fraction is to be added or subtracted ; thus, mm -\ , the apparent sign of the fraction is plus, and indicates that the fraction is to be added to m. 104. The Real Sign of a fraction is the sign of its numeri- cal value, when reduced to a monomial, and shows whether the fraction is essentially a positive or a negative quantity ; thus, in the last fracti6n, "" — '-, let a? = 2, and a = 12 ; then a — X a^ — ax 4-^12x2 —20 _ _ ^. .. = — r- — = — ,rj- = — 2. Hence, the real sign a — X 12 — 2 10 of this fraction is minus, though its apparent sign ispZi^s GENERAL PRINCIPLES OF FRACTIONS. 10«i« Since fractions indicate division, all changes in the numerator and denominator of a fraction will afTect the value and sign of that fraction according to the laws of division ; and we have only to modify the language of the General Prin- ciples of Division (70), by substituting the words numerator^ denominator^ and fraction^ for the words dividend^ diviso?^ and quotient, and we shall express the laws governing the changes in the value and sign of a fraction. CHANGE OF VALUE. PRIN. I. Multiplying the numerator multiplies the frao- Hon, and dividing the numerator divides the fraction. Define 'the apparent sign of a fraction. The real sign. Fractions alvrays indicate what ? Adapt the general principles of Division to frao- tions. Repeat the priiwiples that govern change of value. 7* 78 FRACTIONS. Prin. II. Multiplying the denominator divides the frac- tion, and dividing the denominator multiplies the fraction. PuiN. III. Multiplying or dividing both numerator and denominator by the same quantity does not alter the valu:? of the fraction, 106. These three principles may be embodied in one GENERAL LAW. A change in the numerator pi^oduces a like change in the value of the fraction ; but a change in the denominator pro- duces an orposiTE change in the value of the fraction. CHANGE OF SIGN. 107. Prin. I. Changing the sign of either numerator or denominator, changes the real sign of the fraction. Prin. II. Changing the signs of both numerator and da- nominator at the same time^ does not alter the real sign of the fraction. Prin. III. Changing the ap)parent sign of the fraction changes the real sign. reduction. 108. The Reduction of a quantity is the operation of changing its form without altering its value. CASE I. 100. To reduce a fraction to its lowest terms. A fraction is in its lowest terms when its numerator and denominator are prime to each other. Uah'c 1. Reduce ^^ 2/^2 to its lowest terms. Apply the general law. The principles that govern change of signs, Define Reduction- What is Case I? REDUCTION. «fg OPERATION. Analysis. If we divide both numerator and denominator of this i4 aoc *-0^ fraction by the same number, its 2ld^bc^ oao value will not be changed (105, III) ; and if wo divide by the great- est common divisor, the quotients will be prime to each other (90), and consequently the fraction will be in its lowest terms. By inspec- tion we find labc to be the greatest common divisor; and divi- ding both terms by this quantity, we have o— , the answer. d^x + ax"^ 2 Reduce — r. ;;- to its lowest terms. a' — x^ OPERATION. a^x -f ax^ ax (a -^ x) ax a^ — x^ (a — x) (a + x) a — x Analysis. We first resolve the numerator and denominator into their prime factors, and then cancel the common factor (a -f x), and we have , the answer. a — X From these examples we deduce the following Rule. Divide both rrumerator and denominator by their greatest common divisor. Or, Eesolve the numerator and denominator into their prim^ /actors, and cancel all those that are common, EXAMPLES FOR PRACTICE. 8. Reduce —-7 t^ its lowest terms. Ans, ot. ioao 00 , ^ , 14aVv .. , . . ^ 2at/ 4. Reduce r-. — -^ to its lowest terms. Ans, -o^. zlax^ o Give analyses. Rule. go FRACTIONS. 5. Reduce ^^ ^ , to its lowest terms. Ans, -zr^ — . bSa^xy^ liy 6. Reduce ^r^rrz — pr-r- to ^^s lowest terms. 17a^ — 21fl5 7. Keduce jtt — :— ^ to its lowest terms. 8. Reduce — ^ 77- to its lowest terms. Arts, ., . ,^ a;* — 6* 07^ + 6*^ ^2 1 ^ ][ 9. Reduce • — to its lowest terms. Ans, . xy + y y ex I cot^ 10. Reduce ■ — 7— to its lowest terms. acx + ahx c 4- cr Arts. -^-^. ac + CO 11. Divide a^y^ + a?V by aoc^y + a^y*. ^^«. — . 12. Divide 4a + 46 by 2a' _ 261 Ans. xy I 2 a — b' 8 13. Divide n^ — 2n^ by n^ — 4n + 4. ^ns. -^, 14. Reduce — 5 =- to its lowest terms, 15. Reduce ^ . ^ — ; — 5 to its lowest terms. x^ + 2ca7 + G^ 16. Reduce ^^-^ ~ to its lowest terms. or — a' 17. Reduce , , ,^ to its lowest terms. ar 4- ary^ REDUCTION. 81 CASE n. 11©. To reduce a fraction to an entire or mixed quantity. (lb -\- X 1. Reduce — ^ — - to a mixed quantity. OPERATION. Analysis. Since the value of - . the fraction is the quotient of "^ = a -| t^® numerator divided by the de- ^ ^ nominator, we perform the divi' sion indicated, and obtain a for X the entire part of the quotient, and H- t for the fractional part. Hence, the KuLE. I. Divide the numerator by the denominator as far as possible, for the entire part, II. Write the remainder over the denominator, and annex the fraction thus found to the entire 2^cirt, with its proj^er sign. Note. — If any term be found in the numerator, whose literal part is exactly divisible by some term in the denominator, and having a greater coefficient than this term of the denominator, the reduction will be possible ; otherwise, it will be impossible. EXAMPLES FOR PRACTICE. 19 a^ 4- bx 2. Reduce -^ and to mixed quantities. bx Ans. 2 J and a -i . ^ a 8. Keduce — to a mixed quantity. Ans. 5a -\ • 2a* 26' 4. Reduce r— ■ to an entire quantity. Ans. 2a + 26, Give Case II. Analysis. Rule. P g2 FRACTIONS. 1 5(/S 2.x 5. Reduce — — r-^-- ^ to a mixed quantity. Ans. 3 a — ^-j. . ^ ^ a^ _f_ a7; + 6\ . , ^.^ ^^ 6. Reduce to a mixed quantity. a " yt Ans, a + b A . ^ ^ . 12a^ + 4a — 3c^ • ^ ... ^ 7. Reduce , to a mixed quantity. Ans. 2a -{- 1 — j-. . _ , lOcx + a — 6 ^ . , ^.^ ^^ 8. Reduce ^ to a mixed quantity. ^^ ^ a; , « — ^ Ans. DC H ^ — . n -n ^ a;^ + 2^2/ + !/' + ^ ^ . , ,.^ ^^ 9. Reduce —, — to a mixed quantity. x + y X Ans. X + y -^ ; — ^ x + y X^ — - 6c^cf — — 77i 10. Reduce ^—z to a mixed quantity . X'^ 771 „ Alls. —^—5 2a 2cd 11. Reduce — ^7-7 — to a mixed quantity. \ «>. u. «' + «^' CASE m. 111. To reduce any fraction to the form of an entire quantity. It is evident that if an algebraic quantity be in the form of a fraction, and the fraction in its lowest terms, it will not re- duce to an entire quantity by the last case. But the principle of negative exponents enables us to express the value of any fraction whatever in the form of an entire quantity. 1. Reduce ^ to the form of an entire quantity. Give Case III. REDUCTION. 83 OPERATION. Analysis. It has been shown that a fraction is - = a X -: = a X c-' = ac-' ^^"^^ *^ *^® product of O* C^ its numerator into the -. -. reciprocal of its denomi- nator; or ^ = a X -^ (99). But — = c~^ {SO); whence the ex- pression becomes a X c~' = ac~^. From this example we deduce tbe following Rule. Reduce the fraction to its lowest terms^ and then r)iuUiply its numerator by the reciprocal of its denominator expressed by negative exponents. examples for practice. Reduce the following fractions to the form of entire quan- titles : — 2. a'b c'' Ans. a^bc-^. 3. Ans. m^a^^b^^C^. 4. Sa' 26V Ans. 3 X 2-'a'b-^c-'\ 5. aVcwi ax^cm*' Ans. ax-^nr^. 6. X — y X + y Ans. {X — y)(x + y)-\ 7. a^ + 2( ic -f c* Ans. (a + c) (a — c{''\ 8. ahn — 2am^ -{- m'' Ans. mz (a — m)""*. 9. x^ — ^x'z' + z^ Ans . (X' - 2O (a;^ + sT'P^\ Give Analysis. Rule. g4 FKACTIONS. 112. Since a factor with a positive exponent may be trans- ferred from the denominator to the numerator by making its exponent negative, a factor with a negative exponent may be transferred by making its exponent positive. Hence we have this general conclusion : A factor may he transferred from either term of afrac- Hon to the other, by changing the sign of its exponent, EXAMPLES FOR PRACTICE. 1. Reduce -^7^ to the form of an entire quantity. Ans. a^bc^. 2. Reduce :r—^ to the form of an entire quantity. 7n~ x~ *^ Ans, Soc^y^m^. c(a — — 7yC\ 3. Reduce —7-^^ tAt- to the form of an entire quantity. ^ ^ Ans. c(a' — m'y 4. Reduce _^ _^ to the form of an entire quantity. Ans. ab'^c^, X ~ ^b^ b^z^ 5. Reduce to positive exponents. Ans. — . cz-^ "^ "^ cx^ 'fyi(x — — IJ)^^ 6. Reduce — ^^^ ^^^— to positive exponents. ^ . m Ans, X' — y' Reduce the following fractions to forms having only known quantities in the numerators and unknown quantities in the denominators. i -- Ans '"^"' ■ c-'xz «*z 9 ^^ Ans ^~' How may a factor be transferred from one term of a fraction to the other! REDUCTION. 85 CASE IV 113. To reduce a mixed quantity to a fraction. 1. Reduce 2| to a fraction. ^16 , ^, 16 3 16 + 3 19 Analysis 2 = -^ : and 2| = -g- -{- -g- = — g — = -g-. X 2. Reduce a + ?- to a fraction. OPERATION Analysis. It is evident that a =« ,x__ ah + X y- But y = ah X y, (99) ; also ^ b'^ b X 1 ^^^. 1^^, — =x X 7" ; and ao times r- addea Ob' b to X times r- is equal to a6 + a; times 7-, or (a6 + a;) X 7-, which ab + X ^ IS equal to — 7 — , the answer. The algebraic operation is exactly like the arithmetical, and governed by the same principle. Hence, the Rule. Ilultiply the entire jjart by the denominator of the fraction ; add the numerator if the sign of the fraction he plus, and subtract it if the sign be minuSy and write the result over the denominator . EXAMPLEb FOR PRACTICE. 3. Reduce 7^ and ax -^ — to fractions. ^ A ,a .acx + b Ans. %^ and . ^ c X 4. Reduce 3 — ^ and x^ to fractions. y x^ti — — X Ans. 4 and — ^ . y 5. Reduce y — 1 + = — — ^ to a fractional form. y + l Give Case IV. Analysis. Rule. g5 FRACTIONS. 6. Reduce .c -f V H to the form of a fraction. ^ + y x + y 7. lleduce 4 + 2x + to a fraction. ^ , 4:c + 2cx+h Ans. . c . ^ . r 2x + 6 ^ ,. . 13x — 5 8. Reduce 5a; ^ — to a fraction. Ans. -5 — . 6 o 9. Reduce Ba — 9 —7,- to a fraction, a-i- 6 3 Ans, a + 3 2ax -fa' , . , (x + aY 10. Reduce x A to a fraction. Ans. -. X X & 11. Reduce a + 6 H — -r to a fraction. Ans. ;— 2a:* ft^ 4-rr* 12. Reduce a 4- cc H — to a fraction. Ans. . a — X a — X ax n ' A ^ — 2aa; 13. Reduce a to a rraction. Ans. . a — X a — X CASE V. 114. To reduce fractions to a common denominator. 1. Reduce -, -, and -, to a common denominator. x' y z Analysis. "We multiply both terms of each frao tion by the denominators of the others; that is, the terms of the first by yz^ the terms of tiie second by iC2, a,nd the terms of the third by xy. This process cannot alter the values of the fractions (105, III.,) and it must reduce them to a common denominator, because each new denominator is necessarily the product of all the given denominators. Hence, the iA "What is Case V ? Give analysis. a ayz X b y xyz xbz xyz c z ^xyc xyz REDUCTION. 87 lluLE. Multiply each numerator by all the denominators except its own, for the new numerators ; and all the denomi- nators together for a common denominator. Note. — Mixed quantities must first be reduced to fractious, and entire quantities to tractioual forms by writing 1 for a denominator. EXAMPLES FOR PRACTICE. 8a: 26 2. Reduce -^, x-» and d. to a common denominator. 2a 6c , 9 ex 4ab , Qacd Ans. ^— , ^-, and — — . bac bac bac 8 Reduce — -, — , and — , to a common denominator. m^ mx c *^ acmx bc'^m'^ , m*x Ans, — r — , -v-, and —. — . m^xc 77VXC nrxG 3 2x 2x 4. Reduce 7-, -^-, and a + -— , to a common denominator. 4 o a da Sax ^ 12a'' -4- 24x Ans. ~— , —-^ , and :r- . 12a i2a 12a 5. Reduce , — , — , and -— - — -, to a common deno- . . ^ — y iXi -\- y x^ -{- y^ mmator. :/ ^ i/ ^ if a{a^ -f xy"^ -f x'^y + y^) m(a^ + xy'^ — x^y — y^) z^x"^ — y^) ^ns, 7 7 , 7 , — — . X — 2/* x — y X — y* a "4" c X a 6. Reduce , , and -, to a common denominator. X a — G b . b(a^ — c^) bx^ . axfa — c) Ans. T^^- {, -—. -, and y-7 (. bx{a — c) bx{a — c) bx{a — c) CASE VL 115. To reduce fractions to their least common de- nominator. Since a fraction can be reduced to higher terms only by multiplication, each of the higher denominators it may have Give Rule. Repeat Case VI. 1^ FRACTIONS. must be some multiple of its lowest denominator. Hence, a common denominator for two or more fractions must be a com- mon multiple of their lowest denominators, and the least com- mon denominator must be the least common multiple. C 771/ I. Reduce -r^ and— r; to their least common denominator. ab^ a^b OPERATIOlSr. a^b^ ~- ab^ = a] and c X a = ao, a^b^ -r- a^b ==b] and m X 6 = 6m. c ac ^ m bm ab' ^~aW ' a^ ^'¥^'' Analysis. We find by inspection that a^b^ is the least common mul- tiple of the given denominators ; it is, therefore, the least denomi- nator to which the fractions can be reduced. To ascertain what factor will reduce each denominator to a^i^, we divide this term by each denominator, and obtain a and b. Since the given deno- minators must be multiplied by a and b respectively to reduce them to the required denominator, the corresponding numerators must also be multiplied by these factors for the new numerators ; and we have c X a, or ac^ for the first numerator, and m X b, or biUj for the second numerator, and — — and — r, the answer. a^b d'b Hence the following Rule. I. Find the least common multiple of all the denO" minators for the least common denominator. II. Divide this common denominator by each of the given denominators, and multiply each numerator by the corre- sponding quotient. The products will be the new nur merators, EXAMPLES FOR PRACTICE. Reduce the following fractions and mixed quantities to then least common denominator. * 9 ^ ^ or,^ ^ A h^"cdm acdx _ ahh 2- ac' ¥c' ^""^ ^ ^"«- abVd' ^d^d' ^"'^ a^d- Give analyBis. Kule. REDUCTION. 89 « m c 4- m ^ d . mbc dbc + abm , acd 3. -;„ , and —:, Arts. -,7-, 57 1 and -^r-. a^' ac ah a^hc a'bo a^bo A ^ + ^ g — b a^ iacx^ -f- 4:bcx^ ^o?c — 6a&(? ^ 3a*a? ^ns. ...... , — T7T-0— 1 — , and ^ e — d X , . ac — ad bx _ a^6m 5. — r— , -2, and m. Ans. :pr—, -^, and — ^. ah a^ a^b a^b a^b or o aocu ■ I ■ cc/ c oc ij 6. a + -, — , and ic. Ans. —^ , — , aiid -^. y xy xy xy xy ^ a b 1 <^ 7. , — j — , and x — y' X '\' y' x^ — y^ a(x + y) bfx — y) , c x'^ — y^ * x^ — 2/* ' x^ — y** 8, ^ ^ITi'Sr^T'^'^'i^rzn- . x^ 4- x^ A- x^ + X X* -{- x"^ _ ic* Ane. ,— 3 . -^—^ , and ^— ^. X* ^ a — b , a — b 9. and ao a(a + 6)* a' — ¥ T cfa — 5") Ans, — — yr and —^^ — — r^. ac(a + 0) ac(a -f- 6) 10. .^^-^ and i(l — m^) c( 1 — m) * j_ng ?£ and "^^^ "^ ""^ 8* 00 FBACTIONS. ADDITION. fil6. We have seen (50), that entire quantities may be added wlien they have a common factor, to serve as the unit of addition. In like manner, fractions may be added when they have a common unit; and since the fractional unit is the reciprDcal of the denominator, fractions to be added must have a common denominator. CL C 1 What is the sum of ^ and - ? h Analysts. The fractions have a com- OPERATION. I ^ d Q a 4- c ^^^ ^^^*' tT* ^^ T *^^^^ ^^^^ ^^ taken a times, and in j- it is taken c times ; hence, in the sum of the fractions, it must be taken a plus c times, expressed 2. What is the sum of -, — , and -, — ? on bm OPERATION. J- = J \ Analysis. We first reduce /„ . , , the given fractions to their n rrti f Fractions reduced , " . . _ = ±11. V to a common de- ^^^ commoD denominator, 671 bmnf nominator. and then add as in the first d dn \ example. bm bnin' amn cm dn amn + cm -f dn bmn bmn bmn bmn From these examples we derive the following Rule. I. Reduce the fractions to their lead common de^ nominator. II. Add the numerators^ and write the result over the common denominator. What is a fractional unit. Give analysis of Addition of Fractions. Biile. ADDITION. 91 Notes. 1. When there are mixed quantities, the entire quantities and the fractions may be added separately ; or the mixed quantities may be 1 educed to fractions and added. 2. A tractional result should be reduced to its lowest terms. EXAMPLES F<)R PRACTICE. 3. Wliat is the sum of .-^ and — :, — ? 4. What is the sum of -, — , and -- ? y axi^ X Ans. ba + Sm Ans. 66 1 a + z + y' xy 5. What is the sum of - and ? he X 6. What is the sum of ^, q, and j ? Ans. x + y^. Z O 4: Jim* rj^ 2 \X 7. What is the sum of — ^, — and ~ 1 ^Oo: 14 ' J„s. — 2j— 11 2a 8. What is the sum of r and -r ? Ans. -f a -^ b a — b ' d^ — 6** 9. Add -^- to -^—, Ans. ^l+y]. X -j- y X — y x^ — 2/ 10. Add — -^-^ to — ;:: Ans. — = . SOG ic bo 11. Add :j — j — , = , and = — ; — . Ans, 1 + a' 1 — a' 1 + a 1 — a ' 12. Add — , — , and j-. Ans, r-^-^ . b ob ia I2ab ^^^ 6a6 — 86»— .12acf 166c ^3a — 46 13. Add =-c|- and — ^r— . 126e 36 2a — 6 -4 ns. — -2 . 4c I FRACTIONS. 14. Add 2a:, Sx + x» ^^^ ^ "^ T" ^^' ^^ "*" "45 15. Add 5a; + ' ^ s-i^d 4a7 + 3 **""•" ^ 5^ • ^ns. 9x -A '- — =-^ 15a: ^h 1 16. Add TTT— T-Tx a^^ — ;-?• a — b 17. Add — r— , — r — , and . ab be ac Ans, 0. 18. Add and . . a — X Ans, , X 19. Add — ^--, , and 5-. ^ns. o - y ay da Say ^., , ^,a + b ^ a — b . 2(a^ + b^) 20. Add — "-T to — — T. Ans. \ rr^, . , - a a — 86 ^ a' — 5* — ab 21. Add -7, J—, and r--^ . ^ ^^ ^^'^ , acd^W + a' Ans. J-— . bed 22. Find the sum of r and 7. Ans. —, 7- a -i- b a — b a^ — b' ,2* 1 V X 23. Find the sum of and „ _ Ayis. -^ -,. X + y x^ — y^ x^ — y^ 24 Fmd the sum of = r and v—; — s* ^ws. ^ ^ 1 — a* 1 -f a^ 1 — a^ 25 Find the sum of — hi- and 1 — ( — -, — \ a b \ ab J Ans, L SUBTRACTION. gg SUBTRACTION. 117, We have seen (54), that one entire quantity may be subtracted from another, when they have a common factor to serve as the unit of subtraction. In like mannei, one fraction may be subtracted from another w^hen they have the .^me frac- tional unit, or a common denominator. I. From -' subtract ^. Analysis. The fractions have a com- OPERATION. 2 a d Q d ^ mon unit, -r- ^^ Tt this unit is taken a times, and in -j- it is taken c times ; hence, in the difference of the fractions, it is taken a minus c times, a — c expressed — r — . From this example we derive the following Rule. I. Reduce the fractions to their least common de^ nominator, II. Subtract the numerator of the subtrahend from the nu- merator of the minuend, and write the r^esult over the common denominator. Note. — Mixed quantities must be reduced to fractions before subtract^ ing; and fractional results should be reduced to their lowest terms EXAMPLES FOR PRACTICE. *jx 2x 1 2 From -r- take — ^ — . 21x — 4:x+2 17rr-f2 .^nS, yi ■ = 7^ • Un^er what circunjstance^- can one fraction be subtracted from an- other t (Stiyia an&tysis. Rule. 94 FRACTIONS, 3. From take x^y x-\-y OPERATION. 1 1 ^ + 2/ ^ — y ^y X — y X -\- y x^ — y^ x'^ — i/^ ^^ — 2/** 4. From .^ take -~. Ans. ^ _ _ 2ax . K>ax . Wax 5. From -^ take -^. An%, ^r-. 8 2 b -I ^^ C} q 6. From — ; — -- take -^ ■=* An%, a + 1 a=^ — a + 1 ' 1 4. a'*' . _ 3^.5 . 3.T -~ 10a 7. From -r- take ,7-. -ans. -. 4a Ix 4ax 8. From 7- take rr-, ^ns, — =-x . 4:X oa iZax I 2 3 X 9. From ^ take — — ^. ^ns. -^ ='. X — 1 X + 1 x^ — 1 10. From 2a — 2x + ^ ^ take 2a — 4a; + ^ ^ . a a; a'^ — a;' Ans. 2x + ax 11. From ^^ take — ;;; . Ans. -^^^-r? . oc ic 35c ,„ _ 5x + l,i 210^ + 3 ^ 127^+17 12. From — - — take . Ans. ^1 . 7 4 ^o 13. From --^—^ take ^ ^ . -4ns. — ;;-^- ^. -:a oa 6a 1.1 T7 1 +a\ , l — a^ . 4a' 14. From = ^ take zr— - — j. Am. = ^,. 1 — a^ 1 4- tt*^ 1 — a 15. From x + ^'pL take -^-^. X— '^y SUBTRACTION. 95 16 FrQm — ^ — take — ^- — . 17. From ^^ — ^s-^ take , . -472s. ^, a — a -|- a — o 18. From ^ take , Ana, X — 3 X * ' x^ — 3x* 19. From 6a H —- — take 4a + T" • Ans. 2a + ^ti. 20. What is the value of ^^^^~ ^ ? a^ — ^2 ^ — ^ ^ ^WS, — r-T. a -f O 21. What is the value of l-±^* Irifi' f l — x^ I + x^ 4^ 22. What is the value of 1— X*' a — X a* — x^ „ Ans. X 23. What is the value of ^^ + -=^ — ^- ? oc ac ab o a rt'^ -rrri . . 1 i ^ 3x 2x 5.T _ , 21a; 24. What IS the value K)f -7- + -= ^ r ^ns. -77^. 4 5 o 40 25. From take v ^^^^ —^ — — n n — 1 n — n 26. From -z r, take -z . Ans, 1 — x'^ 1 — X ' 1 -f- x' 27 From 4+^— take , ^i^— r.. (rt — b) {x — a) {ci — 6) (x — 6) -, x -4- c , (^a: — a) (x — 0) 96 FRACTIONS. MULTIPLICATION. CASE I. 118. To multiply a fraction by an entire quantity, 1, Multiply -by c. Analysis. A fraction may be multiplied OPERATION. ^^ multiplying its numerator (105, I) ; ^N-e £j dQ therefore multiply the numerator, a, by c, o and obtain for the required product, -r-- 2. Multiply — hj X, Analysis. A fraction may be multiplied OPERATION. ^y dividing its denominator (105, 11) ; we therefore divide the denominator, xy, by x, mm ^ -— - X ^ = — and obtain for the required product, — . *^y y y Hence, Multiphjing a fraction consists in multiplying its numera- tor, or dividing its denominator. EXAMPLES FOR PRACTICE. c - . cm . 3. Multiply J by m. Ans, -r-, a'^x 4, Multiply -j by ax. Ans, c — df 5. Multiply -j-z by cd. Am, — . a Multiply g| by 7. V Ans. -^• 7 Multiply zrhj a + X. Ans. r-. X — — X X — — X 8. Multiply =- — - — - hj X + y. Ans. -r- ^'' h(^x + y) -" ^ b What is Case I ? Give analysis. Deduction. MCJLTIPLICATIOK. 97 ,. 9. Multiply —:, 5 by m — v. Ans. -— I ^ ^ m' — y^ ' "^ rn + y 10. Multiply — by x^ + !• Ans. --g . iu ■ X Sj —•— X 119. It is usually advantageous to indicate the multiplicap tion, and apply cancellation before obtaining the actual product. 1. Multiply - by x. OPERATION. Analysis. Having indicated t1i« mul- tiplication, we cancel the common factor, ^ ^ ^ _. ^ a;, from both numerator and denominator, X and we have a for the product. 2. Multiply ^ by 6m. OPiCRATlON. Analysis. Having indicated the mul- c X 6m tiplication, we cancel 3m, and obtain 2c — g— — = 2o for the product. Hence, 1. A fraction is multiplied by its own denominator by simply suppressing the denominator. 11. If a fraction be multiplied by its own denominator, or by any multiple of that denominator, the product will be an entire quantity. ^ EXAMPLES FOR PRAOTIOB. 35 3. Multiply — by t/. Ans, x. 4. Multiply -Tj- by 5ft. Ans. Sax. cd} 5. Multiply by a — x. Ans. cd}. a ' X When may cancellation be applied in mnltiplioatlon ? Give fixvt dedutrtioi^. Seodnd. 9 O 98 FRACTIONS. 6. Multiply — by 2x». Ans. 6^V. cc 7. Multiply ~~7r-^ by 20. Ans. 6a — 2x. 10 a — X . 8. Multiply -ggg- by 896. -4ns. a — x. 9. Multiply -2 3 by a + a;, Ans, 10. Multiply -^ by x^ — 1. Ans. 3c(a: + 1). 1 1. Multiply ^5-i^ by a" — 2ab + h\ Ans. a" — h\ CASE n. 120. To multiply an entire or a fractional quantity l)y a fraction. 1. Multiply a by -. FIRST OPERATION. ANALYSIS. It IS evident that the product of two quantities is the same, ^ X_-__ y^ as=^ whichever be taken as the multiplier ; COG h consequently a, multiplied by — , is eqaal to -- multiplied by a ; and, according to Case I, — multiplied . «& by a, IS — . c SECOND OPERATION. Analysis. "We first reduce the multi- 6 , , b — =00 plier, — , to an entire form, 5— by =^. 14. Divide — r — by 42/ 15. Divide ~g- by -^. 16. Divide^ by |5. 17. Divide a; , a; :ri^y2' -4ns u4ns. .4ns. Ans. l — a 2 am Wz 2c * d{my + a?) Ans, Ans, X -{ . c 2a + X 70a; — 15 Ans, Ans, Ans, 10a;— T* 9a; — 3 x 18a; — 21 a;=» — 1 Ans, 12a. 8?/ 18. Divide 5^J-£. by ^ ab be Ans, Ans. Ans, Ans, Ans, 5' 7 2.i' a -f- 1 2 a;— I ox — , y DIVISION. ;i05 19. Divide !^— - by ^^5-+?. Ans, 2m — 2n. 20. Divide -r- ^7 jst- Ans, -,—. o do 2a 21. Divide -77— f- by -7-^. -dns. -tt-;; — . 22. Divide , ^~\,, by ?^^. ^n8. a: + ~ 23. Divide l + \\)yl — \ Arts, x — h Scd x' — 6* a a^ 'a — 1* 2^. Divide (,-i- + ^-) by P^Z- \i + X 1 — xJ ^ (i—xy Ans. :^ ■. 1 4- .^;' 6 + 1 25. Divide (^ + 2.^) by (6 + J--1). ^n. ^^, . 123. The division of one fraction by another, or of one mixed quantity by another, may be indicated in the form of a complex fraction, and the result reduced to a simple fraction, 1. Divide a -| — by x ■{- —^ c z Analysis. We indicate the division OPEBATION. |3y writing the dividend above a horizon- a '^~~- *^^ ^i°6> ^^^ the divisor below. Then, C^ gC2: + bz since the denominator of a fraction will y cxz + cy disappear when the fraction is multiplied ^ • ^ by any multiple of its denominator (119, II), we multiply both numerator and denominator of the complex fraction by cZy the least common multiple of the denominators of the fractional parts, and obtain tho acz -}- bz simple fraction, jT — . Hence, to simplify a complex fraction, Multiply both numerator and denominator by the least com- mon multiple of the denominators of the fractional parts. Explain the process of reducing a complex to a simple fraction. Give deduction. lOf FRACTIONS. EXAMPLES FOR PRACTICE. 2^ 2. Reduce ^ to a simple fraction. Atis. p. 3. Reduce ^ ' " to a simple fraction. _ _ b — i Ans. ^±l'Il^ bdn — en 4. Reduce — 7—^ to a simple fraction. Ans. ^- 60 - + c a 4- Ac 5. Reduce - — ~- to a simple fraction. Ans. ~. x + ^ ^ 4ar + 2z 6. Reduce p to a simple fraction. Ans, a + T ac + 1* 7. Simplify the fraction -, Ans, 1 + ^ »i + wi 8. Simplify the fraction ^ " 1-^ + ^ 1 *" . {a — 3>r^ a{x^ — X +1) 9. Simplify the fraction ^Je:' Ans. l^^l-^ZZ^. ^^ ■_! n^ 10. Simplify the fraction ^"-^^ . Ans. — . 1 1. Reduce ^* to a simple fraction. x^jo^ — x) Jjjj^^Y" EQUATIONS. K ^ ■'^' \\^. 107 LA.,.-- ft U SECTION II. J EQUATIONS. \ IS4. An Equation is an expression of equality between \ two quantities; thus, «: = 4, 5u: = 60, 3x = a + &, are ! equations. j Tlis First Member of an equation is the quantity on the left i of the sign ; and I The Second Member is the quantity on the right of the j sign ; thus, in the equation, a -j- b = 7x — y, the quantity, \ a + &, is the first member of the equation, and the quantity, j 7x — ?/, is the second member. 1 t^fei. An Arithmetical Equation is ope which expresses | the equality of numbers or sets of numbers ; as 10 = 10 ; \ | 4+3 = 6 + 1. '. Y I^H. An Algebraic Equation is one- which contains one or ] more literal quantities ;' as, ox = Ig ; c{a + b) = d. Alge- braic equations serve to express the relation^ between known I and unknown quantities, and to determine the values of the ] unknown quantities by comparing them with some that are j known. / iSI7. A Numeral Equation is one in which all the known J quantities are expressed by numbers ; as, ox -{- 2x = 25. . ] 1^8. A Literal Equation is one^in which some or all the l known quantities are expressed by letters ; as, x + 3ca: = m; | ^2/ + cc^ = 91. ^ ^ 1 ISO. An Identical Equation is one in which the tw^-roem^'^ hS| Define an equation. The members. An arithmetical equation. An Blgebraic equation. A numeral equation. A literal equation. An ideu* tical equation. 108 SIMPLE EQUATIONS. bers are the same, or capable of being reduced to the same ex- pression by performing the operations indicated. Thus, 9^ 1 _ 2x 1") " ' y are identical equations, ox -j- ox = ox J 130. The Degree of an equation is denoted by the highest exponent of the unknown quantity in the equation. Thus, X — ar ^^^ equations of the first degree. CC ~y- OX = C J ' '^ [ are equations of the second degree. ax^ -{- Ox =h) are equations of the third degree, &c. x^ = a\ -{- bx= c) 131. A Simple equation is an equation of the first degree. 13S. A ftnadratic equation is an equaticto of the second degree. \ v 133. A Cubic equation is an equation of the third degree. TRANSFORMATION OF EQUATIONS. 134. The Transformation of an equation is the process of changing its form without destroying the equality of its mem- bers. ■ f Since an equation is only an expression of equality l^etween two quantities, all the changes that can be made in the mem- bers of an equation, by which their values are altered without destroying their equality, are embraced in the axioms (4©), and may be stated as follows : — I. The same or equal quojitiiies may be added to both members, (Ax. 1.) II. The same or equal quantities may bet subtracted from both members, (Ax. 2.) III. Both members may be multiplied by the same or equal quantities. (Ax. 8.) Define the degree of an equation. A simple equation. A quadratie equation. A cubic equation* What is the trauBformation of an equation? State the principles upon which all transformations are based. ONE UNKNOWN QUANTITY. 1Q(J LY. Both membei's may he divided by the same or equal quantities. (Ax. 4.) Y. Both members may be raised, by involution, to the same power. (Ax. 8.) YI. Both members may be reduced, by evolution, to the same root. (Ax. 9.) Note. — As the principal object in transforming an equation is to find the value of the unknown quantity, we present here only those cases necessary to the solution of Simple Equations. I CASE I. 130. To transpose any term of an equation. ^ Traijsposition is tire process of changing a term from one member of an equation to the other, wit^iout destroying the equality. \ 1. In X + a = 6, transpose a to the second member. OPERATION. Analysis. Since the equality « , 7 of the members is not destroyed by taking the same quantity from both (134, 11), we subtract a from ^ X = b — a each member, and obtain for a re- H^ult, x = b — a. The same result may be ob- tained by dropping + « from the Subtract, a = a Or, x + a=b . cc = 6 — # first member, and writing — a inO# the second member, as in the second operation. 2. Inb = X — c, transpose c to the first member. OPERATION. Analysis. We add c to both b = X — c members of the equation (134, Add, c = c I)» and obtain b -\- c = x. The b 4- G~^x same result may be obtained by dropping — c from the second Or, b = x — c member and writing + c in the b -\- c =: X ^^^^ member. What is the object of transformation? What is Case I? Transposi tion? Give analyses. 10 110 SIMPLE EQUATIO^^S. It will be seen that in each of these examples, the term transposed disappears from the member in which it is given, and reappears in the other member with the opposite sign. Hence the following KuLE. Dy^op the term to he transjiosed from the member in which it stands, and write it in the other member with its sign changed, EXAMPLES FOR PRACTICE. In the following equations transpose the unknown terms to the first member, and the known terms to the second. 3. dx + m = b. • Ans. 3x = b — m. 4. 4:x = 2x -\- cd. Ans, 4:x — 2x =^ cd, 5. 3m 4- 12.C -— c = ic + d Ans. \2x — X ^= d — Sm + c, 6. — bc'^d + ax = — bx — m. Ans. ax + bx = bc'^d — m. 7. 4ac'x — od^ = a^d — d^x, Ans. 4:acx + d^x = d^d + ScZ"^ 8 a + b — x — c' = 2c — bx. Ani!^ bx — x = 2g — a — b + c\ 9. a^x — cd = b + a'^x — ax. Arts, a^x — a^x + ax ==b -\- cd. 10. ab — xc = bed — g. Ans. — xc== bed — g — ab, 11. m = ax — Sex + m\ Ans. Sex — ax = m^ — m 12. = ab—Scx — 2ax-^c, Ans, Sex + 2aa; = ab — o. CASE IT. IS©. To clear an equation of fractions. We have seen (119, II), that if a fraction be multiplied by auj multiple of its denominator, the prodiu't will be an entire ,1 Give Rule. Case II. ONE UNKNOWN QUANTITY. m quantity. Ilence it follows, that if several fractions be multi- plied by a common multiple of their denominators, all the pro- ducts will be entire quantities. 1. Transform — -f — = 3 into an equation having no frac- tional terms. OPERATION, Analysis. We multiply every ^erm of ^ the equation by 12, the least common mul- ^. -- = 3 tiple of the denominators, canceling each ^ ^ denominator as we proceed ; thus, 4 is con- 9x -\- 2x = 36 tained in 12, 3 times, and 3 times Zx is 9x ; 6 is contained in 12 twice, and 2 times x ia 2x; and passing to the .second member, 12 times 3 is 86. Hence, the KuLE. Multiply all the terms of the equation by the least common multiple of the denominators, i^educing each frac- tional term to an entire quantity by cancellation. ♦ Notes. 1. If a fraction have the minus sign before it, change all the signs of the numerator, when the denominator disappears. 2. The pupil will readily see that an equation may also be cleared of fractions by multiplying each term by all the denominators. EXAMPLES FOR PRACTICE. Clear the following equations of their fractions : — 2. ^ + ^ = 11. Ans, 3a: + Sx = 132. 3. 5 _ ^= a, / Ans. lOx — 3a; = 12c a -{- c value of a:, VERIFICATION. a -{- c To verify this value, wo substitute it for x in the given ? V. __i£_ j_. h (I) equation, and obtain (i); por- C a + C a + c forming the operations in- ab ab dicated, we have (2), in which cTa^c ^^ a 4- c' ^^' *^^® *^^ members are identi- cal, and the value obtained for X is verified, and the equation Is satisfied. From these examples we deduce the following Rule. To reduce an equation : — I. Clear the equation offraciionSj and perform all the operations indicated. II. Transpose the unknown terms to the first member of the equation, aiid the known terms to the second member, and reduce each member to its simplest expression, factoring, when necessary, xoith reference to the unknown quantity. III. Divide both members by the coefficient of the unknown quantity, and the second member loill be the value required. To verify the result : — Substitute the value found for the unknown quantity in the given equation, and perform the operations indicated. If the result is an identical equation, the value is verified. The three steps in the reduction of a simple equation, con- Give the Rule for reducing an equation. For verifying the result The steps in a rod action. 116 SIMPLE EQUATIONS. taining but one unknown quantity, may be briefly stated as follows : — 1st. Clear of fractions. 2d. Transpose and unite terms. 8d. Divide. Notes. 1. It is often advantageous to transpose and reduce, in part, before clearing of fractions. 2. In the last step, when the coefficient of the unknown quantity is negative, dividing will give a positive result. EXAMPLES rOR PRACTICE. 4. Given bx + 22 — 2x= 81, to find x. Ans, x = S. 5. Given 4a; + 20 — 6 = 34, to find x, Ans. a; == 5. 6. Given 8a; + 12 + 7a; = 102, to find x, Ans. a; = 9. 7. Given 10a; — 6a; + 14 = 62, to find x. Ans. x = 12. 8. Given bx — 10 = 3a; + 12, to find x. Ans. x == 11. 9. Given 8a; — 20 = — x — 4, to find a;. Ans.x=4t 10. Given 4a; + 45 = 7a; — 80, to find x. Ans. x = 25. 11. Given x — 1 = 4a; — 91, to find x. Ans, x = 80. 12. Given 8(a; + 1) +.4(a; + 2) == 6(a; + 3), to find x. Ans. a; = 7. Note. — First perform the multiplications indicated, and then reduca 13. Given 5(a; + 1) + 6(a; -f 2) = 6(a; + 7), to find x, Ans. a; s= 5. 14. Give 7(a; + 3) — 4(3x_16) = 45, tofinda;. Ans. X « 8. 15. Gi^en ax -\- bx= am -f 6m, to find x. OPERATION. Analysis. We factor botli ax + hx ^am-\- bm (i) members of the given equation (a + ft> = (a 4- b)m (2) and obtain equation (2); and ^ dividing bya-r o, the coefncient a; = w (3) of Xt wo obtain a;=m, the anjswer. DNE UNKNOWN QUANilTY. 117 16. Given ex — a? = 5c -^ 6, to find x, Ans, x = b. a — c 17. Given ax+ dx^ a — c, to find x, Atis, x = , . 18. Given ax -f- w = ccc -f n, to find x. n — m a — c 19. Given ax — bx = c + dx — m, to find x, c — m Ans. x = a — b — df 20. Given ^ + 16 = J + J + 17, to find x. OPERATION. o^ XX Analysis. We first drop ^IQ = ^ ^ -^ V7 (i) 16 from both members of the SX XX equation, and obtain (2) = — -f — -f 1 (2) Then, clearing of fractions, transposing and reducing, we 6a; = 4rr + cc + 8 (3) ^ave a; = 8, the Ans. x=S (4) X X 21. Given ^ — 3 + — =5 — 3, to find x. Ans. a; == 6. 22. Given ^ — ~ 4- 2 = 3, to find a;. Ans, x = 12. 23. Given -j + g" — fi" ~ 19' *^ ^^^ ^ ^^^* aj = 2 24. Given -o" + j = -^ + To» to find x. Ans. x = 38. 25. Given - + ^^ + 26 = S&, to find x. ^ ^ ^ 2ab-\'6a Ans. -^r-; . 2 + a 26. Given ^ + 2^ + 11 = j + 17, to find x. Ans. X =a 10. 27. Givfen ^a: + ^x + ^vnd 3&, td find the value of a. XX 'X ^.^ ,+ 3 + 4=39 a) 6x + 4x + 3:r=39x 12 (2) 13x = 39 X 12 (3) ic = 3 X 12 (4) or, aj = 36 (5) 118 SIMPLE EQUATIONS. Analysis. "We multiply both members by 12, the least common multiple of the de- nominators, indicating the operation in the second mem- ber, and obtain (2) ; reducinji;, vre have (3) ; we next divide by 13, observing, in the second member, that a quantity may be divided by dividing any one of its factors, and obtain (4) ; wlienca a? ==36. 28. GiTen \x -f ^x 4- ^x-^- ^x= *11, to find x, Ans, X = 60. 29. In }^ + ^x i-|x = 130, find x. Ans. x = 120. 30. Given ^x + ^oj + y*^^ = ^^y ^^ ^^^^ ^• Ans, X = 120. 31. Given iy + Jz/ + 42/ = ^2, to find y. Ans, y = 84. 32. Given 5a; -f ^x + ^x = 34, to find x. Ans, X = 6, _„ _ 3x x—1 ^ 20x + 13 ^ . _ 33. Given ^r — = 6a: -. , to find x. 4 2 4 ' iFIRST OPERATION. 2>x 57 — 1 ^ 20x + 13 4 2 -•^' ■ 4 U/ 3a; — 2a:+ 2 = 24a; — 20a?- -13 (2) -_3a; = — 15 (3) a;= 5 (4) Analysis. AVe multiply by 4 to clear of fractions. The first term multiplied by 4 is 3a: ; the product of the second term is 2a; — 2 , but since the fraction has the minus sign before it, this quantity must be subtracted, v^hich is done by changing the signs of the terms, thus ■ — 2a; + 2. The product of the first term in the second member is 24a; ; the product of the second term, or fraction, in the second member is 20a; + 13, w^hich being subtracted, as indicated by the sign before the fraction, becomes — 20a; — 13. Ke3ucing, we obtain :ne unknown quantity. 119 Note. — Yv«ung operators are liable to the mistake of omitting to change the signs of all the terms, when a fraction having the minus sign before it, and a polynomial numerator, is reduced to an entire quantity. Thia error may be avoided by the method which appears below. SECOND OPERATION. dx + 20x + l^=:24cx + 2x — 2 (S) Sx = 15 (4) a; = 5 (6) Analysis. We first transpose the fractions having the minus sign, and obtain (2) in v^hich the fractions are positive ; then clearing of fractions and reducing, we have a: ^= 5 as before. 34. x—S 9 x+4 ^ ^ ^ Given X ^ — = o q — » *o find x. Ans. x = 2. 35. x + 2 x — S , ^ x — 1 , ^ , Given — q J [-2 = X ^ — , to find x. 36. Ans. x=7, 4x — 2 3a7 — 5 ., , ^ ^ Given — :pi rr^ — = 1, to find x, , 11 13 Ans, a? = 6. 37. X x — 2 a; , 13 ^ ^ ^ Given g- o — = — 2 "^ T' ^' Ans. X = 10. Find the value of the unknown quantity in each of th« following equations : OQ ^ , ^ , ^ ^ . 12a 2 "*" 3 i" ~ ^^' ^ "^ "IF* a? a; a; - ahc 39. - + - + — = 1. Ans. X = 40. a b c ' ' ac + 6c? -f a5* 1+x 1+a ^ _1 1S.0 SIMPLE EQUATIONS. 41. —^ g ^ = _1. Ans. a?=«13. 42. 1 — 5-^^ + 8 + 1 — 10 = 100^13. o 4 O Ans, X = 120. 15 1 43 — To — Q — ro = ^Tl- -4ns. j;=:2. a; + o 3a; + 9 ' ^ 44 - -j (- ~ = 2, -47W. X = -Ka^ + 6^ + c^. X OL X 45. (a + b) (a — b) (a; .^ m) == fc^m. Ans, x = 11^ — 80 8^ — 5 . 46. g -—- — O. Ans. 37 = 10. An 1 + I + m 12(m + 6) 47. ^ t = 9 ' • ^ns. a; = 20. 48. ^-3 + 2(^-3) = ^-^ + a. OPERATION. Analysis. To sim- Put a; — 3 = 2/ ; (i) ^^^^"^ *^® equation, we represent x — 3 by y, Then y + 2^/ = — ~ + o (2) ^"^ ^^e equivalent equa- ^ 4 tion is (2), reducing 12a "which, we obtain y = y==-BB ^ 12a ^ "oF* -But y represents Or, X — 3 = -^^ (4) x—3 ; and restoring this 35 value in (i), we obtain «. Q r -^^^ ^s <^)» whence, by trans- posing, a; = 3 + -35-. 49. ^+ ^^^+1) = 3(x+ 5)_20. Ans-x^l £0. ^_c + ^ + 5^^IZf} = 2i ^ns. a;= a + 1. .- _. 9a; + 20 4a;_12 a; ^ ^ ^ 51. Given — gg— = ^__ + -, to find x. ONE UNKNOAVN QUANTITY. 121 OPERATION, Ax 9a7 + 20 _ 4^7 — 12 x ~M "5^—4 "^ 4 36(4^ — 12) bx — 4 20(5;r — 4)==36(4^' — 12) 5(5a; — 4) = 9(4:t;~12) — 11a; =—88 x^S 20 = -4 52. 6x + 7 . 7a; — 13 Given — J-, h (1) (2) (3) (4) (5) (6) Analysis. We first multiply by o6, and dropping 9x from both members, we obtain (2). Next, clearing of fractions, we obtain (3) ; dividing by 4, we ob- tain (4) ; performing the multiplication indi- cated, and reducing, we obtain (5) and (6). 2ar + 4 9 6^ + 3 , to find X. 6j;+ 7 9 3 + OPERATION. 7^—13 2a? + 4 6a; + 3 ■ 7a?— 13 ^ ^^Tl ' 21a? — 39 3 2x + 4 = 5 2a7+ 1 21x — 39 = 10a; + 5 11a; = 44 a? = 4 (1^ (2) (3) (4) (5) (6) Analysis. "We first multiply each term by 3, which is done by dividing each denominator by 3, and obtain (2) ; multiplying again by 3, and reducing, we obtain (3) ; clear- ing effractions and reducing, we obtain, finally, (6). Note. — By clearing eqoiations of the simplest denominators tirst, aa iTi the examples just given, we sometimes avoid not ofaly laborious multi- plications, but the involution of the unknown quantity to higher powers. 53. Given 7a? + 16 21 ri h -Q» to find X. 21 Ans, 64. Given 65. Given 7x + 2a X + a 21 4a; — 11 ' 2ar+l 402 — 3a; 29 11 5, to find X. Ans, X = 471 — Qx 12 43a 8a — 21* , to find X, Am, x = 72. J22 SIMPLE EQUATIONS. I ^^ ^. 18a? — 19 .11^ + 21 9^ + 15 ,» , \ 56. Given — ^- + -g^^ = — ^, to find ;.. . I Q I -I Q I 57. Given ^ + 5 + x = — -q— , to find :r. ' Ans. x = %. I 58. Given ~^^ —. — ^ = —-rjr + » ^^ ^^^ ^' ] Ans, X = — • ! PROPORTION. \ i 141. It is often convenient to express the relations of alge- ■ braic quantities in the form of a proportion, and from the \ proportion derive an equation. ■■. For this reason we present here so much of the theory of ] proportion as will enable the pupil to make use of this prac- j tical advantage, reserving the full discussion of the subject for j a subsequent chapter. ] 14S. Ratio is the quotient of one number divided by an* i B i other. Thus, the ratio of -4 to ^ is -r. \ A 143. Proportion is the equality of ratios. i Thus, if -^ = r, or ^ = rv4 ; 1 and —=: Tf or D = rG'y j G J then the four quantities. Ay B, (7, and D, are proportional, j and their proportionality is expressed thus : i A: B :: G : D, I in which A and Z) are called the extremes^ and B and C the I means. \ Define Ratio. Proportion. The extremes. The meana. ^ ONE UNKNOWN QUANTITY. 123 If in the place of B and D we write their values, rA and rCj this proportion will be A : rA : : C : rC ] and we have A X rC = rACy the product of the extremes ; and G X rA = rACf the product of the means. Hence, The product of the extremes is equal to the product of the means. 1 Given 2 : x : : Q : bx — 4, to find x, OPERATION. Analysis. The extremes are 2 and 2 : X : : Q : bx 4 ^^ — 4, and their product is 10a; — 8 ; -I A 8 = 6a: (1) ^^^ means are x and 6, and their pro- A ci ^ duct is 6a:; and since these products are equal, we have equation (i). ^' = -^ (^) Reducing, we have x = 2. Hence, to convert a proportion into an equation, we have the following Rule. Place the product of the extremes equal to the prch duct of the means. EXAMPLES FOR PRACTICE. 2. Given x : 25 : : 60 : 3, to find x. Ans, x = 500. 3. Given a: : x + 6 : : 2 : 6, to find x. Arts, a: = 3. 4. Given x + 2 : a : : 6 : c, to find x, , ah ^ Ans, X = 2. G 5. Given a: + 6 : 38 — x : : 9 : 2, to find a;. Ans. X = 30. 6. Given x + 4 : x — 11 : : 100 : 40, to find x, Ans. X = 21, 7 Given x + a: x — a::c:c?, to find x. Ans. x = -^^ ~. c — d Rule fer changing a proportion to an equation. 124 SIMPLE EQUATIONS. 8. Given x :2x — a : : a : h, to find x. , a/ Ans, X — 2a — ti 9. Given a:b: :2y : dfio find y, Ans. y = ^j, 10. Given a* — ac : ax: :1 : (d — 6), to find x. Ans. x^= (d — h) (a — c). 11. Given cc : 75 — a; : : 3 : 2, to find x. Ans, x == 45 PROBLEMS. 144. A Problem is a question requiring the values of un- known quantities from given conditions. 145. The Solution of a Problem is the process of finding the values of the unknown quantities. We have seen that equations serve to express the relations of algebraic quantities, and to determine the values of the un- known. In Algebra, the solution of problems is generally effected by means of equations. 1. If from five times a certain number, 24 be subtracted, the remainder will be equal to 16 ; required the number. SOLUTION. Analysis. We denote the number \ required by x. By the condition of ' Let X = the number. j,,^ problem, 5 times x minus 24 is I . -^ equal to 16, which, expressed alge- i ' ^ ' braically, gives equation (i). : Dx = ^\J (2) Reducing this equation, we have x \ a: = 8 (3) =8, the number required. ; In this problem, the condition from which the equation is i formed is clearly expressed, and furnishes the equation directly, \ 2, A merchant paid $480 to two men, A and B, and he j paid three times as much to B as to A. How many dollars \ did he pay to each ? \ Define a problem. The solution of a problem. ONE UNKNOWN QUANTITY. 125 SOLUTION. Analysis. We let x :*opre- Let X = the sum paid to A. ^?"' *^^ ^"/^ f'^ *^ ^ ' ^^^ _ ^ . 1 T^ Since he paid 6 times as much ox = the sum paid to B. ^^ j^ ^^ ^^ ^ .3^ ^^i^ represent 4x = the sum paid to both.. the sum paid to B ; and bj ad- T 3"j.Qn dition, \x is the sum paid to * both. But hy one condition X = 120, A's share. (2) ^f the problem, 480 dolhirg ?s 3x = 360, B's share. (3) the sum paid to both. There- fore 4x is equal to 480, which expressed algebraically, is equation (i). Reducing, we find the value of a:, or the sum paid to A, to be $120; and 3 times this sum, or 3 times the value of x, the sum paid to B, $300. In this problem the equation is drawn from an implied con- dition. In the algebraic notation, 4x' is the sum paid to both ; and in the enunciation of the problem, 480 dollars is the sum paid to both ; these two expressions of the same quantity are put equal to each other, to form the equation. From the examples given, we derive the following GENERAL RULE. I. Eepresent one of the quantities whose value is to he de- termined hy some letter or symbolj and from the known rela- tions find an algebraic expression for each of the other quan- tities, if any, involved in the question. ^ II. Form an equation from some condition expressed or implied, hy indicating the operations necessary to verify the value of the unknown quantity, III. Reduce the equation. The three parts of the rule, or the three steps in the solu- tion of a problem, may be named as follows : 1st. The notation; 2d. The equation ; 8d. The reduction of the equation. Note. — By the first two steps, the problem is translated from common into algebraic language; andv>this is called the statement of the problem. Give general Kul%for solution of problems. Steps of the process. 126 SIMPLE EQUATIONS. PROBLEMS. 8. A father is 3 times as old as his son, and the difference of their ages is 24 years ; what is the age of each ? Ans. Son's age, 12 ; father's age, 36. 4. A gentleman purchased a horse, a chaise and a harness, for $230. The chaise cost 3 times as much as the harness, and the horse $20 more than the chaise ; what was the cost of each? r Harness, $30. Alls J Chaise, $90. (Horse, $110. 5. Two men bought a carriage for 86 dollars; one paid 26 dollars more than five times as much as the other ; what did each pay ? Ans. One paid 10, the other 76 dollars. 6. A man had six sons, to whom he gave 120 dollars, giving to each one 4 dollars more than to his next younger brother ; how many dollars did he give to the youngest ? Ans. $10. 7. Three men received 65 dollars, the second receiving 5 dollars more than the first, and the third 10 dollars more than the second ; what sum did the first receive ? Ans, $15. 8. A man paid a debt of 29 dollars, in three different pay- ments ; the second payment was 3 dollars more than at first, and the third payment was twice as much as the second ; what was the amount of the first payment ? Ans. $5. 9. The greater of two numbers exceeds the less by 14, and o times the greater is equal to 10 times the less ; wliat are the numbers ? Ans. 6 and 20. 10. Moses is 16 years younger than his brother Joseph, but 3 times the age of Joseph is equal to 5 times that of Moses ; what are their ages ? Ans. 24 and 40. 11. On a certain day, a merchant paid out $2500 to three men, A , B, and C ; he paid to B $500 less than the sum paid to A, and to C $900 more than to A ; required the sum paid to A. Ans. $700, ONE UNKNOWN QUANTITY. 121 12. There are three nnnibers which together make 72 ; the second is twice as much as the first, and the third is as much as both the others ; what are the numbers ? A71S. 1st is 12; 2d, 24; 3d, 36. 13. A man paid $750 to two creditors, A and B, paying 4 times as many dollars to B as to A ; how much did he pay to each? Ans. To A, $150 ; to B, $600. The last problem would have been essentially the same iu character, if any other sum had been paid out, and if the money had been distributed to the two men in any other ratio. We may therefore make this problem general by stating it as follows : A merchant paid a dollars to two men, A and B, paying n times as many dollars to B as to A ; how much did he pay to each ? SOLUTION. Analysis. Let T , xi • 1 i i ^ denote the sura Let X = the sum paid to A ; ^^^^ ^^ ^ . ^,^^^ nx = the sum paid to B. ^x ^in denote the X + nx= a a) 6""* P^id to B, (l+n)x^a (2) ^"^ ''trJl^ ^ ^ sum paid to both, X = =^-~ — , the sum to A ; (3) which, by the con- 14"^^ dition of tiie pro- na ., . -r. blem, must be nx==^-^^,themmtoB. m ^qual to «, as ex- pressed in (i). Reducing, we obtain the value of x, or A's part, in (3); and multi- plying by n we obtain the value of nx, or B's part, in (4). VERLFICATION. na a -{- na To verify these values, we , add them, and find that their 14- ^1 1-fn 1 -\- n gm-jjj jg ^^ ^\^q whole amount __^ (1 + n)a paid to both, as stated in the ~ 1 -f. itT problem. = a 128 SIMPLE EQUATIONS. 14. My horse and saddle are worth $100, and my horse is worth 7 times as much as my saddle ; what is the value of each ? Ans, Saddle, $12^ ; horse, $87^. 15. My horse and saddle are worth a dollars, and my horse is worth n times as much as my saddle ; what is the value of each ? ^ o. -m ^ 1 ^^ Ans. Saddle, ; horse, = — ■ — . 1 + n ' 1 -f 71 16. A farmer had 4 times as many cows as horses, and 5 times as many sheep as cows, and the number of them all was 100 ; how many horses had he ? Ans, 4. 17. A farmer had n times as many cows as horses, and m times as many sheep as cows, and the number of them all was a ; how many horses had he ? . a , ' *^ Ans, -^ ; horses. i -f ^ + ^^^^ 18. A school-girl had 120 pins and needles, and she had seven times as many pins as needles ; how many had she of each sort ? Ans, 15 needles, and 105 pins. 19. A teacher said that her school consisted of 64 scholars, and that there were 3 times as many in Arithmetic as in Algebra, and 4 times as many in Grammar as in Arithmetic ; how many were there in each study ? Ans. 4 in Algebra ; 12 in Arithmetic ; and 48 in Grammar. 20. There was a school consisting of a scholars ; a certain portion of them studied Algebra, n times as many studied Arithmetic, and there were m times as many in Grammar as in Arithmetic ; how many were in Algebra ? Ans, J— , . 1 + n + rnn 21. A person was $450 in debt. He owed to A a certain ' gnm, to B twice as much as to A, and to C twice as much as to A and B ; how much did he owe each ? Ans. To A, $50 ; to B, $100 ; to C, $300. ONE UNKNOWN QUANTITY 129 22. A person said that he was owing to A a certain sura, to B four times as much, to C eight times as much, and to D gix times as much ; and that $570 would make him even with the world ; what was his debt to A ? Ans. $30. 23. A person said that ne was in debt to four individuals, A, B, C, and D, to the amount of a dollars ; and that he was indebted to B, n times as many dollars as to A ; to C, m timea as many dollars as to A ; and to D, p times as many dollars as to A ; what was his debt to A ? Ans. dollars. 1 + n + 771 4- p 24. If $75 be divided between two men in the proportion of 3 to 2, what will be the respective shares ? Analysis. "We express the greater share by x, and the smaller share by 75 — x. By the conditions of the pro- blem, these shares are in the proportion of 3 to 2, as ex- pressed in the proportion. Converting this proportion into an equation and reducing, we find the respective shares to bo $45 and $30. Analysis. Since the shares are in the proportion of 3 to 2, they may be represented by SOLUTION. Let X : = the greater share ; 75- — X = the smaller share ; JU : 75. — a;;:3:2 2x = 225 — 3x (1) ox = 225 C-^) X == 45, greater ; (3) 75 — X = 30, smaller. (4) Let SECOND SOLUTION. ; = the greater share ; ; = the smaller share ; bx = 75 (1) ^=15 (2) Sx == 45, greater ; (3) 2x = 30, smaller. (4) Sx and The sura of the shares, 5a;, must be equal to $75, the amount divided; whence, by reduction, we find the shares to be $45 and $30. as before. KoTK. — V/hen proportional numbers are required, it is generally most convenient to represent them by only one letter, with coefficients of the given relation or proportion. Thus, numbers in proportion of 8 to 4, may be expressed by Sx and 4x, and the proportion of a to 6 may be expressed by az and bx. This avoids the use of a proportion in the solu- tion of the problem. 130 SIMPLE EQUATIONS. 25. Divide $150 into two parts, so that tlie smaller may be to the greater as 7 to 8. Ans, $70, and $80. 26. Divide $1235 between A and B, so that A^s share may be to B's as 3 to 2. Ans. A's share, $741 ; B's, $494. 27. Divide d dollars between A and B, so that A's share may be to B's as m is to n. Ans. A's share, — - — ; B's, . m -f- n m -\- n 28. Two men commenced trade together ; the first put in $40 more than the second, and the stock of the first was to that of the second as 5 to 4 ; what was the stock of each ? Ans. $200, and $160. 29. A man was hired for a year, for $100 and a suit of clothes ; but at the end of 8 months he left, and received his clothes and $60 in money, as full compensation for the time he had w^orked ; what was the value of the suit of clothes ? Ans. $20. 30. Three men trading in company gained $780, which must be divided in proportion to their stock. A's stock was to B's as 2 to 3, and A's to C's as 2 to 5. What part of the gain must each receive ? Ans. A, $156; B, $234; C, $390. 31. A field of 864 acres is to be divided among three* farmers. A, B, and C, so that A's part shall be to B's as 5 to 11, and C may receive as much as A and B together ; how much must each receive ? Ans, A, 135 acres; B, 297; and C, 432. 32. Three men trading in company, put in money in the fol- lowing proportion ; the first, 3 dollars as often as the second 7, ard the third 5. They gain $960. What is each man's Bhare of the gain ? Ans. $192 ; $148 ; $320 ONE UNKNOWN QUANTITY. X3l 83. A man had two flocks of sheep, each containing the same number ; from one he sold 80, and from the other 20 ; then the number remaining in the former was to the number in the latter as 2 to 3. How many sheep did each flock origi- nally contain ? A'ns. 200. 84. A gentleman is now 25 years old, and his youngest bro- ther is 15. How many years must elapse before their ages will be in the proportion of 5 to 4 ? Ans. 25 years. 85. There are two numbers in the proportion of 3 to 4 ; but if 2i be added to each of them, the two sums will be in the proportion of 4 to 5. What are the numbers ? Ans. 72 and 9G. 3G. A man's age when he was married was to that of his wife as 3 to 2 ; and when they had lived together 4 years, his age was to hers as 7 to 5. What were their ages w^hen they were married ? Ans. His age, 24 years ; hers, 16. 87. The difference between two numbers is 12, and tho greater is to the less as 11 to 7 ; what are the numbers ? Ans. 21 and 33. 38. The difference between two numbers is a, and the greater is to the less as m to n ; what are the numbers ? . ma ^ na Ans. aud . m — n m — n 39. The sum of two numbers is 20, and their sum is to theii' difference as 10 to 1 ; what are the numbers ? Ans. 9 and 11. 40. The sum of two numbers is a, and their sum is to their difference as m to 7i ; what are the numbers ? Ans. Greater, -^^ — — ; less, ^^ — t^ ~ 2m ' '2m 41 A certain sum ot money was put at simple interest, and nO — xiVZQ — xiiSiU (A) Put 120 = a ; then ej — a::a + 6 — ir::8: 14 (B) 8a + 48 — • Sjc = 14a — 14x (1) 6x = 6a — 48 (2) X = a — 8 (3) or ^ = 112 (i) 132 SIMPLE EQUATIONS. iu 8 months it amounted to $120 ; had the interest continued 14 months, the amount would have been $126. What was the principal ? SOLUTION. Analysis. We let Let X = the principal. ^ represent the prin- 120 _ :r = int. for 8 mo. ?^P^H ^^^' '"^^^"^'^ To^ . , /, 1 ^ iiig It from each 126 — X = mt. for 14 mo. ^ ^ . ^^^ amount, we have 120 — Xy the interc-Jt for 8 months, and 126 — Xf the interest for 14 months. But since interest is pro- portional to the time, we have proportion (A). For brevity, we write a for 120, and obtain (b). Converting this proportion into an equation, and reducing, we have x = a — 8, or 120 — 8, or 112. Note. — The artifice, employed above, of representing a numeral by a letter, and restoring the value in the final result, is of much use, and gives true delicacy to algebraic operations. The pupil should be en- couraged in its use. 42. A sum of money placed at simple interest, in 13 months amounted to $113, and in 20 months, to $120. Required, the sum at interest. Ans. $100. 43. A certain number diminished by 45, is to the same number increased by 45, as 1 to 31. What is the number ? Ans. 48. 44. The number 12 is | of what number?' SOLUTION. r^et X = the number. Analysis. We represent the re g quired number by x ; and by the con- ■^ = 12 (1) dition of the question, f of x must be equal to 12, which gives equation (i). Sx = 48 (2) Reducing, we find x = 16, the answer a; = 16 (3) 45. The number a is | of what number ? Ans. ~. 46. The number 21 is f of what number ? Am. 4^. SOLUTION. Let X = the number. a = 130 X , X 6 +7 =^ a) 7^ f 6^ = 42a (2) ISx = 42a (3) X = 3y^a (4) or, a? = 420 (5) ONE UNKNOWN QUANTITY. Igg 47. The number 21 is the — part of what number ? m 48. The number a is the — part of what number ? » , an Ans. — . m 49. If you add together ^ and 4 oi* a certain number, the mm will be 130 ; what is the number ? Analysis. Representing the re- quired number by a:, and the numeral 130 by tty we have by the condition cf the problem, equation (i). Keducing, we have x = 3 ,\a (4) , and restoring the value of «, we have a; = 420, the num- ber sought. 50. A farmer wishes to mix 116 bushels of provender, con- sisting of rye, barley, and oats, so that the mixture may con tain f as much barley as oats, and ^ as much rye as barley ; how much of each kind of grain must there be in the mixture ? Note. — Instead of employing the literal quantity, a, Tve may avoid the labor of multiplying numbers together, by indicating the operation. Analysis. We let x represent the number of bushels of oats. Then, by the conditions, | of x, 5x , 1 , or -=-, must be the bar- 1 A 1 c^^ ^^ ley ; and J of y, or t^, must be the rye. Put ting the sum of ail equal to 116, we have equa- tion (1). Multiplying by 14, indicating the ope- ration in the second Biember, we have (2) ; and reducing we have, finally, x = 56. 12 SOLUTION. Let X = oats ; dx , , y = barley ; bx . + ^f+?i-m a) Ux + 10x+ 507=116 X 14 (2) 29a: = 116 X 14 (3) a?=4x 14 (4) or, X = 56 (5) 134: SIMPLE EQUATIONS. 51 Divide 48 into two such parts, that if the less be divided by 4, and the greater by 6, the sum of the quotients will bo 9 Ans, 12, and H6. 52. A clerk spends | of his salary for his board, and | of the remainder in clothes, and yet saves $150 a year. What is tiis salary ? Ans. $lo50. 58, An estate is to be divided among 4 children^ in the fol- lowing manner : to the first, $200 more than | of the whole; to the second, $340 more than -I of the whole ; to the third. $800 more than ^ of the whole ; and to the fourth, $400 more than ^ of the whole. What is the value of the estate ? Ans. $-l-800. 54. Of a detachment of soldiers, | are on actual duty, ^ of them are sick, i of the remainder absent on leave, and the rest, which is 380, have deserted ; v/ hat was the number of men in tiie detachment ? Aiis. 2280 men. 55. A man has a lease for 99 years ; being asked how much of it had already expired, he answered that § of the time past was equal to | of the time to come. Required the time past and the time to come. Assume a = 99. Ans. Time past, 54 years. 56. It is required to divide the number 204 into two such parts, that -r of the less being taken from the greater, the re- mainder will be equal to -|- of the greater subtracted from 4 times the less. Ans. The numbers are 154 and 50 Put a " 204, and restore value in the result. ( 7. Tn the composition of a quantity of gunpowder, the {litre Vvas 10 pounds more than | of the whole, the sulphur 44 pounds less than J of the whole, and the charcoal 2 pounds less than \ of the nitre. What was the amount of gun- powder ? / =t ONE UNKNOWN QUANTITY. 135 SOLUTION. Let X — the Vs hole ; 2x ~- + IC = the nitre ; o X - — 4| = the sulphur ; ^-r + -^ — 2 = the charcoal. 3 ^6^2T^T ^ '^' ~ 4a; + a:+-y- + ---+21 a) = 6j:. (2) Analysis. Represent- ing the whole composi- tion by 0-', we obtain ex- pressions for the nitre, sulphur, and charcoal, according to the condi- tions of the question ; and the sum of these three ingredients must be equal to rr, the whole quantity (i). Multiply- ing by 6, we have (2) ; reducing, gives ("0 ; mul- tiplying by 7 we have ('!) ; reducing again, gives (5) ; dividing by 3, we have (G) ; and reducing still again, we have a;zz69, the answer. 58. Divide $44 between three men, A, B, and C, so that the share of A may be | that of B, and the share of B j that of C. Ans, A, $9 ; B, $15 ; C, $20. 59. What number is that, to v/hich, if we add its |-, ^, and ^, the sum will be ^0 ? Ans. 24. 60. What number is that, to which, if we add its -g-, -J-, and 1%^V21 =.. (3) 4x + 60 + 21x7=7a:. 60+21 x7=3x. 20+ 7x7= X. Or, 69= X, (4) (5) (6) (7) \, the sum will be a ? Alls. Via 25* 61. In a certain orchard, i are apple trees, \ peach trees, ^ plum trees, 100 cherry trees, and 100 pear trees. How many trees in the orchard ? Ans. 2400. 62. A farmer has his sheep in five different fields, viz. : \ in the first field, |- in the second, ^ in the third, ^V in the fourtli, and 45 in the fifth. How many sheep in all the fields ? Ans. 120 136 SIMPLE EQUATIONS. 63. After. paying out ^ and | of my money, 1 had remain- ing $66 ; liow many dollars had I at first ? Ans, 120 64. After paying away — and — of ray money, I had a dol- lars left How many dollars had I at first ? . amn Ans, . 65. If from ^ of my height in inches 12 be subtracted, ^ of the remainder will be 2. What is my height. Ans. 5 feet 6 inches. 66. A young man, who had just come into possession of a fortune, spent f of it the first year, and | of the remainder the next year, when he had $1420 left. What was his fortune . Ans, $11360. 67. A person at play lost ^ of his money, and then won 3 shillings ; after which he lost ^ of what he then had ; and, on counting, found that he had 12 shillings remaining. How much had he at first ? Ans, 20 shillings. 68. A person at play lost a fourth of his money, and then won 3 shillings ; after which he lost a third of what he then had, and then won 2 shillings ; lastly, he lost a seventh of what ho then had, and then found that he had but 12 shillings remaining. How much had he at first ? Ans. 20 shillings. 69. A shepherd was met by a band of robbers, who plun- dered him of half of his flock and half a sheep over. After- w^ard a second party met him, and took half of what he had left and half a sheep over ; and soon after this a third party met him and treated him in like manner ; and then he had 5 sheep left. How many sheep had he at first ? Ans. 47. 70. A man bought a horse and chaise for 341(a) dollars. If f of the price of the horse be subtracted from twice the ONE UNKNOWN QUANTITY. Ig'J price of the chaise, the remainder will be the same as if ^ of the price of the chaise be subtracted from 3 times the price of the horse. Required the price of each. Arts. Horse, $152 ; chaise, $189. 71. If A can build a certain wall in 10 days, and B can do the same in 14 days, what number of days will be required to build the wall, if they both work together ? Analysis. Let x reprev«ient the number of days both are to work. Since A will build j'g- of the wall in 1 day, he will build X Ypj of it in X days ; and since B will build ^V in 1 day, he will build y-r in aj days. But since the wall is to be comjMed by both working x days, the sum of these two fractions must be equal to unity ; hence equation (i), which, reduced, gives 5|-, the number of days required. 72. If A can do a piece of work in a days, and B can do the same in b days, how long will it take them if they both work together ? ab Ans. days a -{- b 73. A can do a piece of work in 12 days, and B can do the same in 24 days ; how many days will be required, if they both work together ? Ans. 8. 74. A laborer. A, can perform a piece of work in 5 days, B can do the same in 6 days, and C in 8 da^ys ; in wliat time can the three together perform the sam? work ? Ans. 2-X days 12 * SOLUTION. Let X = : the number of days ; X : A'^ 5 part of the work ; X B's 1 part of the work. XX 10 "^14== a) 7x + 5x = 70 (^) 12x = 70 (3) X = ^1 (4) 138 SIMPLE EQUATIONS. 75. A laborer engaged to serve for 60 days on these condi- tions : for every day he worked he should have 75 cents and his board, and for every day he was idle he should forfeit 25 cents for damage and board. At the end of the time, he received $25. How many days did he work, and how many days was he idle? SOLUTION. Analysis. Representing , , .,, the idle days by a:, 60 — x Let X = days he was idle ; ^^^^ ^^ ^^^ ^^^^^.^^ ^^^^^ 60 — a? r=: days he worked. n^d he worked the whole 45 _— a: 25 (^) ^^ days, his wages would have amounted to $45. But X = 20 (2) for every day he was idle, 75 cents, but 25 cents in addition, making $1 a day. In x idle days, therefore, he lost x dollars. Consequently, the amount due him was $45 — 2: dollars. But by the conditions of the problem, the money due him K^as 25 dollars ; hence we have (i), from which we find he was idle 20 days, and worked 40 days. 76. A person engaged to work a days on these conditions; for each day he worked he was to receive b cents, and for each day he was idle he was to forfeit c cents. At the end of a days he received d cents ; how many days was he idle ? . ah — d Ans, -J— — . + c 77. A boy engaged to convey 30 glass vessels to a certain place, on condition of receiving 5 cents for every one he de- livered safe, and forfeiting 12 cents for every one he broke. On settlement, he received 99 cents ; how many did he break ? Ans. 3. 78. A boy engaged to carry n glass vessels to a certain place on these conditions : he was to receive a cents for every one he delivered safe, and to forftfit b cents for every one he broke. On settlement he received d cents ; how many did he Ans, The number represented by -^ — - ^ "^ a -^ b TWO UKKNOWN QUANTITIES. 139 SIMPLE EQUATIONS CONTAINING TWO UNKNOWN QUANTITIES. 14:&. Independent Equations are such as cannot be reduced to the same form, or derived one from the other ; as x-\- Sy := a, and 4x -[- % = ^- Independent equations refer to the same problem, and express diiFerent conditions of the problem. 14:7 • We have seen that, in order to find the value of any unknown quantity in an equation, we separate it from the other quantities, and cause it to stand alone as one member of the equation. But if the equation contain two unknown quanti- ties, the value of neither can be determined by this process. To show the reason of this, let us consider th^ following equa- tion : Transposing y, we have :r rr: 20 — ^, in which x is still undetermined, because its value in the second member of the equation contains the unknown quantity, y. Again, transposing x in equation ^^\ we have y = 20r-x, in which y is still unknown, because its value contains the un- known quantity, x. Hence, Two iinhiown quantities cannot he determined from a sin- gle equation. The equation given above expresses this condition : viz., the sum of two numbers is 20 ; and since there are many pairs or couplets of numbers of which the sum is 20, x and y can have no particular or exclusive values. The equation is satis- Define independent equations. They always refer to Avhat ? Why cannot the va'ues of two unknown quantities be determined from on« equation 1 140 SIMPLE EQUATIONS. fied if we make a? = 1 and y = 19i or, a; == 2 and y =^ 18, etc * for * ' 1 + 19 = 20, 2 + 18 = 20, etc. ' But if we combine another equation with this, as x — 2/ — 4, wliich expresses a different condition : viz., that the difference of the two numbers is 4, then only one Talue for x and one value for y will satisfy both equations, or answer both condi- tions. To find these values we may proceed thus : x-\-y = 20 a) x — y= 4: (2) By addition, 2x = 24, or a: = 12 Subtracting (2) from (i), 2^/ = 16, or ?/ = 8 And 8 and 12 are the only numbers whose sum is 20 and difi'erence 4. From this result we learn that Two uvknoivn quantities can he determined from two independent equations. To effect the reduction, we must derive from the two a new . equation containing but one unknown quantity. This opera- tion is called ELIMINATION. 148. Elimination is the process of combining two or more equations, containing two or more unknown quantities, in such a manner as to cause one or more of the unknown quantities contained in them to disappear. There are three principal methods of elimination : 1st, Bi/ substitution; 2d, By comparison; od, By addition and subtraction. CASE L 149. Elimination by substitution. 1. Given 2x + by = 31, and Sx ■+ 2y = 19, to find the values of x and y. How many equations are required that the values of two unknowr* quantities may be determined? V/hy? Define elimination. Name tho methods. Give Case I. TWO UNKNOWN QUANTITIES. ^4^ OPERATION. Analysis. We transpose 5y in equation (A), and di- 2a; + 52/ = 31 (A) vide by 2, and obtain (i), QX -]- Zy = iv (B) which expresses the alge- ' Qi p, braic value ofa:. Tins value X = ^ (1) of X we substitute for x in QQ - _ ^ (B) ; thus, instead of 3a:, ^ + 2v = 19 (2) "^G write its value, 3 times 2 31— 5y 93—157/ ^ . , 93 — 15i/ -f % = 38 (3) — 2~» ^^ — 2~ ^ '^^^''^ — 111/ = — 55 (4) with the other terms of ^ __ 5 /gN equation (B)written in their ^1 9^ order, gives (2), an equatiou X =s "^ (6) containing only one un- ^ known quantity, y ; there- ' 05 = 3 (7) fore x has been eliminated. Reducing in the usual manner, we have y = 5. Since y is 5, 5y is 25 ; and substituting ^his value in the second member of equation (i), we have (6), which gives the value of x in known terms. Ileducing, we obtain a; = 3. Hence, the following Rule. I. Find the value of one of the unknown quan* titles in one of the given equations, II. Substitute this value for the same unknown quantity in the other equation, EXAMPLES FOR PRACTICE. o n- r to find x and y, tx — y= 6) ^ Ans, X = 7 1 y = 1, Give Analysis. Rule. SIMPLE EQUATIONS. 6 Given | ""^ _^ ^^ = 13 I *^ ^"^ ^ ^^^ ^' Ans. 2/ = 3; a* = 14. 6. Given | ^ __ _Ji [ ^^ find y and ^. -4ns. y / . Given i , . OT r to find x and 2:. C ^ + 5^ = 27 3 Ans. cc = 7 ; 2: = 4. 8. Given ] 4f "n^ — "6 1 ^^ ^^^ 2/ and z. Ans. 2/ ~ 14 ; 2 = 10. CASE n. 130. Elimination by comparison. 1. Given Sx + 2y = 16, and 4x -i- Sy = 23, to find the values of x and y. OPERATION. Sx + 2y== 16 4a? + 3^/ = 23 16 — 2v x = 3 ir = 23 _3y 4 16- ■2y_ 23 -3y 3 4 64- -8y = 69 _9y y = 5 dx 10 = 16 x=2 (A) (B) a) (2) (3) (4) (5) (6) 0) Analysis. Transposing 2?/ in (A), and dividing by 3, we obtain (1). From (B), in like manner, we obtain (2). We have thua found two algebraic values for a:, which must be equal to each other. (Ax. 7.). Equating these values, w^e have (3), an equation containing but one unknown quantity, y; consequently, x has been eliminated. Reducing this equation, we obtain y=5. Since y is 5, 2y is 10 ; and substitutirg this value in (A), we have (6), which reduced, gives x = 2. Give Case XL Analysis. TWO UNKNOWN QUANTITIES. |43 Hence, tlie following Rule. I. Find the value of the same unknown quantity in each of the given equations, II. Form an equation by placing these values equal to each other. EXAMPLES FOR PRACTICE. 2. Given {^l_2yZ 4_\ ^^ ^^^ ^ ^^^ 2/- Ans. X = S] 2/ = 4.. 3. Given i . . ^ r to find y and x, Ans. 2/ = 5 ; x = 1. . _ S2x + 5^ = 29") ^ „ ^ , 4. Given i o ___ i t ^^ ^^^ ^ ^^^ ^' Ans. X = 7 'j 2 = 3. 5. Given ]^ ^ "" ^j, [■ to find z and ?/. L2z — y = 15) Ans. z = lo'j 2/ «= 11. . _ n2x + Uy = 26\ ^ . , . 6. Given -s ^ "^ o f to find cc and y. (. 6x — 2/=2> Ans. cc = 1 ; j^ = 1. ^ ^. (-4^ + 122/ =5) ^ ^ ^ 7. Given i , o i r to find x and y. i x+ 2y = 1) ^ Ans. aj = i ; y = |. (^+^«14) U^ 6 f 8. Given ( > to find x and y. (8'"^6"" ) ^ns. a; = 16; y = 12. 9. Given ( r. ^ /to find a; and y. I , oa; — '^y ___^\ \ 6 / Ans. cc = 3 ; y == 2. Give Rule, 12x + ISy = 96 12x+ 62/ = 48 (1) (2) 12y == 48 y= 4 (3) (4) 144 SIMPLE EQUATIONS. CASE III. 151. Elimination by addition or subtraction. 1. Given 6cc + 9y = 48, and 4x + 2y =^ 16, to find the values of x and y, OPERATION. . Analysis. We multiply equa. tion (A) by 2, and obtain (i) ; wo 6jc + 9i/ = 48 (A) next multiply equation (b) by 3, and 4x -\- 2v = 16 (B) obtain (2). We have thus obtained two equations in which the coefti- cients of x are equal. Subtracting equation (2) from equation (i), mem- ber from member, the terms con- taining X cancel, and we have (3), an equation containing but one un- ~T ~ ^ ZTTfi known quantity, y. Reducing, we d ~ t ^°^ ^^""^ y == 4. Substituting this Ax — (6) value of y in (b) we have (5), which a; = 2 (7) reduced gives x = 2, 2. Given 7x — 2y = 31, and bx + 2y =: 29, to find x and y. OPERATION. Analysis. Since the coefficients 7x _ 22/ = 31 (A) <>f y' !" *« ^'^T ^ff"""'' '''^ 9 9Q numerically equal, and have oppo- Ox + 22/ = 29 (B) gj^g signs, their sum is zero, and y I2x = 60 (1) ^^^^ ^® eliminated by adding equa- X z= d (2) ^^^'^^ ^'^'^ ^^^ ^^^' ^^"^^^^ ^y ^^^^" . ber; this gives equation (i), from 25 + 2t/ = 29 (3) which we obtain x = 5. Substi- 2y = 4: (4) tuting this value of x in (b), we ob- y =z 2 (5) *^i^ ^^)' ^'^^^^ Sives 2/ = 2. From these examples we derive the following KuLE. I. Multiply or divide the equations by such num bers or quantities that the coefficients of the quantity to he eli minated shall be made equal Give Case III. Analyses. Rule. TWO unkinov/n quantities. l45 I. 1/ these coefficients have unlike signs, add the prepared eqvaHons ; if like signs, subtract one equation from the other. Notes. 1. If the given equf^.tions require to be mnltiplied, find the le!i!*t cornm )n multiple of the coefficients of the quantity to be eliminatwl, find , r to find x and y, Lex — my = 26 j ^ . a + 6 a — b Ans. x =^ ;. y ■■ G " m 19. Given ] ' / [-to find x and v. ( aas — by = c J ^ ^ b' + d" b—c Ans. X = —J— — :- ; y = -— — a{b 4- c) ^ b -\- c I \- ny =m + n 20. Given ( ^ , ) to find a? and y. ) mx , nV , , o I Ana, X = mn ; y «= — . n 148 SIMPLE EQUATIONS. TROBLEMS PRODL'CINQ EQUATIONS CONTAINING TWO UNKNOWN QUANTITIES 1«>3. Many of the problems hitherto given require the de- termination of more than one unknown quantity ; but the quantities are so related to each other that they can be ex- pressed algebraically by the use of a single letter. The solu- tion is often rendered more simple by using as many letters as there are quantities to be determined. If two letters are em- l^loyed to represent unknown quantities, the conditions of the problem must furnish two independent equations ; otherwise it will not be capable of solution (147). 1. A man bought at one time 3 bushels of wheat and five bushels of rye for 38 shillings ; and at another time, 6 bushels of wheat and 3 bushels of rye for 48 shillings. What was the price of a bushel of each ? SOLUTION. Analysis. We represent . the prices by a; and y. Since Let j;== price of wheat; 3 b^^hels of wheat and 5 y = price of rye. bushels of rye cost 38 shil- 3^ + by = 38 (A) li"g8, we have equation (A), Q^ , 3^ = 48 (B) furnished by the first condi'. -— — - tioD ; and since 6 bushels of bx+Wy= ib a) ^heat and 3 bushels of rye (>x -f 3i/ = 48 (2) cost 48 shillings, we have •7^ ___ 23 (3) equation (B), from the second condition. Multiplying (A) by (2) to make the coefficients 2/ == 4 (4) Bx -|- 20 = 38 (5) of X equal, equations (A) and ^ = 6 (6) (B) become (l) and (2). Sub- tracting (2) from (1), we have (3), which, reduced, gives y =^ 4. Substituting this value of y in (A), W9 have (5), which gives t/ = Q, 2. A man spent 30 cents for apples and pears, bnying his apples at the rate of 4 for a cent, and his pears at the rate of 5 for a cent. He afterward let a friend have half of hia apphs and one-third of his pears, for 13 cents, at the same rate. How many did he buy of each sort ? TWO UNKNOWN QUANTITIES. 149 SOLUTION. Let X = number of apples, and y = number of pears. Hence, j — cost of apples, and j = cost of pears. By the first condition, ►- 4. |. ,rr 30 ca) 4: O By the second condition, ^ + :^ = 13 (B) o 10 a) Multiplying (b) by 2, | + j| -_ 26 Subtracting (i) from (a) ^ — ^ ^ 4 ,2, Reducing, 3/ = 60 (3) Substituting in (A), + 12 r:::i: SO (4) 4 Reducing, x _-=: 72 (s) 8. What fraction is that, to the numerator of which, if 1 be lidded, its value will be 4, but if I be added to the denomina- tor, its value will be | ? SOLUTION. Let — -- the fraction. y By the first condition, -i—'— == _ (a; y *> X 1 By the second condition, --. = -/ (B) ?/ -f 1 4 Clearing (A) of fractions, ?yx + 3 = t/ (i) Clearing (i: of fractious, ^c — 1 =:= y (2) Siil)tracting (i; from (2;^ x — 4 znz (3) Trans[)()sing, cc r-- 4 (4) From (1) y r- 15 (S) X ^ Hence, tlie fraction is - r^ — (6) y lo 18* 150 SIMPLE EQUATIONS. 4. .A 2:eiitlenian paid for 6 pairs of boots, and 4 pairs of shoes, $41; and afterward, for o pairs of boots, and 7 pairs of shoes, $o2 ; what was the price of eaeh per pair ? Ans. Boots, $6, shoes, fl. 5. What fraction is that, to the numerator of which, if 4 1)6 added, the value is ^;but if 7 be added to its denominator, its value is i ? Ant^. j\, 6. A and B have certain sums of money : says A to B. '*Give me $15 of your money, and I shall have five times as much as you will have leff Says B to A, "Give me $5 of your money, and I shall have exactly as much as you will have left ;" how many dollars has each ? Ans. A has $35; and B, $25. 7. What fraction is that whose numerator being doubled, and its denominator increased by 7, the value becomes § ; but tlie denominator being doubled, and the numerator increased by 1^, the vahie becomes |? Ans. |. 8. If A give B $5 of his money, B will have twn'ce as much money as A has left ; and if B give A $5, A will have thrice as much as B has left ; how murh had each ? Ans. A has $13 ; and B, $11. 9. A merchant has sugar at 9 cents and 13 cents a pound, and he wishes to make a mixture of 100 pounds that shall be worth 12 cents a pound ; how many pounds of each quality must he take? Ans, 25 pouTids at 9 cents, and 75 pounds at 13 cents. 10. A person has a saddle worth £50, and two horses. When he saddles the poorer horse, the horse and saddle are i^'orth twice as much as the better horse ; but when he saddN's the better horse, the horse find saddle are together worth thrt'o timee the other ; what is the value of each horse ? Ans. £40 and £KU. 11. One day a gentleman employs 4 men and ^ boys to labor for him, and ])ays them 40 shillings; tlie next day he hires at the same rate, 7 men and (i boys, for 50 shillings; what are the daily wages of each? Ans. Man's, 5 shillings ; boy's, 2 shillings 6 pence. TWO UNKNOWN QUANTITIES. ][5| 12. A merchant sold a yard of broadcloth and 3 yards of velvet, for $25 ; and, at another, time, 4 yards of broadcloth and 5 yards of velvet, for $85 ; what was the price of each per yard ? Ans. Broadcloth, $10 ; velvet, $5. 13. Find two numbers, such that half the first, with a third part of the second, make 9 ; and a fourth part of the first, with a fifth part of the second, make 5. Ans. 8 and 15. 14. A gentleman being asked the age of his two sons, an- swered, ^' If to the sum of their ages 18 be added, the result will be double the age of the elder ; but if 6 be taken from the difference of their ages, the remainder will be equal to the age of the younger;" what were their ages ? Ans. 30 and 12. 15. A says to B, " Give me 100 of your dollars, and I shall have as much money as you." B replies, "Give me 100 of your dollars and I shall have twice as much as you ;" how many dollars has each ? Ans. A, $500 ; B, $7aO, 16. Find two numbers, such that § of the first and | of the second added together, will make 12 ; and if the first be divided by 2, and the second multiplied by 3, | of the sum of these results will be 26 Ans. 15 and 10|. 17. Says A to B, "^ of the difference of our money is equal to yours ; and if you give me $2, I shall have five times as much as you ;" how much has each l^ Ans, A, $48 ; B, $12. 18. A market-woman bought eggs to the amount of 65 cents, some at the rate of 2 for a cent, and sonio at the rate of 3 for 2 cents. She afterward sold them all for 120 cents, tlieroby gaining a half cent on each egg ; how many of each kind did she buv ? Ans. 50 of one : 60 of the other j^52 SIMPLE EQUATIONS. 19. What two numbers are those, whose sum is a and dif- ference b t SOLUTION. Let X = the greater. And y = the less. From the first condition, X + y=a a) From the second condition, X — y :=b (2) Adding (2) to (i), 2x=^a + b (3) Subtracting c^). from 0), 2y=za — b (4) Ilence, a . b greatei And a b less. In these results, we have the algebraic expression of the fol- lowing general truth : The half sum of any two numbers, a to find x, y, and z, ^ + 3y4-4z = 2lJ OPERATION. Analysis. By tran* posino;, we obtain X -jr y + z = .) (A) equation (i) from (A), X + 2y -i- Sz = lQ (B) (2) from (B), and (3) from X -\- ^y -{■ 4z =21 (C) (C). These equations ~ give us three values ^= ^ — y — 2: (1) ^-^j. ^^ Equating the oc==l6 — 2y — Sz (2) 1st and 2d values, we a: = 21 — 8?/ — 4z (G) have (4), and equating the 2d and 3d values, we have (5). We have thus eliminated x, and obtained two equations with two unknown quantities. By trans- ^ ^^ '^ ^^^ posing terms in (4) and (9) (5) and reducing, we (10) have (6) and (7), giving ^j. two values for//. Equa- ting these values, we eliminate y, and oh^iin (8), from which we fiiwl z = 2. By substi- 9 — y — z = 16 — 2y — Sz (4) U^2y — Sz = 21 -^Sy — 4:z (5) y= 7 — 2z (6) y== ^— z (7) 154 SIMPLE EQUATIONS. tilting the value of z in (7) we obtain ?/ r=r 3 ; and substituting the values of z and y in (1) w^e obtain x ~ 4. In this example we have eliminated bj the method of comimri^on. C2x + 4:y — nz = 22^ 2. Given < 4x — 2i/ + 52 = 18 > to find x, y and z. \Qx + ly — 2 = 63 J Analysts. We multiplj ^a; by 2, and obtain (i). Writing (B) underneath this and sub- tracting, we eliminate x, and obtain (2). Next, multiplying (A) by 3, we obtain (3). W^riting (C) under this and subtracting, we again eliminate a:, and ob- tain (4). Multiplying (4) by 2 gives (5). Writing (2) under this and subtracting (5) from it, we eliminate y and obtain (6), which reduced, gives 2 = 4. Substituting this value of z in (4), and reducing, we have y =: 7 ; and substituting the . values of z and y in (A), and reducing, we find x = 3. lu this example we have eliminated principally by the method of addition or subtraction. From the illustrations given we deduce the follovfing Rule. I. Combine one of the equations with each of the others, eliminating successively the same unknown quantity : the result will he a new set of equations containing one less unknown quantity. II. Combine one of the new equations with each of the others, eliminating a second unknown quantity ; and the re- Give rule for reducing equatians containing three or more unknown quautitiei. OPERATION. 2x +42/— 32 = 22 4x — 2?/+ 52=18 ^x + ly — 2 = 63 (A) (C) 4.r + 82/ — 62 == 44 4x — 2^/4- 52=18 (1) 10?/— 112 = 26 (2) 6x+ 12//— 92 = 66 %x+ ly— 2=63 (3) by — 82 = 3 (4) 10?/— 162= 6 10^ — 112 = 26 (5) 52 = 20 2= 4 y= 7 x= 3 (6) 0) (8) (9) THREE OR MORE UNKXOWN QUANTITIES. ^55 suit will be a new set containing two unknown quantities lesn ■Jian the oj-iginal. III. Continue this iwocess till an equation is found con- taining but one unknown quantity. lY. Eeduce this equation and find the 'jalue of the un - known quantity. Substitute this value in an equation con- taininj two unknown quantities, and thus find the value of a, second. Substitute these values in an equation containing three unknown quantities, and find the value of a third ; and 80 on, till the values of all are found. Note — Instead of combining the first witli each of the others, we may combine the first with the second, the second with the third, and so on ; or we may pursue any other order of combination best suited to the mutual rehitions of the coefficients. loo. It is evident that if there are more unknown quanti- ties than equations, the last resulting equation will contain tv7o or more unknown quantities, and the solution will be im- possible (147). nence the following general law: There must be as many independent equations as there are unknown quantities. Note. — If there are more independent equations than unknown quan- tities, some of them can be dispensed with in reducing the equations. EXAIvrPLES FOR PRACTICE. 1 Given < hx + \y ~~ 22 = 20 \ to find x, y, and z. 1 11^ + 7^ — 62 = 37 J Ans. .T == 2 ; y = 3 ; 2 = 1 r3j;-f 57/ + ^= -C)] 2. Given < G^ 4. S?/ -f 22 = 31 V to find x, y, and z. i.9^ 4_ 4^^^42=, 50 J Ans. ^=2; 2/=3; z=5. What number of equations is required in the solution of a problem ? Wuy? 156 SIMPLE EQUATIONS. Note — AVhr^n several coefficients are unity, or multiples of enoh otlieT, cert.iij) exppdients may be employed to facilitate the calculation, for wliicb uo specitic rules can be given. (x + y + z^n^ 3. Given < x -^ y — 2 = 25 \ to find Xj y, and z. lx — y — z= 9 J Subtract the 2d from the 1st, and '2z = 6, or 2: =3. Subtract the 8d from the 2d, and 2y = 16, or 3/ = 8. A«M the 1st and 3d, and 2x = 40, or x = 20. ill -i- V -\- X -{■ y = 10 \ U-\-V-i-Z+X=ll) ft,,, , « { to find the value of ] ^ ^ ^^ ^-( each. Ju + x + y-{-z = lS\ \v-\-x-\-y + z = l-l] Since in each equation one letter is wanting, assume u + V + X + y -{- z = s Then 8 — z = 10 s — y:=ll 8 — x = 12 ^^^^®' ^y 8 — v^lS substitu- By addition, 56* — « = 60 6-= 15 T I tins: the value of Sy z = b y = i x = S v = 2 u=\ Required, the values of the unknown quantities in the fol lowing equations : rx4-?/ + 2=26^ Cx = 12^ 5. Jx— -2/ —"^r ^^^' \y= ^^ (x — z =6j (2=6, f x — y — z=^ 6^ Cx = ^^, 6 <^ — x — z=12\ Ans. ly=r.2\^ ^ llz-^y—x = 2A) U^12. 'x + iy-lOO^ f^--64, 7. ^y+izr^rlOoV Arts. ly = 12, ^z -1-^x1=: 100 J (2 = 81. {i THREE OR MORE UNKNOWN QUANTITIES. ^57 = 52 \ = 82) ' = 68 \ = 30 ( = 32) z -I- 1^ = 68 \ A718. If; -f- W = 30 ' U '\- X ^x + 81/ = 23 2/ 4- 3^ = 81 X -f 2/ + 2 -f 2ifr r= 39 r 4x 4- 2y — 82 = 4 ^ 10. .^ 3x — 5?/ 4- 22 = 22 V Arts. f i^ + ^!/ + i^ - 22 ] 11. -{ i^c -f 2/ + i^ = 83 V J72s. ^ ^ + ^2/ - A^ == 19 j 12. ^ .?: 4- 2 = 6 V Ans, -(?/== ^(a 4- c — 6), ( 2/ 4- z = c J (2=^64-0 — a). I . . \ / « CJ^ -t y -{- _az = a -\- ac + c j I a; = - , 13. / ^'-^ + i/ + a'^2: = 3«c ( j^^g ) y = ac, ' ( ) c ac^c 4- 2t/ + ac2 = o,' 4- 2ac 4- cM j z == — I { a PROBLEMS PRODUCING EQUATIONS CONTAINING THREE y + ^z = 470. (1) By the conditions, < y + 4:x + 4tzz= 580. (2) 1 z + 5x + 5y = 680. (3> Adding 2.x to (i), 'Sx + Sy + Sz=: 470 - 2x (4) - 87/ to 02), 4x + Ay -\- 4:z — 580 + '^y (5) '' 42 to (3), ( ^ 5x + by 4- bz = 680 + 4z (6; Assume 8 = X + y + z Equation (4) becomes r Ss = 470 + 2x (7) (5) J 4s = 580 + Sy (8) (G) (5.=. 680 + 4z (9) Multiplying (t) by 6, ^1Ss = 2820 + 12x (10)- '' (8) by 4, . Us = 2820 + I2y (11) " (9) by 8, [Vos = 1890 + 122 (12) Adding (lo), (U), and (i2), 49s = 7080 4- 12s (13) S78 = 7080 (14) 8 = 190 (15) Substituting value of s in (T) , Cx = $50, A^s, " " («), 7. Til the solution of the last example, the wages of the wife are found to be 0, which means that she received no wages. The following examples will illustrate negative re- sults : 1. A man worked for a person 10 days, having his wife with him 8 days and his son 6 days, and he received 10 dollars 30 cents as compensation for all three ; at another time he wrought 12 days, his wife 10 days, and son -t days, and he received 13 dollars and 20 cents; and at another time he wrought 15 days, his wife 10 days, and his son 12 days, at the same rates as be- fo'-e, and he received 13 dollars 85 cents. What were the daily wages of each ? Ans, Husband, 75 cents ; w^fe, 50 cents; son, — 20 cents. The sign minus signifies the opposite to the sign plus. Hence the son, instead of receiving wages, was at an e.rjjensc of 20 cents a day, and the language of the problem is thus shown to be incorrect. 2. Two men, A and B, commenced trade at the same time ; A had 3 times as much money as B, and continuing in trade, A gained 400 dollars, and B 150 dollars ; A then had twice as much money as B. How much did each have at first ? Without any special consideration of the problem, it implies that both had money, and asks how much. But on solving the problem with x to represent A's money and y B's, we find a; = — 300 and 2/ = — 100 dollars. 14* L 162 SIMPLE EQUATIONS. That is, they had no money, and the minus sign in this ease indicates debt; and the solution not only reveals the numeri- cal values, but the true conditions of the problem, and points out the necessary corrections of language to correspond to an arithmetical sense. The problem should have been written thus : A was three times as much in debt as B ; A gains 400 dollars, and 13 150 ; A now has twice as much money as B. How much were each in debt ? Ans, A's debt, $300; B's, $100. These results are positive, and show that the enunciation corresponds to the real circumstances of the case 3. What number is that whose fourth part exceeds its third part by 12 ? Ans, — 144. But there is no such abstract number as — 144, and we can- not interpret this as debt. It points out error or impossihility^ and by returning to the problem we perceive that a fourtb part of any number whatever cannot exceed its third part ; it must be, its third part exceeds its fourth part by 12, and the enunciation should be thus : What number is that whose third part exceeds its fourth part by 12, An^. 144. Thus do equations rectify subordinate errors, and point out special conditions. 4. A man when he was married was 30 years old, and his wife 15. How many years must elapse before his age will be three times the age of his wi^e ? Ans, — 7^ years. The question is incorrectly enunciated ; 7i years before the marriage, not after, their ages bore the specified relation. 5. What fraction is that which becomes -| when 1 is add(3d to its numerator, and I when 1 is added to its denominate]'. Ans. In an arithmetical sense, there is no such fraction The algebraic expression, zj|, will give the required results. INVOLUTION. 163 SECTION III. INVOLUTION; OR, THE FOllMATION OF POWERS. I #18. A Power is the product obtained by repeating a quantity several times as a factor. 1«>l>, Powers are indicated by exponents, from which thej take their names. Thus, let a represent any quantity : Its fir Hi power is a=::z a^ Its second power is aa = a" The third power is aaa = a' The /oi/WA power is aaaa = a^ The fifth power is aaaaa = a' In general terras, a to the ??th power is aa^ &c., to n factors, and n may be any number whatever. IGO. The First Power of any quantity is the quantity itself. The Square of any quantity is its second power. The Cube of any quantity is its third power. 101. A Perfect Power is a quantity that can be exactly produced by taking some other quantity a certain number of times as a ftictor; thus, a^ -{-■ 2ab + ^^ is a perfect power, be- cause it is equal to (a -f 6) x (« + b) ; x^ -f- o.i:^ + 8^ + 1 is a perfect power, because it is equal to (x -f 1) (a? + 1) (a^ 4- 1). iv^oTE. — Tt i« thoufiht host to omit qne«tioTis at the bottom of the pngea in the remaining pait of this work leaving the teacher to use such as may be deemed appropriate. 164 INVOLUTION. ICI^, An Imperfect Power is a quantity that cannot be exactly i)ro(Ju('ed by taking another quantity any number of times as a factor; as, «^ -\- by x + oy, and d^ ~{- ab + b'\ IGJI. Involution is the process of raising any quantity to any given power. Involution, in algebra, is performed by suc- cessive multiplications, as in arithmetic. The first power is the quantity itself. The second power is the product of the quantity multiplied bj itself. The third power is the product of the second power by the quantity. The fourth powder is the third power multiplied by the quan- tity, etc. POWERS OF MONOMIALS. 104. In the power of a monomial there are three things to be considered : 1st, the coefficient ; 2d, the exponents ; ^d, the sign. 1st. With respect to the coefficient: Let it be required to raise 'la to the third power: w^e have 2a X 2a x 2a = 2 x 2 x 2a' = 2V = 8al Hence, The coefficient may be raised to the required power separately. 2d. With respect to the exponents : We observe that The second power of aMs a^ x a^ = a^"^^ = a*. The tliird power of a^ is a? x af X a^ = a^^^^^ — al The nth power of a' is a^ x a^ x a^ x etc. = a^+^+' ^etc. Hence, Thp exponent is repeated as many times as there are units in the index of the power. 3d. With respect to the law of signs : It is obvious tliat the re])etition of any positive quantity as a factor must produce a positive result. But the successive powers of negative quantities must have varying signs. POWERS OF MONOMIALS. \Q^ Lei K be required to raise — a to successive powers. We have Second power, — a x — a = -f a^ positive. Third power, + a^ x — a = — a"\ negative. Fourth power, — a^ x — a = -f a*, positive. Fifth power, + a* X — a== — a\ negative. Ilence, 1st. All the powers of a positive quantity are positive, 2d The even powers of a negative quantity are positivey and th4 odd powers negative. 16tS. From these principles we deduce the following Rule. I. Baise the numeral coefficient to the required power. II. Multiply the exponent of each letter by the index of the required power. III. When the quantity is negative, give the odd powers the minus sign. EXAMPLES FOR PRACTICE. to the od power. Ans. a^. to the 4th power. Ans. y^. to the 5th power. Ans. F^'\ to the 4th power. Ans. x^'\ to the 8d power. Ans. y^^. to the 6th power. Ans. x^. to the mth power. Ans. a:'"", to the 8d power. Ans. aV. 9 Raise ab'^x* to the 2d power. Ans. a-7/j;^ 10. Raise cV to the 5th power. Ans. c^y^. 11. Required the od power of Sax^. Ans. 27aV. 12. Required the od power of — 2.t. Ans. ^-8x1 13. Required the 4th power of — B^r. Ans. Slx\ 14. Required the 2d power of Sa'b^. Ans. 6Aa*b\ 15. Required the Bd power of ^x^z Ana. 12^a^z\ 1. Raise x^ 2. Raise y^ 8. Raise F' 4. Raise ^ 5. Raise y^ 6. Raise x'* 7. Raise x" 8. Raise ax"^ 166 INVOLUTION. . Expand the following indicated powers. 16. ( — 2a"7. Ans. — 32a^». 17. (^a'bcy, Ans. a'b'c^\ 18. {6ah/x)\ Ans. 2]Qa''y\t:^. 19. (2cr//c'y. " Ans. IGaW'c'^ 20. (^8a'"6"')3. Ans. — 27g^"&^^ 21 (Dni'''b')\ -—-.-,-_- Ans, 24:3m^'"b^, 22. l — qr')\ Ans. a"^. 23. (26-"*c-")^ Ans. S^-'^^'c"**. POWERS OF FRACTIONS. 166. 1. What is the 3d power of ^ ? OPERATION * /a \^^ a a a a x a x a o' Vc / ~" c c c~'cxcxc~', and y =zG -^ d. / Then (x + yf = x' + Sxhj + Sxy' + 7/. Restoring the values of x and i/, we have (a + by + 3(a + by(c + d^ + 3(a -f 6) (c + dy + (c + (Q«. Now we can expand the binomial quantities contained in parentheses. The method of substitution which we have been obliged to adopt in the last three examples, has been long in use among mathematicians for expanding binomials with coefficients. 174. A direct method of obtaining the coefficients in the expansion of a bhiomial in which the terms have coefficients, has been developed by J. II. French. LL.U., Superintendent of the New York State Map and Atlas Survey, and we present it under the name of 15* J74 INVOLUTION. FRENCH'S THEOREM. Any binomial having coefficients, may be involved by actual multiplication. llequired the 5th power of 2a + ox. 2a + ox 2a + 3x •2d power, 4d'+12ax + dx^ 2a+Sx SaF+2Aa'x + ISax^ V2a' x 4- 36aa;^ - f 27.?^ 3d power, Sa'-i-SQa^x -f- 54aj;^ 4-*-^7x-^ 2a '■\-ox iQa'+f2a'x +108aV + ^^:ax' 24:a'x -f lOSa'V +I(j2ax' + 81 j;* 4ili power, 16a'+9Qa^x +216a-;rH^T6a5^~4^T^ 2a 4- o-r. 82a^ + 192a*^+432aV +432aV4-102a^ 48a*^+ 288aV+648aV-f-648a.:r*+248^ 5th power,"32^^+240a*j;4-T20aV+l080aV+810aj;*+243j;» ITo, By a close analysis of the result, we may arrive at general principles which will enable us to expand any binomial having coefficients with the same facility that Newton's Theo- rem enables us to expand to any power a -\- x. If we examine the result in the same order as we did in (IG8-I7S^), we shall find that there is no difference in the expanded form of (a -j- a;),* and (2a + 3^)^, except in the coefficients. That is, In any power of a binomial, the number of terms, the signs, the letters, and the exponents of the literal part in the several terms are independent of the coefficients of the binomial root We will therefore confine our analysis to The Coefficients. ITG. The rigid demonstration of the law which governs tne formation of the coefficients being too difficult for this place, Is reserved for the University Algebra. Bnt tlie law itself may be exhibited as follows : POWERS OP BINOMIALS. I75 1st coefficient, 32 = 2^, i. e. the coefficient of the leading quantity in the root raised * to the power of the given index. 82 X 5 X 3 2d coef., 240 = , i. e. the product of the co- efficient of the first term, the exponent of the leading quantity in the first term, and the coefficient of the following quantity in the root, divided by the coeffi- cient of the leading quantity in the root. o , c\ocx 240 X 4 X 8 . . ^ . ^ ., od coef. 720 == ^ -^ , i. e. the product of the co- » efficient of the second term, the exponent of a in that term, and the coefficient of X in the root, divided by the product of tlie coefficient of a in the root and the num- ber of the term, counting from the left. 4th coef., 1080 = —^-f^ — , ^'. ^- the product of the co- efficient of tlie third term, the exponent of a in that term, and the coefficient of X in the root, divided by the product of the coeffi- "^^-.,_.^ cient of a in the root and the number of the term. (Or, which is the same thing, the exponent of x m the third term -f 1 ) 176 INVOLUTION. ' 1080 >f 2 y '^ 6th coef., 810 = — -^ i. e. the product of the co efficient cf the fourth te]"]u, the exponent of a in that term, and the coefficient of X in the root, divided by the product of the coefficient of a in the root and the num- ber of the term. 6th eoef., 243 = ^T^TK > ^'- ^- ^^^^ product of the co- efficient of the fifth term, the exponent of a m that term, and the coefficient of X in the root, divided by the product of the coeffi- cient of a in the root and the number of the term. 177. From this example and analysis we may deduce the LAW or THE COEFFICIENTS. I. The coefficient of the first term in any power is ahvoi/a equal to the corresponding power of the coefficient of the leading term in the root, II. The coefficient of the second tei^m is obtained by mnU tip)lying the first coefficient by the exponent of the leading quantity, and this jjroduct by the coefficient of the following quantity in the root, and dividing by the coefficient of the leading quantity in the root. Universally; — The coefficient of any term may be ob- tained hy multiplying the coefficient of the preceding term by the exponent of the leading quantity in that term, or by the number of the term from the last, and by the coefficient of the following quantity in the root, and dividing this i^esultby the product of the coefficient of the leading quantity in the root, multiplied hy the number of the term from the first. POWERS or BINOMIALS. J[77 Note. — Tbe coefficient of the last term in any power is alw&ys equal to the corresponding power of the coefficient of the following term in the root. If 8. There is another class of biuomials that come under a modification of this Theorem, viz. : those having exponents. To illustrate the application, let us write out the fourth power of 2a' + SxK 2a^ + Sx^ 2d power, 4a« + 12a\x^ + 9x* 2a» -f Sx' 8a» + 2iaV + ISa^x* 12a'x^ + 86aV + 27xS 8d power, 8a» -f- oQa^^^ + b4:a'x\ + 27x« 2a^ + ^x' l(ia''+ 72aV + lOSa^^* + 54aV 24aV + lOSa^x' + 162a»x« + 81x» 4th power, 16ai2+ 95^,9^2 j_ 2l6a'x' + 216aV + 81^ On examining the result, we shall find that the exponent of the leading letter is 12(=="* ^ ^) in the first term, and that it di- minishes regularly by 3 in each succeeding term. Also that the exponent of the following letter is 2 in the second term, and increases regularly by 2, in each succeeding term, to the last, where it is 8(=-'^^). Hence the exponents are governed by the law of Newton's Theorem, as shown in (17^), modi- fied by the values of the exponents. The coefficients are the same as the coefficients of (2a + S:v)\ (174), and may be obtained in the same manner, if we keep constantly in mind tlie fact that the first exponent, 12, is the exponent 3 of the leading quantity in the root raised to the fourth power, and that the real exponent which we are to use as a factor of our dividend is the exponent of the leading quantity in any term divided by the exponent of tlie leading quantity M 178 INVOLUTION. in the root. But, as this is liable to be forgotten, we can nse the exponent ol the leading quantity in any term, whatever it may be, as a factor of the dividend, if we write the exponent of the leading quantity in the root, as a factor of the divisor. Observing this direction, and the indicated operations for ob- taining the several coefficients in (2a^ -f- ox^y^ will be as follows : 1st coefficient, 2* = 16 2d coefficient, ^ — '^ — - = 96 3x2 8d coefficient, ^ ^ g ^ ^ = 216 A,x. ^ ' , 216 X 6 X 3 ._ 4th coefficient, -x ^ ^r- =216 3x3x2 p;.!. «, . , 216 X 3 X 3 _^ oth coefficient, —. ?r — — - =81 ' 4x3x2 S79. Examining the indicated operations for obtaining the coefficients of the expanded form of (2a + Sx^ (177), we observe the following facts : IsL Each dividend after the first term is composed of three factors, the first of which is the coefficient of the preceding term, the second, the exponent of the leadiug quantity in the preceding term, and the third, the coefficient of the following quantity in the root. 2d. Each divisor is composed of two factors, the first of which is the number of the preceding term counted from the left, and the second the coefficient of the leading quantity in the root. 3cZ. The second factor of the dividend decreases regularly by 1, and the first factor of the divisor increases regularly by 1, in each succeeding coefficient. ^tJi. The third factor of the dividend, and the second factor of the divisor, are the same in each coefficient, i. e., they are constant. POWERS OP BINOMIALS. X79 We may, therefore, Let a= 1st coefficient of any binomial. 6 = 2d coefficient of any binomial. n == the exponent of any binomial. Then {ax±z hyy any power of any binomial. j^ssume the first coefficient to be Ci, the second C2, the third, C3, &c., and we shall have the following General Formula for Coefficients. Ci = a'^ _ Qvnh a C3 == ^'^^ ^-^)^ C 'la Q,(n — 2)^ oa Q^(n — 3)6 4a &c. &c. Note. We have now carried the investigation of this Theorem as far as the plan and limits of this work admit. It is general in its application, and may be used in the involution of ^i\y binomial whatever. Its full develop- ment, including its application to negative indices and binomial roots, will be found in future editions of the University Algebra. EXAMPLES FOR PRACTICE. 1. Required the 4th power of 2a + 8x. Ans. 16a* + 96a^^ + 216aV + 216aa;^ + 81x*. 2. Expand (2a — bh)\ Ans. 8a' — 60a^6 + ISOaft^ — 125^8^ 8, What is the cube of Ix + 2ay ? Ans. 843a;s + lUx'ay + 84a:ay + 8ay. 4. What is the fifth power of 5a — 2c? Ans, 3125a^— .6250a*c4-5000aV— 2000aV+400ac*— 32e\ 5. Expand (a;' 4- 3 2/^)5. Ans. ^i» + \f)J^y'' + 90xy 4- 270xy -f- 405:ry + 243^1°. 180 INVOLUTION. 6. Expand (2a^ + axf. Am. 8a^ + I2a^x + 6a'x' + 2a'x\ 7. Expand (x — !)«. Arts, x' — Qx' + 15x'^ — 20^8 + 15^=^ — Qx-x-l. 8. Expand (3a; — 5)^ J.ns 21 x" — 135x'^ + 225a" — ■ 125, 9. Expand (Aa'b -— 2c'y, Ans. 2^QaW — bl2aWc' + 'SS4:a%'c' — 128a^6e« + 16aj . 10. Expand (| + ^|)'. Note. — The quantity (^-\- -~) = ( — " + j ^) • 1 ' ^ 15 4 _L 90 , , , 270 , , 405 ^ , 243 . 11. Expand (3 — 2r) to the 6tli power. Ans, 729— 2916r+4860r^^4320r'+2160r*— 576r5 + 64rfi.. . 12. Expand (x + — j to the 7th power. 5 13. Expand (l + o^) ^ ,25 125 , 625 3 . 8125^ , 3125^ Ans. l + -^x + —x^ + -^x"" + -jQ-^ + -g2"^- 3 5 14. Expand (^ — ^xj 81 45 75 ^ 250 3 625 Ans. ^ — —x^-^x ^x +-gj-^. 15. Expand (x' — Sy^. Ans. ^i<'— 15xY+90a;y— 270^y+405^y— 243y^o. ROOTS OP MONOMIALS. j81 EVOLUTION; OR, THE EXTRACTION OF ROOTS. ■^^ 1§0. A Eoot is a factor repeated to form a power ; or, it is one of the equal factors of a quantity. 181. Evolution is the process of extracting tlie root of a quantity. It is tlie converse of involution, and is indicated by the radical sign, v^- 18^. The Index of the root is the figure placed above the radical sign to denote what root of the quantity unde4- the radical is to be taken ; thus, in i/a, 5 is the index of the root, and denotes that the fifth root of a is to be taivcn. 183. A Surd is the indicated root of an imperfect power ; the root thus indicated cannot be exactly obtained or ex- pressed ; thus, v/2 is a surd, because the number 2, not being a perfect square, can have no exact square root. A surd is sometimes called an irrational quantity. ROOTS OF MONOMIALS. 18-4. To discover the process of extracting roots, we must observe how powers are formed, and then trace the operations bach. Thus, to square a, we double its exponent, which makes d\ (^7.) The square of a^ is a^, the cube of a^ is a^, &c. The 4th power of x, is x^ ; the ?ilh power of x^ is ^*"; &c., &c. Now, since multiplying exponents raises simple literal quan' tities to j^owers, dimdiiig exponents must extract roots. Thus, the square root of a* is a^^ = a^ ; the cube root of a^ is a'^' = a\ The square root of a must have its exponent (1 understood), divided by 2, which will give a^ ; the cube root of a in the like manner is a"^, and the exponents, I, ^, I, ^, &c., indicate the 16 X82 EVOLUTION. second, third, fourth, and fifth roots of any quantity whose ex- 5 poneut is 1. The 6th root of ^ is x^. In l!ke manner, | ex- presses the 4th root of the 3d power of a quantity. - Hence the foUowing principles : I. Boots are projjei^ly expi^essed by fractional exponents. II. The numerator shows the power of the quantity, whose root ts to be extracted. III. TJie denominator shoics what root of that power is to he extracted. It is the index of the root. We have seen (1114:) that any power of a positive quantity is positive, and tliat the even powers of a negative quantity are positive, and the odd powers negative. From this it results, that I. lire odd roots of a positive quantify are always p)Osi- tive, and the even roots are either jjositive or negative. II. The odd roots of a negative quantity are negative, and the even roots are impossible or imaginary. Note. — An Imaginary Quantliy ia the indicated even root of a negative quantity, as ^ IT^ or ^Zl2^. 1. What is the square root of 64a* j;^ ? OPERATION. Analysis. Since the power of a monomial is (64aV)^±= d= 8a'j? formed by involving each factor, (165), conversely, VERIFICATION. *^° ^""^ "^^^ ^^ "^'^'"^^ by extracting the root of (+ 8a'^) X (+ ^a^x) = 64a V each factor separately. ( — %d'x) X ( — Sa'^a;) = Qtla^x"^ The square root of 6 1 is 8 ; of a^ is a^ ; of x^ is x ; and the entire root is Sa^ic, to which we give the double sign, zfc, (re a plus or minus), because either + "^a^x, or — %a^x, squared, will pro- duce G4aAf^ as is seen in the verification. 185. From these principles and illustrations we hare the following ROOTS OF MONOMIALS. 1^3 Rule. I. Extract the required 7^oot of the numeral coefficient. II. Divide the exponent of each letter by the index of the root. III. Prefix the double sign, rb, to all even roots, and the minus sign to the odd roots of a negative quantity. Note 1. Under this rule for monomials we shall introduce no numeral coefficient the required root of which will consist of more than one place ; hence the root may be found by trial. EXAMPLES FOR PRACTICE. 1. What is the second root of ^a'^x^if ? Ans, d= Sax^y^, 2. What is the third root of Say ? Ans. 20% 3. What is the fourth root of 81a*:c^^ ? Ans. d= oax^ 4. What is the fi:fth root of 32a^x^y^ ? Ans. 2ax'f, 5. What is i\\Q fourth root of 81a*Z)V^ ? Ans, rtSafe^c^ 6. What is the third root of — 27a^ V ? Ans, — 3a*a;. Find the following indicated roots : 7. {—21a^by, Ans. — 3«.^6, 8. (25xY)i Ans. ± 5xV. 9. v^lGa;*'". Ans, =fc 2x"'. 10. -v/oy. Ans. a^'if. 11. V125a«m». Am, Sa^wi^ or ^a^^^, 12 x/^ -4/15. a^ 18. {xTy'^Y. Ans, xy^, or arV^. L !!' 14. (a^6'0". Ans, a% or 6y^ 15. Find the cube root of 4a*. Note 2 If the coefficient is an imperfect power, it may be treated as a literal factor, and its root indicated. Ans, 4 V, or a^^f. 184 EVOLUTICN. 16. Find the 5th root of 7a^6^^ Ans. l^Jb'. 17. Find the 9th root of — lox^y^, Ans, — 15^x V'^- 18. Extract the square root of — -. Note 3. Since the power of a. traction is formed by involving the nume- rator and denominator separately, the root of a fraction will be obtained by extracting the root of*the numerator and denominator separately. 19. Extract the cube root of — tt-t-.- 7i\ 20. Extract the fourth root of — . 64 a^ 21. Extract the square root of ^-z — ,. SQUARE ROOT OF TOLYNOMIALS. Ans. 3x [ns. - "" 2a V A71S. 1 n ' Ans. 9x 186. In order to discover the process of extracting the sqtiare root of a polynomial, we must observe how the squares of polynomials are formed. If we square a -{ b, we shall ha^ e (a + by = a^ -f 2ab 4- b\ This result, expressed in words, is as follows : The square of (he first terii, plus twice the product of the two terms, plus the square of the second term. The last two terms of the power may be factored as follows; 2ab -f b' = (2a + b)b, which is expressed thus : Twice the first term plus the second, multiplied by the second. SQUARE ROOT OF POLYNOMIALS. 285 1. Extract the square root of a'^ _|. 2ah + 61 OPERATION. Analysis. Reversing; the pro- ^2 j^ 2ah + &H a + & ^^^^ of invclutlon, we extract tiio 2 ~ square root of a\ and obtain a. ^ the first term of the root. The 2a ^ h ) 2a6 + 6^ next term of the power is 2ah =: t;, , , 2 2a X &> or if^^7^■ce the first term cf tlie root multiplied hy ilie second , we therefore divide this terra hj 2a, twice the first term of the root, to obtain 6, the second term of the root. Placing h in the divisor also, at the right of 2a, we have 2a + h, or twice the first term plus the second, which, multiplied by h, gives 2a6 + b\ the last two terms of the power. Again, let us form the square of any polynomial, as a -f 6 + o, in the following nmnner : Assume s = a -\- b, the first part. c = the second part. Then (5 + c)' = s' + 2sc + cK Hence, The square of any polynomial, considered in itco pjarts, is equal to the square of the first part, plus tivice the p)rodiict of the two parts, plus the square of the second p)art. Thus the root of any quantity can be brought into a bino- mial, and the rule for a binomial root will answer for a root containing any number of terms, by considering the root already founds however great, as one term, or one part. 2. Find the square root of a^ -j- 2ab -f ¥ -r^ac + 2bc -f c*, OPERATION. a'' + 2ab + b' + 2ac + 2bc + o\a + b-hc) a' la + b 2ab + b' '2ab -4- ¥ 2a -f 26 4- c 2ac + 26c -f c^ 2aG + 26(? + c^ 16* 186 EVOLUTION. Analysis.- -Proceeding as before, we obtain two terms of the root, a {- b, and a remainder of 2ac + 2bc + c-. We now consider a -\- b as the first part of the required root, and write 2a + 26, or twice the pari already founds for a divisor. Dividing, we obtain c, the next term of the root, which, as before, we phice in both the root and divisor. Multiplying this complete divisor by c, and subtracting the product from the dividend, we have no remainder, and the work i» complete. 18 T. FroiL these illustrations we deduce the following Rule. I. Arrange the terms according to the powers of some letter, beginning with the highest, and write the square root of the first term in the root, II. Subtract the square of the root thus found from the first term, and bring down the next two terms for a dividend. III. Divide the first term of the dividend by twice the root already found, and write the result both in the root and in the divisor, lY. Multiply the divisor, thus completed, by the term of the root last found, and subti^act the product from the dividend, and proceed with the remainder, if any, as before. Note. — According to the principles established in (184), every square root obtained will still be a root, if all the signs of its terms be changed. EXAMPLES FOR PRACTICE. 1. What is the square root of a* + 4a^6 — 4a'^ 4- 46^ — 85 + ^ ? Ans, a' + 2b — 2, 2. What is the square root of l—ib+4:b'-{-2y—4:by-i-7/? Ans. 1 — 2b + y. 3. Wh^i is the square root of 4x*— 4x^ + IS.r^—Gx + 9 >' Ans, 2x^ — x-i-3. 4 What is the square root of 4a:* — 16a;^-4-24a:^ — 16x-f4 ? Ans. 2x^ — 4a; -f 2. 5. What is the square root of 16a;* + 24:x^ + S9x^ -f 60a: 4- 100 '/ Ans. 4x^ -{- Sx + 10. SQUARE KOOT OF POLYNOMIALS. 187 6 What is the square root of 4x* — IG.c* + Sx'^ + lOx + 4 V Ans. 2x2 — 4x — 2. 7. What is the square root of x' + 2xy + y- + 6x2 4- %2 8. What is the square root of a^ — ab + J 6' ? 9 What is the square root of r^ — 2 -f- - ? a b ha Ans. , , or 7-. b a a o 2 11 2 10. What is the square root of x^ — 2x^2/ + 2/^ ^ ^ i i i i ^ns. x-^ — t/-^, or 1/*^ — x-^. 11. What is the square root of 1—4^ + lOz'^— -202^ + 252:^ -.242^ + l^z' ? Ans. 1 — 22-1- 82^ — 4z«. 12. W^hat is the square root of a« — 6a^c -f 15ttV — 20aV 4- 15aV — 6ac^ + c' 't Ans, a« — Sa^c + 3ac^ — d". 13. What is the square root of 22-— 22 + 1-f 22/i-~2/i-f/i2? Ans, 2 + h — 1. SQUARE ROOT OF NUMBERS. 188, In extracting the square root of numbers, the first thing to be considered is the relative number of places in a given number and its square root. This relation is exhibited in the following illustrations : Iloot«<. Squares. Roots. Squares. Ill 1 9 81 10 1,00 99 98,01 100 1,00,00 999 99,80,01 1000 1,00,00,00 From these examples we perceive that a number consisting of one place may have one or two places in the square ; and that in all cases the addition of one place to the root adds two places to the square. Hence, 188 EVOLUTION. If a number he pointed off into periods of two figures each J commencing at the right, the number of full peinods, and the left hand full or jMrtial period will be equal to the number of places in the square root ; the highest period answering to the highest figure of the root, 189* The square of any numeral quantity may be formed after the manner of algebraic squares. For example, let a ~ 40, and 6 = 7; then a + o = il. And since the square of a + 6 will represent the square of 47, W8 have a^ = 1600 2a6= 560 6^= 49 a^ -f 2ab + 6^ = 2200 = (47)1 Hence, the binomial square may be used as a formula for extracting the square roots of numbers. 1. Extract the square root of the number 2209. OPERATION. Analysis. Here are 99 OQ UO -L 7 >- 17 ^^"^ P'^'^^^' indicating ^^,uy i^u -i~ ^ — 4-/ ^^^ places in the root, Q? = 1600 corresponding to tens 2a. = 80 609 ^"^ units. The great- 2a + b = 87 609 ^f ^^";\''«i" ^^ 's 16, ' Its root IS 4, or 4 tens = 40. Hence, a = 40 Then 2>a = 80, which we use as a divisor for G09, and obtain 7 for a quotient. The 7 is taken as the value of 6, and 2a -f 6, the com- plete divisor, is 87, which, multiplied by 7, gives the last two terms of the binomial square, 2a5 -{- 5^ = 609, and the entire root, 40 4- 7 = 47, is found. Arithmetically, a may be taken as 4 instead cf 40, and we may write 16 in hundreds' place, instead of 1600, the ciphers being superfluous. Then 2a will be 8 instead of 80, and in dividing, we say 8 is contained in 60 (not in 609) 7 times. If the given number consists of more than two periods, we obtain the two superior figures of the root from the first two SQUARE ROOT OF NUxMBERS. 1S9 periods, as before, and bring down another period to the remainder. We then consider the root already found as one qaantity, and treat it as one figure. 2. What is the square root of 390424 ? OPERATION. Analysts. Disregarding the local '^o G < cA^por, value of the fl,i!;ures, we have a = G, ;iJ J-t,^4:(po:^ 2fl = 12, and 12 in 39, 3 times, which 36 " gives 5 = 3. We next suppose a = 63, 123 "S94 ^"^ -^ — ^^^ 5 ^"^ 1^^ ^^ ^^-' 2 *^^^'^» o/»rk or the second value of 5 = 2. In the same manner, we would repeat the 1262 25 24- formuI?« of a binomial square as many 25 24 times as we have periods. It is evident ihat we may obtain the divisor 126 from the last complete divisor 123 simply by doubhng its last figure 3 ; and thus the divisors may be derived each from the next preceding, successively. From these examples and Illustrations we deduce the following- Rule. I. Point the given nu7nher off into periods of tivo figures each, counting from the units^ place to the left and right. II. Find the greatest 2^67 feet square in the left-hand pe- riod, and write its root for the first figure in the required root ; subtract the square of this figure from the first joeriod^ and to the remainder bring down the next period for a dividend. III. Double the i^oot already found, and write the result on the left for a divisor ; find how many times this divisor is contained in the dividend, exclusive of the right-hand figure, and place the result in the root and at the right of the divisor. lY. Multiply the divisor thus completed by the last figure of the root ; subty^act the product from the dividend, and to the remainder bring down the next p)eriod for a new divi- dend. Y. Double the right-hand figure of the last complete divi- sor for a new divisor, and continue the operation as before. 190 EVOLUTION. r EXAMPLES FOR PRACTICE. 3. What is the square root of 883G ? Ans. 94. 4. What is the square root of 106929 ? Ans. 327. 5. What is the square root of 47829G9 ? Ans. 2187. 6 What is the square root of 43046721 ? Ans. 6561. : What is the square root of 387420489 ? Ans. 19683. 8. What is the square root of 1209996225 ? Ans. 34785. 9. What is the square root of 6596038656 ? Ans. 81216. 10. What is the square root of 342694144 ? Ans. 18512. U. What is the square root of. 2573733560796 ? .4ns. 1604286. 12. What is the square root of 10.4976 ? Ans. 3.24. 13. What is the square root of 3271.4207 ? Ans. 57.19 -f . 14. What is the square root of 4795.25731 ? Ans. 69.247+. 15. What is the square root of .0036 ? Ans. .06. 16. What is the square root of .00032754 ? Ans. .01809 +. 17. What is the square root of .00103041 ? Ans. .0321. Note. — If both terms of a fraction are perfect squares, or if the frao- tion can be reduced to terms which are squares, the root may be ob- tained b}^ the rule for algebraic fractions. Otherwise, the fraction may be reduced to a decimal. 18. What is the square root of §f ? Ans. |. 19. What is the square root of ^^^j ? Ans. ■^-^, j 20. What is the square root of ^^i£^ ? Ans. ^%\. ^ SQUARE ROOT OF NUMBERS 191 21. What is the square root of yVg ? Observe -j^^^ = ||. Hence, the square root is |, 22. What is the square root of ff §1 ? Ans. -§. 23. What is the square root of f|gi ? Ans, f. 24 What is the square root of | '/ Ans. 866 +. 25. What is the square root of ^ ? ^ns^ .8819 H-. CUBE ROOT OF POLYNOMIALS.* ^ lf^<0. We may derive the method of extracting the cube root of an algebraic quantity in a manner similar to that pursued in square root, by analyzing and retracing the combi- nation of terms in the binomial cube. Forming the cube of a -\- b, we have (a + by =a' + 2>a'b + Sab' + 5^ ; from which we see that I. The first term of the power is the cube of the first term of the root; and II. The second term of the poiveris three times the square 9f the first term of the root multiplied by the second. It is evident, therefore, that to find the first term of the root, we must extract the cube root of the first term of the power ; and to find the second term of the root, we must divide the second term of the power, 3a^&, by three times the square of the first term of the root, 8a'^ ; thus, 3a^& -- 8a- = b. The last three terms of the power may be factored as follows : (Sa^ + 3a6 + b'')b. To reproduce these terms from the divisor already found and the root, we must complete our partial divisor, 3a^, by the addi- * We are Indebted to J. C. Porter, A.M., of the Clinton Liberal Institute, for the valuable method of Cube Root presented here and in the Practical Arithrootia. It is an extension of, and improvement upon Horner's Method, and secures the result with less labor than any other method heretofore presouted. 192 EVOLUTION. tion of Sab + b^, and multiply the divisor thus completed by b. Putting the correction, oab + b"^, under the form of (3a + b)b, we shall have, 3a^ = Trial divisor. 3a -{ b = First factor of correction. oab 4- 6^ = Correction of trial divisor. 3a^ + Sab + b^ — Complete divisor. 1. .Find the cube root of a^ + Sa'b + Sab^ + b\ OPERATION. a' + Sa'b + Sab' + b' {a-hb ^ |3a^ Sa:'b + Sab' -f b' da + b Sab + b' \sa' + 3a6 + ¥ Sa'b + 3a6^ + b' Analysis. Taking the cube root of a**, we obtain a, the first terra of the root. Subtracting the cube of a from the given polynomial, we have Za^b -j- 3a6^ -f b^ for a remainder or dividend. We next write Za'^ at the left of the dividend for a trial divisor. Dividing the first term of the dividend, we obtain b, the second term of the root. We next multiply the former term of the root by 3, and annex the latter term, 6, and obtain 3a -\- b, the first factor of the correction to the trial divisor. Multiplying this by b, we have oab + b'\ the correction. Adding this to the trial divisor, we have 3a^+ 3a6+ 6^ the complete divisor. Multiplying the complete divisor by the last term of the root, and subtracting the result from the dividend, we have no remainder, and the work is complete. Again, let us form the cube of any polynomial, as a 4- 6 + c, in the following manner : Assume s = a + b, the first part ; c = the second part. Then (s + c^ = s^ + Ss'c + Ssc' + c\ The first two terms of the root, regarded as one part, sus- tain the same relation to the third, as the first sustains to the second : and so on. The binomial cube, therefore, furnishes the method of ex- tracting any cube root whatever, by treating the root already found, at each step, as a simple term. CUBE ROOT OF POLYNOMIALS. 193 2. What if5 the cube root of ^ — 40^^^ + 6a^ + 96a; — 64 ? OPERATION. Sa;«+2aj 6ar»+42;g 33;Q+6a^-4 — 12a;«— 24a;+16 3a;* + 6a;»+ 4a;« 6a;»+12a;*+ 8a^ 3x*+12»»+12a;3 _i2a;<— 48a:'+96a5— 64 3a;<+12a;«— 24a; +16 —12ai*—4:Sx^-\-96x—64. Analysis. Since it was shown, in involution, that the exponents of any letter in a power form a regular series, we arransje the terms according to the powers of x. The cube root of a^ is aj^ the first term of the root; subtracting the cube of x^ from the polynomial, and arranging the remainder according to the powers of x, we have 6x^ — 40.r* + 96a; — 64 for a dividend. We next write 3 times the square of a;^ or 3a:*, for a trial divisor ; and dividing 6x*, the first term of the dividend, we obtain 2x for the second term of the root. Having found the second term, we must complete our divisor as in the first exam- ple. Therefore, to 3 times the first term we annex the second, and obtain Zx"^ + 2x, the first factor; and multiplying this by the second term, we have 6ar* -f 4:x^ for the correction to the trial divisor. Add- ing, we have Sa:* -j- 6a:^ -f ^^'^i the complete divisor. Multiplying this by the second term, 2x, and subtracting the product from the dividend, we have for a new dividend, — 12a:* — 48a:^ + 96a; — 64. Now, since the two terms of the root already found, considered as one part, sustain the same relation to the third term, as the first term sustains to the second, the trial divisor to obtain the third term will be 3 times the square of the first two terms, or 3(a;2 -{- 2xy = Z.'c^ -{- 12a:^ -f 12a:2« This quantity is found in the operation by add- ing together 4ic*, the square of the last term of the root; 6x^ -f- 4a:^ the correction ; and 3a;* -f 6a:^ -f- 4a;2, the first complete divisor. Dividing — 12a;*, the first term of the dividend, by 3a;*, the first term of the divisor, we obtain — 4, for the third term of the root. To find a correction of the trial divisor, the first factor will be the last term, — 4, annexed to three times the former terms of the root, or 3a;^ -f 6a; — 4. This quantity is found in the operation by taking the first factor of the last correction, with its last term multiplied by 3, and annexing the — 4. Multiplying this by — 4, we obtaia — 12a;2 — 24a; + 16, for the correction. Adding this to the trial divisor, we have 3a;* + 12a;^ — 24a; + 16, for the complete divisor. Multiplying this by — 4, and subtracting the result from the divi- dend, we have no remainder, and the work is complete. 17 N 194 EVOLUTION. 191. From these illustrations we deduce the foLowing KULE. I. Arrange the polynomial according to the powers of some letter^ and write the cube root of the first term in the root, II. Subtract the cube of the root thus found from the poly^ nomial, and arrange the remainder for a dividend. III. At the left of the dividend write three times the square of the root already found for a tr^ial divisor ; divide the first term of the dividend by this divisor, and write the quotient for the next term of the root, lY. To three times the first term of the root annex the last term, and write the result at the left, and, one line below, the trial divisor ; multiply this binomial factor by the last term of the root, for a correction to the trial divisor ; add the cor- rection, and the result will be the complete divisor. Y. Multiply the complete divisor by the last term of the "^oot, subtract the product from the dividend, and arrange the remainder for a new dividend. YI. Add together the square of the last term of the ro3. To form the cube of a number, let a = 50, and b = 4. Then a + 6 = 54 ; and cubing, we have, a«= 125000 3a^6= 30000 8a&2 == 2400 ___J h^=- 64 (a + 6)8 == a» + Za'^h + Sab^ + 6^ = 157464 = (54)^ Hence, in the cube of a number. The figures of the root, with their local values, have the same combinations as the terms of an algebraic quantity. 1 What is the cube root of 157464 ? OPERATION. Analysis. Pointing off the 157 464 I 54 number, the two periods show -jo;; that there will bo two figures, ' — oT/<7>l *^^^ ^^^^ units, in the root. Sincn 7500 3l464 ^^le highest figure of the root cor- 154 616 8116 32464 responds to the highest period of the power, we find the greatest perfect cube in the first or left hand period, which is 125, and place 5, its root, for the figure of the required root. Subtracting the cube number 125 from the first period, and bringing down the next period, we have 32464 for a remainder or dividend. Since the figures in a cube root, with their local values, have the same combinations in the cube as the terms of an algebraic quantity, we write at the left of the dividend three times the square of the root already found, or 75, with tvro ciphers annexed, for a trial divisor. Dividing, we obtain 4, Tor the next figure of the root. To complete the divisor, we multiply the first figure of the root by 3, and annex the last, and obtain 154 for the first factor of the correction. Multiplying this number by 4, wo have 616, the correction to the trial divisor. Adding, we have 8166, the complete divisor. And multiplying this by 4, and subtracting tb.e product from the dividend, there is no remainder, and the work is complete. CUBE ROOT OP NUMBERS. 197 OPERATION. 12,812,904|234 8 63 189 1200 4812 1389 4167 694 2776 158700 645904 161476 64e5904 2. Wliat is the cube root of 12812904 ? Analysis. The great- est cube in the first pe- riod is 8, and its root is 2, ■which we write for the first figure of the required root. Subtracting 8, and bringing down the next period, we have 4812 for a dividend. Annexing two ciphers to 3 times the square of 2, we have 1200 for a trial divisor. Dividing, we obtain 3, the next figure of tho root. To complete the divisor, we have, bj the same method as be- fore, 63 for the first factor of the correction, and 189 for the correc- tion. Adding the correction, we obtain 1389 for the complete divisor. Multiplying this by 3, subtracting the product from the dividend, and bringing down the next period, we have 645904 for a new divi- dend. As in the algebraic method, we add 9, the square of the last root figure; 189, the last correction; and 1389, the last complete divi- sor ; and annex two ciphers, for a new trial divisor. Dividing, we obtain 4, the next figure of the root. AVe then take the first factor of the last correction, with its last figure multiplied by 3, and annex the last root figure, 4, and obtain 694 for the first factor of the new correction. Multiplying this by 4, we have 2776, the correction. Then completing the divisor, multiplying by the last root figure, and subtracting the product from the dividend, we have no remainder, and the work is complete. From these examples we derive the following Ktjle. I. Point off the given number into periods of three figures each, counting from units'' place to the left and right. II. Find the greatest cube in the left-hand, period, and place its root for the first figure of the required root. Subtract the cube from the first period, and to the remainder bring doivn the next period for a dividend, III. At the left of the dividend, write three times the square of the root already found, and annex two ciphers for 17* 198 EVOLUTION. a Mai divisor; divide the dividend ^ and write the quotient for the next term of the root. lY. To three times the first figure of the root annecb the lastf and place the result at the left, and one live below the trial divisor ; multiply the factor by the last 7^oot figui^e, for a correction to the tr^ial divisor ; add the correction^ and the result will be the complete divisor. Y. Multiply the complete divisor by the last figure of the root, subtract the product from the dividend, and to the remainder bring down another period for a new dividend. YI. Add together the square of the last figure of the root, the last correction, and the last comp)lete divisor, and annex two ciphers for a new trial divisor ; and by division obtain another figure of the root. YII. Take the first factor of the last correction, with its unit figure multiplied by 3, and annex the last figure of the root, for the first factor^ of the correction to the new trial divisor , with which proceed as in the former steps till the work is complete. Notes. 1. If at any time the product is greater than the dividend, diminish the corresponding root figure and correct the erroneous work. 2. If a cipher occur in the root, annex two more ciphers to the trial divisor, bring down another period in the dividend, and proceed as before. EXAMPLES FOR PRACTICE. 3. What is the cube root of 148877 ? Ans. 53, 4. What is the cube root of 571787 ? Ans. 83. 5. What is the cube root of 1367631 ? Ans. 111. 6. What is the cube root of 2048383 ? Ans. 127. 7. What is the cube root of 16581375 ? Ans. 255. 8. What is the cube root of 44361864 ? Ans. 354. 9 What is the cube root of 100544625 ? Ans. 465 10. What is the cube root of 12358-435328 ? Ans. 2312 CUBE ROOT OF NUMBERS. 199 11. What is the cube root of 999700029999 ? Ans. 9999. 12. What is the cube root of 2456 ? Ans, 13.491+. 13. What is the cube root of .004019679? Ans. .159. 14. What is the cube root of 2287.148 ? Ans. 13.175 +. CONTRACTED METHOD. 104. The methods of direct extraction of the cube root of surd numbers are all too tedious to be much used, and seve- ral eminent mathematicians have given more brief and practi- cal methods of approximation. Olc of the most useful methods may be investigated as follows : Suppose a and a + c two cube roots, c being very small in relation to a ; a* and a^ + Sa'-c + Sac^ + c^ are the cubes of the supposed roots. Now, if we double the first cube (a^), and add it to the second, we shall have 8a^ -f Sa'-c -f 3ac' + c*. If we double the second cube and add it to the first, we shall have Sa' + 6a'c + 6ac^ + 2c\ As c is a very small fraction compared to a, the terms con- taining c'' and c^ are very small in relation to the others ; and the relation of these two sums will not be materially changed by rejecting those terms containing c^ and c^, and the sums will then be And oa^ + Qd^c, The ratio of these terms is the same as thg ratio of a + oto a f 2c. c Or the ratio is 1 + a + c But the ratio of the roots a to a + c, is 1 H 200 EVOLUTION. Observing again, that c is supposed to be very small in rela- c c tion to a, the fractional parts of the ratios and are ' ^ a -\- G a both small, and very near in value t<) each other. Hence, we have found an operation on two cubes which are near each other in magnitude, and that will give results very near m proportion to their roots ; and by knowing the root of one of the cubes, by this ratio we can find the other. And as this relation will still exist if one of the roots is a surd, the propor- tion will furnish a method of approximating to values of surds. For example, let it be required to find the cube root of 28, true to 4 or 5 places of decimals. Since 27 is a cube near in value to 28, the root of which we know to be 3, Assume a^ = 27, or a = 3 . (a + cf = 28, or a + c = ^28- Then 27 28 2 2 54 56 Add 28 27 Sums 82 : 83 3 : a -I- / a — 6, together, we have Sav^a — 6, the simplest form of the radical. From these illustrations we deduce the following R.XJLE. I. Separate the factors of the quantity under the radical sign into two groups^ one of which shall contain all the perfect powers corresponding in degree ivith the radical. II. Extract the root of the rational part, and multiply the root, ooejficientj and surd or radical part together. REDUCTION. 205 EXAMPLES FOR PRACTICE. Reduce the following radicals to their simplest form , Ans. a\^hc. 8. v^^^ 4. 2v^a^ 5. 3V50^ 6. ai/TO^. Ans. 2xy\/ X, Ans, s/W. 7. 6v/81m\ 8. '^d'—a'i 9. xy^ x^if- — a^y. 10. Wriah'&, 11. 2av/l47aV?/. 12. 5^125x. 13. 2cv/82^\ 14. (a + 6) s/d? — 2a?h + ah\ 15. {a — h)s/a'h + 2ah^'{-h\ 16. cZ ''/^^t/ — 2a;y + a;?/'. 17. (ISOa;^!/)* 18. (24^2/^2)^ 19. (54am«)*. 20. (aV — a^62') i Ans. av^l — c?. <4ns. ^y V X — 2/. u4ns. 24&cv^2ac. ^ns. 14a^x'v Sa^/. ^?is. 25v'.x. -4ns. 4o^v ac. ^ns. (a2_52)N/a. -4ns. d(x — y) \^xy, Ans. Q>x(pxyy^. Ans. 2xy{^x'^zy\ Ans. 8m"(2a)^. Ans. a^z(a — 6)^. CASE n. SOI, To reduce a rational quantity to a radical, or to introduce a coefficient of i radical under the radical sisfn. 1. Reduce bax^ to the form of the cube root. OPERATION. bax" = {pax?y n= V\2ba?x^ 18 Analysis. "We cuhc each factor of the given quantity separately, and indicate the cube root of tbe result J 206 RADICAL QUANTITIES. 2. Reduce 1c \' x to a radical without a coefficient. Analysis. We laise OPERATION. the coefficient, 2c, tu the (2c)^ = (16c*)* fourth power, and we have (IGC*)*. Multiply. ing this result by re* , we {\<6&y X x^ = (IGc^o;)^ Hence 2c^ x = ^ V^d^x have V^lGc^x-. Hence, the Rtile. I. To reduce a rational quantity to a radical : — Involve it to the same power as the requii^ed index ^ and write the result under the corresponding radical sign. II. To introduce a coefficient of a radical quantity under the radical: — Involve it to the same power as the radical, multiply the radical by the result^ and write the product under the radical sign. EXAMPLES FOR PRACTICE. 8. Keduce ax^z^ to the form of the square root. Ans. -J a?-xH^\ . 4. Keduce 9a^?/ to the form of the cube root. Ans. VlWo>Yy or (729a^y)^. 5. Reduce a + ex to the form of the fourth root. Ans. (a* + 4a3cx + GaVx" + ^a&x"" + c^x^Y . 6. Introduce the coefficient oi a^^/c under the radical sign. Ans. "^ a^c. 7. Introduce the coefficient of 3a >/2a*.;c under the radical sign. Ans. ^ 54a'a:. Reduce the following quantities to equivalent radicals with* cut coefficients : 8. (2a - c)^^4. Ans. (32a»— 48a2c + 24ac'— 4c«)K 9. 4c' v^^ Ans. ^T024ac^^ 10. ax^(a + bxyy, Ans. ^ aV + a'bx'y. REDUCTION. 207 11. Cg6 + x)yd'h- — 'Zahx + x'. Ans. s^^aW— 2d'h'x' + x^. 12 (a^ — ¥)s^. Ans. (a' — 2a'h' + ah^y. CASE m. 20f5. To reduce radicals of diflerent degrees to a common radical index. 1. Reduce a^ and h^ to a common radical index. Analysis. "We have seen (201) OPERATION. that the value of any quantity is i 3 not changed by involving it to a ■= d any power and placing the result 7)3 = h^ under the corresponding radical , J j^ ^ sign or index. That is, the nth a^ = (a^)^, b^ = (&0^ root of the nth power of any Qx' quantity is the quantity itself. I g i_ 6 / T- Now, as the index of a in the given a^ = V a^, 6^ = V i'' example is I, and of h -J-, we may raise the two quantities to any powers that will make the deno- minators of their indices the same. This we do by reducing the in- dices -J and J to equivalent indices having a common denominator, as shown in the operation. Hence, KuLE. I. Reduce the indices to a common denominator. II. Write the numerator of each equivalent index as an exponent of its respective quantity^ and place the result un^ der the common radical sign or index, EXAMPLES FOR PRACTICE. 9 1 3 2. Reduce a^, d'\ and cn to a common radical index. OPERATION. 1. h 3 __ n ~ 4n Sn 18 6n 2 a^ = An a6»i = : V^" = = (a^^L 1 dn = a67t =rr VdF- = = {d'yn 3 r= 18 cen = v;? = = Cc^«>l 208 REDUCTION. 8. Keduce m, (an)^, cxy'^^ and 5 to a common radical index. Ans, (m«)^, {ahi^^^^, (c'xY)\ (125)* 4. Reduce a*j?, (any and v^cZ to a common radical index. Ans, 'V ^i^ V 7^, 'Vd\ ADDITION OF RADICALS. 303. 1. What is the sum of 3 s^ab and 5 \^ab ? OPERATION. o ^~T Analysis. We make the common radical, y— \/a6, the unit of addition ; and adding the co- efficients, we obtain 8 \/a6, the required sum. 2. What is the sum of ^2bi)a' and >/lM? OPERATION. Analysis. We reduce the radi- v^250a^= 5a ^2a'^ ^^^^ *^ their simplest form, and 3 . — ; — 3/- — obtain two similar radicals. Add- vlba = -^ "<^ ing their coefficients, we have . Sum, 7a ^2^ 7a v'2^. From these examples we deduce the following Rule. I. Reduce each radical to its simplest form, II. If the resulting radicals are similar, add their coeffi- dents, and to the sum annex the common radical ; if dis- similar, indicate the addition by the plus sign. EXAMPLES FOR PRACTICE. 3. Add 3 \/2>a^x and a V'48a; together Ans. 7a >/%x. 4 Add VSOm and Vl25m together. Ans. 9 v^5m. 5. Add v/72, n/I28, and ^\ Ans. 16v^2. 6. Add xv/3^, 32/\/3a^, and 2%/3^J^l Ans. ^xy>/Za. 7. Find the sum of >/2a^xy and VWxy. Ans. (a 4- b)\^2xy. SUBTRACTION. 209 8. Find the sum of ^a'-x — d'y and \/'^d\x — y). An^. Sav^x — y. 9. Find the sum of \/80^ and v/245a«6^. _ An^, (4a6+7a^6^)v/5. 10. Find the sum of 3V3a^^, y/Vld^x, and v^36^a:. ^ns. (5a 4- lS)V'6x, 11. Find the sum of >/2^2\/2^^, and v^2F. An$. (a 4- fe)'V2. 12 Find the sum of V^^ and l/Vlha\ _ Ans. 7Va\ 13. Find the sum of V270a'm and VlM0¥m7 A71S, (3a + 66)Vl0^. 14. Find the sum of l^^%VSxY, and ^^ Ans. (x+yy ^ xy, 15. Find the sum of \ / r-^ and \ / ~. ^ris. Va. r—.. V 16 ^16 16. Find the sum of \/^ and Vab'^ xins. a\^b + b Va. 17. Find the sum of \^a^i and V^. 18. Find the sum of 2(Aa''by and (36tt^5)^". Ans. 10 ab- '"^ 19. Find the sum of (aV — aVy)^ and a(x* — o;^?/) ^ . -- ^;is. 2ax(x — y)^, SUBTRACTION OF RADICALS. / " S©4. 1. From \/98a take v/50a^ OPERATION. Analysts. Ileducin;^ the radicals to their simplest form, we obtain the two V 98a = 7v2a similar radicals, 7 v^2^ and 5 V2S. Making \/ ^Oa = 5v^2a the radical part the unit of subtraction, we take the difference of the coefficients, Difference, 2 V2a and obtain 2\/2a. Hence the following 18* o 210 RADICAL QUANTITIES. Rule I. Beduce each radical to its simplest form. II. If the resulting radicals are similar, subtract the coef- ficient of the subtrahend from the coefficient of the minuend^ and to the remainder annex the common radical; if dissimi- lar ^ indicate the subtraction by the minus sign. EXAMPLES rOR PRACTICE. 2. From 3\/5a^c take aV^dc. Ans. 2av^e. 8. From V^162:c*2/ take 4\/8x^. Ans. x^s^^y. 4. Yrom \^'A{)a%^ytfikQ'yba%y. Ans. a%^bhy. 5. From 3 v^l28a'6(? take 4.as/T%^. Ans. 4.a^2bc. 6. From ^375?& take VUab^. Ans. {ba—2b)^^ab. 7. From ^lQaW)^ take 2a(a56')\ Ans. ^a\ab''Y. ^- 8. From VZd'c -\- Qabc + Sb"c take V^126''c. Ans. (a — 6)v^8c-. 9. From v^2a"^c^ take ^2bc'^. Ans, ac^'Za — ^\^2b, \ 10. From \/a^ — a^ take Vah' — b\ -4ns. {a — b)^a — h. 11. From v/Ftake v^f Ans. W'l, ^ 12. From 6^82 take Q'^^S. Ajis. lOv'i 18. From24/32^nake4v/^^ Ans. ^a{^a'—^¥\ MULTIPLICATION OF RADICALS. CASE L SO^. To multiply radicals of the same degree. Since the root of a quantity composed of several factors is obtained by extracting the root of each factor separately {I§,'5), we have (a6>-= a^ _X 5=;_ ' Or, by radical sign, V ab = Va x \^b ; Conversely, Va x ^b = Vab. Hence, if we consider a and b as representing any two quanti- ties, and n the index of any root, we have The 2yroduct of the roots of any two quantities is equat, to the root of their product. MU-LTIPLICATION. 211 I. Multiply 3a v^i^ by 2 v/^. OPERATION. , Analysis Since the pro- duct will be the same, m Ba^'^x X 2%/^/ = 6a\^xy whatever order the factors are taken, we inultiply the coefficients Za and 2, and obtain 6a; and the radical parts \/x and v/'y, and obtain, by the principle enunciated above, y^xi/ ; and the entire product is Q)a\/xy, Hence the following Rule . I. MultijAy the coefficients together for the coeffi' oient of the product. II. Multiply the quantities in the radical parts together ^ and place the product under the common radical sign. III. Reduce the entire result to its simpAest form. EXAMPLES FOR PRACTICE. 2. Multiply 2a^^Sx by 4v^ Ans. 8aV^3^ 3. Multiply b^ac by ^^am. Ans. da^mc. 4. Multiply 3>/5^by 4n/20x. A7is. UOx^^'y^ 5. Multiply 2\/9x^ by s/oxyz. Ans. 6x\^yz. 6. Multiply 2 v^ 14 by 3^4. Ans. 12v'7". 7. Multiply S>/ghj2y/S. Ans. 18. 8. Multiply 3N/2by4v/8. Ans. 48. 9. Multiply V6 by >/T507 Ans. 30. 10. Multiply ^Ihjs/I. Ans. -}V3. + 11. Multiply a + ^bhj '^b. Ans. a\^b + b. 12. Multiply X + v^2/ ^7 ^ — ^V- -^^s* ^^ — V- 13. Multiply v^m + v^n bj v^m — \^n. Ans. m — n. 14. Multiply v^a + >/c by \^a + v^c. 4^ u4ns. a + 2v^ac + c 15 Multiply a(6/ by c(c?)-. Ans. ac{bdy. 16. Multiply 2c(o!'bdf^ by (3a&)^. ^ns. 2ac(3ra)'l 17. Multiply (x'^^/)^ ^7 (p^y^y* ^^^- ^V. 212 RADICAL QUANTITIES. CASE n. 306. To multiply radicals of difierent degrees. 1. What is the product of a- multiplied by 6^ ? OPERATION. Analysis. Letting P represent the pro- 1 \_ duct of the given quantities, we form equa- Y = a b^ (1) |.j(3^ (;^)^ Squaring both members we have p2 _. Q^^ /o) (^)> in which the index of the factor a is 1. pg 3T 2 3 Cubing (2) we have (3), in which both factora J are cleared of their radical indices ; and ex- P = {a^b^y (4) tracting the sixth root, we have (4), in which the product is under a common index. BECOND OPERATION. ANALYSIS. We first reduce the given ra- 1 3 dicals to equivalent quantities having a com- ^" ~ ^2 mon radical index, by (Case III., Reduction), o ^^ J 9/a"x c>bVd -f x, Ans. 12b'Va'\d + xf. 8. Multiply Y I by y J Ans. S/^.. APPLICATIONS. 213 DIVISION OF RADICALS. '' ■ CASE I. SO?. To divide radicals of the same degree. Since iLe root of a fraction, or of the quotient of one quan- tity divided by another, is obtained by extracting the root of e^ch term separately (t §■«!), we have Kb) = 1' 6" Or by the radical sign, \ / -= -— . Conversely, y-^^ = \J - ; hence The quotient of the roots of two quantities is equal to the root of their quotient, 1 Divide 6a^V^'?/ by 2a >/;r. OPERATION. Analysis. Dividing the coefficients, we have 3a for the new coefficient. (Sa^\^ x y And, by the principle stated above, \/xy divided by '^^=\ / " — = v^2/ ' the entire quotient, therefore, is oa\/ y. Hence the following Rule. I. Divide the- coefficient of the dividend hy the coefficient of the divisor. TI. Divide the quantity in the radical part of the dividend by the quantity in the radical part of the divisor ^ and place the quotient under the common radical sign, III. Prefix the former quotient to the latter j and reduce the result to its simplest form. 214 RADICAL QUANTITIES. EXAMPLES FOR PRACTICE. 2. Divide 4\^a6c by \lV ac. Ans. 2\/5. 3. Divide ^\'Zba}'xy by v^5a^;r. Ans, [m^y. 4. Divide 2v/20iW by \/2m. Ans. 2Qm>/m. 5. Divide v/lGO by n/8. .4?is. 2x^5". 6. Divide v^M by \/6 u4ns. 3. 7. Divide 8 n/ 72 by 2 v/ 6. Ans. 8n^3." 8. Divide 3\/l0 by \/T5. ^7?5. \/6. 9. Divide (a'b'cy by (a&)^. Ans. a(bcy . 10. Divide 12(xY)* by S(xy)K Ans. ixij(y")K 11. Divide (a'b'cd')^ by (a25cZ)K u4ns. d(b'c)K^ 12. Divide ^^a"^ — a*6 by Va. Ans. aVl — a6. '"t 13. Divide ^a^ — &^by ^oTT. J^ns. \^'a^'b 1-1. Divide (a^ — Jfy by (a — 6)i Ans. (a + ^>)i 15. Divide sy-l by \/^- ^^s- — >/ab^. t CASE II. S©8. To divide radicals of different degrees. 1. What is the quotient of (aby^ divided by a^ ? OPERATION Analysis. Letting Q represent tho 1 quotient, we have equation (i). Jlais- /-x (^' bj ^^^ ing both members to the sixth power, V we have (2) an equation without a radical index. Dividing a^b'^ by €^ Q6 __ ^ ^ we have (3) ; and extracting the sixth <^' root and reducing the index of b to Q^^ z=z b^ (3) its lowest terms we have, (4) tho ro- Q = &^ = b^ (4) ^^'^^^^ ^^^"^^- APPLICATIONS. 215 SECOND OPERATION. Analysis. We first re- X 3 , duce the dividend and di- a'^ =z a = (a^^ visor to the same radical in- (a^b^y' -^ (a^y = b^ — 6* dex, by (202), and then di- viding as in Case I, vre have tne same result as before. Hence, KuLE. I. Reduce the radical parts of the dividend and divisor to a common radical index. II. Divide the rational and radical parts separately, as i?% Case L EXAMPLES FOR PRACTICE. 2. Divide (ax-y by (xyy, Ans. y — r • _2_ 3. Divide \^x by \^x, Ans, x'^^- 4. Divide 30 by Vb. Ans. Wb. 5. Divide lQx{a + cf by 5(a + c)\ Ans. 2x(a + cy'\ 6. Divide (p? — x"") (m + 2/)" ^1 (« + x)(m + yy, Ans. (a — x) (in + ^z) "'"• 7. Divide %/f by v/f. Ans. 2^-^^^^. 8. Divide -^-=. by —y= . Ans. x-~. - V ax ^x '^d PRINCIPLES RELATING TO THE APPLICATION CP INVOLUTION AND EVOLUTION. ^I^9, The application of Involution to the solution of radical equations, is governed by the principles illustrated by the three following examples. ]. Kaise Va to the second power. Analysis. Placino; the pro- OPERATION. ^^^t ^f tj^g .lg^^gj.g ^^^^^ ^^^ >/a X v^a = v^o^ = a common mdical sign, (205), wo have \/a^ — a. 216 RADICAL QUANTITIES. 2i Raise >^'a + 6 to the second power. OPERATION. Analysis. Since the _ _ given quantity is a bi- ( \^a + by = a + 2b v^a + b^ nomial, we write a, the square of the first term ; 26 %/«, twice the product of the two terms ; and h'\ the square of the second term. 3. Kaise \^a + "^b to the second power. OPERATION. Analysis. Since the _ quantity is a binomial, ( ^/a + \^by = a + 2 \^ab + b we write a, the square of the first term ; 2 \^ab, twice the product of the two terms ; and b, the square of the second term. These three examples establish the following principles : I If a radical quantify be involved to a power cor- responding to the radical index^ the radical sign will be removed. (1.) II. If a quantity containing both radical and rational terms be raised to any power, the radical sign will not be removed. (2). III. If a quantity consisting of two radical terms of the second degree be squared, the result will contain but a single radical term. (3). SIO. The application of Evolution to the solution of equa- tions above the first degree, is governed by the principles illus- trated in the following examples : 1. Extract the nth root of a". OPERATION. Analysis. We divide the exponent of the power n/-^ = a ^^ ^^^® given quantity, by the index of the required root (185), and obtain 1 for the exponent of the root, which is omitted in the written result. SIMPLE EQUATIONS. 217 2. Extract the nth. root of a" + 6. Analysis. By the principles of involution, every power of a mo- nomial consists of one term only ; and the powers of a binomial con- sist of at least three terms. And since a" -f 6, the given quantity, has more terms than any power of a monomial, and a less number of terms than any power of a binomial, it cannot be a perfect power, and we therefore indicate its root, thus, ^a" + b. 3. Extract the square root of a^b^ + 2a^b'^ + a'^b\ OPERATION. Analysis. "We find by \/^= a'^b (1) ^^'^^^' ^^^ ^^^ ^^^' ^^^^ *^^ y— — ol the given terms are per- , V a'b^ = ab (2) ^^^^ squares, and that twice 2 x a'b Xab^ 2a 'b^ (3) ^^e product of their square y/a'^b'^ + 2a^¥ + a^b^ = a^b + ab roots is equal to the other term of the given quan- tity (3). This answers the condition of a binomial square (186), and we have a^b + ab for the required root. The principles illustrated by these three examples may be stated as follows : I. The exponent of a quantity will be removed by extract* ing the root whose index corresponds to tJie exponent. (1). II. The root of a binomial is necessarily a surd^ and a binomial always becomes a radical by evolution. (2). III. A trinomial is a perfect square when two of its terms are perfect squares and positive^ and the remaining term is twice the product of the square roots of the others , and either positive or negative. (3). SIMPLE EQUATIONS CONTAINING RADICAL QUANTITIES. 211, 1. Given 4 + \/ic-— 3 = 7 to find the va*.. ^ x. OPERATION. Analysis. We first tra<>spose 4 to the second member, so that the radical may stand alone, and ob- tain (2). We next square both 4,+ s/x — S = 7 (1) V' j; — 3 = 8 (2) X — 3 = 9 (3) members to clear the equation of a: = 12 (4) the radical sign (209, 1), and ob- 19 tain (8). Keducing^, wehave» = 12. 218 RADICAL QUANTITIES. 2. Given s/ x — 2 + \/ x + 6 = 4 to find x, OPERATION. Analysis. In order to x/ JZ2 + n/^T6 = 4 (1) ^^°'f *•>* involution of x / y ^ ' to the second power, we Vx-^1^\ — N/^_+_b (2) transpose one of the radi- ir~2=16— .8v/a;+6 + iC + 6 (3) cals to the second mem- \/^-|.d==3 (4) ber, and obtain (2). In- ic-f-6=:9 (5) volving both members to ^_.3 /gx the second power, we have (3), an equation containing cnly one radical. Transposing and reducing, we have (4), in which the radical stands alone. Squaring both members, we obtain (5), and reducing, cc = 3. From these examples, we derive the following Rule. I. If the equation have hut one radical term con- taining the unknown quantity, transpose the terms so as to make the radical stand alone, as one member ; then clear the equation of the radical sign by involution, and reduce as usual. II. If there be two or more radical terms, clear the equation of radicals by successive involutions, adjusting the terms at each step according to the principles enunciated in (209). Note. 1. Equations which appear to be higher than the first degree, sometimes become simple equations by reduction of terms. EXAlVrPLES FOR PRACTICE. 3. Given \^x + 5 = 9, to find x, Ans. x = 16. . 4 4. Given v ^ — 3 = "y - , to find x. Ans. a; = 7. ▼ X ■ o 5. Given ^^4 + (a? — 2) ^ = 3, to find x, Ans. x == 27. 6. Given x — -^x^ + 6 — — 2, to find x. Ans. x==i. 7. Given x + v'a?^ — 7 = 7, to find x. Ans. a? = 4. ^ Given ^^x + 12 = 2 -f y^x, to find x. Ans. x = 4. SIMPLE EQUATIONS. 219 9 Given 2 + (Sxy = v/5^ + 4, to find x, A71S. X = 12. 10. Given a; + 2 = ^4 + a7V^64 + ^to find .r. Ans, a; = 6. 11. Given a; — ^ v^a? = v^a;^ — x^ to find a?. Ans, a; = ^|. 12. Given Va; — 32 = 16 — v^^, to find a;. Note. 2. For brevity, put a = 16, and restore the value of a in tha inal, or reduced equation. Ans, X = 81. 13. Given \/3 + a; = y., , , to find x. Ans, a? = 3 Vd 4- X _6_ 14. Given Vx — 16 = 8 — Vx, to find a;. Ans, X = 25, 15. Given Va; + 8a = 2 -v^a, to find x, Ans, x = a. . 2a a 16. Given ^cx + a = ~v^, to find x, Ans. x = ~» Vza c 17. Given \^x + 2a = Vl^+ >/x — 2a, to find x. Ans, cc = — , A 1 m« 18. Given .-„ - = .-- — = ■ — a:, to find x. Ans. aj=-/m^ — 1. v^^+28 -^^+38 19. Given -7= = -p= , to find x, Va;+ 4 va;+ 6 Note. — Place y/x = i/j then find y -4w« aj = 4. 2a SO Given ^a;+>/ci + ic = (a + x)i* 220 QUADKATIC EQUATIONS. SECTION IV. QUADRATIC EQUATIONS. ^13. A ftuadratic Equation is an equation of the second degree, or one which contains the second power of the un- known quantity ; as ic^ = 9, or x^ + 3x = a. Quadratic equa- tions are divided into two classes, pure and affected. 31 3* A Pure Quadratic Equation is one which contains the second power only, of the unknown quantity ; as, x^ = 25, or x^ -f- 8a6 = 2c. jloTE. — A pure equation, in general, is an equation whicli contains only one power of the unknown quantity. 914. An Affected Quadratic Equation is one which con- tains both the second and the Jlrst powers of the unknown quantity ; as, x^ + Sx = 10. 215. The Eoot of an equation is such a value as, when substituted for the unknown quantity, will satisfy the equation. PURE QUADRATICS. S16. Since a pure quadratic equation contains only the second power of the unknown quantity, the unknown terms may always be united into one by miking the unknown quan- tity the unit of addition. Hence, Uvery pure equation of the second degree can he reduced to the form of ax^ = b] in which a and b are supposed to represent any quantities whatever. Thus the equation 3^2 — 21 = 7 — x\ becomes, by transposing and uniting terms, 4x^ = 28. ^ .^^U^^^4^^,+■ Z^ h Ir, uH^ LiAJ^ Z. <^«iA^<^ ^7w^'- + 2(3^ = 6«,'^+4i, t? Z-W^ J^ . ^^1 ir. VuhJL. Lu/ fjL. 0:t/x-\- An^. X = zizl. 17. Given x >/a^ + x' = a^ — x^, to find x. _ Ans, x= dhaV^. 18. Given ^x^ — a^ = a v^m — 1, to find x. _^ Arts, ic = zb a Vw. PROBLEMS PRODUCING PURE QUADRATIC EQUATIONS. ^IT. A problem may often furnish either a pure or an affected quadratic equation, according to the notation assumed. 1. Find two numbers whose difference is 6, and whose pro- duct is 40. SOLUTION. Let re — 3 = the less number ; ic + 3 = the greater. o:^ — 9 = 40 a) x' = 49 (^) X = 7 (3) X - — 3 = 4, the less number ; ■x + 3 = 10, the £?r( ?ater. Analysis. We repre- sent the less number by X — '3, and the greater by a; 4- 3, thus making the difference of the num- bers 6, according to the first condition of the prob- lem. Multiplying x — 3 by a; + 3, we obtain x^ — 9, the product of the two numbers, which we put equal to 40, according to the second condition of the problem, and we obtain (i), a pure equation, lleducing, we find a; = 7, if we Uijo only the plus sign. Then x — 3 = 4, the less ; and oj + 3 = 10 the greater. NcTR. — If we take a; = — 7, in the aoove problem, we shall have z — 3 ~: — 10, the less number, and x -\- S = — 4, the greater. This result, considered in an algebraic sense, satisfies the conditions of the problem, for —4 — (—10) =r 6, the dilference, and (—4) x (—10) = 40, the product. In general, however, such a value will be taken for the ua- kuowu quantity, as will satisfy the conditions arithmetically. //i 224 QUADRATIC EQUATIONS. This problem will give rise to an affected quadratic, if we assume the following notation : Let X = less number, 00 + Q = greater, Then x''+ Qx = 40, an equation containing both powers of x^ and for the solution of which rules will be given hereafter. In the problems which follow, an affected quadratic may be avoided by proper no- tation. 2. The sum of two numbers is 6, and the sum of their cubes is 72 ; what are the numbers ? SOLUTION. Let 2x = difference ; 3+07 = greater number ; 8 — X == less. (3 + xy = 27 + 27x + dx" + ar^, cube of greater; (3 — xy = 27 — 27a; + 9x^ — ar^ cube of les s. ^ I +18a;2=72 a) X^==l (2) X =1 (3) S + X = 4f greater number ; 3 — x = 2, less. Analysis. We let 2x represent the difference of the two numbers, 3 4-ic, half the sum plus half the difference, the greater (153); and 3 — Xf half the sum minus half the difference, the less; and as the Bum of these quantities is 6, this notation satisfies the first condition of the problem. Cubing each, and adding the results, we obtain for the sum of the cubes, 54 + 18x^ which we put equal to 72, accord- ing to the second condition of the problem. Reducing, we obtain 3 -f a; = 4, the greater number ; and 3 — x = 2, the less. 3. A and B distributed 1200 dollars each among a certain number of persons. A relieved 40 persons more than B, and B gave to each individual 5 dollars more than A ; how many were relieved bv A and B ? PURE QUADRATICS. £25 SOLUTION. Let ^ + 20 = th^ number relieved by A. ; X — 20 = the number relieved by B. rru ^-^^ . ^ 1200 ^^^^ ^-+20+ ^ = ^=20 ''' 240 240 DmdiDg(i)by5 ^^^20+ ^ = ^^120 <*' Assume a = 20 and h = 240 Eq. (2) becomes — ; — + 1 = (s) x-\-a X — a Keducing (3) x^ = a(26 + a) (4) Restoring values of a and 6, a-} = 10000 (5) By evolution x = 100 (6) ' a; + 20 = 120, number A relieved ; Hence, ' ^> |^_20== 80, " B 4. Divide tlie number 56 into two such parts^ that their product ehall be 640. Ans. 40 and 16. 5. Find a number, such that one third of it multiplied by one fourth of it, shall produce 108. An^, 86. 6. What number is that, whose square plus 18 is equal to half its square plus 30^ ? Ans, 5. 7. What two numbers are those, which are to each other as 5 to 6, and the difference of whose squares is 44 ? Note. — Let 6x = the greater, ^nd hx = tlie less. An%, 10 and 12. 8. What two numbers are those, which are to each other as 3 to 4, and the difference of whose squares is 28 ? AuB. 6 and 8. 9. What two numbers are those, whose product is 144. and the quotient of the greater divided by the less is 16 ? An^, 48 and 8. P 226 QUADRATIC EQUATIONS. 10. The leDgth of a certain lot of land is to its breadth as 9 to 5, and its contents are 405 square feet. Kequired the length and breadth in feet. Aiis. 27 and 15. 11. What two numbers are those whose diiTerence is to the greater as 2 to 9, and the difference of whose squares is 128 ? Ans. 18 and 14. 12. Find two numbers in the proportion of ^ to §, the sum of whose squares shall be 225 ? Note. — Multiplying the fractions J and | by 6, or reducing them to a common denominator, we find their ratio to be 3 to 4. Ans. 9 and 12. 13. There is a rectangular field whose breadth is | of the length. After laying out ^ of the whole ground for a garden, it was found that there were left 625 square rods for mowing. Required the length and breadth of the field. Ans. Length, 80 rods; breadth, 25. 14. Two men talking of their ages, one said that he was 94 years old. *^ Then," replied the younger, ^' the sum of your age and mine, multiplied by the difference between our ages, will produce 8512." What was the age of the younger ? Ajis, 18 years. 15. A fisherman being asked how many fish he had caught, replied, " If you add 11 to the square of the number, 9 times the square root of the sum, diminished by 4, will equal 50." How many had he caught ? Ans. 5. 16. A merchant gains in trade a sum, to which 320 dollars bears the same proportion as five times the sum does to 2500 dollars ; what is the sum ? Aiis. $400. 17: What number is that, the fourth part of whose square being subtracted from 8, leaves a remainder of 4 ? Ans. 4. 18. There is a stack of hay, whose length, breadth and height are to each other as the numbers 5, 4 and 3. It is worth as PURE QUADRATICS. 227 many cents per cubic foot as it is feet in breadth ; and the whole worth, at that rate, is 192 times as many cents as there are square feet in the bottom of the stack. Required the dimensions of the stack. / SOLUTION. Let 5x = length ; 4.r = breadth ; 8x = height ; 6x X 4a; X ox = cubic feet in stack ; 5x X 4x = square feet in bottom ; 5a: X 4a; X ox X 4x = cost ; 192 X 5x X 4a; = cost. ba; X 4x X 3a; X 4x = 192 x 5a; x 4a; a) 3a; X 4a; = 192 (2) a;^= 16 (3) a; = 4 (4) Ans. Length, 20 feet ; breadth, 16 ; height, 12. Analysis. According to the given proportions, we let 5.x% 4x, and 3a; represent the three dimensions. Then bx X 4x X 3a;, their indi- cated product, will be the solid contents ; and 5a; X 4a;, the area. According to the conditions of the problem, we multiply the cubic contents by 4x, the breadth, and the square contents of the bottom by 192, and obtain two values for the cost, which being put equal to each other, give (i). canceling the factors 5a; and 4a; from both members, we have (2). Again canceling 3 X 4, or 12, from both members, we have (3), which gives a; = 4. Hence, 5a; = 20 feet, 4a? = 16, and 3a; = 12, the required dimensions. Note. — The advantage of keeping the factors separate, as in the solu- tion just given, has been fully illustrated in the former part of the book- The pupil may apply the same method to some of the examples which follow. 19. A man purchased a field, the length of which was to its breadth as 8 to 5. The number of dollars paid per acre was equal to the number of rods in the length of the field : and the number of dollars given for the whole, was equal to 13 19* 228 QUADRATIC EQUATIONS. times the number of rods round the field. Required the length and breadth of the field. Arts. Length, 104 rods ; breadth, 65. 20. There are three numbers in the proportion of 2, 3, and 6 ; and their product is equal to 108 times their sum. Re- quired the numbers. Ans. 12, 18, and 30. 21. It is required to divide the number of 14 into two such parts, that the quotient of the greater divided by the less, may be to the quoticLt of the less divided by the greater, as 16 i 9. Ans. The parts are 8 and 6. 22. What two numbers are those whose sum is 12, and whose product is 35 ? A7is. 7 and 5 Note. — For notation, see 2d problem. 23. The difference of two numbers is 6, and the sum of their squares is 50 ; what are the numbers ? Ans. 7 and 1. 24. The difference of two numbers is 8, and their product is 240 ; what are the numbers ? Ans. 12 and 20. AFFECTED QUADRATICS. 218. Since an affected quadratic equation contains both the first and second powers of the unknown quantity, the equa- tion will contain two unknown terms, and only two, after the coefficients of each power are united. Thus the equation, 3x2 — 12x = 180 — x' + 4x, becomes, by transposition, and reduction of terms, 4x2 — 16^ = 180, Or, x^ — 4x = 45. ^nd if we represent — 4 by 2a, and 45 by b, we have x^ + 2ax = b (A). Hence, Every affected quadratic equation can be reduced to the form of x^ + 2ax = 6, in which 2a and b are supposed to represent any quantities whatever, positive or negative. AFFECTED QUADRATICS. 229 SIO. Since the first member of the general equation (a), is a binomial, its root is a surd (!BiO, II), and the equation in that form cannot be reduced by evolution. We observe, however, that x^, the first term of (A), is a perfect square, and 2ax, the second term, contains x, the root of this square ; and it only requires that another square be added, such that twice the pro- duct of the two roots shall be equal to 2ax, the second term, to constitute this member a perfect square (^10, III). The square to be added will evidently be a\ giving x'' + lax + a^ = 6 -\- o} (b), 1 nAnflR- '^ in which the first member is a perfect square. But a, whose square, (a^), we have added, is half the coeffi- cient of X in the second term. Hence, If the square of half the coefficient of the first power of the unknown quantity he added to both members of a quad- ratic equation in the form of x"^ -\- 2ax = 5, the first member will become a perfect square. Note. — The term, a^, is added to the first member to complete the square; and to the second member to preserve the equality. (Ax. 1.) EXAMPLES FOR PEACTICE. Complete the square in each of the following equations : Ans. ir^ + 4x + 4 = 96 + 4. Ans, x^ — 4x + 4 =as 49. Ans. x"" — Ix + 4„9 ^ 8_i, Ans. sc" + 2x + lr=. 16. Ans, x^ + 12x + 36 = 64. Ans, cc* + 6x + 9 = 25. Ans. x" — 15a; + 2f 5 = f Ans. a;« — |ic + i = ^g^ 1. ir» 4- 4x = 96. 2. x« — 4x = 45. 8. x' — lx=^ 8. \. x' + 2x = 15. 5. ic* + 12a; = 28. 6 a;« + 6x == 16. 7. a;^.— 15x = — 54. 8. 20 230 QUiBRATIC EQUiTIONS. 9. x'—^x^ J. Ans. x'^ — -^ + M- = tA' a c , o « a^ c , a^ SS^. Reducing the equation, x''+2ax-i-a''=:b+a\ By evolution, we have x-j-a =ziz\^b+a^. Transposing a, x = — a+\/6+a^ 1st root. Or, X = — a — >/6+a^,2d root Hence, Every affected quadratic equation has two roots, unequal in numerical value. find the -, r.. , 4 — 32:c 8(1 + ^) , Tp 1. Given a;2 . = _X___ — L j^ 16, to values of x. OPERATION. .'-1 _ 32^ __ 8(1 + 0.) 3 ~ 3 "^ a) 3x»— 4 + 32x = 8 + 8x + 48 (2) ^x" + 24:^ = 60 (3) x" + 8:^; = 20 (4) x^ + 8x + 16 = 36 (6) X -1- 4 = ±6 (6) x^2 (7) Or, :c = _ 10 (8) Analysis. Clearing of fractions, we obtain (2). Transposinpi; and uniting, we have (3). Dividing by 3, we have (4), an equation in the form of (A). As 8 is the coefficient of x^ we add 16, the square of one half of 8, to both members, and obtain (5), in which the first member is a perfect square. Extracting the square root of both members, we have (6). Transposing 4, and uniting — 4 with -f 6, the plus value of the root, we have .r = 2, the first value of ar; uniting the — 4 with AFFECTED QUADRATICS. 231 <— 6, the mimig value of the root, we have x = — 10, the second value of X. Hence, for the solution of an afiected quadratic equa- tion, we have the following general KuLE. I. Beduce the given equation to the form of x^ + 2ax = h. 11. Complete the square of the first member^ by adding to Loth members of the equation the square of one half the co- efficient of the second term, III. Extract the square root of both members, and reduce the resulting equation, EXAMPLES FOR PRACTICE. Reduce the following equations : 2. x' + 4x = 96. Ans. a; = 8, or — 12. 3. x^ — 4:x = 45. A71S. a; = 9, or — 5, 4. x^ — 7x = 8. Ans. x = 8, or — 1, 5. x^ '\- 2x = 15. Ans. a; = 3, or — 5. 6. x' + 12;r = 28. Ans. a? = 2, or — 14. 7. x"^ + Qx ==i 16. Ans. cc = 2, or — 8. 8. x"^ — 15;r = — 5^. Ans. ^ == 9, or 6. 9. ^2 __ 1^ ^ i| 3, ^r^s. a: = 7, or — \/. 10. x'^ — |x = J. Ans. ^ = 1, or — J. Notes. 1. The ten preceding examples are all in the form of 8;2 -|- 2az = b, and require no application of the first step of the rule. 2. If in the preparation of the equations which follow, the square of the unknown quantity appears with the minus sign^ make it positive bj changing ail the signs of the equation. 12. 3^7^ -- 25^7 = — 72 + bx. Ans. a; = 6, or 4. 13 2x'' + 100 = 32^ — 10. Ans. x = 11, or 5. 14. %x — 300 = 204 — ^x\ Ans. x = 12, or — 14. 15. bx^ + 80 = — 505 — llO^r. Ans. X = — 9, or — 13. 16. 2x^ — 9j7 = — 4. Ans. a? = 4, or ^, 232 QUADKATIC EQUATIONS. 17. 1 — .r = 5 ~rc. Ans. ic r= 4, or — 14. y. nr. -X. A 2x 76 18. 3 — x^ = — ^ . Ans. a? = 6, or — 5f . 19. x' + | = ^-J + if. Ans. a) = l,or-2i 20. J— 30 + a; = 2x — 22. Ans. a; = 8, or— 4. ^ 21. | — I 4- 7i = 8J. Jns. x = 1|, or — g. Aj^ 22. J_15 = ^ — 14|. ^ns. cc = 3, or— J. 2^-;^8-£l^- ^n.. = 12,or_2. 24. cc — 1 + ——: == 0. Jns. ic = 3. or 2 X — 4 ^_ 22 — X 15 — a? . or^ io 25. ^^ = ^. ^ns. cc = 36, or 12. 20 X — 6 2x — 7 cc + 1 . . , 26. rr- = - — V-o. % Ans. aj = 4, or — 1. X — 1 ^x + o 27. x^ + 2ax = 3al ^ns. a; == a, or — 3a 28. a;2 — 4cx = 4cc?— 2d!x — 3c' — d\ Ans. x=Zc — dy or c — d. 29. x^ — 2aa; = m* — o?. Ans. x= azhm. 30. x^ — 2cx = 4m — c^ u4ns. x = c it 2 >/m. SECOND METHOD OF COMPLETING THE SQUARE. 3^1. It frequently happens in the reduction of a quadratic to the form of x^ + 2ax = 6, that some or all of the terms become fractional, and render the solution complex and diffi- cult. In such cases, it will be sufficient to reduce the three parts of the equation to the simplest entire quantities, by uniting terms, and dividing the equation by the greatest common AFFECTED QUADRATICS. 233 divisor of the two members. The equation will then be in the form of ax^ -\-hx = c (A), in which a, 5, and c, are entire quantities, prime to each other. To render the first member of equation (a), a binomial square, we may make its first term a perfect square, by multi- plying the equation by a, and afterwards complete the square by the ezlQ already given. The operation will appear as follows : ax^ + bx = c] (1) Multiplying (i) by a, aV + bax = ac ; (2> Putting y = ax ] And (2) becomes y^ + by = ca ; (S) b^ b^ Completing the square, ^' + ^2/ + r = ^^ + j- W The first member of equation (4) is a binomial square ; but one of the terms is fractional, a condition which we are seek- ing to avoid. The denominator of the fraction is the square number, 4 ; and if the equation be multiplied by 4 to clear it of fractions, the first member will still be a square, because it will consist of square factors. Hence, multiplying the equa- tion by 4, we obtain, 4i/2 + 4:by +b^ = 4:ca + b' ; (S) Restoring value of y^ 4:a'^x^ + 4:abx + &^ = 4ca + b\ (6) Factoring this result, and comparing it with the primitive equation^ ^.^us, Primitive equation, ax^ + bx = c ; (i) Square completed, 4a(a^^ + bx) + 6^ = 4a(c) + b^. (6) we perceive that equation (6) may be obtained by mu-Itiplying equation (i) by 4a, and adding b^, to both members of the result Hence, 20* 234 QUADRATIC EQUATIONS. If a quadratic equation in the form of ax^ + 6^ = c, he multiplied by 4 times the coefficient of the second power of the unknown quantity^ and the square of the coefficient of the first power be added to both sides, the first member will be- come a perfect square. 1. Given 5x^ + 4xc = 204, to find the values of x. OPERATION. bx" + 4x = 204 (1) lOOa;^ + 80x + 16 = 4096 (2) 10x+ 4 = ±64 (3) lOx = 60 or — 68 (4) x=: 6 or — 6| (5) Analysis. To complete the square, we multiply equation (i) by 4 times 5, and add the square of 4 to both members, and obtain (2). Extracting the square root of both members, we have (3). Trans- posing 4, we have (4); and dividing by 10, we obtain a; = 6, or — 6|. 2. Given x^ — 5x = — 6, to find the values of x. OPERATION. Analysis. In this example, 4 times the coefficient of x^ is 4: a;2 — 5x = — 6 0) ^Yq therefore multiply by 4, and 4x' — ( )4-25 = l (2) add the square of 5 to both 2x — 5 = d= 1 (3) members, and obtain (2). Re- jc = 3 or 2 (4) ducing by evolution and trans- position, we have a: = 3, or 2. As the second term of a binomial square is dropped in extracting the square root, we may place ( ) in the equation, preceded by the proper sign, when we complete the square. From these examples we derive the following KuLE. I. Reduce the equation to the form of ax^ + bx «= c, in which the three terms are entire, and prime to each other. II. Multiply the equation by 4 times the coefficient of x\ ard add the square of the coefficient of x to both members. III. Extract the square root of both members, and y^educe the resulting equation. NcTE. — If the coefficient of a:^ is 1, the second method will be applied "with advantage, provided the coefficient of x is odd; but if it is even, the first rule is preferable. AFFECTED QUADRATICS. 235 EXAMPLES FOR PRACTICE. 3. Gi ix» 2x — 5x = 117, to find the yalnes of x. Ans. cc = 9, or — C^. 4. Given Bx^ — 5x = 28, to find the values of x, A71S. x = 4:j or — |. 6. Given 2x^ — a; = 70, to find the values of x, A71S, a: = 5, or — y. t) Given bx^ + 4x = 273, to find the values of x. Ans. cc = 7, or — 7|. 7. Given 2x* + 3ic = 65, to find the values of x. Alls, a: = 5, or — 6^. 8. Given Sx^ •{• bx = 42, to find the values of x. Ans. a: = 3, or — 4-|. 9. Given 8x* — 7a; 4- 16 = 181, to find x. Ans. cc = 5, or — 4|-. 10. Given lOx* — 8x + 8 = 320, to find x. Ans. a = 6, or — 5 J. 11. Given Sx^ + 2x = 4, to find x. Ans. x = — ^±:^VW. 12. Given bx^ 4- 7x = 7, to find x. Ans. a7=~/^±^3^\/2i. 13. Given ~ + j%== ^£^, to findx. Ans. x = ^^'^^^^^ 14. Given 12 + ic = ^^ , to find a;. ^yi.s. cc = 5, or — 12. Find the approximate roots of the following equations : 15. a^^ — 5x = — 2. Ans. X = 4.5615 +, or .4384 +. 16. 2a;^_Sx==12. Ans. X = 3.3117 +, or — 1.8117 -f . 17. %x^ — x^\. Ans. X = .7675 +, or — .4342 +. 236 QUADRATIC EQUATIONS. Ans. X = 1.6180 +, or — .6180 +. 19. ^x' + 3x = 5. Ans. X = .8042 +, or — 1.5542 +. 20. ic« — 7a; = — 11. Ans. X = 4.6180 +, or 2.3820 + HIGHER EQUATIONS IN THE QUADRATIC FORM. Q22. Any equation is in the quadratic form, when it eon tains but two powers of the unknown quantity, and the index of the higher power is twice the index of the lower. Such equations are reducible to one of the following forms : a^" + ax** = 6 ; or, ax^" + 6x" == c ; and may therefore be reduced by one of the rules for quadratics. ] . Given x^ — 4x* = 621, to find the values of x. OPERATION. Analysis. To sam- plify the application Put y =a^ of the rule for quad- y^ = oc^ ratios, we put y=a^ y^ — ^y = 621 m ^°^ 2/' = ^. The •^ _ ( ) + 4 = 625 (2> Siven equation then ^ 9 Hh 25 (3) ^^^^^^ ^^^' ^ quad- ^ ; I 27 or -23 (4) '-«c in the general ^ form. Solving m the or, ar* = 27 or — ^ (5) ^.^^^^ manner, we a? = 3 or >/ — 23 (e) obtain (i). Restor- ing the ra^ue of y, we have (5) a pure equation ; and extracting the cube root of both members we have a; = 3 or \/ — 23. Note 1. It will be remembered that the odd roots of a negat»»'^ quan- tity are realy while the eveJi roots are imaginary (184). Hence, by ex- tracting the cube root of — 23. and prefixing the minus sign, we Snd the approximate value of the secona root m the example above. Thus, ^ir23 = 2.84+. 2. Given x* - ^x- = 56, to find X, OPERATION. Put V- y y' = .? = a:^ 15 or — .7 y' — y ( ) + i 22/ — 1 = 56 = 225 = zfc: = 8 (1/ (2) (3) (4) or, x'' = 8 or — .7 (5) 1 = 2 or(- ■7) ff (6) X = 4 orr- •7) ^(7) HIGHER EQUATIONS IN THE QUADRATIC FORM. 937 Analysis. If wo represent x- by y, cc* will be 2/^ and the equation, (l), takes the usual quadratic form. Reducing by the second rule, we have (4). Restoring the value of ?/, we obtain (5). Extract- ing the cube root of both members,we ob- tain (6); and squar- ing both members we N. ^ - 2 have a;=4 or ( — 7)^. NoTB 2. The expression ( — 7)^ signifies the cube root of the second power of — 7 ; or, \/49 = 3.65 +. EXAMPLES EOR PRACTICE. 3. Given x* + 2x^ = 24, to find the values of x. OPERATION. aj* -f 2x' = 24 a) Completing square, a;* + ( ) + 1 = 25 (2) Extracting root, cc^ -f 1 = db 5 (3) Transposing, cc' = — 1 zfc 5 (4) Uniting terms, x* = 4, or — 6 . (5) Extracting root, x = it 2, or db >/ — 6 (6) 4 Given a* — Zx^ = 550, to find the values of x. Ans. a; = db 5, or db ^— 22. ] 5. Given 3x — x"^ = 44, to find the values of x. \ Arts, X = 16, or 13|- 6. Given x* — 7x' = 8, to find the values of x. 1 Ans, X = 2, or — 1, \ 7. Given x* — 6x' = 567, to find the values of x. \ AriB. X = 8, or — 2.758 -f - i 238 QUADRATIC EQUATIONS. POLYNOMIALS UNDER THE QUADRATIC FORM. ^S3. When a polynomial appears under different pow<»-re or fractional exponents, one exponent being twice the other, we may represent the quantity by a single letter, and apply one of the rules for quadratics, as in the last article. 1. Given (x^ + 2x)2 + 2(x' + 2x) = 80, to find the values of X, OPERATION. Assume y^ = (x^ ^ 2xy ; And 2/ = ^^ + 2a; ; Then, 2/' +22/ = 80; Completing square, 2/^ + ( ) + 1 = 81; By evolution, y + 1 = =h 9 ; Reducing, y = S^ ov — 10. Restoring the value of y, we have two equations containing x ; Thus, x^ + 2a? = 8, OTx' + 2x = — 10 ; Whence, a;^ + 2^ + 1 = 9, or ^'-^ + 2^ + 1 = — 9 ; And, O! + 1 =db3;or^ + l =± 3^^^^; Therefore, a; = 2 or — 4 ; or, a? = 3\/ — 1 — 1 ; or, -- (1 + 3 v/I=T). EXAMPLES FOR PRACTICE. 2. Given (x + 3) + 2(x + 3)'^ = 35, to find the values of X, OPERATION. (a^ + 3) + 2(a; + 3)* = 35 d) Completing square, (^ + 3) + ( ) + 1 = 36 (2) Extracting root, v'aj + 3 + l=:it6 (3) Transposing and uniting terms^, ^ x + S=^d or — 7 (*) Squaring both members, a; + 3 = 25 or 49 (5) Transposing and uniting, ar = 22 or 46 (6) FORMATION OF QUADRATIC EQUATIONS. £39 3. Given (\j'' + lyy + Kv'' + ^2/) = 96, to find one value of y, Ans, y = 2. 4. Given 10 + cc — (10 + a:)^ = 12, to find one value of X, Ans. x = Q 5. Given (- + 2/)'+ (- + 2/) = 30, to find y, Ans. 2/ = 3 or 2, or — 3 ± n/3. 6. Given (x + 12)* + (x + 12)^= 6, to find the valuea of X, Ans. x = 4:j or 69. 7. Given 2x' + 3a; + 9 — 5v^2x^ + 3x + 9 = 6, to find the real values of x. Ans, x=S OT — 4 J, or a? = — | =b |- ^ — 55. 8. Given (x + of + 2b(x + a)* = Bb\ to find the valuea of X. Ans, x = b^ — a, or 816* — a. FORMATION OF QUADRATIC EQUATIONS. S34. The Absolute Term of an equation is the term or quantity which does not contain the unknown quantity. ^^5. The roots of quadratics possess certain properties which enable us to reconstruct the equation when its roots are known. Let us resume the general equation, x'^+2ax=h ; (a) Completing the square, x^+2ax+a^=a^'\-b ; By evolution, x+a=±^ a^+b ; Hence, x= — a+\^a^+by 1st root; And, ir=— a— Va'+6, 2d root. Adding these two roots, we have — a+ ^oT+l) — a—Va'-i-b Sum, T— 2a 240 QUADRATIC EQUATIONS. Multiplying them together, we have . a + \^a' + b — a— \^o? + h a^ — aVo? + h + a Va^ + h — {o? + h) Product, — h. If we transpose the absolute term of equation (a) the equa- tion will appear as follows : x^ + 2ax — 6 = (B) Comparing the sum and product now obtained, we con- clude that in every equation in the form of x^ + 2ax — b = 0> I. The sum of the two roots is equal to the coefficient of the unknown quantity in the second tervfiy taken with the contrary sign, II. The product of the two roots is equal to the absolute, term taken with its proper sign. 1. Form the equation whose roots are 4 and — 12. OPERATION. Algebraic sum of roots, 4 — 12 = — 8 Product of roots, 4 x ( — 12) == — 48 Hence, a;^^ 8a? — 48 = Analysis. The algebraic sum of the roots is — 8 ; and their pro- duct is — 48. Hence, from property (I), we have 8 for the coeffi cient of the first power of the unknown quantity ; and from (II), we have — 48 for the absolute term in the first member; hence the equation is x^ -^ Sx — 48 = 0. From these principles and illustrations we have the following Rule. I. Write the second power of the unknown quan- tity for the first term. FORMATION OF QUADRATIC EQUATIONS. 241 II. Take the algebraic sum of the roots, with its sign changed, as the coefficient of the unknown quantity in the secand term. III. Write the product of the roots with its proper sign for the third term, and place the whole result equal to zero EXAMPLES FOR PRACTICE. 2. Form the equation whose roots are 10 and — 7. Ans, x^ — 3a; — 70 = 0. 3. Form the equation whose roots are 12 and — 5. Ans. «2~7x — 60 = 0. 4. Form the equation whose roots are 6 and — 15. Ans. a;2 + 9x — 90 = 5. Form the equation whose roots are 1 and — 2. Ans. ^2 + X — 2 = 0. 6. Form the equation whose roots are 4 and 13. Ans. x" — 11 X + 62 = 0. 7. Form the equation whose roots are — 5 and — 3. Ans. x^ + 8a; + 15 = 0. 8. Form the equation whose roots are 4^ and 5^. Ans. x^ — lOx + 24.*75 = 0, SECOND METHOD. SS6. Let us suppose that a and h represent any quantities, and find the product of the binomials. aj — a and a: — ^ 6 ; thus, X — a X — b Product; a:^ — (a + h)x + ah. Now by placing this product equal to zero, thus x^ — (a + h)x + a6 = 0, we form the equation whose roots are a and h ; because the coefficient of a; — (a + 6), taken with the contrary sign, is 21 Q 242 QUADRATIC EQUATIONS. the sum of a and 6, (S3«5, I) ; and the absolute term, a6, is the product of a and h (S^^, II). Hence Every quadratic equation in the form of x^ + 2.ax — h = ^ is composed of two binomial factors, of which the first term in each is the ujiknown quantity j and the second term, the two roots with their signs changed. To illustrate this by a numerical example, take the following Ciination : x^ + 4a^ — 60 = (1) Transposing, x^ -{- 4:x ==: 60 (2) Completing squp.re, x^ + ( ) + 4 = 64 (3) By evolution, x 4- 2 = zb 8 (4) Yv^hence, x = 6, 1st root ; And, X = — 10, 2d root. Connecting these roots with x, with their signs changed, and multiplying, we have X — 6 X + 10 . x"^ — 6x lOx — 60 x^+ 4x — 60 = Thus we have reconstructed equation (i). Hence, iluLE. I. Connect each root, with its contrary sign, to an unknown quantity. II. Multiply together the binomial factors thus formed^ and place the product equal to zero, EXAMPLES EOR PRACTICE. 1. Find the equation which has 3 and — 2 for its roots. Ans. x^ — X — 6 = 0. 2. Find the equation which has 5 and — 9 for its roots. Ans. x^ + 4x — 45 = 0. 3. Find the equation which has 7 and — 7 for ilts roots. Ans. cf' — 49 as 0. FACTORING TRINOMIALS. 243 4. Find the equation which has 8 and — 12 for its roots. Ans. x^ -{- 4:x — 96 = 0. 5. Find the equation which has — 5 and — 7 for its roots. Ans. x" + 12x + 35 = 0. 6. Find the equation which has a and a — 6 for its roots. Ans, x^ — (2a — h)x -f a? — a6 = 0. - 7. Find the equation which has h and — a for its roots. Ans, x^ + (a — l))x — ah =■ 0. FACTORING TRINOMIALS. ^fM7* The principle established in the last article, enailes US to resolve any trinomial, in the form of x^ -{- ax + b, into two binomial factors, either exact or approximate. I. Resolve cc^ -f 5x + 6 into two binomial factors. OPERATION. Analysis. Since the given a;2 _|. 5x -f 6 = — quantity is in the form of a x^ 4- dx = 6 quadratic equation reduced ^2i/^\i95_. 1 *^ ^^^ member, we place it ^^^ ^ p^ __ _, -t equal to zero, and solve the ^x 4- D — d= 1 resulting equation. Then, •'^ = — o, or A changing the signs of the (x 4- 3) (x + 2), factors. roots, and connecting them Or, 2 X 3 =-6, absolute term, ^'^^^ ^' ^'® ^^^^ (^' + ^') 2 + 3 = 5, coefficient of x, (^ + ^), the factors that com^ And( 4- 3Wx + '^Wactors pose the tr,nomiaU226). Or, Ana 1^^ ^ 6) (^x+ ^), tactois. ^^ ^^^ ^^^ ^^ inspection, two factors of 6, the absolute term, whose sum is equal to 5, the co- efficient of X in the middle term, and thus form the binomial factors sought. Hence, the following FtULE. I. Place the trinomial equal to zero, and reduce the resulting equation. II. Connect each root, with its sign changed, to the lowest flower of the literal quantity, and the result will he the binomial factors required. Or, Find by inspection two factors of the absolute term, whose hum is equal to the coefficient of the middle term; and con* nect each ivith its proper sign, to the literal quantity 244 QUADRATIC EQUATIONS. EXAMPLES FOR PRACTICE. Factor the following trinomials. 2. x'^^x — 20. 3. a' — 7a + 12. 4. i^ — 7a — S. -18. -X' Ans, (x — 5) (x 4- 4) Arts, (a — 3) (a — 4) A71S. (a — 8) (a + 1) Ans, (x — 6) (cc + 5} Ans. (x + 9) (x — 2) Ans. (x + 14)(ic — 3) 8. x^ + 2x — 5. Ans. (x — 1.449 + ) (x + 3.449 + ) 9. x^ — Sx + 8. Ans. (a; — 6.828 + ) (^ — 1.172 +) x:' x^ + 7x- X' THE FOUR FORMS. 928, In the general equation, x^ + 2ax = b, 2a, as we have seen, may be either positive or negative ; and b msiy be either positive or negative ; therefore, for a representation of every variety of quadratic equations, we have the four general formi; x^ + 2ax = 6 (1) cc* — 2ax = b (2) x^ — 2ax = — b (3) x"^ -f 2ax = — 6 (4) A reduction of these several equations gives for the values of X as follows: x= + a± Va" + b cc = + a db V^a'- — b x = — a db v^a^ — b a) (2) (3) (4) REAL AND IMAGINARY ROOTS. 229. By examining the roots of the four forms given in the last article, we find that in the first and second forms, fl^ -f 6 , the quantity under the radical is positive ; its root can, PROBLEMS. 245 therefore, always be taken, either exactly or approximately. But iu the third and fourth forms, a^ — 6, the quantity under the radical will be negative when the term h is greater nume- rically than d^ ; in which case, the root cannot be extracted (184), and must be imaginary. Hence, I. In each of the first and second forms, both roots are always real. II. In each of the third and fourth forms, both roots are imaginary when the absolute term is numericalli/ greater than the square of half the coefficient of the unknown quan- tity in the second term. Note. — Imaginary roots of an equation furnished by a problem, indi- cate that the conditions of the problem are impossible or absurd. PROBLEMS PRODUCING QVADRATIC EQUATIONS. S30. 1. If four times the square of a certain number be diminished by twice the number, it will leave a remainder of 80 ; what is the number? Ans. 3. Note. — The number 3 is the only number that Tvill answer the required conditions ; the algebraic expression — 4 will also answer the conditions ; but the expression is not a number in an arithmetical sense. 2. A person purchased a number of horses for 240 dollars. If he had obtained 3 more for the same money, each horse would have cost him 4 dollars less ; required the number of horses. Ans. 12. 3. A grazier bought as many sheep as cost him 240 doL lars, and after reserving 15 out of the number, he sold the rs- mainder for 216 dollars, and gained 40 cents a head on the number sold ; how many sheep did he purchase ? Ans. 75. 4. A company dining at a house of entertainment, had to pay 3 dollars 50 cents ; but before the bill was presented two of them had left, in consequence of which, those who remained, had to pay each 20 cents more than if all had beej present ; how many persons dined ? Ans. 7. 21* 246 QUADRATIC EQUATIONS. 5. There is a certain number, which beini:^ subtracted from 22, and the remainder multiplied by the number, the product will be 117 ; what is the number ? Ans. 13 or 9. 6. In a certain number of hours a man traveled 36 miles; if he had traveled one mile more per hour, it would have taken him 3 hours less to perform his journey ; how many miles did he travel per hour ? Ans. 3 miles. 7. A man being asked how much money he had in his purse, answered, that the square root of the number taken from half the number would give a remainder of 180 dollars ; how much money had he ? Ans. $400. 8. If a certain number be increased by 3, and the square root of the sum be added to the number, the sum will be 17 ; what is the number ? Ans. 13. 9. The square of a certain number and 11 times the num- ber make 80 ; what is the number ? Ans. 5 or— 16. 10. A poulterer going to market to buy turkeys, met with four flocks. In the second flock, were 6 more than 3 times the square root of double the number in the first ; the third contained 3 times as many as the first and second ; the fourth contained 6 more than the square of one third the number in the third ; and the whole number was 1938. How many were in each flock ? Ans. 18, 24, 126, 1770. Note. — Let 2x'^ equal the number in the first. Also see (223). 11. The plate of a mirror 18 inches by 12, is to be set in a frame of uniform width, and the area of the frame is to be equal to that of the glass ; required the width of the frame. A71S. 3 inches. 12. A square courtyard has a rectangular gravel walk round it The side of the court is two yards less than six times the width of the gravel walk, and the number of square yards in the walk exceeds the number of yards in the periphery of the court by 164 ; required the area of the court, exclusive of the walk. Ans. 256 yards. TWO UNKNOWN QUANTITIES. 247 13. A and B start at the same time to travel 150 miles ; A travels 3 miles an hour faster than B, and finishes his journey- s' hours l)efore him ; at what rate per hour does each travel ? A71S. 9 and 6 miles per hour. 14. A company at a tavern had 1 dollar 75 cents to pay ; but before the bill was paid two of them left, when those who remained had each 10 cents more to pay ; how many "/^ere m ib3 company at first? Ans. 7. 15. A set out from C, toward D, and traveled 7 miles a day. After he had gone 32 miles, B set out from D toward C, and went every day y*^ of the whole journey ; and after he had traveled as many days as he went miles in a day, he met A ; required the distance from C to D. Ans. 76 or 152 miles ; either number will answer the con- ditions. QUADRATIC EQUATIONS CONTAINING TWO UNKNOWN QUANTITIES. ^31. In general, two equations essentially quadratic, In- volving two unknown quantities, depend for their solution on a resulting equation of the fourth degree. A solution may be effected, however, by the rule for quadratics, if the equations come under one of the three following cases : 1st. When one of the equations is simple, and the other quadratic. 2d. When the equati-ons are similar in form, or the unknown quantities are involved and combined in a similar manner. 3d. When the equations are homogeneous. We give illustrations of the three classes in succession. 1st. Simple and Quadratic. 1 . Given 1^, ^ 2/ == S3 ]* ' ^ ^""'^ "^ ""^^ ^ 248 QUADRATIC EQUATIONS. OPERATION. X +2y =.9 (I) ^ x" + 2y^ = SS (2) From (1), cc = 9 — 2y :z^ Sqnaring (3), a;^ = 81 — S6y + ^f (4) From (2), ic2 ^ 38 — 2y^ (6) From (4) and (5), 81 — S6y + iy^ = 33 — 2y^ (6) Reducing, y^ — 6?/ = — 8 (7) "Whence, 2/ = 4 or 2 (8) And from (i), a; = 1 or 5 (a) 2d. Similar Equations. 2. Given | / ^ 9i | > to find a; and y, operation. X ^- 2/ == 10 0) xy =24 (2) Squaring a), x" + 2^:2/ + 2/^ = 100 (3) Multiplying (2) by 4, 4:xy = 96 (4) Taking (4j from (3), x" — 2a;2/ + 2/' = 4 (5) Extracting square root of (5), X— y = ±2 (0) But in (1), x+ 2/ = 10 (7) Adding (6) to G), 2a? = 12 or 8 (SI Taking (6) from (7), 2y = 8 or 12 (0) Whence a; = 6 or 4 (10) And y = 4 or G (11) 8. Given 1 ^ i. /i _ rq | » *o ^^^ ^ ^^^ 2/- TWO UNKNOWN QUANTITIES. £49 OPERATION. X + y = 10 a) x" -\- y^ = 58 (2) SquariQg a), nc^ + 2^2/ + 2/' = l^^ (2> Taking (2) from (S), 2^2/ = 42 (4) Taking (4) from (2), a;^ — 2xy + y^ = IQ (6) Extracting square root of (5), x — y = zt: 4: «) But in (1), ^ + y =10 (7) Whence, ar = 7 or 3 (8) And 2/ = 3 or 7 (9) 4. Given | ^ ;J; ^3^ 05 1 , to find x and y, OPERATION. X + y = d a) ^3^2/^== 35 (2) Cubing (1), rr* + 3:^^?/ + ^^y' + y'=12^ (3) Taking (2) from (3), ^x^'y + 3x2/^ — 90 (4) Factoring (4), Sxy(x + y)=90 (s) Dividing (5) by (i), Bxy = 18 (6) Or, xy = 6 (V) Combining (i) and (7), | ic = 3 or 2 (8) as in 2d example, J 2/ = 2 or 3 (o) . 5. Given { J, Z ^ Z 728 } ^ ^^ ^""^ ^ ^""^ 2/- OPERATION. CC — 2/ = 8 (1) a:8^y3=,r728 (2) Dividing (2) by a), x^ -{- xy -i- y^ = 91 (S) Squaring (i), cc^ — 2xy + ^7" = 64 (4) Taking (4) from (S), 3:7-2/ = 27 (5) Or, xy = d (e) Adding (6) to (3), x^ -f 2xy + i/' = 100 CO Extracting square root of (7), x + y = ±:10 (8) But in (1), x>^y = S (9) Whence, 'x== 9 or — 1 (lo) And 2/ — 1 or — 9 (H) 250 QUADRATIC EQUATIONS. 8d. HOMOGENEOUS EQUATIONS, OPERATION. 2x^ — xy = Q (Ij 2y' + Sxy = 8 (2) Assume x = vy (3) Substituting vy in a) 2uy — vy^ == Q (4) And in (2) 2y'' + Bvy' = 8 (6) From (4) y' = ^— 2'i;=^ (6) From (5) y'-2-+^ <^> -, . , 8 6 jCiquauiiJg (."; tiuu v/ 2 + Sv 2v"^ — V («) Reducing (9) 8i;^ — • ISv =6 (9) Whence, V =:2 (10) Substituting v in (7) 2/^-1 (11) Whence, 2/=:tl (12) From (3) and (i2) a; =r±2 (13 Note. — For simplicity, only one value of -y was taken in equation (10), S33. Referring to the three classes, we find that, 1st. When the equations are simple and quadratic^ they may he solved by ordinary elimination. 2d. When the equations are similar, they may he solved by taking advantage of multiple forms, and of the relations ex- isting between the sum, difference, and iDvoduct of the un- known quantities. ^d. When the equations are homogeneous, they may he solved by the use of an auxiliary quantity. TWO UNKNOWN QUANTITIES. 251 EXAMPLES FOR PRACTICE. Find the values of x and y in the following equations. (2x^y^l2\ I X = 7 or — 11, ^- 4x^ + 2^ = 53r ^''^' l2/=.2or— 31. ( X —^ = 1\ . j'a: = 4or— -5, 2- 1 .'_3j' = 13|- ^«*- iy = lor_2. 5 or 7, ( £C2/ = 35 3 ( 2/ = / or 5. rx_2/==-l| r:« = 6or-7, 5 jo: +2/ =1125) r. = 568, 2/ fa: — v= 4) . fa: = 5or — 1, i X — 2/ = 3 - (2/ = 2or — 5. AuB. 2 or 3, 3 or 2. Note.— Put — = P, and ~ = 0. 9. 1^+2/^ = 1^2) ^^^ j:. = 5or3, U 4-y =8 3 • • (2/ = |.« + ,« . , = 78) ^^. 1^ = 5 f2x' — 3xy=50) "• \ v^ — 2f =50 3 • ■■{ 2/ = 3 or 5. 9 or 3, 3 or 9. . cc = 10or — 5v^~.2, 252 QUADRATIC EQUATIONS. PROBLEMS PRODUCING QUADRATICS WITH TWO UNKNOWN QUANTITIES. Note. — In several of the examples which follow, the pupil may have the choice of using two symbols, or one, in the solution. It will be use- ful to solve by both methods. S3S. 1. Divide 100 into two such parts, that the sum of their square roots may be 14. Ans. 64 and 36. 2. Divide the number 14 into two such parts, that the sum of the squares of those parts shall be 100. Ans. 8 and 6. 3. Divide the number a into two such parts, that the sum of the squares of those parts shall be b. Ans, i(a ± V26^ir^^. 4. It is required to divide the number 24 into two such parts, that their product may be equal to 35 times their dif- ference. Ans, 10 and 14. 5. The sum of two numbers is 8, and the sum of their cubes 152 ; what arc the numbers ? Ans. 3 and 6. 6. Find two numbers, such that the less may be to the greater as the greater is to 12, and that the sum of their squares may be 45. Ans. 3 and 6. 7. What two numbers are those, ^"^hose difference is 3, and the difference of their cubes 189 ? Ans. 3 and 6. 8. What two numbers are those, whose sum is 5, and the sum of their cubes 35 ? Ans. 2 and 3. 9. A merchant has a piece of broadcloth and a piece of silk.s The number of yards in both is 110 ; and if the square of the number of yards of silk be subtracted from 80 times the num- ber of yards of broadcloth, the difference will be 400. How many yards are there in each piece ? Ans, 60 of silk ; 50 of broadcloth. 10. A is 4 years older than B ; and the sum of the square? of their ages is 976. What are their ages ? Ans. A's age, 24 years ; B^s, 20 years TWO UNKNOWN' QUANTITIES. 9/33 11. Divide tlie number 10 into two such parts, that the square of 4 times the less part may be 112 more than the square of 2 times the greater. Ans. 4 and 6. 12. Find two numbers, such that the sum of their squares may be 89, and their sum multipUed by the greater may pro- duce 104. A 118. 5 and 8. 13. What number is that which, being divided by the sum of its two digits, the quotient 6| ; but when 9 is sub- tracted from it, there remains a number having the same digits inverted ? Ans. 82. 14. Divide 20 into three parts, such that the continued pro- duct of all three may be 270, and that the difference of the first and second may be 2 less than the difference of the second and third. Ans. 5, 6, and 9. 15. A regiment of soldiers, consisting of 1066 men, forms into two squares, one of which has four men more in a side than the other. What number of men are in a side of each of the squares ? Ans. 21 and 25. 16. A farmer received 24 dollars for a certain quantity of wheat, and an equal sum for a quantity of barley, but at a price 25 cents less by the bushel. The quantity of barley ex- ceeded the wheat by 16 bushels. How many busliels were there of each ? Ans. 32 bushels of wheat, and 48 of barley. 17. A kborer dug two trenches, one of which was 6 yards longer than the other, for 17 pounds 16 shillings, and the dig- ging of each trench cost as many shillings per yard as there were yards in its length. What was the length of each ? Ans. 10 and 16 yard^i. 13. A and B set out from two towns distant from each other 247 miles, and traveled the direct road till they met. A went 9 miles a day, and the number of days at the end of which they met, was greater, by 3, than the number of miles which B went in a day. How many miles did each travel ? Ans. A, 1 17 miles ; and B, 130. 22 254 QUADRATIC EQUATIONS. 19. The fore wheels of a carriage make 6 revolutions more than the hind wheels, in going 120 yards ; but if the circum- ference of each wheel be increased 1 yard, the fore wheels will make only 4 revolutions more than the hind wheels, in the same distance ; required the circumference of each wheel. Ans. 4 and 5 yards. 20. There are two numbers whose product is 120. If 2 bo added to the less, and 3 subtracted from the greater, the pro- duct of the sum and remainder will also be 120. What are the numbers ? Ans. 15 and 8. 21. There are two numbers, the sum of whose squares ex- ceeds twice their product, by 4, and the difference of their squares exceeds half their product, by 4 ; required the num- bers. Ans. 6 and 8. 22. What two numbers are those, which being both multi- plied by 27, the first product is a square, and the second the root of that square ; but being both multiplied by 3, the first product is a cube, and the second the .root of that cube ? A71S. 243 and 3. 23. A man bought a horse, which he sold, after some time, for 24 dollars. At this sale he lost as much per cent, upon the price of his purchase as the horse cost him. What did he pay for the horse ? Ans. He paid $60 or $40; the problem does not decide which sum. 24. What two numbers are those whose product is equal to the difference of their squares ; and the greater number is to the less as 3 to 2 ? Ans. Iso such numbers exist. 25. What two numbers are those, the double of whose pro- duct is less than the sum of their squares by 9, and half of their product is less than the difference of their squares by 9 ? Ans. The numbers are 9 and 12. ARITHMETICAL PROGRESSION. 255 SECTION V. ARITHMETICAL PROGRESSION. 934, An Aritlimetical Progression is a series of numbers or quantities, increasing or decreasing by the same difference, from term to term. Thus, 2, 4, 6, 8, 10, 12, the square and reducing, we n' 4- 6n = 567 (S) t ^' „H()+ 9 = 576 <7l"^^Iir^''' '"'''''"'- n + 3 = 24 (7) n = 21 (8) 260 ARITHMETICAL PROGRESSION. Hence, for the determination of any required parts in an arithmetical progression, we have the following HuLE. Substitute, in the formulas (a) aiid (b), the known quantities given in the example, and reduce the resulting equations. PROBLEMS. 8. Find seven arithmetical means between 1 and 49. Note. — If there are 7 means, there must be 9 terms; hence, n = 9, a = 1, and L = 49. Ans, 7, 13, 19, 25, 31, 37, 43. 4. The first term of an arithmetical series is 1, the sum of the terms 280, and the number of terms 32 ; what is the com- mon diderence, and what the last term ? Ans. d= \,'h = 16^. 6. Insert three arithmetical means between ^ and -J. Ans. The means are f, Z^, ^\< 6. Insert five arithmetical means between 5 and 15. Ans. The means are 6f , 8^, 10, llf, l^. 7. Suppose 100 balls be placed in a straight line, at the distance of a yard from each other ; how far must a person travel to bring them one by one to a box placed at the dis- tance of a yard from the first ball ? Ans. 5 miles 1300 yards. 8. A speculator bought 47 building lots in a certain village, giving 10 dollars for the first, 30 dollars for the second, 50 dollars for the third, and so on ; what did he pay for the whole 47 ? Ans. $22090. 9. In gathering up a certain number of balls, placed on the ground in a straight line, at the distance of 2 yards from each other, the first being placed 2 yards from the box in which they were deposited, a man, starting from the box, traveled 11 miles and 840 yards ; how many balls weie there ? 'ins. 100. PROBLEMS. 261 10. How many strokes do the clocks of Venice, which go on to 24 o'clock, strike in a day ? Ans. 300. 11. In a descending arithmetical series, the first term is 730, the common difference 2, and the last term 2 ; what is the number of terms ? Ans. 365. 12 The sum of the terms of an arithmetical series is 280, the first term 1, and the number of terms 32 ; what is the common difference ? A7is. ^. 13. The sum of the terms of an arithmetical series is 950, the common difference 3, and the number of terms 25 ; what is the first term ? Ans. 2. 14. What is the sum of n terms of the series 1, 2, 3, 4, 5, Ans. S = 7^(1 4- n), Li PROBLEMS IN ARITHMETICAL PROGRESSION TO WHICH THE PRECEDING FORMULAS, (A) AND (B), DO NOT IMMEDIATELY APPLY. ^4^. Since the sum of the extremes, in an arithmetical series, is equal to the sum of any two terms equally distant from the extremes (SS8, I), we have the following special properties : I. When three quantities are in arithmetical progression^ the mean is equal to half the sum of the extremes, II. When four quantities are in arithmetical progression, the sum of the means is equal to the sum of the extremes. Take, for example, any three consecutive terms of a series, fliS a + 2(i, a + 3c?, ci + 4cf, and we perceive, by inspection, that the sum of the extremes is double the mean, or the mean is half the sum of the extremes. 262 ARITHMETICAL PROGRESSION. . Take four consecutive terms, and Tve have a+ 2d, a+ -od, a + id, a + 5d, (a + 2d) + (a + dd) == 2a + 7d, sum of the extremes: (a + od) + (a + M) = 2a + 7d, sum of the means To facilitate the solution of problems, when three terms are in question, let them be represented by (x — ?/), x^ (x + y), y l^eing the common diiference. When four numbers are in question, let them be represented by (x — 3?/), (x — y), (x+y), (x + ^y), 2y being the common difference. Such notation will often secure the formation of similar equations; and in applying the principles enunciated on the preceding page, the common difference ivill disappear by addition. 1. Three numbers are in arithmetical progression ; the pro- duct of the first and second is 15, and of the first and third is 21 ; what are the numbers ? SOLUTION Let y = common difference, X — y = first term, X = second term, X + y = third term. Analysis. We represent the common difference by y, and the first term by x — y. Then a; must be the second term, andar-fy, the third term. From the given conditions we have (1) and (2). Dividing (2) by (1) we obtain (3), which re- duced to find the value of x, gives (4), Substituting this value of X in (2), we have (5), and reducing this equation, we find y = 2, Hence, from (4), we get :c = 5, &c. x{x — y) = 15 (1) x' — y' = 2l (2) x+ y 7 ic ~ 5 (3) (i) Y-y- =' (5) 2ly'= 21 X 4 (6) 2/'= 4 (7) y= 2 (8) X = 5, 2d term, (9) X — y = S, l&t term, (10) X + y= 7, 3d term, (11) PROBLEMS. 263 2. There are four numbers, m aritlimetical progression ; the Rum of the two means is 25, and the second number, multiplied by the common difference is 50 ; what are the numbers ? Ans. 5, 10, 15, and 20. 3. There are four numbers, in arithmetical progression ; the product of the first and third is 5, and of the second and fourth is 21 ; what are the numbers ? Ans, 1, 3, 5, and 7. 4. There are five numbers, in arithmetical progression ; the sum of these numbers is 65, and the sum of their squares 1005 ; -what are the numbers ? Note. — Let x = the middle term, and y the common difference. Theu z — 2y, a; — y, x^ z -\- y, z -\- 2y, will represent the numbers. Ans, 5, 9, 13, 17, and 21. 5. The sum of three numbers in arithmetical progression is 15, and their continued product is 105 ; w^hat are the num- bers ? Ans. 3, 5, and 7. 6. There are three numbers, in arithmetical progression; their sum is 18, and the sum of their squares 158 ; what are the numbers ? Ans. 1, 6, and 11. 7. Find three numbers, in arithmetical progression, such that the sum of their squares shall be 56, and the sum arising from adding together once the first and twice the second, and thrice the third, shall amount to 28. Ayis. 2, 4, 6. 8. Find three numbers, such that their sum may be 12, and the sum of their fourth powers 962 ; and the numbers have equal difi'erences in order from the least to the greatest. Ans. 3, 4, 5. 9. Find three numbers having equal differences, and such that the square of the least number added to the product of the two greater, may make 28, but the square of the greatest number added to the product of the two less, may make 44. Ans, 2, 4, 6. 264 ARITHMETICAL PROGRESSION. 10. Find three numbers, in arithmetical progression, such that their sum shall be 15, and the sum of their squares 93. Ans, 2, 5, and 8. 11. Find three numbers, in arithmetical progression, such that the sum of the first and third shall be 8, and the sum of the squares of the second and third shall be 52. Ans. 2, 4, and 6. 12. Find four numbers, in arithmetical progression, such that the sum of the first and fourth shall be 13, and the dif- ference of the squares of the two means shall be 39. Ans. 2, 5, 8, and 11. 13. Find seven numbers, in arithmetical progression, such that the sum of the first and sixth shall be 14, and the pro- duct of the third and fifth shall be 60. Ans. 2, 4, 6, 8, 10, 12, and 14. 14. Find five numbers, in arithmetical progression, such that their sum shall be 25, and their continued product 945. Ans. 1, 3, 5, 7, and 9. 15. Find four numbers, in arithmetical progression, such that the diflference of the squares of the first and second shall be 12, and the difference of the squares of the third and fourth shall be 28. Aiia. 2, 4, 6, and 8. GEOMETRICAL PROGRESSION. 265 GEOMETRICAL PROGRESSION. S4I* A Geometrical Progression is a series of numbers increasing or decreasing by a constant multiplier. Thus 2, 6. 18, 51, 162, &c., is a geometrical series, in which the first feerm is 2, and the multiplier is 3. The series 9, 3, 1, |, J, ^V? ^^-y ^^ ^^so a geometrical series, in which the first term is 9, and the multiplier is \. When the multiplier is greater than 1, the series is ascend- ing , and when the multiplier is less than 1, the series is descending. ^4^. The Ratio is the constant multiplier. Note. — The words, means and extremes, as defined in Aritlimetical Progression, have the same application in Geometrical progression, or in any series. CASE I. S43. To find the last term. If we represent the first term of the series by a, and the ratio by r, then a, ar, ar^, ar^, ar*j &c., \^ will represent the series ; and we observe, 1st. The first term is a factor in every term. 2d. The exponent of r in any term is one less than the number of the term ; thus, in the second term it is 1 ; in the third term, 2 ; in the fourth term, 3 ; and in the nth term it must he n — 1. Therefore, if n represent the number of terms in any series, and L the last term, then L = ar""-^ (A). 23 266 GEOMETKICAL PROGRESSION. 1 Hence, the following I Rule. I. Raise the ratio to a power whose index is one j less than the number of terms. \ II. Multiply the result by the first term. ' | Note. — "^Ylien any two successive terms in the series are given, it is j evident that the ratio may be found by dividing any term by the pre- * ceding term. | EXAMPLES FOR PRACTICE. 3 1. The first term of a geometrical series is 3, and the ratio \ is 2 ; what is the 6th term of the series ? \ Ans. L == (2)s X 3 = 9G. \ 2. The first term of a series is 5, and the ratio 4; what is "\ the 9th term ? | Ans. 327680. j 8. The first term of a series is 2, and the ratio 8 ; what is \ the 8th term ? i Ans. 4374. • j 4. The first term of a series is 1, and tlie ratio | ; what is ] the 5th term? Ans. -ij\. \ 5. The first term of a series is 2, and the ratio 3 ; what is \ the 7th term ? \ Ans. 1458. J 6. The first term of a geometrical series is 5, and the ratiG j 4 ; what is the 6th term ? I Ans. 5120. 1 I 7. The first term of a series is 1 and the ratio 2 ; what is \ the 10th term ? \ Ans. 1024. ] 8. The first term of a series is 1, and the ratio | ; what is j the 8th term ? | Ans. ri\%%. I ^ GBOMETRICAL PROGRESSION. 267 CASE IL 844. To find the sum of the series. Let S represent tlie sum of any geometrical series chen we have S = a + r/r 4- ar^ + ar^^ &c., to ar^^, (l) Multiply this equation by r, and we have rS = or 4- or^ + «^, &c., to ar'*"* -f ar^. (£) Subtracting (i) from (2), we have (r — 1)S = ar' — a, (») But from Case I. we have L = ar^-\ And multiplying by r, rL = ar". Therefore, by substituting the values of ai^ in (s), (r — l)S = rL — a. Or. S = -^^-y. (B) Hence, the following Rule. Multiply the last term by the ratio, and from the product subtract the first term, and divide the remainder by the ratio less one. EXAMPLES FOR PRACTICE. 1. The first term is 5, the last term 1280, and the ratio 4 5 what is the sum of the series ? Ans. S = mj^A^' = 1705. 2. The first term is 2, the last term 486, and the ratio 8 ; what is the sum of the series ? Ans, 728. 268 GEOMETRICAL PROGRESSION. ^ 3. The first term is 12, the last term 7500, and the ratio 5 ; \ what is the sum of the series ? Ans. 9372. \ \ Note. — If the last term is not given, it may first be found by Case I. \ \ 4. What is the sum of 8 terms of the series, 2, 6, 18, &c. \ Ans, 05()0. -l 5. What is the sum of 10 terms of the series, 4, 12, 36, &c. ? I Ans. 118096. ] 6. What is the sum of 9 terms of the series, 5, 20, 80, (fee. ? ] Ans, 436905. 1 7. What is the sum of 5 terms of the series, 3, 4i, Q^, &c. ? Ans, 39 /g. \ 8. What is the sum of 10 terms of the series, 1, f, |, / ax^ 272 GEOMETRICAL PROGRESSION. APPLICATIONS. 948. In a geometrical progression there are five parts, as follows : 1st. The first term ; 2d. The ratio ; 3(1. The number of terras ; 4th. The last term ; 5th. The sum of the terms ; Since the independent equations, symbol a. a r. a n. ti L. ii B. L=ar^-' (A) S--L-a (B) contain all of the parts, any two may be determined when the other three are known. Note. — Since n enters into tbe above formulas only as an exponenty the process of determining it requires a knowledge of logarithms, and must be omitted here. 1. The sum of a geometrical progression is 468, the number of terms is 4, and the ratio 5 ; what is the first term ? Analysis. From the example, we have 468, 4, and 5, for the values of S, ?2, and r, respectively ; ii — 1 is 3, and 7'""^ is, therefore, equal to 125, which substituted in formula (A), *:^ives (1). The value of r, substituted in formula (B), gives (2). To eliminate L we substitute its value, 125a, in i2\ and obtain (3), which reduced, gives a = 3. 2. The sum of a geometrical series is 847, the ratio 3, and the number of terms 5 ; what is the first term ? Ans. 7. OPERATION. S = 468 n = 4 r=5 L = 125a 0) 408 ^^Y" (2) ^gg^625«-a 4 (3) 156a = 468 (4) a = 3 (5) PROBLEMS. 273 3. The sum of a geometrical progression is 6220, the ratio 6, and the number of terms 5 ; what is the last term ? Ans, 6184. 4. The first and last terms of a geometrical scries are 2 and 162, and the number of terms 5 ; required the ratio. 1 /L\«-i Note. — From formula (A) we nave r^^l-~) Ans. 3. 5. The first term of a geometrical series is 28, the last term 17500, and the number of terms 6 ; what is the ratio ? Ans. 6. 6. The first term of a geometrical series is 32, the last term 4000, and the number of terms 4 ; what is the ratio ? Ans. 5. 7. Find two geometrical means between 4 and 256. Note. — Two means will require /owr terms. Hence w =4, from which data r may be found, as above. Ans. 16 and 64. 8. Find three geometrical means between 1 and 16. Ans. 2, 4, and 8. PROBLEMS IN GEOMETRICAL PROGRESSION TO WHICH THE FORMUIAS (A) AND (B) DO NOT DIRECTLY APPLY. 249, The general representation of the terms of a geome- trical progression is made thus : ^f ^y, ^y\ ^¥\ &c-» in which x is the first term, and y the ratio. But when par- ticular relations are given, it may become necessary to adopt a different notation. 274 GEOMETRICAL PKOGRESSION. When three terms are considered in the problem, they may DC represented thus : Or, x\ xy, y\ Since, in each case, the product of the extremes is equal to the square of the mean (5S4G the numbers, First condition, x'y' = 64 0) Second condition, ^-f r^^y^ 4- y^ == 584 (2i Adding (i) to (2), x-«-f 2.ry + / = 648 ( = 324-2) (3) Extracting the square root, ^ + y» = 18 n/2 (4) Taking 3 times (i) from (2), .r^— ■ .2.r^?/-H f = 392 (= 196 2) (6) Square root, y^ — a:^=14v/2 (6) Equation (4), 3/S 4- ^ = 18n/2 (7) From (6) and (7), x^ = 2%/2 (8) Squaring (8), a^==% (9) Cube root, a;' = 2, 1st term, m From (6) and (7), t/3 = 16v/2 (11) Squaring (ii), 2/^=512 (12) Cube root, t/^ = 8, 3d term, (13) From (10) and a3), a-y == 16 (14) Square root, xy ~ 4, 2d term. (15) 3. Of three numbers in geometrical progression, the sum of the first and second is 90, and the sum of the second and third is 180 ; what are the numbers ? Note. Represent the numbers by x, a:y, and ary*. Ans. 30, 60, and 120. 4. The sum of the first and third of four numbers in geo- metrical progression is 20, and the sum of the second and fourth is 60 ; what are the numbers ? Ans. 2, 6, 18, 54. 5. Divide the number 210 into three parts, such that the last shall exceed the first by 90, and the parts be in geometrical progression. Ans. SO, 60, and 12Q 276 GEOMETRICAL PROGRESSION. 6. The sum of four numbers iu geometrical progression in 80, and the last term divided by the sum of the mean terms is 1^ ; what are the numbers ? Ans. 2, 4, 8, and 16. 7. The sum of the first and third of four numbers in geo- metrical progression is 148, and the sum of the second and fourth is 888 ; what are the numbers ? Ans. 4, 24, 144, and 864 8. The continued product of three numbers in geometrical progression is 216, and the sura of the squares of the extremes is 328 ; what are the numbers ? Ans. 2, 6, 18. 9. The sum of three numbers in geometrical progression is 13, and the sum of the extremes being multiplied by the mean, the product is 30 ; what are the numbers ? Ans. 1, 3, and 9. 10. Of three numbers in geometrical progression, the sum of the first and last is 52, and the square of the mean is 100; what are th-e numbers? Ans, 2, 10, 50. 11. There are three numbers in geometrical progression; their sum is 31, and the sum of the squares of the first and last is 626 ; what are the numbers ? Ans. 1, 5, 25. 1 2. It is required to find three numbers in geometrical pro- gression, such that their sum shall be 14, and the sum of their squares 84. Aris. 2, 4, and 8. 13. Of four numbers in geometrical progression, the second number is less than the fourth by 24, and tlie sum of the ex- tremes is to the sum of the means, as 7 to 3 ; what are the numbers ? Ans. 1, 3, 9, and 27. 14. The sum of four numbers in geometrical progression is equal to their common ratio + 1, and the first term is y'^ ; what are the numbers? Ans. -j^^, j^^, y^^, f J. 15. How much will $500 amount to in 4 years, at 7 per cent, compound interest ? Note. — Since we have the principal at the commencement, and the trst year's amount at the end of the first year, the number of terras is 5. The ratio is ^1.07, equal to the amount of $1.00 for one year. Ans. $655,398 PROPORTION. 277 PROPORTION. 2o0. Two quantities of the same kind may be compared, and their numerical rehation determined, by seeking how many times one contains the other. The rehition of four quantities may be determined by comparing the relation of two of like kind, with tv^o others of like kind. These comparisons give rise to ratio and proportion. 2^1. Ratio is the quotient of one quantity divided by an- other of the same kind, regarded as the standard of comparison. There are two methods of indicating ratio. 1st. By TVTiting the divisor and dividend with two dots be- tween them ; thus, a : b is the indicated ratio of a to 6, in which a is the divisor, and b the dividend. 2d. In the form of a fraction ; thus, the above ratio becomes a 3f>S. Proportion is an equality of ratios ; thus, if two quantities, a and b, have the same ratio as two other quantities, c and d, the four quantities, a, b, c, and d, are said to lye pro- portional. Proportion is written in two ways ; thus, a : b : : c : df which is read, a is to 6, as c is to cZ ; or thus, a :b ~ c : d, which is read, the ratio of a to 6 is equal to the ratio of c to d. Note. — The second is the modern method, and more fitly expresses the nature of proportion. 24 ■» 278 PROPORTIOIS. fSSS» A Couplet is the two quantities which form a ratio. S34, Tlie Terms of a proportion are the four quantities which are compared. 25^. The Antecedents in a proportion are the first terms of the two couplets ; or the first and third terms of the pro- portion. S^o6. The Consequents in a proportion are the second term.. of the two couplets ; or the second and fourth terms of tlie proportion. ^^7. The Extremes in a proportion are the first and fourth terms. S58. The Means in a proportion are the second and third terms. Sof^. A Mean Proportional between two quantities is a quantity to which tlie first of the two given quantities has the same ratio as the quantity itself has to the second ; thus, if a : 6 = 6 : e the quantity, 6, is a mean proportional between a and c ; and the three quantities are said to be in continued pro2:)ortion ^HO. A Proposition is the statement of a truth to be de- monstrated, or of an operation to be performed. 361. A Scholium is any remark showing the application or limitation of a preceding proposition. 3©S. If in the proportion a : h = c : d tie 2d method of indicating ratio be employed, we shall ha"ve b d - = - (A) a c GENERAL PRINCIPLES OF PROPORTION. 279 which is the fundamental equation of proportion. And any proposition relating; to proportion will be proved^ when shown to be consistent with this equation. GENERAL PRINCIPLES OF PROPORTION. PROPOSITION I. ^6S« In every proportion, the product of the extremes is equal to the product of the means. Let a:h =^ cidj represent any proportion ; Then by formula (a), — = - ; ■ a c Clearing of fractions, he = ad. That is, the product of h and c, the means, is equal to the product of a and c?, the extremes. SCHOLIUM. — From the last equation, we have ad ^ The first mean, h = G The second mean, ^ ~ "T he' d I ha The first extreme, a — •, a (1). (2). The second extreme, d = a Hence, 1st. Either mean is equal to the product of the extremes divided hy the other mean. (i). 2d. Either extreme is equal to the product of the means divided hy the other extreme. (2). 280 PROPORTION. PROPOSITION n. 2S4. Conversely. If the product of hvo quantities be equal to the product of two others, then two of them may he taken for the means, and the other two for the extremes of a p)roportion. Let be = ad (i) Dividing by c, 6 = — (2) b _ad c b a _d ~ c a :6 = Dividing (2) by a, - = - (3) Hence by formula (A), a : b = c : d in which the factors of the first product, cb, are the means, and the factors of the second product, ad, are the extremes. PROPOSITION III. @llo. If four quantities he in proportion, they will be in proportion by alternation ; that is, the antecedents will he to each other as the consequents. Let a : b — c : d Then by formula (a), - = - (i) a c be Multiplying (D by c, — = d! (2) Dividing (2) by 5, = - (3) Hence, a : c = b : d in which a and c, the antecedents of the given proportion, are proportional to b and d, the consequents of the given pro- portion. GENERAL PRINCIPLES OF PROPORTION. 281 Scholium. — A proportion and an equation may be regarded as but different forms for the same expression, and every equation may be converted into a proportion under various forms. For example, Let, xy = a{a + h) Then, x : a = a -}- b : y Or, xy : a = a -^ b : I Or, a : X = y : (a -i- b) PROPOSITION IV. 266. 1/ four quantities be in proportion, they will be in proportion by inversion ; that is, the second will be to the first, as the fourth to the third. Let a : b = c : d Then by formula (a), - = - (i) •^ a c Clearing of fractions, be = ad (2) Hence by Prop. II, b : a : : d : c. PROPOSITION V. 307. If three quantities be in continued proportion, the product of the extremes is equal to the square of the mean. Let, a : b — b : c By Prop. I, ac = hb = ¥ Scholium. — Extracting the square root of the last equation, vre have b = V ac\ hence. The mean prop)ortional between two quantities is equal to the square root of their product 24* 282 PROPORTION. PROPOSITION VI. 268. Quantities which are loroportional to the same quantities are j^rojooriional to each other. If a : 5 = P : :Q (A) And c :d=P :Q (B) We are to prove that a :b=c ; id From (A) Form (B) d Q Hence, (Ax. 7), b _d a '^ c Or, a : b = c : d PROPOSITION VII. 269. If four magnitudes be in proportion, they must be in proportion by composition or division ; that is, the first is to the sum of the first and second, as the third is to the sum of the third and fourth ; or, the first is to the difference between the first and second, as the third is to the difference between the third and fourth. If a lb =^ c : d (A) We are to prove that aia±b=^G\cdcid From (A), b _d a"' c (1) Take 1 == 1 C2) Adding (i) to (2), i + Ui + 1 a c (8> Subtractmg (i) from (2), a c (4) From (3), a + b c -h d a c (6) From (4), a — b c — d a c («) Hence, from (5), a I a ■■{- b = c : G + d And from (6), a I a — b = c : c — d. GENERAL PRINCIPLES OF PROPORTION. 283 Scholium. — This composition may be carried to almost any extent, as we see by the following iiivestigatiou ; Take the equation, b _d a c a) Multiplying by m = VI And vib md a "" c (2) Adding, * n = n And n -f m6 . md — = n H a c (8) Reducing, 71 a a mb nc -}- md c (4) Hence, a : 7ia -f mb = c : nc -f md Or, subtracting (2) from n = w, and proceeding as before, we shall have a : na — mb = c ; nc — md PROPOSITION VIII. S70. If four quantities be in proportion, the sum of the two quantities which form the first coujjlet is to their dif- ference, as the sum of the tivo quantities which form the second couplet is to their difference. If a : b = c : d (A) We are to prove that a -{- b : a — 6 = c-f(/:c — d From (A), by Prop. YII, a : a + b = c : c -j~ d (i) Also, a : a — 6 = c : c — d (2) From 0), — ■ — = — — (3) a c ^ a — be — d From (2) = (4) a c Tx. . T ,. ^ — be — d Dividmg (4), by (S), — — - = — — - (51 a -j- c -j- a Whence a + b : a — b = c + d : c — d 284 PROPORTION PROPOSITION IX. !I71. If four quantities he in proportion, either couplet may he muliiplied or divided hy any number whatever, and the quantities will still he in proportion Let a : b = c : d h d Then, - = - (i;. a c Multiplying both numerator and denominator of either of these fractions by any number, n, (105, III), We have, Also, Hence from (-2), na : nb = c : d And from {^\ a : b = nc : nd] in which, if n represent a whole number, the couplets are mul- tiplied; and if 71 represent a fraction, the couplets are divided. PROPOSITION X. S7S. If four quantities be in proportion, either the ante- cedents or the consequents may be multiplied by any number and the four quantities will still be in pi^oportion. Let a:b =c id Then, b d Take Multiplying (i) by (2), (105, I), nh d = (2) na G h nd — = : (3) a no Dividing :i) by (2), (lOti, II), Hence from (3\ And from (4), a = c (1) m = m (2) mb md ), — —— (3) /' a c h d - — - : (4) "ilia mc a : mb c : md ma : b = mc : d GENERAL PRINCIPLES OF PAOPORTION. 285 PROPOSITION XI. ^7*1. IJ four quantities he in proportion j like powers or roots of the same quantities will he in proportion. Let a ih =^ c : d Then, -= - (i) ' a ^ ' nih power of d), ^= ^» (2) 1 I 5» d^ nth root of a), —= ~ {^ a« en Hence, from (2), a" : 6" = c" : d"" And from (3), a" : 6" = c^ : (Z« PROPOSITION XIL fSl^^:. If four quantities in proportion he multiplied or divided term hy term hy four others also in proportion^ the 2:)roduct or quotient will still form a proportion. If a I h = G : d rA) And X : y = m : n (B) We are to prove that ax : hy = cm : dn And a b G d X ' y ~~ m ' n From (A), ad— ho a) From (B), xn = ym (2) Multiplying (i) by (2), {ax') (dn) = (h^J) (cm) (3) Dividing (i) by (2), \x J \nJ \y / \m) (4) From (3) by Prop. II, ax : by =?= cm : dn And from (4), a b G d X ' y m ' n 236 PROPORTION. PROPOSITION XIII. 275. If any number of proportionals have the same ratiOj any one of the antecedents will be to its consequent as the sum of all the antecedents is to the sum of all the conse- quents. Let a : b = a : b (A) iVlso, a : b= G : d (B) a : b =ni : n [fi) &c. = &c. We are to prove that a : b == {a -r c -}- m) i (b -{- d '^- ri) From (A), ab = ab From (B), ad = cb From (C), an^= mb By addition, a(b -f c? -f- n) = b(a -f c -j- in) By Prop. II, a :b= (a + G-i-m') : (b + d + n) PROBLEMS IN PROPORTION. 2711. 1. Find two numbers, the greater of which is to the less as their sum to 42, and the greater to the less as their difference is to 6. SOLUTION. Let X = greater ; y == By the conditions, ] = X + y : 42 (1) X — y : Q (2) Prop. YI, X + y : 4:2 = X — y : 6 (3) Prop. Ill, X + y : X — y = 42 : Q (4) Prop. YIII, 2x:2y == 48 : 36 (5) Prop. IX, X : y = 4 : S (6) From (1), by Prop.YI, 4 : S = x -}- y : 42 a) From (2), by Prop. YI, 4 : '6 = x — ^ : 6 (8) From (7), by Prop. I, cc 4- y == 56 (9) From (8), by Prop. I, x — y = 8 (lo) Hence, cc = 32 And y = 24 PROBLEMS. 287 2. Divide the number 14 into two such parts, that the quo- tient of the greater divided by the less, shall be to the less divided by the greater, as 100 to 16. Let SOLUTION. X = the greater ; y = the less By the conditions, From'(i), by Prop. IX, Prop. XI,' Hence, from (4), But Therefore, a; = 10 2/ = 4 3. Find three numbers, in geometrical progression, whose sum is 13, and the sum of the extremes is to the double of the mean as 10 to 6. Let SOLUTION. x, xy, xy'^^ represent the numbers. ^ ^.,. ( x ■\- xy -{■ xri"^— 13 By tne conditions, \ v j ix^f + .T : 2:rt/ = 10 : 6 (1) (2) From (2), by Prop. IX, y^ -}- I : 2y = 10 : 6 (S) From (3),byPr.TIII, (2/^+22/4-1): (2/'— 2y+l)x= 16 : 4 4) Prop. XI, t/ + l:2/ — 1 = 4:2 (6) And 21/ : 2 = 6 : 2 (6) Hence, y == 3 And X == 1 288 PROPORTION. 4. The product of two numbers is 35, and the difference of their cubes is to the cube of their difference as 109 to 4 ; what are the numbers ? SOLUTION. Let X = :the greater, and y = the less. By tne conditions, {.- f : xy = 35 109: :4 (1) (2) Dividing 1st couplet ) of (2) hj x — y, ) »' + ^2/ + y' :(^ — ^// = 109; :4 (3) Or, x' + xy + y^'. ix'^ - 2xy + 2/^ = 109; :4 (4) From (4) Prop. VII, Sxy :x'- ^2xy + y' = 105 :4 (5) But, (1) oxy = 105 Hence, from (5) (x_yy = 4 (6) And X — y = , 2 (7) From (1) and (7), a: + 2/ = 12 (8) Hence, x = :7 And 2/ = 5 5. What two numbers are those, whose difference is to their sum as 2 to 9, and whose sum is to their product as 18 to 77 ? Arts. 11 and 7. 6. Two numbers have such a relation to each other, that if 4 be added to each, the sums will be to each other as 3 to 4 ; and if 4 be subtracted from each, the remainders will be to each other as 1 to 4 ; what are the numbers ? Ans. 5 and 8. 7. Divide the number 16 into two such parts, that their pro- duct shall be to the sum of their squares as 15 to 34. Ans, 10 and 6. 8. There are two numbers whose product is 320, and the difference of their cubes is to. the cube of their difference as 61 is to 1 ; what are the numbers ? Ans. 20 and 16, APPROXIMATE ROOTS. 289 APPROXIMATE ROOTS OF HIGHER DEGREES. 2? 7. A root of any number is evidently one of the equal factors which compose the number. Thus, the square root of 25 is one of the two equal factors, 5x5, which produce 25. The fifth root of 32 is one of the five equal factors, 2x2x2 X 2 X 2, which produce 32. If a number is not composed of as many equal factors as there are units in the index of the required root, the number is a surdy and the required root can only be obtained approxi- mately. This may be done, as we have seen (195j, by ex- tending the general method of extracting the root, to decimal periods. Another method, and the one we are now about to consider, is to decompose the number into factors nearly equal, and average the result, $578. If the number a be composed of 3 factors, each equal to X, then g — ^ — V a. In like manner, if the number a be composed of three fac- tors, Xf 2/, z, nearly equal to each other, it is obvious that 1 = v^a, nearly. l|p To illustrate this principle by numeral examples, we have 2 X 3 X 4 = 24 A A 2 4-3 +4 And q = ^ But v^24 n= 2.88 +. That is, one third of the sum of the three unequal factors, 2, 3 and 4, which compose 24, is 3 ; while the cube root of 24 is 2 88 -f , a number less than 3 by only .12. 25 T 290 APPROXIMATE ROOTS. Again, tako 5 X 7 X 8 = 280 And ''■l'-' = 6M + But v/280 = 6.54 + Difference, .12 4-. Similarly, we shall find that if uxyz = a, Then u+x+y+z Hence, in general terms. If a number be decomposed into n factors, nearly equal U} each other, the nth part of their sum will be nearly equal to the nth root of the number, ^79. If a number does not consist of factors nearly equal, or if it is a prime number, it may be decomposed into approxi* mate factors. Thus, if xyz = a, Then x = — , yz in which, if y and z be assumed, x may be computed, i L Find the cube root of 100. OPERATION. Let xyz = 100 Assume y = 4 And z = 5 Then X 100 100 - yz ~ 2Q - Adding A^alues of x+y + z, 4+5+5 = 14 Hence, dividing by 3, VlOO = 4.66 + , 1st approx. } 2/ = 4.641 2 = 4.642 X = 4.64176 + APPROXIMATE ROOTS. 291 As the root songbt is greater than 4.6, and less than 4.7, it must evidently be some number between these two. We will therefore IS'ext, assume 2/= ^-^ And, z= 4.7 T,e„. . = l^=i^= 4.62503 + * yz 21.62 Adding the new values, x + y -\- z = 13.92503+ Hence, dividing by 3, ^JFO = 4.64167 + , 2d approx. Next, assume And, Then, dividing as before, Adding the new values, oc + y + z = 13.92476 + Dividing by 3, VWO = 4.64158 +, 3d approx. which is correct to tire last decimal place. 2 Find the square root of 3. OPERATION. Let xy = 3 Assume ^ = 1.6 Then, a: = A = 1.875 + l.o Adding, x + y = 3.475 + Dividing by 2, n/3 = 1.7375, 1st approximation. Next, assume y = 1.732 Then, x = T-|ri= 1.7321016 + Adding, x + y= 3.4641016 + Dividing by 2, \/3 = 1.7320508 + , 2d approx. which is correct to the last decimal place. Hence, to obtain the approximate 7?th root of a number, we have the following KuLE. I. Assume n — 1 factors as near the required root as may be found by inspection, II. Divide the given number by the product of the assumed factors, and the quotient will be the remaining factor. 292 APPROXIMATE ROOTS. III. Divide the sum of the n factors thus obtained, by n, and the quotient ivilt be the first approximation to the required root, lY. Assume a second set of.n — 1 factors, and proceed as before ; and thus continue^ till the desired approximation is obtained. Note. 1. By inspecting the two examples given, it will be seen that the second approximation is less than the lirst, and the third less than tho second. Hence, every approximation is greater than the true root, and this principle will govern the selection of the factors to be assumed iu each step. 2. Also, by inspecting the two preceding operations it will be seen that each approximation after the first obtains one or more correct decimal figures in the required root. EXAMPLES FOR PRACTICE. 3. Kequired the 4th root of 18. OPERATION. Let uxyz = 18 {" u = 2 = 2 Assume, ' = 2 Then by (II), z = 2.25 Adding, u + x + y + z = 8.25 Dividing by 4 (III), v/l8 = 2.0625, 1st approx. 2^06 Next, assume -^ ^ = 2.06 .!/ = 2.05 Then, as before, z = 2.069113 + Adding, u-^- x + y +_z = 8.239113 + Dividing by 4, v/l8 = 2.059778 +, 2d approx. XoTR. — It will bo observed, in the above example, that 16, a number near 18, is a perfect 4th power, whose root is 2. Hence, ti, x and y were assumed each as 2. 4. Required the cube root of 130. 5. Required the 4th root of 260. 6. Required the 4th root of 640. 7. Required the 5th root of 2. 8. Required the 5th root of 38. APPROXIMATE ROOTS. 293 QSO. When tlie numbers are large, or the index of tlie root is higher than the 3d or 4th, the following table will be of use in assuming the factors for the 1st approximation. It will be seen that the first column contains certain numbers, taken at intervals between 1 and 270C0 ; the second column contains the squares of the numbers in the third ; the third column contains the cube roots of the numbers in the first ; the fourth contains the 6th roots ; and the fifth, the 9th roots. A aI Ai A* Ai 1 1 ... ... 1 ... ... 1.0000000 ... ... l.OOOOOC 8 4 ... ... 2 ... ... 1.4142136 ... ... 1.259921 27 9 ... ... 8 ... ... 1.7320508 ... ... 1.442250 64 . 16 ... ... 4 ... ... 2.0000000 ... ... 1.587401 125 25 ... ... 5 ... ... 2.2360680 ... ... 1.709976 216 . 36 ... ... 6 ... ... 2.4494897 ... ... 1.817121 843 . 49 ... ... 7 ... ... 2.6457513 ... ... 1.912933 512 . 64 ... ... 8 ... .... 2.8284271 ... ... 2.000000 729 . 81 ... ... 9 ... ... 8.0000000 ... ... 2.080084 1000 . 100 ... ... 10 ... ... 3.1622777 ... ... 2.154435 1331 . 121 ... ... 11 ... ... 3.3166248 ... ... 2.223980 1728 . 144 ... ... 12 ... ... 3.4641016 ... ... 2.289428 2197 . 169 ... ... 13 ... ... 3.6055513 ... ... 2.351335 2744 . 196 ... ... 14 ... ... 3.7416574 ... ... 2 410142 3375 . 225 ... ... 15 ... ... 3.8729833 ... ... 2.466212 4096 , 256 ... ... 16 ... ... 4.0000000 ... ... 2.519842 4913 , 289 ... ... 17 ... ... 4.1231056 ... ... 2.571282 5832 , 324 ... ... 18 ... ... 4.2426407 ... ... 2.620741 6859 . 361 ... ... 19 ... ... 4.3588989 ... ... 2.668402 8000 , 400 ... ... 20 ... ... 4.4721360 ... ... 2.714418 9261 . 441 ... ... 21 ... ... 4.5825757 ... ... 2.758923 10648 , 484 ... 22 ... 4.6904158 ... ... 2.802039 12167 . 529 ... ... 23 ... ... 4.7958315 ... ... 2.843867 13824 , 676 ... ... 24 ... ... 4,8989795 ... ... 2.884499 15625 625 ... ... 25 ... ... 5.0000000 ... ... 2.924018 17576 . 676 ... ... 26 ... ... 5.0990195 ... ... 2.962496 19683 729 ... ... 27 ... ... 5.1961524 ... ... 3.000000 21952 784 ... ... 28 ... ... 5.2915026 ... ... 3.036589 24389 841 ... ... 29 ... ... 5.3851648 ... ... 3.072317 27000 900 ... ... 30 ... ... 5.4772256 ... ... 8.107232 25* 294 APPROXIMATE ROOTS. 1. Find the 7tli root of 2211. Referring this number to the table, we find the number nearest to it in column A is 2197, whose Gth root (A^), is is 3.6 -f , and whose 9th root (A^), is 2.3 +. Therefore, the 7th root of 2211 must be nearly equal to 3, and we will tako this number for each of the six factors to be assumed. Thus, OPERATION. Let 3 X 3 X 3 X 3 X 3 X 3.2; = 2211 3)2211 3)"737 3) 245.666666 3) 81.88888 8 3) 27.296296 3) 9.098765 37032921 = 0?. Hence, ^22n = 3 + 3 + 3 + 3 + iT"3T^:^2921 7 _ 21.03 2921 ~" 7 = 3.004703, 1st approximation. For the second approximation, we may assume 3.004 for each of the 6 equal factors. As the method above used would be somewhat tedious, we may avoid the labor, by an application of the binomial theorem. For this purpose let it be observed that the powers of a frac- tion are less than the fraction itself. Thus, (.iy = .oi (.1)3 = ,001 (.1)*=.0001 We have found that the 7th root of 2211 is greater than 3, by a small fraction. Let this fraction be represented by x ; end we shall have the following equation : (3 + xy = 2211. APPROXIMATE ROOTS. 295 Expanding the first member, indicating the powers of 3, we have 2187 + 7(3^^ + 21(3)^0:2 + 85(3)^:c^ +35(3yx* -f- 21{Syx' + 7(3>c« + x'^ = 2211. Now as a: is a small fraction, the powers of x will be still Bmaller. And if the terms containing x^j and all the higher powers of x be omitted, the equation will still be approxi- mately true ; and we shall have 2187 + 7(Syx + 21(3)5x'^ == 2211 a) Transposing, 7(3)«:c + 21(3)V = 24 (2) Dividing (2) by 3, 7(3/^ + 2l(3yx^ = 8 (3) Expanding (3), 1701a: + 1701a;'' = 8 (4) Dividhig (4) by 1701, x' + x = .0047081 (5) Completing the square, 4x2+ () -f 1 = 1.0188124 (6) Extracting square root, 2x + 1 = 1.00936 (7) Reducing (7), x = .00468 Hence, v^22Ti == 8 + a: = 3.00468, 2d approx. 2. Find the 7th root of 2412. Referring to the table, we find this number in column A, between 2197 and 2744; and by examining the 6th and 9th roots of these numbers, we find that the 7th root of the given number must be a little greater, or a little less, than 3. Hence, OPERATION. Let 3 -f a; = required root. Put a s= 3 Then, (a + xf = 2412 EKpanding, a;+7a«a;+21alx'+35a*a-» + &c. = 2412 Or, approximately, a' + 7a^x = 2412 (i; Restoring value of a in (i), 2187 + 5103x = 2412 (2) Transposing 5103.C = 225 (3) And X =.044 -f Hence, S + x = 3.044 + 296 APPROXIMATE ROOTS. That is, the seventh root of 2412 is very nearly 3.044; but this is a little too large. If we would find the root more ac- curately we will place 3.04 for a, or (3 + m) in equation (i). Then that equation will become (3 + m)^ + 7 (3 + my x = 2412. Taking the higher terms of (3 + m)', and (3 + my ip separate columns, we have 3^ = 2187 8« = 729 7(3>i = 204.12 6(3)»m= 58.32 21(3ym« = 8.1648 15(3)*m^ = 1.944 85(3)*m» = .18144 20(3)»m»= .03456 S5(3)»m* = .0024192 15(3)''m* = .0003456 789.2989056 Adding, 2399.4686592 + 7(789.2989056)a; = 2412 Whence, from the above equation, we have approximately 2399.4686592 + 7(789.2989056)a; = 2412 ; ^ 12.5313408 _ „^__ ^ Q^-^= 7(m2989056) = '^^^^^^+> Hence, ^2412 = 3.042268 +. EXAMPLES FOR PRACTICE. 8. What is the cube root of 14000 ? 4. What is the 5th root of 812 ? Am. 3.81893. 5. What is the 8th root of 1340 ? 6. What is the 7th root of 9150 1 MISCELLANEOUS EXAMPLES. 297 MISCELLANEOUS EXAMPLES. 1 What is the value of — , when a = 4, 5 = 5, and m = 20 ? m^b--(l'— m)a A71S, -J. 2. What is the sum of (a + b)s^ax + 12(a + b)^^ax -— 7(a -f b)^ax 4n(a -f 5)v^ax? ^ns. fQ(^a + &) + 4n(a + 6)") >/a~x ; Or, (6 + 4n) (a + 6)>/ax. 8. What is the sum of (A + B)^ (x + y) and (A — B)*- (a; + 2/) ? Ans. 2{A.' + B'O (x + y\ 4. What is the sum of ba'b + Sa-^fe^c, 6a*6 + 2a-262(?, 4- 10 ab, and 9a*6 — 8a-=^6^c — 10a6 ? ^ Ans, 20a'b— •^. 5. From v^x^ — ^/^ + 4(0- + y) — Sy^a + a;, subtract 3(a; -♦- 2/) — 2(:r^ — y'y -h 13(a + a^)l ' ^ns. Sv^ic^ — 2/^ + 03 + 2/ — 16v^a + ir» 6. From 17ax* + Say + 10a, subtract 7ax'^ + 4ay + 12a — 2ba, Ans, (lOx^ _ y -^ 2 + 26)a. 7. What is the value of (2x') (j^2xy) (—2mx) (—2xy^) (2m*xy) ? Ans, — (2mxyy 8. Multiply a» — a'6 + afe^ — ft^ 13^ ^ ^ j -<4ns. a* — b\ 9 Multiply as*" -- a;**-^ ^as^'V — 'f^hj x + y. 208 MISCELLANEOUS EXAMPLES. 10. Expand (m — z-- 1) (m + 1) (m' + 1). Ans. m* — m^z — m^z — mz — z — 1. 11. Expand (ox'^i/ — oxy^) (Sx^y — Sxy^), A71S. ^xY — ISxy + 9^y, 12. Expand (c*" + c") (c^" + c"). Ans, c^"' 4- 2c'"+" 4- cr^. 13. Find two factors of x^ + 3/*. ^?i6*. {x + y) and a;* — xhj -{- x^y^ — xy^ + y\ 14. Find two factors of x^ — y^. Ans. (xT- +2/") and ix' — 2/') ; Or, {x — y) and x^ -\- x^y 4- ct-?/' + a:'^/^ + 1/*. 15. Multiply (a' — ah + a6^ — Jh' + Jh' — h") by ( a* + &). Prod a^ _ h\ 16. Find the factors of a^ — h\ Ans, (a'—b) (a^+a'b + a'b'+b^), or (a' + b') (a'—b'). 17. Find the greatest common factor of (a* — 1), (a^ 4- a^), (a« 4- 1). Ans, 0} + 1. 18. Find the greatest common divisor of a" — 5ah 4- 46'' and a^ — ab + ^a¥ — Sb^, Ans, a — b, 19. Find the sum of the following fractions; indicating that sum by S, and condensing it to a single term. a^ h ^ ab , , -, and — -TT-o- -^^s. 1. (a + 6)" a + b' (a + b) 20. What is the sum of the following fractions ? _^^, ^_, and -i-. Ans.^-^L±q^±l. x^ — y^ X '\- y X — y x^ — y^ 21. Divide 51a'(c — 1)"*+" by ITa^c — 1)"^-". Ans, 3(c — 1)*~. 22. Divide 5(a — 1) 4- 6(1 — of by 1 — a. Ans, 1 — 6a. MISCELLANEOUS EXAMPLES. 299 23. Divide 771 — ?i by v'm — >/ n, Ans. v^wi -1- v^n. 24. Divide a' + ab + b^ hj a -{• ^^a'b + b, Ans, a — v^a6 + b 25. Factor a;^y--a:y'—-a:y + iP—y. 26 What is the greatest common divisor of ia^z — \2amz^+ %abcz and ba^nix — Ibm'^xz + 10 bcmx ? Ans. a^ — Smz + 26c. 27. What is the least common multiple of Sd^x, 4:axy, and 16ax^ ? Ans. 4:SaVy. 28. What is the least common multiple of ac — om% o?c + Zacnv and a^c — 9cm* ? Ans. a^c — 9ac7M*. s/aFb '^ai/ 29. Reduce — ^r-, — ; — r., to its lowest terms. a^ — 2ab + b^ ^ab Ans. --. _ ^ — ^ Ttx( ^^ CL — — ^^ b "^ SO. Reduce — -^=z =t— ,-^ to an entire form. Ans. m(a + 6) — 2m s/ob, 81 Find the value of ^^ - (^^ + ^). Ans. ^^. 82. From i + i + |=-J, subtract \ + ^^^. 33. Divide 1 -\ -— :r, by 1 — ^. u4ns. n. n + 1 "^ 71+1 J5^0TE. — The product of the divisor and quotient must be equal to the dividend. Let Q = the quotient, then «c-j^l)-'+:-^- 300 MISCELLANEOUS EXi^MPLES. 34. Divide-^ ;> by (a; ; — ). Ans. . or — x^ \ a + x/ a — x 35. Simplify the fraction, "t! "^ : -^. Ans. ^^^ tax a — x a-^x 86. Divide 2 , by — :: — . 87. What is the value of ^?L±^|:z;(^g! y Ans. a — b' 88. What is the value of ^^^IJt^tt^^I ? (1 + by — (1 — by 39. Raise v^a + 6 to the 4th power. Ans. (a + 6)'. 40. Raise a V — 1 to the 3d power. Ans. d\/ — 1. 41. Raise 2a + 36 to the 9th power. Ans. h\1a^ + 256-27a'6 + 512-81a'6'' + 256-567a'6' + &c. Reduce the following equations : 42. {x + Va)' — (ar — \/a)' = 2a>/^. A «^^ 43. m + 2c 44. 6 v^a 6 -f v'a a? aj x/a6> u4ns. X = by^a. 45. 4 a: + 1 ar 4- 1 ar — 1^ 4 "^ 4 • Ans. a; = d= 3. 46. -g — — rr a=x 5a? — 5. An». aj ==3 zfc 1.224744 -f. MISCELLANEOUS EXAMPLES. gOt 47. -5 = = _ZL_. jins. X == ±*n/2 ±:2v/5 48. vsr=9 = :;7=fg. ^n«. X = 9. 49. (1 + xy + (1 — a?)'' = 242. _ Ans. ic = dh 2 or it v^ — 6. 60. Find the 6th power of \^2 -f "^3, or expand (v^2i -f ^3y)®, and afterward suppress all the powers of x and y, on the supposition that each is equal to unity. Ans, 8 + 24>/6-f 180 + 120\/6 4-270 + 54 v/6+ 27. ifl 9 51. Given - — -—^ = 7 — v, to find the value of y, Ans, 2/ = 5. 25-;p« 9 b2. Given -= — -—^ =7 — 5x, to find the value of x. oa? + 3 u4ws. X = 1. 63. Given |-+ 2 = |(^- 1 ) + ^^^Y^f to find the value of X. Ans, x = o. 54. Given x» + 2x» + cc = (a» + 3a:) (a; — 1) + 32, to find the value of x. Ans. x =z S. 65. Given (x + ly — (x — !)« = l344x. Ans. a; = it 3, or d= 4/— -^^ _^ (ox + ey = 2aV 7 f x = ac^. Ccx + ay =s ac^c' + a')> Cy == ^^« Tcc+y — 2=»4^ rx==3. 58. <2x + 2z — y==d> Ans. < y =^ 2. I 3t/ + 32: — X « 6 3 ( z « 1. 26 ^Q2 MISCELLANEOUS EXAMPLES. lu + x+y=^^a — 6m \ I u = a '-- m. ^, \u 4- X + z = Sa — 1m { , ]x ^ a — 2m, )u + y + z=oa — 8m ( ) 2/ = « — ^^^• \x + y + z = Sa — 9m / [z = a — 4 ;?i. ^"- l4a;»_42/ = 28i ^"'- 1 2/ = 2, or 6J-. f ck' -r 37/' = 28 •> f a; = 5, or v/2i. ^" l a;' + 2ccv = 35 i ^"*- U = 1, or iv'21. 62. 63. + 2x2/ = 35 ) <- 2/ + \^ xy = 15 fx + N/^^=15| ^^^^ (a; = 9. (. J/ 4- Va-j/ = 10 ) (. 1/ = 4. (a;^ + 2/^=106| ^^^ (a. = 3,or3>/=T 64. Two men started from two towns, A, o,nd B, and tra- veltid toward each other. The first went ^, and the second, f of the distance between the two towns, when the men were found to be 16 miles apart; required the distance from A to B Ans, 60 miles. 65. The sum of two numbers is 80 ; and if their difference be subtracted from the less and added to the greater, the re- sults will be as 1 to 7 ; what are the numbers ? Ans. 80 and 50. 66. What number is that whose fourth part exceeds its fifth part by ^j^^ ^ ^^^- h 67. There is a number whose 3 digits are the same ; and if from the number, 4 times the sum of the digits be subtracted, the remainder will be 297 ; required the number. Ans. 333. 68. A certain number increased by 1, is to the same number increased by 4, as the square of the number is to its cube* what is the number ? Ans, 2 MISCELLANEOUS EXAMPLES. 303 69. Fifty gallons of wine are to be put into casks of two sizes ; and 10 of the smaller casks and 2 of the larger, or 5 of the smaller and 6 of the larger, may be used ; required the capacity of a cask of each size. Ans. Smaller, 4 gal. ; larger, 5 gal. 70. Two men have the same income ; one saves one tenth of bis, the other spends $150 per annum more than the first, and at the end of five years finds himself $100 in debt ; what is tht income of each? Aiis. $loOO. 71. A man has 2 equal flocks of sheep ; from one he sells a gheep^ and from the other b sheep, and then he has 3 times as many remaining in the latter flock as in the former ; how many did each flock originally contain ? A7is. ^(Sa — b) sheep. Note. — For a and b, take uny niimher at plensure, such that Sa — h shall be divisible by 2, and form a definite problem. For instance, as- sume a = 12, and 6 = 2, then the number in each flock will be 17. In this manner numeral problems are formed. 72. The sum of two numbers is 72, and the sum of their cube roots is 6 ; what are the numbers ? Ans. 64 and 8. 73. The sum of the squares of two numbers, multiplied by the sum of the numbers is 2336, and the difference of their Bquares, multiplied by the difference of the numbers is 576 ; what are the numbers ? Ans, 11 and 5. 74. The product of two numbers multiplied by their sum is 84 ; and the sum of their squares multiplied by the square of their product is 3600 ; what are the numbers ? Aiis. 4 and 3. 75. A market man bought 15 ducks and 12 turkeys for 105 shillings, and he obtained 2 more ducks for IS shillings, thar. turkeys for 20 shillings ; what were the prices ? Ans. Ducks, 3 shillings ; turkeys, 5 shillings. 76. The sum of three numbers is 12 ; one third of the sum of the first and second is equal to one fifth of the sum of the Becond and third ; and the second minus the first is equal to thf5 third minus the second ; required the numbers ? An&. 2, 4, and (I, gQ4 MISCELLANEOUS EXAMPLES. 77. There is a square tract of land containing 10 times as many acres as there are rods in the fence inclosing it ; how large is the square ? Ans, 20 miles square. 78. A general wishing to draw up his regiment into a square, found by trial that he had 92 men over ; he then increased each side by 2 men, and wanted 100 men to complete the square ; how many soldiers had he ? Ans. 2301. 79. Some bees had alighted upon a tree ; at one flight the square root of half the number went away ; at another | of them ; and two bees then remained ; how many alighted upon the tree ? Ans. 72. 80. A May-pole is 56 feet high. At what distance above the ground must it be broken, in order that the upper part, clinging to the stump, may touch the ground 12 feet from the foot? Ans. 26 1 feet. 81. Divide the number 20 into two such parts that the square of the greater diminished by twice the less, shall be equal to twice the square of the less. Ans. 12 and 8. 82. Three numbers are in arithmetical progression ; their sum is 27, and the product of the extremes is 77 ; required the numbers. Ans. 7, 9, and 11. 83. A gentleman has a garden 10 rods long and 8 rods wide ; he would lay out half the way round it a graveled walk of uniform width and to contain ^ of the area of the garden. How wide shall the walk be laid out ? Ans, 13.4684- rd. 84. Two numbers are in the proportion of a to 6, and when c is added to each, the proportion is as 5 to 6 ; what are the numbers ? ^ «^ -. ho Ans. -p-^ ;^, and 66 _ 6a' 56 — 6a* 85. A man sold a horse for 144 dollars, and gained as much per cent as the horse cost him. What did the horse (^ost him ? Ans. $80. MISCELLANEOUS EXAMPLES. 305 86. Two numbers are in the proportion of 5 to 8, and if 200 be added to the first, and 120 to the second, the sums will be to each other as 5 to 4 ; what are the numbers ? Ans, 50 and 80. 87. A person bought two cubical stacks of hay for £41, each of which cost as many shillings per cubic yard, as there were yards in a side of the other, and the greater stood on more ground than the less by 9 square yards. What was the value of each stack ? Ans, £16 and £25. Let X = a side of the greater stack in yards ; And y = a side of the other ; Then x" —y^ = 9 = a-, 0) And ar'y + xf = 41-20 = 820 = 6 (2) From (2) x^ + y^ = — (8) xy Squaring (3) x* + 2j?y + 2/* = ~r^ (« X y Squaring (i) x^ — 2x'^y'^ -{- y^ = a^ Diff 4:xY =~— a\ X y 88. A farmer had 3 more cows than horses. He bought 2 more cows and sold 3 horses ; and he then had 5 times as many cows as horses. How many had he at first ? Ans. 5 horses and 8 cows. 89. Some boys on a frolic incurred a bill of $12. If there had been two more in the company each would have been charged 30 cents less. How many were in company ? Ans. 8. 90. A person residing on the bank of the Ohio, 15 miles above Cincinnati, can row his boat to the city in 2^ hours, but it requires 7i hours to return. With what force can he ruw his boat in still water, and what is the velocity of the river ? Alls. Man rows 4 miles per hour ; stream flows 2 miles. 2x 2x 91. What is the value of a; H ^— divided hj x when X = 5i ? ^,^ ^ 26* u 806 MISCELLANEOUS EXAMPLES. . . 92. I deposited $200 in a savings bank, which paid 6 per cent, on deposits, interest payable semi-annually. How much did my money amount to in 5 years, the interest being added to the principal at the end of every 6 months ? Note. — The principal is the first term; $1.03, the semi-annual amount of $1 at the rate per cent, is the ratio ; and the number of terms minus 1 is 2 times the number of years. Ans. $268.'78 + . 93. What is the value of the expression -_ — 4- 94. Find the geometrical mean between 2a^h'^ and 24:a^b'^x'^ , 12 8 — Ans. 4a^VSa x Vb^ x Vx, 05. From a bag of money which contained a certain sum, was taken $20 less than its half; from the remainder, $30 less than its tliii'd pai't ; and from the remainder, $40 less than its fourth part, and then there was nothing left. What sum did the bag contain? Ans. 1080. 96. Four numbers are in arithmetical progression ; the pro- duct of the first and third is 27, and the product of the second and fourth is 72. What are the numbers ? Ans. 3, 6, 9, and 12. 97. A merchant gains the first year, 15 per cent, on his capi- tal ; the second year, 20 per cent, on the capital at the close of the first ; and the third year, 25 per cent, on the capital at the close of the second ; when he finds that he has cleared $1000.50. Required his capital. A7is. $1380. 98. The sum of three numbers in arithmetical progression is 15, and their product is 80. Required the numbers. Ans. 2, 5, and 8. 99. Find three numbers in arithmetical progression such that the sum of their squares shall be 2900, and the product of the extremes shall be less than the square of the mean by 100. Ans. 20, 30, and 40. MISCELLANEOUS EXAMPLES. 307 100. What number is that which, if 4 be subtracted from it, ^ of the remainder will be 7 ? Ans. 25. 101. What number is that to which if 1 and 11 be added se- parately, the sums will be to each other as 1 to 3 ? Ans. 4. 102. Given - — --7-3 + ~ — r-r^ == ^y ^^ ^^^^ ^^^ ^^^^^ ^^' ^• (a + bf (a + by . ah Ans. X == — ,-7. a + b 103. What number is as much below 40, as three times that number is below 100 ? Ans. 30. 104. Divide 400 into two such parts, that the sum of their square roots shall be 28. A^ns. 256 and 144. 105. A man sold a horse for a dollars, and gained as much per cent as the horse cost him ; what did the horse cost him ? Ans. (v^aHh~25J 10 — 50 dollars. 106. The product of three numbers in geometrical progres- sion is 1728, and the sum of the first and third is 40 ; what are the numbers ? Ans. 4, 12, and 36. 107. Two quantities are to each other as m to n, and the difference of their square is d^ ; what are the quantities ? md nd Ans. / ~~ :, , -=r^ Vm? — n vm- — 1^ 108. The sum of four numbers in geometrical progression is 85 ; and the sum of the first two is to the sum of the second two as 1 to 16 ; what are the numbers ? Ans. 1, 4, 16, and 64. 109. Th.e base of a right angled triangle is 20 rods, and the perpendicular and hypotenuse are to each other as 5 to 7 ; what is the length of the perpendicular and what the area of the triangle ? Ans. Perpendicular, 20.4124 f rods; area, 204.124 -fs(^. ^-ods. 80.8 MISCELLANEOUS EXAMPLi5S. 110. The extremes of a geometrical series are 2 and 3*744, and the ditference of the first and second term is to the difference of the third and fourth as 1 to 9. liequired the sum of the Beries. -^^^s. 5615. 111. The sum of two numbers added to the sum of their squares is 18, and 10 times their product is 60 ; what are the numbers ? Ans. 2 and 3, or v^3 — 3, and — Vo — 3. 112. The sum of two numbers is to their difference as 4 to 1, and the sum of their cubes is 152 ; what are the numbers ? Ans. 3 and 5. 113. Insert 9 arithmetical means between 6 and 36. Ans. 9, 12, 15, &c. 114. At what rate per cent, will a dollars gain as much in 4 years at simple interest, as in 2 years at compound interest ? Ans. 200 percent. 115. How many terms of the series, .034, .0344, .0348, &c., . will amount to 2.748? Ans. 60. 116. How much will $230 amount to in 12 years, at 6 per cent., simple interest ? Note. — The number of terms will evidently be 1 greater than the number of years. Alls. $395.60. 117. What is the sum of n terms of the series, 3, 3^-, ^, &c. ? Ans. (n + 17)^. 118. I lent a certain sum at 7 per cent., simple interest, and at the end of 5 years received, in principal and interest, $317.79 ; what was the sum lent ? Ans. $235.40. 119. If 6 be the first term of a geometrical series, and 4374 the 7th term, what is the ratio, and what are the other terms ? Ans. Ratio 3 ; whence 18, 54, &c., the other terms of the series. MISCELLANEOUS EXAMPLES. 3()9 120. Sold a horse for $175.50, taking a uote drawing interest at 6 per cent. Not needing the money, I did not collect the note until the end of 6 years ; what amount did I collect? Ans. $238.68. 121. The sum of the extremes of 4 numbers in geometrical progression is 35, and the sum ©f the means is 30 ; what are the numbers ? Ans. 8, 12, 18, 27. 122. I own a mortgage of $875 on a farm, due in 6 years, at 6 per cent, interest, payable annually. If no part of the moit. gage or interest is paid until the end of the 6 years, how muck will be the auiountdue at compound interest ? Ans. $1241.20 -f. 123 Three times the product of two numbers is equal to their difference multiplied by the difference of their squares. Also, 45 times the square of the product is equal to the difference of their 4th powers multiplied by the difference of their squares ; what are the numbers ? Ans. 4 and 2. 124. The principal is $300, the time 3 years, and the rate 6 per cent., compound interest, semi-annually; what is the amount ? Note. — The number of terms is 1 more than 2 times the number of years; and the ratio is 1.03. Ans. $358.2156. 125. The length of a plat of ground is 4 rods more than its breadth ; and the number of square rods in its area is equal to the number of rods in its perimeter. Required the length ard breadth. (Length, 6.8284 + rods. iBreadth, 2.8284 + rods. 126. Find the side of a cube which shall contain as many solid units as there are linear units in the distance between its two opposite corners. j^^g^ ^/g^ 127. Find two numbers, such that their product shall be equal to 4 times their difference ; and the difference of their squares shall be 9 times the sum of the numbers. Ans, 3 and 12. glO MISCELLANEOUS EXAMPLES. 128. A and B together carried 90 eggs to market, and sold at different prices, each receiving the same sum. Had A taken as many as B he would have received 32 cents for them. Had B taken as many as A, he would have received 50 cents for them ; how many did each take to market ? Ans. A 50, B 40. 129. Find two numbers, such that the difference of tlicir scjuares shall be 12 ; and 3 times the square of the greater minus twice their product shall be 32. Ans. 4 and 2. 130. The product of 5 numbers in arithmetical progression is 945, and the sum of the numbers is 25 ; what are the num- bers? Ans. 1, 3, 5, 7, 9. 13 1. A man has two unequal measures. If he lay out a plat of ground having the greater measure for its length, and the less for its breadth, it will contain 40 square feet ; but if he lay out a plat having twice the greater measure and once the less for its length, and once the greater and twice the less for its breadth, it will contain 432 square feet. How many feet in length is each measure ? Ans. Less, 4 ; greater, 10. 132. A carpenter agreed to live with a farmer during the winter, on the condition that for everyday he worked he should receive $1.50, and for every day he was idle he should forfeit 65 cents. At the expiration of 129 days they settled, and the carpenter received nothing ; how many days did he work, and how many was he idle ? Ans. He worked 39 days, and wa«; idle 90 days. 133 Find three numbers such that the product of the first and second shall be 6 ; of the first and third, 8 ; and of the second and third, 12. Ans. 2, 3, and 4 134. In a plane triangle the base is 50 feet, the area 600 feet, and the difference of the sides 10 feet ; required the sides and perpendicular. Ans. Sides, ^0 and 40 ; perpen. 24 feet. MISCELLANEOUS EXAMPLES. ^H 135. There are four numbers such that the product of tho first, second, and third is a ; the product of the first, second, and fourth is b ; the product of the first, third, and fourth is c; and the product of the second, third, and fourth is d, Required the numbers. , ^^ abed -^^abcdi -^ abed \^abcd Ans. — -r^— , , — -— , . d b a 136. A is a traveler 11 miles in advance of B, and travels 4 miles per hour ; B starts to overtake him, and travels 4-^ miles the first hour, 4| the second, and 5 the third, increasing his rate :^ of a mile per hour ; how many hours before he will overtake A ? ^?7S. 8. 137. The base of a right angled triangle is 6, and the three Bides are in arithmetical j)rogression ; what are the sides ? Ans. 6, 8, 10, or 4i 6, 7 J. 138. Two squares contain together an area of 52 inches ; and twice the difference of their area is equal to the number of inches in the perimeters of the two. Required the con- tents, of each. . (Greater, 86 inches; I Less, 16 inches. 139. A certain number, consisting of two digits, is equal to twice the product of its digits ; and if 27 be added to the number, its digits will be inverted. Required the number. Ans, 36, 140. The sum of three numbers in arithmetical progression is I, and the sum of their reciprocals is 7^- ; what are the numbers ? Ans. ^, \, and |. 141. Find three geometrical means between -^ and f. Ans. ^>/6', ^, and ^v^6. 142. There are three quantities related as follows : the sum of the squares of the first and second, added to the first and second, is 18 ; the sum of the squares of the first and third, added to the first and third, is 26 ; and the sum of the squares gio MISCELLANEOUS EXAMPLES. of the second and third, added to the second and third, is 32 Required the quantities. r 1st, 2, or — 3 ; Ans, < 2d, 3, or — 4 ; I 3d, 4, or — 5. 143. The compound interest of a certain sum of money for 8 years was $864 ; and the interest for the first year was to the interest which accrued the third year as 25 to 36. Required the sum at interest. Ayis. $500. 144. Find 3 numbers in arithmetical progression such that their sum shall be 36, and 4 added to the product of the ex- tremes shall be equal to the square of the mean. Ans, 10, 12, and 14. 145. Find three numbers in geometrical progression whose sum shall be 52, and the sum of the extremes to the square of the mean as 10 to 36. Ans, 4, 12, and* 36. 146. There are three numbers in geometrical progression ; their continued product is 1, and the difference of the first and second is to the difference of the second and third as 1 to 3 ; what are the numbers ? Ans. ^, 1, and 3. 147. There are three numbers in geometrical progression, such that three times the first, twice the second, and once the third, taken in order, form an arithmetical series ; and also the first, the second increased by 8, and the third, taken in order, form an anthnietical series. What are the numbers ? Am, 4, 12, and ?6. ► ••^/ V • • ^ ». •!*' •• ^, • . . v^-* .'Ta^- •*• !• P THTS BOOK IS DUE ON THE LAST DATE ^ STAMPED EET.OW T ..^■'■M :^-^'i AN INITIAL PINE OF 25 CENTS S WILL BE ASSESSED FOR FAILURE TO RETURN 1 THIS BOOK ON THE DATE DUE. THE PENALTY g WILL INCREASE TO SO CENTS ON THE FOURTH B DAY AND TO $1.00 ON THE SEVENTH DAY K OVERDUE. B r.;vv >.: :a42 V- • •/ • ,■• '. -J- K^H LD21-100m-7,'40 (6936s) 1 f^- m^. /s h sir ^^-■ ^^^i^k:^M.0^ i^' wm IWI :\s^