^ IN MEMORIAM FLORIAN CAJORI Digitized by the Internet Archive in 2008 with funding from IVIicrosoft Corporation http://www.archive.org/details/elementarypractiOOdoddrich ELEMENTARY AND^EACTICAL ALGEBB,i^^^^v^ OF IN WHICH HAYE BEEN ATTEMPTED IMPROVEMENTS IN GENERAL ARRANGEMENT AND EXPOSITION; AND IN THE MEANS OF THOROUGH DISCIPLINE IN THE PRINCIPLES AND APPLICATIONS OF THE SCIENCE. BY JAMES B. DODD, A. M., Morrison Professor of Mathematics and Natural Philosophy in Transylvania University. FIFTH EDITION. NE W - Y ORK : PUBLISHED BY PRATT, WOODFORD & CO. 18 5 4. Entered, according to Act of Congi-ess, in tne year 1852. Br JAMES B. DODD, In the Clerk's Office of the District Court of Kentucsy. CAJORl T. P. JONES & CO., STEREOTYPEaS, 183 William stbskt. PREFACE. The following work is designed to furnish a practicable course of Algebra for tbe younger classes of students, witbout tbe omission of any tbing important to a tborougb education in tbe subjects wbicb it embraces. It aims at tbe most metbodical arrangement^ tbe clearest expositions^ tbe best elementary exercises, and tbe most varied and useful applications : — in all tbese respects presenting some new features, wbicb bave been adopted as improvements in tbe metbod of teacbing tbis science. All tbat is appropriate to an Algebraic treatise in a general course of mathematical studies, or necessary in preparation for tbe bigber works of tbe course, bas been introduced, witb tbe exception of a few subjects wbicb are more exclusively pre- liminary to tbe Differential and Integral Calculus. Tbese, witb whatever else may be considered useful in a larger work, will shortly be added to the present treatise. Between this work and tbe author's Arithmetic there will be found a mutual correspondence in many respects, though each is entirely complete in itself The two are commended to the consideration of Teachers of Mathematics, and the Guardians and Friends of Education, as containing a practi- cable, progressive, and thorough course of study, for Schools, on these connected and important branches of science. Transylvania University, ) July 20th, 1852. ^^ Hf^H^ ■^ REMARKS ON THE METHOD OF USING THIS WORK, AND CONDUCTING EXAMINATIONS IN ALGEBRA. The following remarks may be useful to the less experienced Teacher using this work, who would make it fully efficient for the purposes intended. 1. The definitions and propositions numbered (1), (2), (3), &c^ and the Rules I, II, in, Ac, should be accurately memorized and recited by the Student. 2. The accompanying examples, illustrations, or demonstrations, should be requii-ed of the Student, and discussed with him on the paii of the Teacher, with reference to the principles involved in them. 3. The oral exercises in the eai-lier pai-ts of the work, should be exacted ; and the Student should often be examined on the exercises under the Rules, with his book closed. 4. In the solution of Equations and Problems, he should explain each part of the operation, as exemplified in different parts of the work. 5. The Analysis of Contents (see the next page) will be convenient for reviews on the theory of the science ; and such reviews should be fi-equent. The Student will thus become familiar with the phraseology, principles, and order of the science. 6. The Student's acquisitions wiU depend veiy much on the exactness, as well as on the frequency, with which he is examined. Tlie requisitions made on him should be adapted to his capabilities, — ^which, it should be remembered, are liable to be sometimes overrated, and sometimes underrated, by Authors and Teachers. ANALYSIS OF CONTENTS. This Analysis is designed to be used in oral examinations, in reviews. The Teacher will name the topic as presented in this table ; the Learner will respond according to his knowledge of the subject. For example : the Teacher will say, " Science and Art ; " the Learner will re- spond, " Science is knowledge reduced to a system ; Art is knowledge applied to practical pui-poses." CHAPTER I. Preliminary Definitions and Exercises. — Page 1 ... 8. Science and Art, (1). — A Unit — Numbers. (2). — Quantity — Whether Numbers are quantities^ (3). — Mathematics — Its most general Divisions — Arithmetic — Geometry, (4). — Algebra, (5).— Symbols of Quantities, (6). — Symbols of Operations, the Sign + plus, (7). — The Sign — minus, (8). — The Sign X into— A point (.) betw^een Quantities — Quantities in juxtaposition^ (9). — The sign ^ by — Division otherwise denoted — An Integral Quantityi (10). — Use of the Parenthesis or Vinculum, (11), — Factors and Constant Product, (12). — Powers of Quantities, (13). — Roots of Quantities, (14). — The Coefficient of a Quantity, (15). — The Exponent of a Quantity, (16). — What an Integral Coefficient indicates — an Integral Exponent, (17). — Simi- lar and Dissimilar Quantities, (18). — An Algebraic Monomial, (19). — An Algebraic Polynomial — A Binomial — A Trinomial, (20). — Whether the Value of a Polynomial is affected by changing the Order of its Terms, (21). — A Polynomial arranged by powers^ (22). — A Homogeneous Polynomial — Dimensions and Degrees of the Terms of a Polynomial, (23). — Positive and Negative Quantities, (24). — Effect of a Negative Quantity in an Expression or a Calculation — Occasional Use of the Positive and Negative Signs, (25). yJ analysis op contents. CHAPTER II. Addition. — Subtraction. — Multiplication. — Division. — 9 ... 26. Algebraic Addition, (26). — How to Add Similar Terms with like signs j (27). — Sum of Two Equal Similar Terms with cordrary signs, (28). — How to Add unequal Similar Terms with contrary signs, (29). — Sum of Two oi more Dissimilar Terms, (30). Rule I. For the Addition of Algebraic Quantities^ (31). Calculations on the same Polynomial in two or more ( )'s, (32). Algebraic Subtraction. (33). — How to Subtract a Monomial from another Quantity, (34). — How to Subtract a Monomial from a Dissimilar Quantity, (35). Rule H. For the Subtraction of Algebraic Quantities^ (36). How to denote the Subtraction of a negative Monomial — of a Polyno- mial, (37). — Change of signs in a Polynomial without affecting its Value, (38). Algebraic Multiplication — When the Multiplier is positive — When the Multiplier is negative^ (39). — Product of Two Monomials, (40). — Exponent^ in the Product, of a Letter occurring in both the Monomials multiplied to- gether, (41) — Sign of the Product — Reason for this when both the Quanti- ties are negative — When One of them is positive and the Other negative (42).— Product when Either of the Two Factors is 0, (43). Rule HI. To Multiply a Monomial into a Polynomialj (44). Rule IV. To Multiply a Polynomial into a Polynomial^ (45). Algebraic Division, (46). — How to find the Quotient of Two Monomials, (47). — Value of any Quantity with exponent 0, (48). — Sign of the Quotient, and Principle which determines it, (49) — Quotient of divided by any Quantity, and of any Quantity divided by 0, (50). Rule V. To Divide a Monomial into a Polynomial, (51). Rule VI. To Divide a Polynomial into a Polynomial, (52). How the indicated Product of Two or more Factors may be divided, (53). ANALYSIS OF CONTENTS. ^j CHAPTER III. Composite Quantities. — Common Measure. — CoxMmon Multiple. — 27 ... 40. A Composite and a Prime Quantity, (54). — DecompoBition of a Quantity — Divisor and Quotient as Factors of a Quantity, {55). — Wliat the Differ- ence of Two Quantities will divide, (56). — What the sum of two Quantities will divide, (57). — Product of the Sum and Difference of two Quantities, (58). — Square of the Sum of two Quantities, (59). — Square of the Difference of two Quantities, (60). — Case in which a Trinomial may be resolved into Two unequal Binomial Factors, (61.) — One Quantity a Measure of another — A Common Measure of Two or more Quantities, (62). — Greatest Common Measure of Two or More Quantities, (63). — Composition of the Greatest Common Measure, (64). — Principle on which depends the Rule for the Greatest Common Measure, (65). Rule VII. — To Find the Greatest Common Measure of Two Quantities^ (66). Whether the Signs in a Common Measure may be changed, (67). — Of a Factor which is common to all the Terms of the Dividend, (68). — Of a Factor which is contained in all the Terms of the two Polynomials, (69). — One Quantity a Multiple of another. — A Common Multiple of Two or more Quantities, (70). — Least Common Multiple of Two or More Quantities, (71). — Composition of the Least Common Multiple, (72). — Relation of Least Common Multiple to Product and Greatest Common Measure, (73). Rule VIII. To Find the Least Common Multiple of Two or More Quantities f (74). CHAPTER IV. Fractions. — 41 ... 66. An Algebraic Fraction, (75). — Of a Quantity with & negative Exponent, (76) — Two Methods of representing the Quotient when the Divisor is not a Factor of the Dividend, (77). — How any Factor may be transferred from the yjii ANALYSIS OF CONTENTS. Numerator to the Denominator, and vice versa, (78). — Reciprocal of a Quantity, (79). — Reciprocal of a Fraction, (80). — Constant Value of a Fraction, (81). — \yhen a Fraction is positive, and when negative, (82). — Signification of -4- or — prefixed to a Fraction, (83). — Changes of Signs without affecting the Value of a Fraction — How a Polynomial may be changed from Positive to Negative, (84). — A Fraction reduced to lower Terras — How a binomial Common Measure may often be discovered, (85). Rule IX. To Reduce a Fraction to its Lowest Terms, (86). When Two or more Fractions are said to have a Common Denominator. —How Fractions may be reduced, mentally, to a Common Denominator, (87). Rule X. To Reduce Two or More Fractions to a Common Denominator, (88). An Integral Quantity, (89). — A Mixed Quantity, (90). — An Improper Fraction, (91). Rule XL To Reduce an Integral or a Mixed Quantity to an Improper Fraction, (92). Rule XIL To Reduce an Improper Fraction to an Integral or a Mixed Quantity, (93). By what means the Sum of Two or More Fractions is found, (94). Rule XIII. For the Addition of Fractions, (95). By what means the Difference of Two Fractions is found, (96). Rule XIV. For the Subtraction of Fractions, (97). Product of Two or more Fractions, (98). — Effect of Multiplying by t> Fraction, (99). — Compound Fractions, (100). — Equivalent of Multiplying Two or more Fractions together, (101). Rule XV. For the Multiplication of Fractions, (102). How a Fraction is Multiplied by its own Denominator, and what Can- eellations may be made in the Multiplication of Fractions, (103). — Quotient of Two Fractions, (104) — Complex or Mixed Fractions, (105). Rule XVI. For the Division of Fractions, (106). ANALYSIS OF CONTENTS. Jjj CHAPTER ¥. Simple Equations. — 67 ... 90. An Equation— The First Member—The Second Member, (107).— For what purposes Equations are employed — Ho'w applied to the Solution of Questions, (108). — The Solution of an Equation — Verification of the Value found for the Unknown Quantity, (109). — A Simple Equation — A Quadratic Equation — A Cubic Equation, (110). — A Numerical Equation — A Literal Equation — An Identical Equation, (ill). — Transformation of an Equation, (112). — An Axiom — Axiom first, second, &c., (113). — How the Value of the Unknown Quantity is found — Transformations necessary, (114). — How to clear an equation of Fractions — How by means of Least Common Multiple — Advantage of this Method, (1 1 5). — How any term may be Transposed from one Side of an Equation to the other, (116). — Change of the Signs in an Equation, (117). Rule XVH. For the Solution of a Simple Equation containing but one unknown quantity, (118.) A Problem, and in what its Solution consists — General Method of form- ing the Equation of a Problem, (119). — Solution of Problems with Two or more Unknown Quantities — Independent Equations, (120). — General Method of solving Two Equations, (121). — Elimination by Addition or Subtraction, (122). — Elimination by Substitution, (123). — Elimination by Comparison, (124).— Solution of Three Equations— Of Four Equations, (125).— Of Problems in which there are Three or more Required Quantities, (126). CHAPTER VI. Ratio. — Proportion. — Variation. — 91. ... 110. Ratio of one Quantity to another, (127). — -Sign of Ratio, (128). — How the value of a Ratio may be represented, (129). — Direct and Inverse Ratio, (130). — Compound Ratio, (131). — Ratio of the first to the last of any Num- ber of Quantities, (132). — Duplicate and Triplicate Ratios, (133). — Equi- multiples and Equisubmultiples, (134). — Ratio of Equimultiples and Equi- submultiples, (135). — Proportion, (136). — Four Quantities in Proportion, (137). — Three quantities in Proportion, (138). — Direct and Inverse Proper- ^ ANALYSIS OF CONTENTS. tion, (139). — Sign of Proportioiij (140). — Inverse Converted into Direct Pro portion, (141). — Variation — Variation direct — Variation inverse, (142). — Product of Two Quantities varying inversely with each other, (143). — Varia- tion, an Abbreviated Proportion, (144). — A Theorem — A Corollary, (145). — Ratio of two Fractions having a common Term, (146). — How the value of a Fraction varies, (147). — Corresponding Equalties and Inequalities between the Antecedents and Consequents of a Proportion, (148). — A Proportion con- verted into an Equation, (149). — A Fourth Proportional, how found, (150). — Product of the Extremes when Three 'Quantities are in Proportion, (151). — Mean Proportional, how found, (152). — An Equation converted into a Pro- portion, (153). — Ratio of the j^r^^ to the third of Three Proportional Quanti- ties, (154). — Proportion by Inversion, (155). — Proportion by Alternation, (156). What Multiplications may be made in a Proportion, (157). — What Divisions may be made in a Proportion, (158). — Proportion by Composition, (159). — Proportion by Division, (160). — Proportion between the Sum of two or more Antecedents and that of their Consequents, (161). — A Proportion derived from Two other Proportions in which there are common Terms, (162). — Proportion between the Sums and Differences of the Antecedents and Consequents, (163). — Products of the Corresponding Terms of Two or more Proportions, (164). — Proportion between the Powers or Roots of Pro- portional Quantities, (165). — Substitution of Factors in a Proportion, (166). — General Solution of a Problem, (167). — Two Numbers found from their Sum and Difference, (l68). — An Algebraic Formula, (169). — Of a Propor- tion occurring in the Solution of a Problem, (170). — Percentage — Ratio of Percentage — Basis of Percentage, (171). — Amount of Percentage, how found, (172). — Interest — The Principal — The Amount, (173). — Amount of Interest, how found, (174). CHAPTER VII Arithmetical, Harmonical, and Geometrical Progression. — 111. . . 120. An Arithmetical Progression, (175). — Last Term of an Arithmetical Progression, equal to what, (176). — Common Difference of the Terms, (177). — Sum of the two Extremes, (178). — Arithmetical Mean, (179).— Sum of all the Terms, (180). — Formulas in Arithmetical Progression, (181). — An Harmonical Progression, (182). — An Harmonical converted into an Arith- ANALYSIS OF CONTENTS. XI metical Progression, (183). — Harmonical Mean, (184). — A Georaeterical Pro- gression, (185). — Last Term of a Geomitrical Progression, (186). — Power of the rati(f found from the First and last Terms, (187). — Product of the two Extremes, (188). — Geometrical Mean, (189).— Sum of all tlie Terms, (190) — Sum of an infinite number of Decreasing Terms, (191). — Formulas in Ge- ometrical Progression, (192). CHAPTER VIII. Permutations and Combinations. — Involution. — ^Binoaiial. — Theorem. — Evolution. — 121. ... 146. Permutations, (193). — Number of Permutations, how found, (194). — Combinations, (195). — Number of Combinations, how found, (196). — Invo- lution, (197). — A Higher Power found from lower Powers of the same Quan- tity, (198.) — Powers of unity, (199). — Powers of Monomials, how found. (200).— Powers of Fractions, how found — Powers of a mixed Quantity, (201). Sign to be Prefixed to a Power, (202). — Powers of Binomials, or of any Po- lynomials, (203). — Binomial Theorem — Exponents in any Power of (a±6) — Coefficients — Signs, (204). — Formula for the development of (a+6") — At what Term the development will terminate, (205). — Evolution — Extract- ing the Square Root, in what it consists — The Cube Root, (206). — Roots of vmity, (207). — Roots of Monomials, how found, (208). — Roots of Fractions, how found — Roots of a mixed Quantity, (209). — Of a Root whose Exponent is resolvable into two Factors, (210). — What denoted by the Numerator and Denominator of a Fractional Exponent, (2 1 1) . — Root of a power of a Quantity, (212). — Equivalent Exponents, (213). — Sign to be Prefixed to an odd Root of a Quantity, (214). — Sign to be Prefixed to an even Root, (215). — Of an even Vx^ooi oi Q. negative Quantity, (216). — How the Roots of Polynomials may be discovered, (217). Rule XVIII. To Extract the Square Root of a Polynomial. (218.) Principle for determining the Number of Figures in the Square Root of a Number. (219). — Square of any two Parts into which a number may be di- vided, (220). — Periods to be formed in Extracting the Square Root of a Deci- imal Fraction —Why the last Period must be complete — Number of Decimal Figures to be made in the Root, (221). X:ii ANALYSIS OF CONTENTS. Rule XIX. To Extract the Cube Root of a Polynomial, (222.) Principle for determining the Number of Figures in the Cuhe Root of a Number, (223). — Cube of any two Parts into which a Number may be di- vided, (224). Rule XX. To Extract the Cube Root of a Number, (225). Periods to be formed in extracting the Cube Root of a Decimal Fractiwi — Why the last period must be Complete — Number of decimal Figures to be made in the Root, (226). — How any Root whatever of a Polynomial might be extracted, (227). Rule XXI. To Extract any Root of a Polynomial, (228). CHAPTER IX. Irrational or Surd Quantities. — Imaginary Quantities. — 147. . .166. Perfect and Imperfect Powers, (229). — A Rational Quantity — An Irrationa.. or Surd Quantity — Radical Quantities, (230). — Radical Sign — How this Sign may always be superseded, (231). — Similar and Dissimilar Surds,(232). How a Rational Quantity may be expressed und;: ;; CHAPTER I. PRELIMINARY DEFINITIONS AND EXERCISES. Science and Art. (1.) Science is knowledge reduced to a system. — Art is know- ledge applied to practical purposes. The Rules of Art are founded on the Principles of Science. Numbers. — Quantity/, (2.) A UNIT is any thing regarded simply as one; and numbers are repetitions of a unit. Thus the numl)ers tivo, three, &c., are repetitions of the unit one. (3.) CluANTiTY is any thing which admits of being measured. Thus a line is a quantity, and we express its measure in saying it is so many feet or inches long. Time, weight, and distance are also quantities. Numbers are quantities; for every number expresses the measure of itself in units ; and numbers are used to express the measures of all other quantities. Thus we express the measure of Time by a number oi days, hours, &c Mathematics. (4.) Mathematics is the science of quantity. Its most general divisions are Arithmetic and Geometry. 2 PRELIMINARY DEFINITIONS AND EXERCISES. Arithmetic is the science of numbers; or, when practically ap plied, the art of Calculation. Geometry is the science which treats of Extension — in lengthy breadth, and height, depth, or thickness. , (Siy-'AmEBRA i^' a method of investigating the relations of both Arithmetical and Geometrical quantities, by signs or symbols. Symbols of Quantities, (6.) Quantities are represented in Algebra by letters, — known quantities usually by the first, and unknoion or required quantities by the last letters of the Alphabet. Thus the quantity a or b, that is, the quantity represented by a oi b, will generally be understood as known in value ; while the quan- tity X ox y will be unknown or required. Cluantities represented by letters are called literal quantities, in contradistinction to numbers or numerical quantities. Symbols of Operations. (7.) The sign + plus prefixed to a quantity, denotes that the quan tity is to be added, or taken additively. Thus a-{-b, a plus b, denotes that the quantity b is to be added tc the quantity a. (8.) The sign — minus prefixed to a quantity, denotes that the quantity is to be subtracted, or taken subtr actively. Thus a—b, a minus b, denotes that the quantity b is to be sub- tracted from the quantity a. (9.) The sign X into between two quantities, denotes that the two quantities arc to be multiplied together. Thus axb, a into b, denotes that the two quantities a and b are to be multiplied together. A point (.) between two literal quantities, or a numerical and 3 literal quantity, also denotes that the two quantities are to be multi- plied together. a.b denotes a into b, and 3. a denotes 3 into a, or 3 times a. The Product oi several numbers may be denoted by points between them ; thus 1.2.3 denotes 1 into 2 into 3, the same as 1x2x3. PRELIMINARY DEFINITIONS AND EXERCISES. 3 Quantities in juxtaposition, without any sign between them, are to be multiplied together. Thus ab denotes a and b multiplied toge- ther ; and axy denotes a, x, and y multiplied together. (10.) The sign ^ by between two quantities, denotes that the quantity before the sign is to be divided by the one after it. Thus a-^b, a hy h, denotes that a is to be divided by b. Division is also denoted by placing the dividend over the divisor^ with a line between them, after the manner of a Fraction ; - denotes a divided by b, the same as a-f-S. An integral quantity, in Algebra, is one which does not express any operation in division, whatever may be the numerical values which the letters represent. (11.) A parenthesis ( ) enclosing an algebraic expression, or a vinculum drawn over it, connects the value of that expression with the sign which immediately precedes or follows it. Thus {a-\-b).c, or {a-\-b)c, a plus 6 in a parenthesis into c, denotes that the sum of a and b is to be multiplied into c. The same thing would be denoted by a-{-bxc, a plus b under a vinculum into c. — In a-\-bc, only b would be multiplied into c. The vinculum, and the expression affected by it, are sometimes set vertically. Thus a +b —c X is equivalent to {a-i-b—c)x or a+b—cxx. PRELIMINARY EXERCISES. In the elementary oral Exercises which are occasionally inserted, the Student should write down the quantities as they are read to him. m^ Suppose the letters a, b, c to represent the numbers 3, 4, 5, respectively; then what is the numerical value of a+d — c1 - What is the value of ab-{-c ? Of abc—bc-\-a ? What is the value oi ac^bl Oi ab -\- b —ac ? What is the value o^bc^al Of abc-\-ac—bc ? What is the value of {a-\-b) c, a plus 5 in a parenthesis into c? What is the value of {ab-{-c) a ? 0^a-\-b+ c) c What is the value of ( a +b)-^c ? Of {ab-\-b—c) a ? What is the value of (bc—a)-^b ? Of {bc—a-\-b) b ? The Teacher may propose other Exercises of the same nature, should he deem it necessary. 4 PRELIMINARY DEFINITIONS AN,P EXERCISES. Factors. — Constant Product. (12.) Two or more quantities multiplied together, are called the factors of their product ; and the Product is the same i?i value, in whatever order its factors are taken. Thus a and x are the factors of the product ax ; and this product is the same in value as xa. So a, b, and c are the factors of the product abc, or acb, or bca, &c. It is most convenient to set literal factors according to the order of the same letters in the Alphabet ; thus ax ; abc. To understand why the product ax is equal to xa, consider a and x as representing 7iumbers, and that the product of two numbers is the same, when either of them is made the multijMer. For example, 25 times 7 is equal to 7 times 25. For 25 times 7 must be 7 times as many as 25 times 1 , which is 25 ; that is, 25 times 7 is equal to 7 times 25. D^^ Prove that 14 times 9 is equal to 9 times 14. Prove that 31 times 11 is equal to 11 times 31. Prove that 23 times 15 is equal to 15 times 23. Prove that 47 times 18 is equal to 18 times 47. Powers and Roots. (13.) The frst poiver of a quantity is the quantity itself; thus the first power of 5 is 5, and the first power of a is a. The second 'power, or square, of a quantity, is the product of the quantity midtiplied into itself. Thus the second power, or square, of 5 is 5 X 5, which is 25 ; and the second power of a is aa. The third power, or cube, of a quantity is the product of the quan- tity multiplied into its second power, or square. Thus the third power, or cube of 5 is 5Y.5y.5, which is 125 ; and the third power of a is aaa. IIP^ What is meant by the fourth power of a quantity ? What is meant by the fifth power of a quantity ? By the seventh power of a quantity ? What is the square of 3 ? The cube of 4 ? The fourth power of 2 ? The square of 7 ? The cube of 6 ? The fourth power of 10 ? ,^^' PRELIMINARY DEFINITIONS AND EXERCISES. 5 (14.) The second root, or square root, of a quantity, is that quan- tity whose square is equal to the given quantity. Thus the square root of 9 is 3 ; and the square root of aa is a. The third root, or cuhe root, of a quantity, is that quantity whose third power, or cube, is equal to the given quantity. Thus the cube root of 8 is 2 ; and the cube root of aaa is a. [CF^ What is meant by the fourth root of a quantity ? What is meant by the ffth root of a quantity ? By the ninth root of a quantity ? What is the square root of 16 ? The cube root of 27 ? The fourth ;oot of 16 ? The square root of 81 ? The cube root of 1000 ? What is the square of the square root of 4 ? The cube of the cube root of 125 ? The square of the cube root of 64 ? The cube of the square root of 1 6 ? Coefficients and Exponents. (15.) The coefficient of a quantity is any multiplier prefixed to that quantity. — In a more general sense, the coefficient of a quantity is di,ny factor forming a product with that quantity. Thus, in 3a, 3 is the coefficient of a, and denotes 3 times a. In 5ax, 5 is the coefficient, denoting 5 times ax. In -Jrc, ^ is the coefficient of a;, and denotes one-half oi x. When no numerical coefficient is prefixed, a unit is always to be understood. Thus a is \a, once a, and ax is lax, once ax. (16.) The exponent of a quantity is an integer annexed to it, to denote a power, or a fraction annexed to denote a root, of that quantity. Thus a"^, a with exponent 2, denotes the second power, or square, of a ; x^ denotes the third power, or cube, of x ; and so on. The fractional exponents ^, -J, ^, and so on, denote, respectively, the square root, cube root, &c., of the quantity to which they are annexed. a", a with exponent -J-, denotes the square root of a ; cfi denotes the cube root of a; ; and so on. When no exponent is annexed to a quantity, a unit is alvra-ys to be understood. Thus a is a^, the first power of a. (13.) 5' PRELIMINARY DEFINITIONS AND EXERCISES. An exponent is assigned to the product of two or more factors, by affecting such product with a parenthesis, or a vinculum, and the exponent. Thus (ax)^ or ax^, ax in a parenthesis, or under a vinculum, with exponent 2, denotes the square of the product ax ; whereas ax^ denotes a into the square of x. (17.) An integral coefficient indicates the repeated addition of a quantity to itself; while an integral exponent indicates the repeated multiplication of a quantity into itself. Thus 3? (28.) Two equal similar terms with contrary signs, when added together, mutually cancel each other; that is, their sum is 0. Thus 3a — 3a is ; that is, is the simplest expression for the value of 3a — 3a, and is therefore the sum of the two terms. So 5a-\-x—5a is equal to x; for 5a and --5a cancel each other, and X is therefore the sum of the three given terms. ] ADDITION. D;#^ What is the Sum of 2ax and —2ax, that is, the simplest ex- pression for the value of 2ax—2ax'l What is the Sum of 7b and -76? Of a, 3x, and -a? Of Sy^, -3y^, and 5b^ ? Of ax^, 5, and — 5 ? Of —abc, -\-a^x, and abc ? (29.) Two unequal similar terms with contrary signs, are added together by taking the difference of their coefficients, annexing the common literal factor, and prefixing the sign of the greater term. Thus la—Za is 4a; just as 7 cents — 3 cents is 4 cents ; that is, the sum of la and —3a is 4a. And 3a— la is —4a. For —7a is equal to —Sa—Aa, (27); then 3a — 7a is equal to 3a — 3a — 4a; 3a cancels — 3a, and leaves — 4a. (28). ITF^ What is the Sum of 5ab and —Sab ? Of 3a2c and —a^c ? Of 9ax^ and -4ax^ ? Of -a^x and 3a^x ? Of 5ac and -5ac ? What is the simplest expression for the value of 2b— 5b, that is, the Sum of 2b and —5b1 How do you reason in finding that sum 7 What is the Sum of 3a^ and — Sa^ ? and how do you reason in finding it? What is the Sum of ax"^ and —2ax'^ ? and how do you reason in finding it? What is the Sum of — lla^ and Sa^ ? and how do you reason in finding it ? (30.) The Sum of two or more dissimilar terms can only be indi- cated by arranging them in a Polynomial, so that each term shall have its given sign prejixed to it. Thus the Sum of ax and bc^ can only be indicated as ax-\-bc'^ ; and the Sum of aa: and —bc^ as ax—bc^. What is the Sum of 3a and 5b ? Of 5a and -h ? Of 3a2 and 2*2 ? Of ax and by ? Of -36 and -Ay ? What is the Sum of 3b, 5b, and 6c? Of a^, bx, and 3a2 ? Of 4c and -3.y ? Of —ab, 3xy, and —2c ? Of 3^2, 4b, and 2c ? Of 5a, — 5, and 5b ? The preceding principles, and the following Rule, provide for all the cases in Algebraic Addition. ADDITION. 11 RULE I. (3L) For the Addition of Algebraic Quantities. 1 . Find the positive and the negative Sum of similar terms, sepa rately, (27), and then add together the similar sums, (28) and (29). 2. Connect the results thus found, and the dissimilar terms, in a Polynomial, prefixing to each term its proper sign. (30). EXAMPLE. To add together ^a'^-\-2hc—xy, 2a^—3bc-{-5y, bc—a^-\-3xy, and cy—2a^ — 5bc. Aa^ -\-2hc—xy 2a^-3bc-\-5y —a^-\- bc-^3xy —2a^—5bc-\-cy 3a^—5bc-\-2xy+cy+5y The sum of the positive terms 2a ^ and 4a ^ is 6a^, and the sum of the similar negative terms —2a^ and — a^ is 3^2^ ^27). Adding together these two similar sums, 6^2 and — Za^, the result is — Sa^, (29). The sum of the positive terms be and 25c is 3bc, and the sum of the similar negative terms — 5bc and — 3bc is — 8bc. Adding together these two similar sums, 3bc and —8bc, the result is —5bc. The sum of the similar terms 3xy and —xy is 2xy. The results thus found, namely, 3a^, —5bc, and 2xy, and the dissimilar terms cy and 5y, are connected in a Polynomial. EXE RCISE s. 1. Add together ab-\-3c^—2x, 3ab—c^ + 5x, 5c^—2ab+y, and 4ab—c^-\-x. Ans. &ab-{-^c^-{-4:X-{-y. 2. Add together 4^2 — 66+3, 3a^ + 65— 7, 5b+a^—bc, and a''^ — bc-\- 10. Ans. 9a^ + 5b-^6 — 2bc. 3. Add together 2b^-\-bc+x, 3b^—2bc+y, bc-3b^ + 5x, and b^+bc+3y. Ans. 3b^-{-bc-\-6x+4y. 4. Add together 5a—6b^ + 3, 2b^—Aa—l, 7a—b^+c, and a-f 362 + 2C+4. Ans. 9a-5b^ + 3c+3b^. 5. Add togeihei a^b —50+ xy, 4a^b-{-8c—xy, aH+c-\-3xy, and aa+Sc-lS. Ans, ^aH-^lc-\-3xy-\'a^-l3. 12 ADDITION. 6. Add together a3 + 2a2— 33c, 2a^—a^-\-bc, —a^—d^+h^, and 3a3 + 3a2 + 362. Ans. 5a^-{-3a^—2bc+4b^. 7. Add together 2z+ab^-37j^, 2ab^-3x+y^, -ab^-{-5x-y^, and 5x-\-ab^—y. Ans. 9x-\-Sab^ —3y^ —y. 8. Addtogether 4a— 352+ca;2, b^—3a+3, ,2a-i-3b^—cx^, smd 2b^ — 2a—9, Ans. a + 3b^ — e. 9. Add together?^— 2^2—2:2/, 5^2 _66+3a;y, b — 3a^-{-4, and a^—b—3xy. Ans. b+a^—xy-\-A. 10. Add together 3b^-2a^ + 13, 3a^-2b^-5, 4ab+7a^-3, and 262— a3+«3. , Ans. 3b^-\-7a^-{-5 + 5ab. 11. Add together aa;2—2?/4-5, 22/+2aa;2— 33, Aax^—y—b, and 26— 3aa:2 4-3. .Am. 4aa:2— «/. 12. Add together 2c^-^a^ + 3bc, 5c^-3a^-2bc, c^ + 2a^-bc, and b-\-bc—3. Ans. ec^-\-bc+b—3. 13. Add together ab+a^c—5, '3ab—3a^c-^7, 2a^c-2ab—3, and ab-{-a^c-\-5. Ans. 3ab-\-a^c-\-4:. 14. Add together 3b''—2a^x-{-b, —b^ + 3a^x—3b, b^—a^x-\-c, and 332+6— 3c. ^?zs.- 662— 6 — 2c. 15. Add together 2a24-3— ac, 3a2_74-ac, 3ac—5a^-}-9, and 2ac+4:—a^. Ans —a^ + 9 + 6ac. 16. Add together b^c+2-y^, y^-3b^c-10, 2b^c-3-\-2y^, and 62—3/2 + 5. Ans. —6-\-y^-\-b^. 17. Add together a3+6c— Jc, 2^3- 6c+fc, 3a3 + 36c— :^c, and a3+6c+c. Ans. 7a^+4bc-\-c. 18. Add together 63-3a2+|c2, 262 + 5^2+^2^ ^,2_^c+2, and 562 + 3a2_2c+32/. Ans. 63 + 5a2 + lic2 + &62-2^+2+33^. (32i) When the same Polynomial is enclosed in two or more paren- theses,. (11), this polynomial enters into a calculation iu the same man- ner as the common factor of similar monomials. Thus the Sum of 3(a+6) and 5(a+6), is evidently 8(a+6). 19. Add together 2(a— 36) + c, 3(rt-36)— 3c, — 4(a— 36)+c, and 5(a— 36)+3c. A?is. 6(a— 36) + 2c. 20. Add together 3a2 + 2(a— c2), a2_3(^_c2)^ 4a2 + 5(a_c2) and — 7a2 + (a— c2). Ans. a^-\-5{a—c^). 21. Add together 5(a+6-c)+362, 3(a+6-c)-262, 2(a+6-c) and 362 + 2?/, +62. Ans. 10(a+b—c)-{-5b^ + 2y. 22. Add together 2a62 + 3a(6+c2), ab^-2a{b+c^), a(6+c2), and -a62-3a(6+c2). Ans. 2ab^-a(b-{-c^). 23. Add together 5+i(c— c?+w)+26, l-\-^(c—d+m)-b-\-2, nn^ 2(c-d-\-m)+ib. A?is. 8+3J(c-(/+w) + li6. SUBTKAOTIOM. 13 SUBTRACTION. (33.) Algebraic subtraction consists in finding the difference between the two quantities ; that is, in finding what quantity added to the quantity subtracted, will produce that from which the subtrac- tion is made. Thus the difference found by subtracting 3c— 2 + ?/. Ans. a-{-4:—y. 12. From the Sum of az^ — 6b^-i-3ay^, and 2ax^ + eb^ —ay^, subtract —3az^-{-b^-i-2ay^—5. Ans. Qax^—b^ + d. 13. From the Sum of 3ab-\-bc—5x^, and a5—3^c+a;2— 3, sub- tract 5Z'C—3a;2— 4+2/2. Ans. 4:ab—7bc—X2-{-l—y^. 14^From the Sum oi 5x—3xy-{-7y^, and 5a;?/— 3a;— 4y2_ 9, sub- tract —x-\-xy—3y^-{-7. Ans. 3x-{-xy-{-6y^ — 16. 15. From the Sum oi2a—b-\-3cd, 5a-{-3b—cd—3, and a-\-2b — cd, subtract 4a—5b—cd. Ans. 4a+9^ + 2c(i— 3. 16. From the Sum of 3x^+y—5xy, 2x^—3y-{-xy, and 4:y—3x^ —xy+5,&\ihtv3icix^-i-5y—xy—3. A?ts. x^—3y—4:xy+8. 17. From the Sum of 2ay^ '^2b-{-3ax, 2ay^-\-b—ax, and 3i— ai/2 + 3— y, subtract 3ay^ +4:b—ax-^y. Ans. —2b-\-3ax-{-3^2y. 18. Fi;pm the Sum of 3a^ -\- xy"^ ^2by , 5a^ -{-3xy^ —3by, and 3xy^-\-4:a^-\-by, subtract 2a^ —xy^ —5by-{-5. Ans. \0a^-\-8xy^-\-by—5. 19. From the Sum of a^b^-3y^-\-5xy, 3a^b^ + 3y^ -2xy, and 5y^-]-3a^b^—xy-}-6, subtract a^b^-i-xy—y^ —3. Ans. 6a^b^-{-6y^-\-xy-{-9. 20. From the Sum of 5y^~3ax—2bc, 4:ax—2y^-i-5bc, and 3ax-\-y^—bc-^5, subtract ax—bc-\-3y^ — l. Ans. y^-{-3ax+3bc+6. 21. From the Sum of a^y—x^-]-bc—d, 3x^—3bc-{-5d, and a^y-\-3x^—4.bc—d, subtract 5x^—2bc-\-7d. Ans. 2a^y—4:bc—4d. 22. From the Sum of 5b^ + 3ax+2y, 3b^—ax—y-{-l, and 4a« — 62 _52^_^ 2, subtract 3b^-\-2ax—y+10. Ans. 4:b^-{-4:ax—3y—7. 23. From the Sum of 7a+6x^ —y^-{-h x^-{-3y^—c—3, and 7x^+y^-3c+5, subtract 5a+2a;2-2?/2 + 5c+2. Ans. 2a+12aj2 + 5?/2— 9c+l. 24. From the Sum of 2 {a—x)-\-y, 3{a—x)-\-2y, and 6{a—x)—y^ subtract 4{a—x)-{-y—2, (32). Ans. ^{a—x)-\-y+2. 25. From the Sum of 25(a+c)2 + 3, i(a+c)2 — 1, and 3b{a-\-cy ~2, subtract the sum of ^(a 4-^)2 -j-^» and 5b(a-{-cY —h. Ans. 0. 15 ADDITION AND SUBTRACTION. • (37.) It is sometimes expedient merely to indicate the subtraction of a quantity, without performing the operation. To denote the subtraction of a positive monomial, nothing more is necessary than to place the sign — before it ; thus a—b denotes that b, that is, positive b, is to be subtracted from a. The subtraction of a negative monomial, will be denoted by en- closing the quantity, with its negative sign, in a ( ), and prefixing the sign — to the parenthesis. Thus a—{—b) denotes that negative b is to be subtracted from a. When the subtraction is performed, the expression becom|^ a-\-b. The subtraction of a poly7iomial will be denoted by enclosing the polynomial in a ( ), and prefixing the sign — to the parenthesis. Thus a—{b-\-c-\-d)] in which the sign — denotes that the en- closed polynomial js to be subtracted. When the subtraction is performed, the expression becomes a—b—c—d. (38.) The value of a Polynomial is not afiected by changing the signs of any or all of its terms, enclosing those terms in a ( ), and prefixing the sign — to the parenthesis. Thus a+i— c is equivalent to a—(—b-\-c)yOxa—(c—h)', for if (c—b) be subtracted, as is required by the sign — prefixed to it, the result will be a+Z>— c. 26. Under what difierent forms may the value of the polynomial db -\- 2cd —3x-\-5 be expressed ? 27. Under what different forms may the value of the polynomial ax—3y-{-4:b^—5c—7 be expressed? MULTIPLICATION. j -7 MULTIPLICATION. (39.) Algebraic multiplication consists in finding the Product of one quantity taken as many times, additively, or subtr actively, as there are units in another quantity. The quantity to be multiplied is called the multiplicand, and the multiplying quantity the multiplier: both together are called the factors of the product, (12). When the Multiplier is positive, the multiplicand is repeated positively, or is repeatedly added. Thus 5a X 3, 5a multiplied hy positive 3, is 5a-\-5a-{-5a, equal to 15a ; the multiplicand 5a being repeatedly added. When the Multij)lier is negative, the multiplicand is repeated negatively, or is repeatedly subtracted. Thus 5a X— 3, 5a multiplied by negative 3, is ~5a—5a—5a, which is equal to —15a; the multiplicand 5a being repeatedly swi- tracted Multiplication of Monomials. (40.) The Product of two monomials consists of the product of their coefficients multiplied into all their literal factors. For example, 3ac X 2a; is equal to ^acx ; for the factors may be taken in the order 3 X 2acx, which becomes 6aca; by substituting 6 for 3x2. dp* What is the Product of 3a X 4Z> ? and how do you reason in finding that product? What is the Product of 5a2 5x2a;? and how do you reason in finding it ? Of lac"^ Xxyl Of xy^ x 5 ? Of 3 x 7a2Z»2 ? (41.) When the same letter occurs in both the monomials multi- plied together, its exponent in the Product will be the sum of its expo- nents in the two factors. Thus Sa^xx2ax, or 3a2.T^ x2a^a:^, is equal to 6a^axx, (40), and this product becomes Ga^x^ by substituting a^ and x^ for their equiva- lents a^a and xx, (13) and (16). 2 18 MULTIPLICATION. Observe that the exponents of a and x in the product ^a^x^, are the 5um^ of the exponents of the same letters, respectively, in the two monomials which are multiplied together. []^=- What is the Product of 4^3 X2a2 ? Of 3axx5aa:2 ? Of Iax2a^b1 Of 4acxac6? Oi a^b^x^bl Of Qa^chxcl Of anxaH^I Sign of the Product. (42.) The Product of two quantities is positive when they both have the same sign, and negative when they have contrary signs ; — in other words, like sig7is produce +, and U7ilike signs produce — , in multiplying. When both the quantities are positive, their product is evidently positive, — as in common Arithmetic. Thus 3a X 2 is 3a+3a, equal to 6a. When both the quantities are negative, their product is positive, because such product results from repeatedly subtracting a negative quantity, (36). For example, in — 3ax —2 the multiplier —2 denotes that —3a is to be taken twice subtractively ; and since the sign of a quantity is changed in subtracting it, the product is 3a+3a, or 6a. When one of the quantities is 'positive and the other negative^ iheir product is negative because it results from repeatedly subtracting a positive, or adding a negative quantity. Thus in 3aX — 2, the 3a is to be taken twice subtractively, and the product is therefore —3a— 3a, equal to —6a. LkF* What is the Product of — 2a; X — 3 ? and how do you reason on the sign oi the product? Of — 3a5x— 2? Of — Sa^x— 3? Of-a;X-5? What is the Product of 5ax X — 4 ? and how do you reason on the szVti of the product? Of4a2cX— 3? Of 7a;2 x — 5 ? Oib^x—^, , What is the Product of —2ab X 3 ? and how do you reason on the s/o-Tz of the product? Of— 3a2a;x4? Of— 3a2x2? Of— c^xSJ? What is the Product of 4rc2 x — 7 ? and how do you reason on the sign of the product 1 Of 6ay X — 3 ? Of — 3a X —4a 1 Of 66 X-5? MULTIPLICATION. 19 When the Multiplier or the Multiplicand is 0. (43.) When either of the two factors multiplied together is 0, the product will be ; for it is evident that repeated any number of times, produces only ; and the product is the same when the multi- plicand and the multiplier are taken ihe one for the other. Thus 5x0 is 0; axOisO; (a+J— c)xO is 0. The preceding principles and the two following rules, embrace the subject of Algebraic Multiplication. RULE III. (44.) To Multiply a Monomial into a Polynomial. Find the product of the monomial into each term of the polynomial, separately, and connect these products in a polynomial with the proper sig}t prefixed to each, (40), (41), (42). EXAMPLE. To multiply Sab -\- bc^ —2xy— 5 hy 2a^x Sah-\-bc^—2xy—5 2a^x 6a^bx-\-2a^bc^x—Aa^x^y—\0a^x It will be most convenient to multiply from left to right. The first term of the multiplicand is positive, and the multiplier being also jiositive, the first term, ^a^bx, of the product i& positive. In like man- ner the second term, 2a^bc^, of the product is positive, &c. EXERCISES. 1. Multiply 2a2_3J-|-c— 2?/by 3. Ans. 6a2_95+3c-6y. 2. Multiply 3^2 —a-?/— 2/2 + 4 by 2. Ans. 6x^—2xy—2y^-^S. 3. Multiply 45+6-2— 3?/+ 1 by —y. Ans. —4J)y—c^y-\-Sy^—y 4. Multiply — a+36 — 2a;2 — 3 by ~2x. Ans. 2ax—^bx+^x'^ + Qx. 5. Multiplya52— ^»3_f_2c_y by —5. Ans. —5ab^ + 5b^ — 10c-{-5y. 6. Multiply —3 +ax — 1 + byhy —a. Ans. 3a—a^x-\-a—aby. 7. Multiply 7x+by^ — 4:C+d^ by 4. Ans. 28x+4.by^ — IQc+Ad^. 8. Multiply a2^_c2_^a._2 by 3a. A7is. Sa^b—Sac^ + Sax—Qa, 20 MULTIPLICATION. RULE IV. (45. ) To Multiply a Polynomial into a Polynomial. Multiply one of the polynomials by each term, separately, of th* other polynomial^ and add together the several products. EXAMPLE. To multiply 2a'^-{-^ac—c^ by 3a— 5c 2a^-}-4tac—c^ 3a —5c ea^-\-12a^c-3ac^ -10a2c-20ac2 + 5c3 6a3+ 2a26--23ac2 + 5c3. We multiply the first polynomial by 3a, and then by —5c, accord- ing to the preceding Rule. The two products thus obtained are set with similar terms one under another, and added together. The correctness of this Rule, as well as of the preceding one, will be evident upon considering, Jirst, that if each part of a quantity be multiplied, the whole will be multiplied ; and, secondly, that if one quantity be multiplied by each part of another quantity, the former will be multiplied by the whole of the latter. EXERCISES. 9. Multiply «2_2a+5 by a+3. Am. a^+a^—a+15. 10. Multiply 2:?;24_3^__i ]t)y x—5. Ans. 2x^-7x^ — lQx-\-5. 11. Multiply <22 — 2aa;+^^ by a-f-cc. Ans. a^—a^x—ax^-{-x^. 12. Multiply a:;2-|-3a:?/+ 2/2 by cc— 7/. Ans. x^-{-2x^y—2xy^—y^. 13. Multiply ^>2_35c—c2 by 25+c. Ans. 2b^ — 5b^c—5bc^—c^. 14. Multiply 2^2 + 3fl^c—c2 by a— 2c. Ans. 2a^—a^c—'7ac^-\-2c^. 15 Multiply 2b^—2bx+x^ by 2b— x. Ans. 4b^ — e>b^x-i-4.bx^—x^. 16 Multiply 63^_^2_j_^l3y^3^^2, jif^s. b'^ + 2b^-\-2b^-\-b^. 17 Multiply 0^2— 2c2+2?/by c2 + 2/. Ans. c^x^ — 2c^-{-x^y-\-2ij^. 18. M-xxlii^ly a'^-\-bx-\-yhy a^—bx. Ans. a^-\-a^y—b^x^—bxy. DIVISION. 21 19 Find the Product of (2a^-Aab-{-2b^){3a-2b). Ans. Q>a^ — I6a^b+Uab^ —U^. 20 Find the Product of {3x^ —2xy-{-2y^)(2x^ + 3xy). Ans. Gx"^ -{-5x^y—2x^y^ + 6xy^ . 21 Find the Product o^2a^ + 3ab-b-){a^ —ab-\-b^). Ans. 2ci*-\-a^b-2a^b^+Aab^-b^. 22 Find the Product of (3x^ —2xy-{-6)(x^ + 2xy—3) Ans. 3x^-\-^x^y—^x^ —ix^y^ -i-16xy— 15. 23. Find the Product of the three polynomials (a+b), if^—b), and {a^ [-ab^h^) , Ans. a^+g,^b-ab^-~b*. DIVISION. (46.) Algebraic division consists in finding a factor or (Quotient, which, multiplied into a given divisor, will produce a given dividend. Thus &a^x^2a, ^a'^x divided by 2a, gives the quotient 3ax, be- cause this factor, multiplied into the divisor 2a, produces the dividend Division of Monomials. (47.) The duotient of two monomials will be found by dividing the coefficient of the divisor into that of the dividend, and subtracting the exponents in the divisor from those of the same letters, respectively, in the dividend. Thus lQa'^x'^y-^5a'^x gives the quotient 2axy, because this quo- tient, multiplied into the divisor 5a^x, produces the dividend lOa^x^y, and 2axy is found by dividing 5 into 10, and diminishing the expo- nents of a and x in the dividend by the exponents of a and x in the divisor. The dividend being the product of the divisor and quotient, the exponents in the dividend are the sums of the exponents of the same letters in the divisor and quotient, (41); hence the exponents in the quotient will always be found by subtracting as above. HF^ What is the Gluotient of 6a^x^y~3ax1 and how do you prove that quotient true ? What is the Quotient of 4ab^c-^b ? and how do you prove that quotient true ? What is the Quotient of 8a^bx-^2a ? and how do you prove that quotient true 1 22 DIVISION. When any Exponent is reduced to 0. (48.) Any quantity whatever with exponent is equivalent to unity ; and a. factor with this exponent may therefore be canceled. For example, d^ -^a'^ gives the quotient a^, by subtracting the ex- ponent of the divisor from that of the dividend. But any quantity contains itself once, and therefore a^-^a^ also gives the quotient 1 ; and these two quotients, a^ and 1, must be equi- valent to each other. Since a in the preceding illustration may represent any quantity we please, any quantity whatever with exponent 0, is equivalent to 1, or is a symbol of unity. To divide lOaH by 5a^. Subtracting the exponent in the divisor from the exponent of a in the dividend, we find the quotient 2a^b. The factor a^ is equivalent to 1, and may therefore be canceled without affecting the value of the quotient. The quotient of 10a ^5 -f- 5a ^ may therefore be given under the two different forms 2a^b and 2b. I (1^^ Under what two different forms may the Cluotient of ^a?x -f-2a2be represented? Of8a62^262? Ofl0a5c-^5a? Of 9aa;2^ a:2 ? Of labc^-^lc^ 1 Of 5ax-^ax 1 Of Zahf-^ay^ 1 Of 10a:2 ^ a;2? Sign of the Quotient. (49.) The duotient of two quantities is positive when they both have the same sign, and negative when they have contrary signs ; — in other words, like signs produce +, and unlike signs produce — , in dividing. Thus -\-ax-^-\-a gives -\-x, because -\-aX -\-x is -\-ax ; —ax-. a gives -\-x, because —a X -\-x is —ax ; -\-ax-. a gives — x, because — aX — x is -\-ax', and — aa;-^+a gives — a:, because +aX— a? is — ax\ (42.) From these examples, it will be seen that the sign of the quotient must be such that the quotient multiplied into the divisor shall pro- duce the dividend. It thus appears that the nde for the sign of the quotient, is the same as that for the sign of the product, of two quantities. The learner must be careful not to apply this rule in finding the sum, or difference, of two quantities. DIVISION. 23 tt^ What is the Quotient of 4:a^oc^21 and how do you reason on the sign of the quotient? Oi6ac^-^2cl Oi 9x^y^3xy'i Of a^ -^a ? What is the Quotient of —7a^-. a^ 1 and how do you reason on the seV/i of the quotient? Of — 12c-2-i-— 4 ? Oi -a^b~—abl Of What is the Quotient of 8a^ -. a^ 1 and how do you reason on the sign of the quotient? Of 10ab-^-2b1 Of ^a^c^-al Of Da:^-^ 2 ? What is the Gluotient of —la^-^la 1 and how do you reason on the sign of the quotient? Of —I2ac^'^a1 Of —^a^b-^^1 Of -20«4-f-4? What is the Quotient of 10a^-:-5a? and why is that quotient true ? What is the Quotient of 12a;2 -, 3 ? and why is that quotient true? What is the (Quotient of —^a^by^2a^1 and why is that quotient true ? What is the Quotient of —20a^x^ -. 5ax ? and why is that quotient true ? What is the Quotient oi —lOOaxy-. xy '^ and why is that quotient true ? When the Dividend or the Divisor is 0. (50.) The quotient of divided by any quantity, is ; but the quotient of any quantity divided by 0, is infinitely great. First, let a represent any quantity we please ; then 0-^ a gives the quotient 0, because this quotient multiplied into the divisor a, produces the dividend 0, (43). Secondly, a divisor may be taken so small as to be contained a great number of times in any given dividend, represented by « ; if the divisor be still diminished, the quotient will be increased ; and if the divisor be diminished without limit, the quotient will be increased %vithout limit. If therefore the divisor were dimished to 0, the quotient would be unli^nited, that is, infinitely great. The character oo is the sign oi infinity ; we have then 0-^a equal to ; and an-O equal to oo, infinity. The preceding principles, and the two following Rules, embrace the subject of Algebraic Division, when the Quotient is an integral quantity. 24 DIVISION. RULE V. (51.) To Divide a Monomial into a Polynomial . Find the quotient of the monomial divided into each term of th* polynomial, separately, and connect these quotients in a polynomial, with the proper sign prefixed to each. (47), (48), (49). • EXAMPLE. To divide 20a^x—15So^ -{-30axy^ —Sax by 5ax. 5axJ20a^x—15ax^ + 30axy^ — 5ax 4a — 3a? + 6y' Dividing into the first term of the dividend, we find the quotient term 4a ; dividing into the second term of the dividend, we find tha quotient —3a: ; and so on, through the dividend. EXERCISES. 1. Divide 3a;3-f6a;2-f3aa;— 15a? by 3a;. Ans. x^-\-2x-{-a—5. 2. Bivide 2ab—6a^x+8a^y—2ahy 2a. Ans. b—3ax-^4:a^y—l 3. Divide 14a2—7ai+21«a?— 21a by 7a. Am. 2a—b-^3x~3. 4. Divide a^ x-\-3a^ x^ — 6ax^ -\-3axhy a. Ans. a^x-{-3ax^ —6x^-\-3x. 5. Divide 5b^~10b^ + 5b^y-\5b^ hj 5b. Ans. b-2b^-\-b^y-3b^. 6. Divide6a;2-f2.x3— 8ca;4+7a?5 by a?2. Ans. b-\-2x—Scx^-{-7x^. 7. Divide4a*— 8^3— 4a2i-[-8aby 2a. Ans. 2a^ —Aa^ —2ab+4., 8. DiYidQ ay ^ + a'^y^—a^ y'^— ay hy ay. Ans. y^-\-ay^ —-a'^y—l. 9. Divide —y-\-by^—5y^-{-3y^ by —y. Ans. l—by+5y^—3y^. 10. Divide 363-9^*2 + 125-15 by + 3. Ans. b^-3b^+4.b-5. 11. Divide — c^ + Sc*- Gc^+c^ by -c^. Ans. c^ —3c^-{-^c—l. 12. Divide 8?/^— 4a2/+a22/2— 3?/ by 3/. Ans. Sy^ —4:a-{-a'^y—3 13. Divide — 10 + 20a — 15a2 + 20 by —5. Ans. 2— 4a-f 3a2_4. 14. Divide a^b—a^b^ + a^b^—a^b hy ab. Ans. a^-~ab-{-a^b-~a^. 15. Divide x*y^-\-x^y^—x^y^—xy'byxy. Ans. x^y^-\-x^y^—xy—l. DIVISION. 25 RULE VI. (52.) To Divide a Polynomial into a Polynomial. 1 . Arrange the divisor and dividend according to the ascending or the descending ^oz^ers of the same letter, (22). 2. Divide the first term of the divisor into the Ji7'St term of the dividend ; multiply the whole divisor by the quotient term ; subtract the product from the dividend ; divide into the remainder, as before, and so on, — observing to connect the several quotient terms in a poly- nomial, with the proper sign prefixed to each. EXAMPLE. To divide <6a^x^'ia'^x-'—\a'^x^^x^ by 2ax-\-x'^ 2aa;-f a;2 ) ^a^x"^ ^'^a'^x^ —^a'^x^ ^x^ ( ^a'^x—2ax'^-\-x^ 6a4a;2+3a3a;3 — 4a2a:*4-a;^ —\.a'^x^ — 2ax^ 2ax^^x^ 2ax^-{-x^ The divisor and dividend are here arranged according to the ae* sccnding powers of «, or the ascending powers of x. The first term 2ax of the divisor, divided into the first term Qia^x'^ of the dividend, gives the quotient term oa'^x. Multiplying the whole divisor by this quotient term, and subtracting the product from the dividend, the remainder is —^a'^x^-\-x^. We next divide the first term of the divisor into the first term, — 4«2a?*, of the remainder, and find the quotient term — 2ax^ ; &c. By this Kule the divisor is multiplied by each fart of the quotient^ and the successive products subtracted from the dividend. When the dividend is thus exhausted, the product of the divisor and quotient is equal to the dividend ; and the quotient is thus proved to be correct. The Divisor, when a polynomial, is sometimes set on the left of the dividend, and the quotient under the divisor. This arrangement is thought to afibrd greater facility in multiplying the divisor by the quotient term. 26 DIVISION. EXERCISES. 16. Divide a^—2ax-{-x^ hy a— X. Ans. a—x. 17. Divide a;^— 3aa;2-|-3a2a;— a^ by a;— a. Ans. x^ — 2ax-\-a^. 18. Divide 6a^ + 9a^ — 15a by 3a^—3a. Ans. 2a^ + 2a-^5. 19. Divide a'* +«^^^+^* l^y «^—«^+^^- Ans. a^ -\-ax-\-x'^. 20. Divide 2a;3 — 19a;2 + 26a;— 16 by x—S. Ans. 2x^—3x+2. 21. Divide 4a;*— 5a?2c2+c* by 2a;2— 3a;c+c2. Ans. 2a;2 + 3a:c+c2. 22. Divide «* — 2a2a;2+a?* by «2_j.2aa;4-a;2. Ans. a^ — 2ax-[-x^. 23. Divide 12—4:a—3a^+a^ by 4-^2. ^^^s. 3_a. 24. Divide 4.7/^—9y^-i-ey—l by 2?/2 + 3i/— 1. Ans. 2y^—3y+l. 25. Divide 2a;4 —32 by a;— 2. Jl?zs. 2a:2 + 4a;2 4-8a;+16. 26. Divide 4a^— 64a by 2a— 4. Ans. 2a*-\-4:U^ + Sa^ + lGa. 27. Divide 6^6 _96^2 ^y 3?/— 6. Ans. 2y^+4:y'^-^8y^ + 16y^. 28. Divide a^+Ax* by a2_2aa;+2a;2. ^7^s. a2^2aa;+2a:2. 29. Divide a^ — x^ hy a^^a^x-\-a^x^-\-ax^-\-x*. Ans. a — x. 30. Divide«/4+43/22;2_322;4byy+22;. Ans.y^—2y^z-{-8i/z^ — 16z^. (53.) The indicated Product of two or more factors is divided by any quantity, when either of the factors is divided by that quantity. Thus to divide 3a'^x[x^-\-ij'^) by 3a:, we divide the factor 3a2a?, and find the quotient a^ix^-^-y^). In the following exercises, the first of the given /actors may be divided by the given divisor. 31. Divide {a^^2ay+y^)[h^-cd^ + 3) by a-^y. Ans. (a+?/)(53-cc?2+3). 32. Divide (2a;3_6aa:2 + 6a2a:-2a3)(c2 + 3c?/-2/2) by x-a. Ans. (2a:2— 4«a;+2a2)(c2 + 3c?/— 3/2). 33. Divide {a^+Ay^) (3a;i/2-5?/3 + 3^4-4) by a^-2ay^2y^. Ans. {a^-\-2ay+2y^){3xy^—5y^^3y^~A:.) 34. Divide [Sa^—2a^x—l3a^x'^—3ax^){y^ + 2xy){y^ + 3x^y^—x^) by 4a2 + 5aa;+a:2. Ans. {2a^ —3ax){y'^ -^2xy){y^ -{-3x^y^ —x^). 27 CHAPTEE, III. COMPOSITE QUANTITIES.—COMMON MEASURE.— COMMON MULTIPLE. Composite and Prime Quantities, (54.) A composite quantity is one which is the product of two fac- tors, each differing from unity; and d. prime quantity is one which is not the product of such factors. Thus 3^2 is a composite, while a+5 is d^ prime quantity. H?^ Is ah a composite or a prime quantity ? Is «+5 a composite or a prime quantity? a^ % Ix^ 1 2a-\-'Zh% 6a — Sc^ 1 Decomposition of Quantities. (55.) Decomposing a composite quantity consists in resolving it into its factors. Any divisor of the quantity, and the corresponding quotient, are two factors into which it may be resolved, (46.) Thus if the binomial 3x-{-6ax be divided by 3a:, the quotient will be 1 + 2a ; the binomial may therefore be resolved into the factors 3x(l-\-2a), 3x into the binomial l + 2«. Uesolve 2a-{-4:ax—Qa^x^ into component factors. Resolve a^x—3ax^-}-8aij^ into component factors. Resolve Aa^-^-a^b — 5a^y into component factors. Resolve 2a^—3ax-\-7a'^y into component factors. Resolve 5ax-{'5a^x — lOa^x^ into component factors. Resolve 7ab—14:ab^ — 14iabx into component factors. A monomial factor may generally be found by mere inspection, and a composite polynomial be resolved into a monomial and a polynomial factor, as above. No general Rule can be given for resolving a polynomial into the polynomial factors of which it may be composed. But there are par- ticular Binomials which have well known binomial divisors, by means of which such Binomials may be resolved into two factors, {65). The following divisions will be found, ou trial, to terminate with- out remainders. 28 COMPOSITE QUANTITIES. (56.) The Difference of two quantities will divide the difierence of any j7owers of the same degree of those quantities. Thus a—b will divide a'^—b^, or a^—b^, or a^—b^, &c. (57.) The Sum of two quantities will divide the sum of any odd powers or the difference of any even powers, of the same degree, of those quantities. ' Thus a-\-b will divide a^-\-b^, or a^-\-b^, or a^ -{-b"^ , &c. ; also a-\-b will divide a^—b"^, or a^—b*, or a^—b^, &c. ; The factors of certain Binomials and Trinomials may also be known from the manner in which the products and squares of binomiaU are formed. This will be seen in the following propositions. (58.) The Product of the sum and difference of two quantities ii equal to the difference of the squares of those quantities. Thus (a+5) (a—b) is equal to a'^—b'^ ; and this last binomial may therefore be divided by either a-\-b ox a — b. This proposition, it will be observed, is included in the two pre- ceding ones, (56.) (57.) \!C^ What is the Product of (a-\-x)(a—x) ? What is the Pro- duct of (a+5) («-5) ? Of (3+2/) (S-^/) ? Of (a;-l) (^c+l) ? (59.) The Square of the sum of two quantities is equal to the sum of the squares plus twice the product of the two quantities. Thus {a-\-b) [a-\-b), that is, the square oi a-\-b, is equal to a^-^b^ + 2ab or a^ + 2ab-\-b^. This trinomial may therefore be divided by a-{-b. \!CF^ What is the Product of (a+a?) (a+x), or the square of «+a;? What is the Square of a+3/? Ofa:+3? Ofa+c? Of 2/4-5? (60.) The Square of the difference of two quantities is equal to tha sum of the squares minus tvrice the product of the two quantities. Thus (a—b) (a—b), that is, the square oi a—b, is equal to a^j^h2_2ab or a^—2ab+b^. This trinomial may therefore be divided by a— 5. []:F^ What is the Product of (a—x)(a—x), or the square of a-xl What is the Square of a— 2/? Oiy-21 Ofb—xl Ofl— T/t ^ COMPOSITE QUANTITIES. 39 ^^ The preceding principles will be found applicable to the following EXERCISES. 1. Resolve a?—x^ into component factors. Ans. {a—x) {a-\-x). 2. Kesolve a^-{-y^ into component factors. Ans. (a-{-y){a'^—ay-\-y^). 3. Hesolve a^—x^ into component factors. Ans. (a~x){a'^ -{-ax-^-x"^). 4. Resolve a^—y^ into component factors. Am. (a^+y^){a-\-y){a—y), 5. Resolve a^—8x^ into component factors. Ans. {a — 2x)(a^-i-2ax+4x^). 6. Resolve l — 8y^ into component factors. Ans. (l-2i/)(l + 2y+4?/2) 7. Resolve 14-27a;^ into component factors. Ans. (l + 3x)(l—3x+9x^). ft Resolve 8a^—27y^ into component factors. Ans. {2a—3y){4:a^ + 6ay+92j^). 9. Resolve a^x^-{-c^y^ into component factors. Ans. (ax-\-cy)(a^x^—axcy-{-c^y^). 10. Resolve a^ — \6x^ into component factors. Ans. (a^+4x^) {a+2x){a-2x). 11. Resolve a^-\-2ax-{-x^ into component factors. Ans. {a-{-x) [a-\-x). 12. Resolve a^ — 2ay-{-y^ into component factors. Ans. {a-y)(a-y). 13. Resolve a^-^Aax+Ax^ into component factors. Ans. {a+2x){a+2x). 14. Resolve 9a^ — 6«?/+2/^ ^^^^ component factors. Ans. (3a— y) {3a— y), 15. Resolve 4:a^-\-12ax-{-9x^ into component factors. Ans. {2a+3x){2a-{-3x). 16. Resolve a^x^ — 2ax-\-l into component factors. Ans. {ax—\)(ax-^l) 17. Resolve l-\-4:xy-\-4x^y^ into component factors. Ans. {\-{-2xy)\{ + 2xy). 18. Resolve 4:a^x^ — \2abxy-\-9b'^y'^ into component factors. A71S. {2ax—3by)(2ax — 3hj), 19. Resolve 9a^x^-\-2'ia'^cx'^y-\-l^c^y'^ into component factors. J.?^s. (3a'^x^ + 4cy) {3a^x^-\-4cy). 20. Resolve IGa^iC^ — 32a^ciC?/^4-16c22/^ into component factors, Ans. {4:a^x—4Lcy^)(4ta^x—4:cy^). ^ COMPOSITE QUANTITIES. (61.) A Trinomial may be resolved into two binomial factors, whenever one of its three terms is a square, another the sum of the products of the square root of that term multiplied into any two quan- tities, and the remaining term the product of the same ttvo quantities. For example, let it be required to decompose the trinomial a2— a — 12. The first term is the square of a ; the second term is the sunt of the products of a multiplied into 3 and —4 ; and the third term is the product of 3 and — 4. Now, from the manner in which the product of two binomials is formed, it is evident that «2_a_i2 is equal to (a+3)(a— 4). In like manner the following Trinomials may be decomposed. 21. Resolve a^ -\-l a-\-12 into component factors. Ans. (a+3)(a+4) 22. Resolve a^^-fa; — 30 into component factors. Ans. (a;+6)(a:-5) 23. Resolve a^ — 8(2 — 20 into component factors. Ans. (a — 10)(a+2) 24. Resolve y'^ — lQy—^^ into component factors. Ans. (y-13)(y+3) 25. Resolve a^x'^-\-^ax-\-\S into component factors. Ans. (ax-\-^)[ax-\-2) 26. Resolve a'^y'^ — Hay -{-2Q into component factors. Ans. {ay—1) {ay— A) 27. Resolve 2>ah^ -\-2\ab-\-2>^a into component factors. The given quantity may be resolved first into 3a(i2_|_7^_|.12); and the trinomial factor thus obtained may be resolved into (J+3) (6+4) Ans. 3a(i+3)(5+4). 28. Resolve 5aj2— 5a;— 60 into component factors. Ans. 5{x-A)[x^-Z). 29. Resolve 2ah^ -{-\^ah-\-2t^a into component factors. Ans. 2fl(^>+6)(5+3). 30. Resolve ^ax^ -\-lQax^ -\-^ax into component factors. Ans. tta;(2a;+3)(2a;+2). COMMON MEASmRE. gj COMMON MEASURE. (62.) One quantity is called a measure of another, or is said to measure another, if it will divide the other, without a remainder ; and A common measure of two or more quantities is any quantity that will divide each of them, without a remainder. Thus 5a is a common measure of 10^6 and 5a^-\-15ab^. D;^^ Name a Common measure of Aa^b and 6ab^. Oi 9a^ and 3a^-i-a^c. Of 5abc+3a^c and 4ac. Of a^o? and a^x-{-a^^. » Greatest Common Measure. (63.) The greatest common measure of two or more quantities, is the greatest quantity that will divide each of them, without a remainder. Thus 3a^ is the greatest common measure of 6a^x-\-3a^i/ and 3a^b — 6a^cd, since it is the greatest quantity that will divide each of these binomials, without a remainder. Here it may be remarked, that when we speak of algebraic quan- tities as being greater or less, we have reference to the coefficients and exponents of the same letters, and not to any particular values which the letters may be supposed to represent. Thus a^ is, algebraically, always greater than a ; though its nu- merical value would be less tHan that of a, if the latter should be taken to represent a, proper fraction, as ^, f, &Ci 0:^^ What is the Greatest Common Measure of Sa^x-^-ia^x and a^y~5a^y^1 Oi 5a^c-\-\^a^c^ and lOa^b—da^bxl 0£ 2ax^ — 6a^i/ and 4:a^x-{-8a7/^ — 2a 1 When two quantities have no common measure but unity, they are said to be prime to each other, (64.) The greatest common measure of two or more quantities is composed of all the factors which are common to those quantities, that is, all the factors which are found in each quantity. For example, the common factors in 3axy^ and 6^2^22; are 3, a, and y^ ; and 3ay^ is the greatest common measure of the two quan- tities. 32 COMMON MEASURE. But as no general Rule can be given for resolving a Polynomial into the iiolynoniial factors of which it may be composed, we make the Rule for finding the greatest common measure depend on the fol- lowing proposition ; — (65.) The greatest common measure of two or more quantities, is the same as that of the least of those quantities and the remainder, or remainders, if any, after dividing the least into the other, or each of the others. We may suppose any two quantities to be represented by a and na-\-h ; n being some integral number, and h less than a. It is plain that any measure of a will also measure na, n times a ; and, measuring na, if it measure na-\-h, it must also measure h. Hence there can be no common measure of a and na-\-h which is not a common measure of a and h ; the greatest common measure, there- fore, of a and ^, will be the greatest common measure of a and na-^h. But h is the remainder after dividing a into wa+^; hence the proposition is true for two quantities ; and in like manner it may be proved for three or more quantities. RULE VI I. (66.) To find the Greatest Common Measure of two Quantities. 1. Divide one of the quantities into the other, and the remainder into the divisor, and so on, until there is no remainder : the last divisor will be the greatest common measure required. — But, 2. When any factor is contained in each term of the divisor, with- out being contained in each term of the corresponding dividend ^ — such factor must be canceled, before dividing. — And 3. When the first term of the divisor is not a measure of the first term of the dividend — multiply the several terms of the latter by some quantity which will thus render its first term divisible, without a remainder. Note — This Rule might be readily extended to three or more quantities ; but we seldom or never have occasion to find the greatest common measure of more than two algebraic quantities. COMMON MEASURE. 33 EXAMPLE. To find the greatest common measure of the polynomials 4:a^x—5ax^-\-x^ and 3a^ —3a^x-\-ax^ —x^. A:a^x- -5ax^ + x^ J' - 3a^x-i- ax^— x^ 4a2 - -5ax -\-x^ > 12^3 - -12a^x-\- Aax^— 4:X^ {Sa -15a^x+ dax^ 3a^x-{- ax^— 4a:' 4 \2a^x-{- 4aa;2 — 16a:3(3a? I2a^x-I5ax^+ 3x^ 19aa;2 — 19a:3 19aa:2 — 19a;3 ; 4^2 —5ax-\-x^ ( 4a— a? 4a2_4aa3 ■ax-\-x^ •ax-{-x^ In this example the factor x is contained in each term of the divisor ^a^x—bax^-\-x^, without being contained in each term of the dividend ; we therefore cancel this factor, and take ^a^—5ax-{-x^ for a divisor. Then, the first term 4a2 of this divisor not heing a measure of the first term 3a^ of the dividend, we multiply the dividend by 4, which renders the first term 12a' divisible, without a remainder. We also multiply the remainder 3a^x-{-ax^ —^x^, still regarded as a dividend, by 4, to make the first term divisible, without a remain- der. In the next remainder, 19aa;2 — 190?', the exponent of a, the letter of arrangement, being less than in the first term of the divisor, — we divide this remainder, after canceling the factor I9a:2. into the former divisor, which now becomes the dividend. The divisor a—x completes the operation, and is the greatest com- mon measure required. i 34 COMMON MEASURE. It remains to elucidate the Rule. The greatest common measure of the two polynomials, is the same as that of the divisor Aa^x—5ax^-\-x^ and the remainder after the first division, (65). On the same principle the greatest common mea- sure of the first divisor and remainder, is the same as that of the first remainder and the second remainder ; and so on, to the last remain- der, which becomes the last divisor. Hence, the last remainder, or divisor — ^being the greatest common measure of itself and the preceding divisor, and so on to the first re- mainder and divisor — is the greatest common measure of the given polynomials. Again ; since the greatest common measure is composed of all the factors found in each of the two polynomials, (64), it is not afiected by canceling a factor, — as x in the first divisor, — which is found in only one of them ; nor by introducing a new factor, — as in multiplying the first dividend by 4, — into only one of them, — since these expedients do not interfere with the original common factors. The suppression of a factor, as x in this example, which is peculiar to the divisor, is necessary ; for, otherwise, the dividend must be mul- tiplied by this factor to render its first term divisible, — and this would introduce a new common factor into the two polynomials, and thus increase the greatest common measure, (64). 1 In the preceding Example, we might have taken the first remain- der, 3a2a:+aa;2— 4ar^, for a divisor, and the first divisor for the divi- dend. The operation from that point would be as follows ; — 3a^x-\-ax^—Ax^ Aa^— 5ax-\- x^ 3a^ -i-ax —4.x^ }— 15ax-^ 3a:2(4 12^2 + 4aa:— 16a:2 — 19ax-\-19x^ — 19ax-]-19x^ — a + X jsa^-^- ax—4.x^{—3a—4:X / 3a^—3ax 4aa;— 4a;2 4aa?— 4a;2 COMMON MEASURE. 35 The greatest common measure is here found to be — (i-\-x, which is the one before determined, with its signs changed. The effect of changing the signs in the divisor, would only be to change the signs in the quotient resulting from the division. Hence (67.) A Common measure of two or more quantities may have all its signs changed, without ceasing to be a common measure of those quantities. The Rule which has been given, directs that one of the two quan- tities be divided by the other, &c., without distinguishing the divisor from the dividend. We therefore remark here, that the same common measure will be found by taking either of the two quantities for the divisor, and the other for the dividend. If the divisor and dividend in the preceding Example, be inter- changed with each other, the operation will be as follows ; — 3a^ — Sa^x-{-ax^—x' ■3a^x^—ax^-\-4:X^ ) 4a^x— 5ax^-\- 3a 12a^x--15a^x^ + 3ax^ ( 4.r 12a^x—12a^x^-{-4:ax^ —4a;* — 3a^x^— ax^-}-4:X* J3a^- 3a^x-{- /3a^+ a^x~ 3^2. ^ax +4x2 J3a3_ 3a^x-[- ax^ — 4aa;2 — 4:a^x-{- 5ax^ — x^ 3 -'\2a'^x-{-\5ax^— 3a;3 ( 4a? — \2a^x— Aax^ + lQx^ 19aa:2_19a;' t9aa;2 — I9a:3 ) a—x /---3a2_ ^a;-l-4a;2 (— 3a+4a: .3a'^-\-3ax — 4aaj+4a;2 --4aaj4-4a:2 36 COMMON MEASURE. In this case the factor x is contained in each term of the dividend, without being (;ontained in each term of the divisor. This factor, therefore, does not enter into the composition of the greatest common measure, (64), and might be canceled before dividing ; and this would simplify the operation. Hence (68.) When any factor is common to all the terms of the dividend, and not to those of the divisor, such factor may be suppressed, without affecting the greatest common measure of the two quantities. It follows also from proposition (64), that (69.) When the same factor is contained in all the terms of two polynomials, their greatest common measure may be found by multi- plying this factor into the greatest common measure found for the po- lynomials without this factor. For example, to find the greatest common measure of a^x—x^ and a^x-^ax^ — 'Zx^. The factor x is contained in all the terms, and the two polynomials vnthout this factor are a^—x^ and a'^-\-ax—2x'^. The greatest common measure of these will be found to he a—x', then, multiplying x into a— a;, we find ax—x^ for the greatest common measure of the given polynomials. EXERCISES. 1. Find the greatest common measure of a^—x^ and a^—x^. 2. Find the greatest common measure of a^—ax~2x^ audi a^ —3ax-\-2x^ . 3. Find the greatest common measure of 3a^—2a—l and 4^3 — 2^2— 3a+l, 4. Find the greatest common measure of x^-\-2ax-\-a^ and ic^ — a^x. Ans. a—x. Ans. a—2x. Ans, a—1. Ans. x-{-a. 5. Find the greatest common measure of 2a;3-16ic-6 and 3x3-24a;-9. Ans. a:'~Sa;— 3. COMMON MULTIPLE. 37 I 6. Find the greatest common measure of Ans. a— a;. 7. Find the greatest common measure of a^—2ax-\-x^ diV^di a^—a'^x—ax^-\-x^. Ans. a^ — 2ax-{-x^, 8. Find the greatest common measure of 6a^-\-'7ax—3x^ and 6a--{-llax-^Sx^. Ans. 2a+3x. 9. Find the greatest common measure of 7x^—23x7/-{-6y^ and 5x^ — 18x^y-\-nx7j^—(Jy^. Ans. x—3y. 10. Find the greatest common measure of a*+a2y2_j_^4 ^iXidi a^—2a^y-\-2ay^—y^. Ans. a^—ay-\-y^, 11. Find the greatest common measure of 3a5 4-6a^a:+3a3a;2 ^nd a'^x-{-3a^x^-\-3ax'^-]-x^. Ans. a^ + 2ax-^x^ . 12. Find the greatest common measure of x^-ax^-8a^x-\-6a^ and x^-3ax^-8a^x^ + 18a''x-Sa^. Ans. x^-\-2ax—2a^. COMMON MULTIPLE. (70.) One quantity is called a multiple of another, if it can be measured by the other, that is, divided by it, without a remainder ; and A common multiple of two or more quantities is any quantity that can be divided by each of them, without a remainder. Thus 8a^x^ is a common multiple of 2a^x and 4a;2. [C^ Name a Common Multiple of 3a5 and 5c^. Of arc^ and 2ay, Of 3a2 and ^x^. Of 56c and «2. Oiy^ and 5x^. Least Common Multiple. (71.) The least common multiple of two or more quantities, is the smallest quantity that can be divided by each of them, without a re mainder. Thus 6aa;2 is the least common multiple of 6a and Ga;^. 38 COMMON MULTIPLE. 03^ What is the Least Common Multiple of 2ab and 3^2 ? Qf 4a; and 4?/2 ? Of 5a and 3a25 ? Of ab and 2dc2 1 Of lax and aa;2 ? Of 3a^ and 6<2?/2 ? Of axy and 5x^y^ ? (72.) The least common multiple of two or more quantities is com- posed of the S9nallest selection of factors that includes the factors of each given quantity. For example, take the quantities 3ah^c and Qa^bxy, Resolving these two quantities into iheir prime factors, we have 3abbc and 3x2 aab xy. If we take 3x2 aa bbc xy, we shall have the smallest selection of factors that includes the factors of each of the two given quantities. Then the product ^a^b'^cxy is the least common multiple, because it is the smallest quantity that each of the given quantities will divide, without a remainder. (73.) The least common multiple of two quantities, is equal to their product divided by their greatest common measure. For since the greatest common measure of two quantities, is com- posed of all the factors which are common to those quantities, (64), these factors will enter twice in the product of the quantities. Tf, therefore, the product be divided by the greatest common mea- sure, the quotient will contain only those factors which are common to the two quantities, and those which are peculiar to each of them ; and these are the factors of the least common multiple, (72). RULE YIII. (74.) To find the Least Common Multiple of Two or more Quantities. 1. Set the quantities in a line, from left to right, and divide any two or more of them by any prime quantity, greater than unity, that will divide them, without a remainder, — placing the quotients and the undivided quantities in a line below. 2. Divide any two or more of the quantities in the lower line, as before ; and so on, until no two quantities in the lowest line can be so divided. The product of the divisors and quantities in the lowest line, will be the least common multiple required. 3. If no two of the given quantities can be divided as above, tW, 'product of all the quantities will be their least common multiple. COMMON MULTIPLE. 39 EXAMPL E. To find the least common multiple of 2ax^, 6a^y, and 3y— 4?/^ 3 J 3ax^ (ja^y 3y-Ay^ a) x^ 2a2y 2ay 2a Sy-iy^ y) 3y—4:y^ 3—4^2 In the first operation we divide 3ax^ and 6a^y by 3, and set down 3y — 4y^ without dividing it ; and in like manner in the second opera- tion. In the third operation we set down x^ without dividing it. Then 3ayx^ x2ax{3-~^y^), equal to 18a^x^y—24:a^x^y^, is the least common multiple of the three given quantities. This Rule depends on proposition (72) : the divisors and quantities in the lowest line, are the smallest selection of factors that includes the factors of each given quantity. EXERCISE s. 1. Find the least common multiple of ax^, 2a^y, 4:y-\-y^, and ax'^-\-^x^, Ans. 8a^x^y^ + 2a^x^y^ + 32a^x^y-}-8a^x^y^, 2. Find the least common multiple of 2ay^, 4:ay^, 2x—4lX^, and x^-\-ax^. Ans. 4:ax^y^ —Sax^y^ -\-4:a^ x^y^ ~8a^ x^y^ . 3. Find the least common multiple of 3a^, ax^, 3a-\-6a^, and x^ — 3x^. A?is. 3a^x^ + 6a^x^—9a^x^-'18a^x^, 4. Find the least common multiple of 4^^, 2uy^, 5a—5ab, and 10a — 5. Ans. 40^2^/3 _40a2%3 — 20«3/3 + 20«5?/3. 5. Find the least common multiple of 5a, lOab, 3y-{-3y^, and 6y^+3y^. Ans. 90abij^ + 60aby* + 30ahy^, •40 COMMON MULTIPLE. 6. Find the least common multiple of 4, 4a;2, 83/— 8, and 2x^—ax^. 7. Find the least common multiple of 10, 5ax, 4a— 8?/2, and 2x~{-6x^. Ans. 20a^x—4:0axy^^-\-Q>0a^x^ — 120ax^y^. 8. Find the least common multiple of 32/2, 12y, 5a;2 — 10, and Ay—Qy^. Ans. 60a;23^2_i20«/2 — 120a;2y4 + 2402/* 9 Find the least common multiple of 14, 7a, 2^2 _4^ and 28+7fly. A7ts. 56a7j^ + 112a+Ua^y^—2Sa^y. 10. Find the least common multiple of a^—x^, and a^—a^x—ax^-\-x^, (73). Ans. a^—a*x—ax*-]-x^. 11. Find the least common multiple of a;2 + 2^>2;+^»2 andx^-b^x. (73). Ans. X* -\-bx^ —h^x^ —h^x. 12. Find the least common multiple of a2-3a5+262 smd a^-ab-2b^. (73). A71S. a^-2a^b-ab^ + 2b''. 13. Find the least common multiple of 4a3_2a2— 3a4-l and 3^2 — 2a— 1. (73). Ans. 12a*-2a3-lla2 + l. I 41 » CHAPTER IV, FRACTIONS. (75.) An algebraic fraction represents the Quotient of its nume- rator divided by its denominator. Thus — represents the Quotient of a divided by 3a?. In reading and algebraic Fraction, it will often be necessary to use the terms numerator and denominator, to avoid ambiguity in refer- ence to the division which is expressed. Thus if the Fraction a-\-x y be read, «+ a: divided by y, it might be Understood that only x is to be divided by y. But the true sense would be conveyed by saying, nume- rator a-{-x, denomitiator y. A Fraction is thus employed to represent the (Quotient, when the divisor is not a factor of the dividend. The quotient in this case may also be represented by means of Negative Exponents. (76.) Any quantity with a negative exponent is equivalent to a unit divided by the same quantity with tlie sign of its exponent changed. Thus a~^, a with exponent —2, is equivalent to —^ . For a'^a''^ is equal to a^, (41) ; and by dividi^ig each of these equals by a^, we find a~2 equal to -tt or — - , (48). a- a^ ^ O;^ What is the fractional equivalent of a~^ 1 and how is it proved 1 Of x~^ ? and how is it proved 1 Of y~^ ? and how is it proved ? Of («-f :r)-2 ? or {.c-y^ ? k 42 FRACTIONS. • (77.) When the Divisor is not a factor of the dividend, the Quotient may he represented hy a fraction (75), or hy the dividend multiplied into the divisor with the sign of its exponent changed. a^x^ will give the quotient ax~^, hecause this quotient multiplied into the divisor x^, produces ax^, (41), which is equal to a, the divi- dend, (48). We have, therefore, ax~^ for an integral, and —^ for a fractional I form of the quotient oi a^x^. [I^ "What is the integral form of the Cluotient of a-^b^ 1 and how is it proved? Of x-^y* ? and how is it proved 1 Of l~a^ ? and how is it proved? Ofa^-f-a?? Oib—ac^'^ Ofa-f-5? Transfer of Factors. (78.) Any factor may he transferred from the denmninator to the numerator, and vice versa, hy changing the sig7t of its exponent. For example, if we divide a hy x^y, fractionally, we have a-^x^y equal to —^r~ . x^y If we divide a hy the factors x^ and y, separately, we shall find a-^x^ equal to ax~^, (77), and ax~^-^y equal to y Hence a divided by x^y gives -2 a ax or — x^y y These two quotients being necessarily equal to each other, we aee that x^ may be transferred from the denominator to the numerator, by changing the sign of its exponent. If we also transfer the factor y, we shall have ' equal to or ax~^y~'^, (77). x^y ""- 1 If we transfer the factor a from the numerator a, or la, to the de- nominator, we find a , 1 - — equal to x^y a ^x^y [C^ Under what different forms may the duotient ofa-^bx^ be represented ? Of Sa-^x^ ? Of 2c-^3a* ? Ofab-^x^y^ ? FRACTIONS. 43 Reciprocals of Quantities. (79.) The reciprocal oi a quantity is a unit divided by the quantity, Thus the reciprocal of a^ is — or a~^ (76). (80.) The reciprocal of a Fraction is equivalent to the fraction in- verted ; that is, with its numerator and denominator taken the one for the other. CL 1 iK Thus the reciprocal of -^ , or ax""^ (77), is — ^j, which becomes — by transferring x~'^ from the denominator to the numerator, (78). [[^ Under what different form^ may the Reciprocal of — be ex- X pressed ? The reciprocal of ^ ? The reciprocal of — ? Of — — ? Constant Value of a Fraction. (81.) The value of a Fraction remains the same when its numera- tor and denominator are both multiplied, or both divided, by the same quantity. For example, if we multiply both terms of — by y, we shall have —r equal to — -— . x^ x^y For the j^rs^ of these fractions is equal to ax~'^ ; and the second is equal to ax~'^yy~^j (77). But yy~^ is equal to y^, (41), and by cancelling this factor, (48), we find the second fraction also equal to ax~^ . Hence the two frac- tions are equal to each other. c'^m^ [T^ Prove that -7- is equal to —7 — . That -— - is equal to — bx x^ X Sig-ns of Fractions. (82.) Since a Fraction represents the quotient of its numerator di- vided by its denominator, a Fraction is positive when its numerator and denominator have the same sign, and 7iegative when they have contrary signs, (49). 44 FRACTIONS. [IF^ Say whether the Fraction — is positive or negative. Say - , a . . . . r* t ^ ^ • • • "whether — — is positive or negative, bay whether — is positive or negative. Say whether — — is positive or negative. (83.) The sign + or — prefixed to a Fraction, — not the sign of either the numerator or denominator, — shows whether the fraction enters additively or subtractively into a calculation. Thus -I L— denotes that the fraction — i— , which is in itself o o negative^ (^2), is to be added, in the calculation into which it enters. "When no sign is prefixed to a fraction, + is always understood. (84.) The signs of both the numerator and denominator may he ehanged, or the sign of either of them with the sign prefixed to the Fraction, — without affecting the value of the fraction. Thus -;- is equivalent to — , since both these fractions are b —b positive, (82). Also — is equivalent — — y— . For the fraction — — is negative, its numerator and denominator having contrary signs ; this nega- tive fraction becomes positive, when subtracted, as required by the sign — prefixed to it, and it is then equivalent to the first fraction. A Polynomial is changed from 'positive to negative, or from nega- tive to positive, by changing the sign of each of its terms. For example, if a—h is positive, a must be greater than h ; then, changing the signs, b—a will be negative. The Fraction is therefore equivalent to . x—y ^ y—x {i^ What other, changes may be made in the signs, without af- ectinjr the value, of this fraction ? fractions. 45 Fractions Reduced to their Lowest Terms. (85.) A Fraction is reduced to loiver terms by dividing its numera- tor and denominator by the same common measure. This simplifies the fraction, without altering its value, (81). A monomial common measure may usually be known by inspection. Thus to reduce the Fraction 4a2^ Q>ac—Qa'^x It is obvious that we have only to divide its numerator and deno- minator by 2a. This gives us the equivalent Fraction, 2ah 3c—4:ax ' A binomial common measure may often be discovered from the principles which have been established for the decomposition of Poly- nomials. Thus to reduce the Fraction a^-b^ a^-\-2ab-\-b^' By proposition (57) we can divide the numerator hy a-{-b; and by (59), we can divide the denominator by the same quantity Thus dividing we find the equivalent Fraction a—b a+b' In all cases in which a Fraction admits of being reduced, we may apply RULE IX. (86.) To Reduce a Fraction to its Lowest Terms. Divide the numerator and denominator by their greatest common measure, (66) : the quotients will be the lowest terms of the given fraction. exercises. ^ . 3«a:2 . ^ 'r.2 1 . Reduce -— rr^ — r— r to its lowcst tcrms. Ans Za^b—^a^ ' ' ab~3a^ 8a ^ -r. , 2«27/-f-4a?/2 . , . ay-\-2y^ 2. Reduce ^ ^ to its lowest terms. Ans. . ^ 46 FRACTIONS. 3. Reduce 3 ^ to its lowest terms. (57). a—h An&. 4. Reduce — — — to its lowest terms. CC Co 5. Reduce — to its lowest terms. Q 3 CLIJ 6. Reduce ~-z — - — - — i. to its lowest terms. 7. Reduce -^ — -— ^ — — - — to its lowest terms. a^-ab+h^ Ans. — x—a A71S. — , — x-{-a Ans. "'-^^ a-\-y A:ns. a^—2ax-{-x^ _ _ . 2ax^—a^x—a^ . ^ o. iteauce ^ _ , ^ ; — r- to its lowest terms. 2x^-\-3ax-\-a^ ax — a^ Ans. x-\-a ^ _ , a^-\-2a^x-\-Sa^x^ y. Reduce — - — — — — — to its lowest terms. 2a* —oa^x—5a^X" a-\-2x-\-3x^ Ans. 2a^—3ax—5x^ ' ..rx -n n 6a^ -\-7ax—3x^ . , lU. Reduce ^ _ . ^ ^ -^r to its lowest terms. 6a^-i-llax+3x^ Sa—x Ans. -. oa-^x 1 1 . Reduce — — — -■ to its lowest terms. a'^-i-a^x—ax^—x^ a^—ax-\-x^ ^ns. — -. CL X 10 -D 4 5a^-\-\^a^y+5a^y^ . , 12. Reduce — — , ^ - , ^ ^ — t^—t to its lowest terms. a^y+ 2a^y^ + 2ay^ -i-y* 5a^ + 5a^y Ans. 2 I .,3 a^y-\-ay'^~\-y fractions. 47 Fractions Reduced to a Common Denominator. (87.) Two or more Fractions are said to have a common denomina- tor, when they have the same quantity for a denominator. Thus and have a common denominator. a-\-b a-\-o Two or more Fractions may often be reduced, very readily, to a common denominator, by multiplying both the numerator and denomi- nator of one or more of them, so as to make the denominator the saine for each. For example, to reduce to a common denominator the Fractions a , he and 2a— 2x 3a— Sx' We have only to multiply the terms of the first fraction by 3, and those of the second by 2. This gives the equivalent Fractions '^ci 2bc — and . (81). 6a— 6x K)a—&x Observe that this reduction does not alter the values of the given Fractions. When this method cannot be obviously applied, we adopt RULE X. (88.) To Reduce two or rtiore Fraction?, to a Common Denomi- nator. 1. Multiply each numerator by all the denominators excerpt its oivn, for new numerators ; and multiply all the denominators together, for a common denominator. 2. If tlie Least Common Denominator be required, — Find the least common inultiple of the given denominators, for the Common denominator. Divide this Multiple by the denominator of each given Fraction, and multiply the quotient by the numerator, for the new numerators. EXAMPL E s. 1 . To reduce -^r- ^ -r , and to a common denominator. 2x G y+^ For the new numerators, we have a . 6 . (y+2), equal to ^ay-\-\2a ; h.2x.{y-\-2), " 2hxy-\-Ux; and c. 3a:. 6 " l^cx. And the common denominator is 3a? .'6 . (2/+2), equal to ISxy+^Qx. 49- FRACTIOxVS. The given Fractions are thus reduced to 6a?/+12a ^bxy-\-^hx iScx . , — ^ , , —7- , respectively. 18a;i/+36a? \Sxij+'Z^x \Qxy+'^(Sx ^ ^ 2. To reduce the same Fractions to the least common denominator. The least common multiple of the denominators Sx, 6, and y-\-2 will be found to be (jxy-\-12x, (74), ^^/hich is the lequiied denominator. Dividing this Multiple by each given denominator, and multiplying the quotients by the given numerators, respectively, we find the new numerators, 2ay-\-4:a, hxy-\-2hx, and 6ca?. The Fractions reduced to their least common denominator, are then 2ay 4- 4a bxy + 2bx &cx &xy-\-\2x ^xy+l2x 6xy-{-l2x respectively. Each of the given Fractions should be in its lowest terms, before proceeding to find their least common denominator ; otherwise, the de- nominator found will not, in all cases, be the smallest by means of which the values of the several Fractions may be expressed. In finding a Common Denominator as above, the numerator and denominator of each given fraction are multiplied by the same quantity. a Thus in the first example, — has both its terras multiplied by 6 and «/+2, — producing the new terms ^ay-\-l2a and lSxy-\-oQx. Hence the values of the given fractions are not altered in reducing them to a common denominator, (81). EXERCISES. 1. Reduce -tt • ;^ > a-nd to a common denominator. 3 2x x—4: 2ax^—8ax 3bx-12b 6cx Ans. Qx^—2ix ' %x^—2^x' 6a:2_24a; ^ _ , 2 a _ a-\-c - . 2. Heduce -;r , — ^ , and to a common denommator. a^ 2/ 1— 2/ 2y^ — 2y^ a^—a^y a^y^-{-a^cy^ jM FRACTIONS. 49 »i. Jteduce — , r— , and — ^r— to a common denominator. ^a'^y^ cxy^ 3axy-{-15xy ^ _ , ax a ^ b — 1 - 4. Iteduce -^,77-, and 7 to a common denommator. 2 3a; 1+6 J±'^S. — 6x-{-6bx ' 6x+6bx ' 6a?+66a; a common denominator. 2a^x^y bxy^ 2ax^y-\-2bx^y 5. Reduce — , -—- , and to a common denominator. y 2x^ xy Atis ■ 2x3^2 ' 2xY^ ' 2x^y^ r> T> J a b ab 6. iteduce -— - , - , and -r to the least common denominator. 2x'^ y x-^-\-x^ ay-\-axy 2'bx'^ -\-2hx^ 2aby 2x'^y+2x^y ' 2x^y + 2x^y' 2x^y-\-2x^y Dinmon denominator. 3a;2y2 4ay 2a — 2a; ''*• T2^ ' 127/2 ' 122/2 8. Reduce — ^ , - , and — to the least common denominator. y2 5 y^—y^ 5ay—5a hy^ —bij^ 5c Ans. 7 Reduce -—,—-, and — — tt to the least common denominator. 4 3y Q>y^ Ans. It (L ft 2 9. Reduce - , -—. , and to the least common denominator. 4 2x^ 2-\-x Ans. 5y^ — 5y^ ' 5y^ — 5y^ ' 5^/^ — 5y^ ) the least common denominator. 2x^y-^x^y 4:a-\-2ax ^a'^x^ 8a;2+4a;3 ' 8^2^P4^ ' 8^M^4^ ' iB— 1 Cb C 10. Reduce — - — , — - , and — -;r to the least common denomi- 3 32/ 67/— 3?/2 nator. 2xy—xy'^—2y-\-y^ 2a'^—a^y c ^y-'iy^ ' Oy-32/2 ' &y-^y^ 50 fractions. Integral and Mixed Quantities Reduced to Improper Fractions. (89.) An integral quantity is one which does not contain any fractional expression ; as 'iax^, or 2ab-~5xy. (90.) A mixed quantity is partly integral and ^d^xily fractional ; as 3aa;2-|--— or 2ah . 5c \-\-y (91.) An improper Fraction is a fraction whose value may be ex- pressed by an integral or a mixed quantity. Thus — is an improper fraction whose value is 2a4-— - . 6a 3a RULE XL (92.) To Reduce an Integral or a Mixed Quantity to an Impro- per Fraction. 1. Under an integral quantity, regarded as a numerator, set 1 for a denominator. Or multiply the integral quantity by any proposed denominator ; the product will be the numerator. 2. In a mixed quantity, multiply the integral part by the denomi- nator annexed ; add the numerator to the product when the sign be- fore the fractional part is +, but subtract the numerator when this sign is — ; and place the result over said denominator. examples. 1. To reduce 3a ^ to a Fraction whose denominator shall be a—2x. 3^2 is the same as —-— ; and by multiplying both terms of this fraction by the proposed denominator, we have 3fl2 equal to ^_^^ , (81). 2. To reduce 3a ^ — — to an improper Fraction. Multiplying 3a ^ by the denominator 2, and subtracting the nume- rator a— a: ^ from the product, — observing to change the signs of the numerator, (36), — we have 3a=-?^ equal to ^-^i::^. FRACTIONS. 51 The reason of this operation will be evident, if we consider that, by multiplying the 3a^ by 2, we reduce the integral part to a common denominator with the fraction annexed to it, according to the first part ' the E-ule. The operation then consists in subtracting numerator from numerator, and placing their difference over the common denomi- nator. Another view to be taken of the preceding operations, is, that the integral term 3a^ is multiplied, and the product is then taken to be divided, by the same quantity, namely, the denominator. The value therefore remains the same. EXERCISES. 1. Reduce Aax^ to a Fraction whose denominator shall be y-{-2. Am. ^f- 2. Reduce a^-\-ox to a Fraction whose denominator shall be 2y^. 2«27/2_f_ 6^7/2 3. Reduce 5xy to a Fraction whose denominator shall be 3-{-x^. 16xy+(5x^y . _ ^-- 3+.^ 4. Reduce y^—6 to a Fraction whose denominator shall be 3ax^. Ans, ^ — 5. Reduce a+5 to a Fraction whose denominator 'shall be a—h. Ans. a—b 6. Reduce fyay^ to a Fraction whose denominator shall be a^ —b"^ 5a^y^-~5ab^y^ Ans, 2 ,2 1 . Reduce oj^ + l to a Fraction whose denominator shall be 3a'^y. 3a^y 8. Reduce ab^~x to a Fraction whose denominator shall be 1—y Ans. ^-J— ^ 1-2/ 52 FRACTIONS. 9. Reduce a^-{-x-^- to an improper Fraction. y 10. Reduce ax^-\-—— to an improper Fraction. Ans. ^— — y+1 /7 I rv" 11. Reduce 5a+35 -— to an improper Fraction. o 140+94-X Ans. 12. Reduce 2a—^c-\ -^— to an improper Fraction. 11a— 20(f-2/2 ^ws. ^ o 1—7/2 13. Reduce 2-\-y^ -j— to an improper Fraction. Ans.l+X 14. Reduce a^-\-x^— ■ to an improper Fraction. Ct — it/ ax^~a^x Ans. a—x Ans. ^ X 15. Reduce a—x-\ to an improper Fraction. X A 02 A 16. Reduce \-^2x — to an improper Fraction. a:+10a:2+4 Ans. ox 5y2 17. Reduce 2x-[-2y+ ' to an improper Fraction. ^X — oy Ax^—2xy-~y^ Ans. „ ^ to an imnroDer Fraction a'^—ax-\-x' 18. Reduce a-{-x ; — „ to an improper Fraction. 2x^ Ans. -, a'^—ax+x' fractions. 53 Improper Fractions Reduced to Integral or Mixed Quantities. By reversing the preceding Rule, we have RULE XII. (93-) To Reduce an Improper Fraction to an Integral or a Mixed Quantity. Divide the numerator by the denominator for the integral part; and set the denominator under the remainder, if any, for \he fractional part, of the result. Connect the fractional to the integral part by the sign + ; or change the sign of the numerator or denominator; and connect it by the sign — . example. To reduce to an integral or a mixed quantity the Fraction ^a^c-[-2ah—2b-\-y 3a Dividing the numerator by the denominator, we find the integral quo- tient to be Zac-\-h, and the remainder —2b-\-y. Setting the denominater under this remainder, and connecting the fraction so formed to the integral quotient, by the sign +, the result is 6a Or, changing the signs in the numerator — 2h-\-y, and connecting the fraction by the sign — , the result, under a somewhat simpler form, is ^ac-\-b -— ^. 6a This form is simpler than the preceding, as it dispenses with one sign in the numerator of the fractional part. The reason of the preceding operation is evident from the conside- ration that every Fraction is equal to its numerator divided by its de- nominator, (75). After obtaining the quotient '5ac-\-b, — the divisor 3a not being contained in the remainder —2b-\-y, the division of these terms is indicated by setting the divisor under them. The fractional part of the result must evidently be added to the integral part ; and this addition is indicated by placing the sign + before the fraction. But the value of the fraction annexed, will not be affected by changing the sign + before it to — , if at the same time we change the signs in the numerator, (84). (54 FRACTIONS. The Fraction formed of the divisor and remainder, will be in its loivest terms, or not, according as the improper fraction reduced, is, or is not, in its lowest terms. For, if the dividend and divisor have any common measure^ the divisor and remainder will have the same common measure, (65). EXERCISES. ^3 — ^3 1 . Reduce to an integral or a mixed quantity. Cb — 2/ ^ ^ , 10x^—5x+3 . , . , . 2. iteduce • to an integral or a mixed quantity. ox 3 Ans. 2:c— 1+-^ • ox 3. Reduce to an integral or a mixed quantity. a-y Ans. a^ -\-a'^y-\-ay^ -\-y^ . 4. Reduce ^ — 5~~2 ^^ ^^ integral or a mixed quantity. X — 6y Ans. x"^ 5. Iteduce ^ to an integral or a mixed quantity. x^—3y^ ' ,ntity. Ans. 2x^ --^^ . n -n ^ 9ax^-2x-{-3 . , . :, . D. Jxeduce to an integral or a mixed quantity. « o 2x—3 A?ts. 3x^ ~ — . 3a a^-\-y^ 7. Reduce to an integral or a mixed quantity. a+y ^ ^ Ans. a^—ay-\-y^. x^ — y^-^4: 8. Reduce to an integral or a mixed quantity. oc+y 4 Ans. x—y-\ . 9. Reduce —^ to an integral or a mixed quantity. 2a~y ° H J' Ans. 2y-\-- . '^^2a-y FRACTIONS. 55 ADDITION OF FRACTIONS- (94.) The Sum of two or more Fractions is found by means of a common denominator. ihus the Sum oi - and - is . XXX For it is evident that a divided by x, added to b divided by x, makes the sum of a and b divided by x. In other words, if each of the parts a and h be divided by a, the ivhole a-\-b will be divided by x. Hence we have RULE XIII. (95.) For the Addition of Fractions. 1 . If the fractions have not a common denominator, reduce them to a common denominator. 2. Add the numerators together, and place the Sum, as a numera- tor, over their common denominator. 3. Mixed quantities may be added under the form of improper fractions; or the integral and the fractional parts may be added sepa- rately. EXAMPL E. h c To add together 2a-{ — and 3a . X y Reducing these mixed quantities to improper fractions, they be come 2ax-[-h , 2ay—c and — . X y Reducing these fractions to a common denominator, we have 2axy-^hy Zaxy—cx ' and . xy xy Placing the sum of these numerators over the common denomina tor, the result is, 5axy-\-by—cx . ^ , by— ex , equal to Da-\ . xy ^ %y 56 FRACTIONS. Otherwise, by adding the integral and ihe fractional parts sepa- rately. — There will be less liability to error, if we change the sign be- fore the fraction in the second quantity to +> and change the sign of its numerator ; thus + ^i , (84). Then, reducing — and — to a common denominator, they become by —ex — and . xy xy Adding these fractions together, and adding together the integral parts 2a and 3a, we obtain _ , hy—cx ^ „ oa-\ , as beiore. xy Improper fractions in the results obtained by this Rule, should be reduced to integral or mixed quantities ; and proper fractions, to their lowest terms. E XERCISES. 1 . A.dd together -7- > — . and — - — (i2 2a2 2. Add together 2x+ —- , and 4a; H — - 4 5 2x X 3. Add together y^+ ~ , and 3y^— — 4. Add together -— , 2a^, and ^ — 2x 5 Ans. 2a+ —^ Ans. 6x-{- -^ . 5x Ans. 4:y^+ — Ans. 2«2-j- _ 5. Add together 3x^-^ ^ and ^- lOrc Ans. 3x^-\-a-{- — . 4 FRACTIONS. 57^ ^2 1 6. Add together x^y, — , and x^ij+ — . 2^24-3 Ans. 2x^y-\ — . n oc ,a^-3x 7. Add tojrether 2a ^ and — - — . ° 4 2 , 2a2-7ir ilTZS. 2a2+ . , a^ 3a ^a^-4 8. Add together — , — , and — ^— . y y y a^y+3ay+a^-4: Ans. — ^ . y^ , x—l ^ ,3x+4: 9 Add together — — - , 2, and — - — . 3 o 14CC+7 Ans. 2+ —TF- 10. Add together 2y-\ — — and y —- . Ans. 3y- ^^ ^ 3x ,a2_}_i 11. Add together 5a^ , —^ , and ^^ - Ans, 5a^-\ • — — /r X 12. Add together 10, — -— , and 3— — . o ^ . -, , , a^-\-x - a^—x-{-l 13. Add together — - — and 8a;^ Ans. 13—-—- 15 7a2 + a:+3 y Ans. 14. Add together 2y^ and 12 3 ^ 3+y^-x^ a 5 15—3a—ay^-\-ax^ 40 FRACTIONS. SUBTRACTION OF FRACTIONS. (96.) The Difference of two Fractions is found by means of a com- vnmi denominator. Thus — subtracted irom — leaves . For the last fraction, , added to — the one subtracted, pro- X X duces — the one from which the subtraction is made. X "We have therefore RULE XIV. (97.) For the 2,uhtr action of Fractions. 1. If the fractions have not a common denominator, reduce them to a common denominator. 2. Subtract the numerator of the fraction to be subtracted from the other numerator, and place the Difference, as a numerator, over the common denominator. 3. A mixed quantity may be taken in subtraction under the form of an improper fraction; or the integral and the fractional part may be taken separately in subtracting. EXAMPLE. From 8a; H to subtract 3a; . y ^ Reducing these mixed quantities to improper fractions, they become Bxy-\-a , ^xw—b — and . y ^ Reducing these fractions to a common denominator, — subtracting the second, of the resulting numerators from the first, — and placing the difierence over the common denominator, we find 5xyw+aw-\-by ' ^ , aw-\-hy — , equal to 5x-\ ~ . yw yw FRACTIONS. 59 Otherwise, by taking the integral and the fractional parts sepa- rately. — To diminish the liability to error in adjusting the signs, we change the fraction in the second quantity to + — , (84). Then, reducing — and to a common denominator, they become aw , —by — and . yw yw Subtracting the second of these fractions from the first, and 3a; from 8a;, we find the difference of the given quantities to be 6x-{- — , as before. yw In all subsequent exercises, improper fractions in the results should be reduced to integral or mixed quantities ; and proper fractions, to their lowest terms. EXERCISES. , ^ a2-f2 , b-S 1. From — - — subtract — — O tiClf 2. From subtract r y 2/2 ^2 ^ 3. From — - — subtract l-fic l—x h c 4. From Sa-\ subtract 2a . X V 6. From 2x^ subtract x^ X y 2a^+4:a—3b+9 Ans. ba y2 a^—a^x—b—bx A?ts. r 1—x^ by+cx Ans. a-\- — XV Ans. x^ xy 60 FRACTIONS. 6. From subtract x—y x^—y^ *^ From 3y2 subtract 2?/2 -| — y ^ 8. From 2a-\-h-\- ^ subtract 3h— ~ . ^3 9. From ^x^—y—~ subtract y-\- Ans. - ^"^ ^ X^~y^ Am. 2/ — 2 ax-\-by xy Ans. 2a-2i+^4^ Ans 4:X^ — 2y 10. From 2J2 + ^±1 subtract b^ - %— . Ans. b^-\ — — b 11. From 5^34- -5 subtract Sa^-j . 2/+1 2/-1 I JL 2 12. From 2x^— ^^-— subtract — a;2 4- -^ Ans. ^x'^ 7/2-1 3a+35+2a2 13. From 2^3- ^J- subtract b^ -— . 3 4 Ans. 253_62_i+^ 14. From 6-\ • subtract l-\ ^ — . y y Am. 4H \ y^ FRACTIONS. 61 MULTIPLICATION OF FRACTIONS. (98.) The Product of two or more Fractions is equal to the product of their numerators divided by the iJroduct of their denominators Thus „ multiphed by -^ produces -^-r . x^ 2/ ^ y For the first of these fractions is equivalent to ax'"^ , and the second to hy~^, (77) ; and the product of these two equivalents, is a.hx-^y-'', (40). And by transferring the factors a;"^ and y"^ to the denominator, we have ab -t:—^ , ab divided by x'^y^. From the preceding it follows, that (99.) Multiplying by a Fraction finds sitch a part of the multiph- cand as is expressed by the multiplier. Thus a X ^ or — X ^ produces -— , -J of a. Compound Fractions. (100.) A Fraction multiplied by a fraction, or divided by an inte- ger, may be expressed by a Compound fraction, that is, a fraction of a fraction. For example, — x^ is | of — , (99) ; also — -^2 is 4- of — . X X X X Hence (101.) Multiplying two or more Fractions together is equivalent to reducing a compound to a simple fraction. Thus — X 4 is 4- of — , equal to — . X ^ ^ X ^ 2x From proposition (98), we have the following Rule. 62 FRACTIONS. RULE XV (102.) For the Multiplication of Fractions, 1 . Multiply the numerators together for a numerator, and the de- nominators together for a de^wminator . 2. An integral quantity and o. fraction are multiplied together, by multiplying the numerator, or dividing the denominator, by the integer. 3. A mixed quantity may be taken in multiplication under the form of an improper fraction ; or the integral and the fractional part may be taken separately in multiplying. It may also be remarked that (103.) A Fraction is multiplied by its own denominator, by merely canceling the denominator. — And equal factors may he canceled in a numerator and its own or the other denominator, without altering the Product of the two fractions, (81). To multiply The first numerator is equal to (a+x) {a—x), (58), and the second denominator, to (a-{-x) {a-\-x), (59), By canceling the factor a-\-x from these terms, and the factor y from the other numerator and denominator, the operation is reduced to a—x 2 _ , , 2a— 2x X — — ; which produces I EXAMPLE . «2. by 2y 3 a2 + 2aa;+a;2 3?/ a-\-x 3ay-\-3xy EXERCISES. 3 1. Multiply together a2 _^2 ^^d . Ans. 3a + 3a;. Sa^y 2. Multiply together a^y and ic+f . Ans. a^xy-\ j- . «,»-,., , «4-^ , 2 ^ 2a-{-2x 3. Multiply together — -— and . Ans. 7— . ^ '' ° 3 a—x 3a— 3a; a . 5ay^ 4. Multiply together 5y^ and y^— -^ . Ans, 5y* -— . FRACTIONS. (53 . , , a-\-b , a—b , a—h 5. Multiply together ^^ and -^-^ . Ans. -^ . 3b , &a^b — 9bx 6. Multiply together 2a^ —3x and . Ans. . ^ -^ ° y—1 2/ — 1 ace . ax a^x^ 7. Multiply together a-\ and x , — . Ans. -^ . ^^ ^ a—x a-\-x a^—x^ a^-b^ a2 a^ + a^b^ 8. Multiply together — —7- and -7 — j^ . Ans. • . 9. Multiply together 6y^ 2 and 6y^-{- —. Ans. 25y* . ^ -.^ , . , -, ox^—5x , 7a . Sax — 5a 10. Multiply together — ^^ and ^^^-g^ . Ans. ^-^-^ . DIVISION OF FRACTIONS. (104.) The Quotient of two Fractions is equal to the dividend mul- tiplied by the reciprocal of the divisor y (80). _- ■, r^ . „ a ^. ,^ . . X . . a y Thus the (Quotient 01 — divided by — is equal to -7- X — . b y b X For since the value of the dividend is not altered by multiplying each of its terms by both terms of the divisor, (81), the quotient is equal to axy X bxy ' y ' And by dividing the numerator of this dividend by the numerator of the divisor, and the denominator by the denominator, we have the ay ... . . a y quotient y- , which is equal to — X — . uX X Observe that the (Quotient multiplied into the divisor, produces the dividend, (46.) n^. ay X . axy . ^ a Thus T^ X — produces ^ , equal to -^ • bx y ^ bxy ^ b 54 FRACTIONS. Complex or Mixed Fractions. (105.) "When the dividend or the divisor is a Fraction or a mixed quantity, the dividend over the divisor, vrith a line between them, forms a complex or 7nixed fraction. a Thus -, — '-{x-\-y) equals — ; — , numerator -7- , denominator x-\-y, \ -^f ^ x-\-y From the nature of Division, and the proposition before demonstra- ted, (104), we have RULE XVI. (106.) For the Division of Fractions. 1. Divide the numerator of the divisor into the numerator of the dividend, and the denominator into the denominator ; or multiply the dividend by the reciprocal of the divisor. 2. A fraction is divided by an integral quantity, by dividing the numerator, or multiplying the denominator, by the integer. 3. An integral quantity is divided by a fraction, by dividing the integer by the numerator, and multiplying by the denominator ; or by 'multiplying the integer by the reciprocal of the fraction. 4. A mixed quantity may be taken in division under the form of an hnproper fraction : or the integral and the fractional part may be divided separately EXAMPLE. 10c2 ^ 5c3 a^—2ax-\-x'^ a^—x^ ' a^—x^ The dividend must be multiplied by - ^ — , which is the recipro- ml of the divisor. If we cancel the factor o— a; from a^ — 2ax-\-x^ , (60), and from a^—x^, {58); and also cancel the factor 5c^, the operation will be, (103), 2 a-\-x , . , , 2a+2x X ; which produces . a — X c ac — ex By thus cancelmg common factors, we find the Quotient in its lowest terms. 1. Divide 20^- 2y^ by FRACTIONS. EXERCISES. 5cx G/) 2. Divide -^ ^^Y' — Z a^—x^ a—x _. ., ax+b , bx—a 3. Divide by — 7— a a^ + 2ab-\-b^ . a+h 4. Divide ^^, by^^ 6a^b ^ 3^2 5. Divide — — ^ . . ;^ by x^~2xy-\-y^ " x—y c4-x^ ^ 6. Divide 3a^ + — ^ by a^ — y 2x^ X '7. -Divide — -^ by a^+a;^ x+a 8. Divide ^ ^ by -^ ax^ x^ 9. Divide ^Z^x ^y ^^ + ^^- 10. Divide 2a^-\-Aay-\-2y^ by — fl^4-^ Ans. lOacx—lOcxy Ans. —■ — a-\-x Ans. abx-\-b^ abx—a^ a-\-b Ans. Ans. 2ax _2b__ x—y Ans. 3+ Ans. Am. y 9x^-\-3c 6a^-2x^ 2x x^—ax-\-a^ 22/2-1 Ans. a^-ah+b^ 2b+3bx Ans. Aa-{-iy 66 FRACTIONS. 11. Add — to -' , and divide the Sum by — . Ans, — — — 3 4 -'4 9y 12. Add '^ to ^, and divide the Sum by —^. Ans.^-^. 5 4 Mo 2x^ 1 Ans. x^ 2x^ 1 a;* 4 13. Add — —to — - — , and divide the Sum by — - — 3(a;2 + 2) 14. Add —--to — - — , and divide the Sum by-^ — 5~~^. ,d o o 1 ^^s- 2(ax-\-2) nj ofu 2oc 15. Subtract -| from ~, and divide the Remainder by -r-. „ 3 2 -^ 3 3ay—2y . Ans. 4x 16. Subtract ^ from — -, and divide the Remainder by -— — -. 4 4 2 Ans. — ^- X Q/ Q, I X 17. Subtract — from — , and divide the Remainder by —^r-' Do o Ans. — ^-- — -' 5 ^ ,_ , . , 2ab—Z .5a ,,...,_, ,5 18. Multiply by — , and divide the Product by — . « » ^ 2a^b-Sa Ans. • X 2 2 I i^ 19. Multiply ^^ by ^, and divide the Product by ~-. 3a2 a-\-o 20. Multiply —rr- by — - — , and divide the Product by ^. ^•^53 9 a^—x^ 21. Multiply --J - by / , and divide the Product by — -^ Ans. 2y Ans. 27+9y+3y^-^y^, 67 CHAPTEE V. SIMPLE EQUATIONS. (107.) An Equation is an expression denoting the equality of two quantities by means of the sign =, equal to, placed between them. The quantity on the left of the sign ■= is called the first rriember, or side, and that on the right the second member, or side, of the Equation. Thus ?>x-\-ah=.^x-{-^cl — 9 is an equation, in which ox-\-ah is the first member, and 5a:+8^— 9 is the second. (108.) Equations are employed in the solution of particular mathe- matical questions, or in the investigation of general mathematical principles. In the solution of questions, the unknow7i or required quantity is represented by a letter, usually x, or ?/, &c., and an Equation is then formed which expresses the relation between this and the known or given quantities. To give a simple example ; — Suppose we wish to find a number the tJiird ^xi^ fourth of which shall together make 35. Let X represent the number to be found, and the Equation will be 4 + -T = 35. 3 4 (109.) The solution of an Equation consists in finding the value of the unknoicn quantity in the equation. The value found for the unknown quantity is verified, or the Equa- tion satisfied, when this value, substituted for its symbol in the equa- tion, makes the^rs^ member the same as the second. The value of x in the preceding equation is 60, since this number, substituted for x, satisfies the equation ; thus \ '^ V^° -35 The mode of solution will vary with the Different Degrees of Equations, (110.) A svmiile Equation, or an equation of thej?rs^ degree, is one which contains no power of the unknown quantity but \i^ first "power. (58 SIMPLE EQUATION?. ox-^ax — 4 = 20 is a simple equation. A quadratic Equation, or an equation of the second degree, is one in which the highest power of the unknown quantity is its second power or square. 2a;2 + 3(2a;+5=r30 is a quadratic equation. A cubic Equation, or an equation of the third degree, is one in which the highest power of the unknown quantity is its third power, or cube; and so on. Equations are also distinguished as Numerical^ Literal, and Identical Equations. (111.) A numerical Equation is one in which all the known quan- tities are expressed by nuinbers. 2a:+5a;=:25— 3 is a numerical equation. A literal Equation is one in which some or all of the known quan- tities are represented by letters. 2a;+<3ra:= 25 — 35 is a literal equation, in which a and b are supposed to represent quantities whose values are known. An identical Equation is one in which the two members are the same, or become the same by performing the operations which are in- dicated in them. Thus 'ix—oabzzz.o(x — ah) is an identical equation. Transformation of Equations. (112.) The transformation of an Equation consists in changing its form, without destroying the equality of the two members, — for the purpose of finding the value of the unknotcn quantity, or of discovering some general truth or principle. These transformations depend, for the most part, on the following Axioms. (113.) An Axiom, is a truth which is self-evident, — neither admit- ting nor requiring any demonstration ; such as, 1 . Things which are equal to the same thing, are equal to each\ other. 2. If equals be added to equals, the sums will be equal. 3. If equals be taken from equals, the remainders ivill be equal. 4. If equals be multiplied by equals, the jyroducts will be equal. 5. If equals be divided by equals, the quotients will be ec[ual. 6. Any like 'powers or roots of equal quantities, arc equal SIMPLE EQUATIONS. gg Solution of Simple Equations Containing but one Unknown Quantity. ^ (114.) The value of the unknown quantity is found by making: its iiymbol stand alone on one side of the Equation, so as to be equal to known quantities on the other side. In order to this, the following transformations may be necessary, or at least may be expedient. 1. Clearing the Equation of Fractions. 2. The Transposition and Addition of Terms. 3. Changing the Signs of all the Terms in the Equation. 4. Dividing the Equation by the Coefficient of the Unknown quantity. We shall apply each of these transformations to the solution of the same Equation. Clearing an Equation of Fractions. (115.) An Equation is cleared of fractions by multiplying each nu- merator into all the denominators except its own — regarding each inte- gral term as a numerator, — and omitting the given denominators Let the Equation be 4 3 "^ 3 Multiplying the numerator of each fraction by the denominators of the other two, and the integral terms 7 and x by all the denominators, we obtain 27a:-252 = 36a;-336+ 12^. The equality of the two members is not destroyed in thus clearing the Equation of fractions, because each of the terms connected by the signs -f- and — in the two members, is thus multiplied by all the denominators. (103) (113. ..4). An Equation may also be cleared of fractions by multiplying its two members by the leai^t common tnultiiile of the denominators ; — observing that a fractional term will be multiplied by multiplying its numerator into the quotient of said multiple -^ the denominator. In the given Equation the least common multiple of the denomina- tors is 12. Multiplying by 12. we find 9a; — 84 = 12:c — 1124-4a:. The advantage of this method is, that the new equation is found in its lowest terms. 70 N SIMPLE EQUATIONS. Transposition and Addition of Terms. - (116.) Any term may be transposed from one side of an Equation to the other by changing its sign. — All the similar terms may thus be placed on the same side, and then added together. In the last Equation 9a;— 84= 12a; — 112 -f- 4a;, by transposing —84 to the second member, and 12a; and 4a; to the first, we have 9a; — 12a;— 4a;i= — 112 + 84; And by adding together the similar terms, -7a;=-28. The equality of the two members is not destroyed by transposing a term with its sign changed from one side to the other, because this is equivalent to adding the term with its sign changed to both sides. Thus by adding 84 to both members of the equation 9ic— 84 = 12a; — 112 + 4a;, the term — 84 is canceled in the first member (28). In like manner by adding — 12a; to both members, 12a; is canceled in the second mem- ber ; so also with 4a;. (113... 2). From the preceding principles it follows, that Two equal terms with like signs on opposite sides of the sign =, may be at once suppressed from the Equation. Change of the Signs in an Eqaation. (117.) All the signs in an Equation may be changed, + to — and — to +, without affecting the equality of its two members. This follows from the principle of Transposition, (116), since in transposing all the terms, the signs would all be changed, but the two members would still be equal. In the Equation already found -7a;=-28, we shall have, by Transposition, 28=:7a:, or 7a;z=28. The only Transformation which remains towards finding the value of X in the equation at first assumed, is that of dividing by the coeffi- cient of ic, the unknown quantity. (113. ..5). Dividing both members of the preceding Equation by the coefiicient of a;, we find -28 , 28 X— — y =4 ; or a;= y =:4. "We have thus found the value of a; to be 4. This value may be verified by substituting it for x in the original equation. SIMPLE EQUATIONS. -JT^ We may now give RULE XYII. (118.) For the Solution of a Simple Equation containing hut one iinktimvn quantity. 1. Clear the Equation o{ fractions, if it contains any. 2. Transpose all the terms containing the unknoivn quantity to one side, and all the known terms to the other side, of the equation. 3. Add together all the similar terms in each member. 4. Divide both members by the coejjicieiit of the unknown quanti- ty ; — observing that when the unknown quantity is found in two or more dissimilar terms, its coefficient will be the sum of its coefficients ill those terms. Note. — When the sum of the terms containing the unknown quantity, after transposition, is negative^ it will generally be expedient, though it is never necessary, to make it positive by changing all the signs in the equation. EXAMPLE. Q> 3 X a?-l-l9 Given h ■— =20 — , to find the value of a?. 2 3 2 Clearing the equation of fractions, by multiplying it by the least common multiple of the denominator, which is 6, we have ^ 3a^ — 9 + 2a;==:120 — 3a: — 57. p By transposition, 3a;+2a;+3a:=: 120-57 + 9. Adding similar terms, 8a:=:72. Dividing by the coefficient of x, 72 ^ .= -=9. Remark. — The student is apt to err in Clearing an Equation of its frac- tions, when, as in this Example, a fraction preceded by the negative sign has a polynomial numerator. The sign — before the fraction in the second member above, de- notes that the fraction is to be suhtracted. When this fraction is mul- tiplied by the 6, the product 3a;+57 is subtracted by changing its signs. This gives the terms — 3a: — 57 in the new equation. 72 SIMPLE EQUATIONS. EXERCISES. Numerical Equations, 1. Given 4a:— 8 = 13— 3a; to find the value of a;. 2. Given 7a:+l7r=10a;— 19 to find the value ofx. Ans. a;=12. 3. Given 8a?4-6=36— 7a; to find the value of a;. Ans. x=z2 4. Given 59 — 7a;=4a:-f 26 to find the value of a;. Ans. x=o 5. Given 20 — 4a:— 12 = 92 — 10a; to find the value of a;. A?is. a: = 14 6. Given 8 — 3a:+12 = 30 — 5a:4-4 to find the value of a:. Ans. a; =7 cc * 3a? 7. Given [-24=—- to find the value of x. 4 2 8. Given -— + -— =13 -to find the value of a?. 2 3 4 Ans. a;=19|-. A?2S. a; =12. ^ ^^. x — 5 „ 284— a; p , , , „ 9. Given \~ox=: — to find the value of x. 4 o Ans. x=0 10. Given — 1 ;— =16 — to find the value of x. Ans. a;=13. 2x 5 X 11. Given 3a:— —^^ — =a:-|- - + 13|- to find the value of a:. O D Ans. a:=10. 12x4-26 a;+3 3 2. Given ^ 2a;=15 to find the value of a:. o 3 Ans. a;=12. 13. Given — — - — =x-\ — to find the value of a;. Ans. x=6. ^^ „. 6a:-10 18-4a; ,,..,, , . 14. Given ^ — = — f-^ to nnd the value of x b 3 Ans. a: =4. _ ^. 3a;-f4 7a:-3 ^-16 15. Given — — = to nnd the value of x. 5 2 4 Ans. x=2. i SIMPLE EQUATIONS. 73 16. Given — ; 16= — to find the value of x. 3 5 4 An^. 03=41. 17. Given — - — =a? — 3 ~ — to find the value of a;. Ans, a: =8. 18. Given -^— — - =l-j — to find the value of x. An?,. x=l. 19. Given x =3a;— 14 to find the value of £c. 5 ^ 20. Given ^ _ ^ -f 5= 5^^^ to find the value oi x. /i 3 8 Ans. 0:=-^. Ans. a; =6. ^ 4^ 2 21. Given — +6a;r= - — — to find the value of a?. 3 o Ans, x=z—fy. r^'^ r^- 3x—5 ^^ 2a:— 4 ,. , , , ^ 22. Given x — =12 -; to nnd the value ol x. (it o Arts. ir=65. 2J^— 3j: 2(2a.-|-3)_^ S.t+I ~1~ A?is. x—3. An Equation in which the unknown quantity is found in every term, with different expone?its in difierent terms, may often be reduced lo a simj)le Equalion hy (iiy'i&nig it hy ^oixiQ power of the unknoum quantity. (113... 5). Thusif2a'3 = 10a;2,— by dividing by ic^ we have 2a;=:10 ; hence a; = 5. x"^ 3,^2 24. Given |-3:?;=7a: to find the value of ic. 5 5 A71S. x=5. o^ n- 3x 2x a;2 — 10a; 25. Given = — to nnd the value oi x. 4 5 2 . Ans. a:=10j^. 4 5 2 26. Given h t;- = -^ — ttt- to find the value of x. 3x 2x a;2 — 10a? Ans. a:=10if. 74 SIMPLE EQUATIONS. Literal Equations. x—h 27. Given ax—c=i to find the value of a?. a-\-c Clearing the equation of fractions, hy multiplying it by the deno- minator a+C) we have a'^x—ac-\-acx—c'^ =zx—b. By transposition a^x-]-acx—x=zac-{-c^ —h. Dividing by the coefficient of x, ac-\-c^—b a^-\-ac—l ax dx 28. Given x-\ =.h to find the value qS.x. c c ^^ Ans. x= — ; — — ; . a-\-c-\-d 29. Given be z=z—d to find the value of a?. 0? X ab-l Ans. X— -^ — -— , . bc-\-d 30. Given '^x—az=ix — to find the value of a;. o 362 + ^ Am. X— -TTT-r . K)-\-0 ^. .-TO '^ax'^ — 2bx-\-ax ^ , , , ^ 31. Given \(ibx^— to find the value of x. a — 2b Ans. X: 12ab—3a «^ ., . x(a—b) , ab ^ , r ^ ^i i r 32. Given -^^— — - =«+ x ~ T value oi x. Ans. .= '<'+'^ 6{a-b)-\-4: Remark. — In an identical Equation the unknown quantity has no determinate value, since any quantity whatever may be substituted for it, and the equation will be satisfied Thus in the equation 3a;— 5 = 3a;— 5, the two members will be equal whatever be the value of .t. (113... 3). SIMPLE EQ,UATIONS. 75 PROBLEMS In Simple Equations of one unknown Quantity. (119.) A Prohlem is a question proposed for solution ; and the solu- tion of a problem by Algebra consists in forming an Equation which shall express the conditions of the problem, and th^ solving the equation. The general method oi forming the Equation of a problem, is, to represent a required quantity by x, or y, &c., and then to perform or indicate the same operations that would be necessary to verify the value of a? or y^ supposing that value to have heen found, EXAMPLES AND EXERCISES. 1. What number is that to the double of which if 13 be added, the sum will be 75 ? Let X represent the required number ; then 2x will represent twice the number ; and, by the conditions of the problem, the equation will be 2a; 4-13=75. The value of a? in this equation is the number required. Am. 31. 2. Find a number such that if it be multiplied by 5, and 24 be subtracted from the product, the remainder will be 36. Ans. 12. 3. What number is that to -J- of which if 25 be added, the sum ob- tained will be equal to the number itself mmws 39 ? Ans. 96. 4. Find a number such that \i\ of it be subtracted from three times the number, the remainder will be 77. Ans. 28. 5. Find what number added to the sum of one half, one third, and one fourth of itself will equal 4 added to twice the number. Ans. 48. 6. Divide the number 165 into two such parts that the less may be equal to -^ of the greater. Let X represent the less fart ; then 165— a; will represent the greater ; and the equation will be 165— a? Ans. 15 and 150. 7. Divide the number 100 into two such parts that six times the less may be equal to twice the greater. Ans. 25 and 75. 76 SIMPLE EQUATIONS. 8. It is required to divide 75 into two such parts that 3 times the greater may exceed 7 times the less by 15. Ans. 21 and 5 i. 9. What sum of money is that to which if $100 be added, f of the amount will be $400 1 Ans, $500. 10. A prize of $100 is to be divided between two persons, — the share of the ^st being ^ of that of the other. "What are the shares ? A71S. $43f ; %(^(\\. 11. A post is ^ of its length in the mud, -J- of it in the water, and 15 feet above the water. What is the length of the post ? Am. 36 feot. 12. Find a number such that if it be divided by 12, the divisor dividend and quotient together shall make 64. Ans. 48. 13. In a mixture of wine and cider, |- of the whole plus 25 gallons was wine, and ^ part minus 5 gallons was cider. What was the whole number of gallons in the mixture ? Ans. 120. 14. After a person had expended $10 more than \ of his money, he had $15 less than ^ of it remaining. What sum had he at first? Ans. $150. 15. Divide the number 91 into two such parts that if the greater be divided by their difference, the quotient may be 7. Ans. 49 and 42. 16. A and B had equal sums of money ; the first paid away $25, and the second $60, when it appeared that A had twice as much left as B. What sum had each 1 Ans. $95. 17. After paying away ^ of my money, and then \ of what was left, I had $180. What sum had I at first? Ans. $300. 18. A line 37 feet in length is to be divided into 3 parts, so that the first may be 3 feet less than the second, and the second 5 more than the third \ what are the parts? Ans. 12, 15, and 10 feet. 19. A can perform a piece of work in 12 days, and B can perform the same in 15 days. In what time could both together do the Avork ? Let X represent the number of days. Then since A could do ^^ of the work, and B -fs of it, in 1 day, — will represent the part of the work A could do in x days, — will represent the part of the work B could do in x days. 15 rjn Qn The equation is — + — =1, the entire work. Ans. 6| days. 12 15 SIMPLE EQUATIONS. 77 20. If A could mow a certain meadow in 6 days, B in 8 days, and L <^ 5 days, in what time could the three together do it ? Ans. 2j^ days. 21. Out of a cask of wine, which had leaked away a third part, 20 gallons were afterwards dra.wn, and the cask was then found to be but half full ; how much did it hold? Ans. 120 gallons. 22. It is required to divide $300 between A, B, and C, so that A may have twice as much as B, and C as much as the other two to- gether. Ans. A $100, B $50, C $ 1 50. 23. A gentleman spends -| of his yearly income in board and lodging, |- of the remainder in clothes, and then has $20 left. What is the amount of his income ? A?is. $180. 24. A person at the time he was married, was 3 times as old as his wife, but 15 years afterwards he was only twice as old. What were their ages on their wedding day ? Ans. 45 and 15 years. 25. Two persons, A and B, lay out equal sums of money in trade ; the first gains $126, and the second loses $87, and A's money is now double of B's; what did each lay out? Ans. $300. 26. A courier, who travels 60 miles a day, had been dispatched 5 days, when a second is sent to overtake him, who goes 75 miles a day, in what time will he overtake him ? Ans. 20 days. 27. An island is 60 miles in circumference, and two persons start together to travel the same way around it : A goes 1 5 miles a day ; and B 20 ; in what time would the two come together again ? Ans. 12 days. 28. A man and his wife usually drank out a cask of beer in 12 days, but when the man was from home it lasted the w^oman 30 days ; how many days would the man alone be in drinking it ? Let X be the number of days ; then — is the part that he would drink in 1 day ; and since the woman would drink -^ of it in 1 day, the equation will be }-—- = —-, the part both would drink in 1 day. oc oO 12 A7ts. 20 days. 29. If A and B together can do a piece of work in 9 days, and A alone could do it in 15 days, in what time ought B alone to accomplish the work ? Ans. 22^ days. 30. The hour and the minute hand of a clock or watch are exactly together at 12 o'clock ; when are they next together ? Ans. 5^j minutes past 07ie 78 SIMPLE EQUATIONS. 31. It is required to divide $1000 between A, B, and C, so that A shall have half as much as B, and C half as much as A and B to- gether. , Ans. $222f ; $444|^ ; S333i 32. A person being asked the hour, answered that the time past noon was f of the time till midnight ; what was the hour ? Ans. 48 min. past 4. 33. It is required to divide the number 60 into two such parts that their product shall be equal to 3 times the square of the less ; what are the parts? Ans. 15 and 45. 34. How much wine at 90 cents a gallon, and how much at $1.50 a gallon, will be required to form a mixture of 20 gallons which shall be worth $1.25 a gallon ? Ans. 8^ gal. ; llj gal. 35. A cistern is supplied with water by three pipes which would severally fill it in 4, 5, and 6 hours. In what time would three pipes running together fill the cistern ? Ans. 1|^ hours. 36. If $1000 be divided between A, B, and C, so that B shall have as much as A and half as much more, and C as much as B and half as much more, what will be the portion of each ? Ans. $2101$; $315if ; $473}f. 37. A person has a lease for 99 years, and f of the time which has expired on it is equal to | of that which remains. Required the time which remains on the lease. A?is. 45 years. 38. A merchant bought cloth at the rate of $7 for 5 yards, which he sold again at the rate of $11 for 7 yards, and gained $100. How many yards were thus bought and sold ? Ans. 583J yards. 39. A and B together possess the sum of $9800 ; and five-sixths of the sum owned by A is the same as four-fifths of that owned by B. What is the sum owned by each ? Ans. $4800 ; $5000. 40. The assets of a bankrupt amounting to $5600 are to be divided among his creditors A, B, and C, according to their respective claims. A's claim is J of B's, and C's is f of B's ; what sum must each of the creditors receive? Ans. $1292^^; $2584^^; $17233^. simple equations. 79 Simple Equations Containing Two or more Unknown Quantities. (120.) It is sometimes necessary to employ two or more unknown quantities in the solution of a Problem ; and in this case there must be formed as many independent Equations as there are unknown quanti- ties employed. Two equations are said to be independent of each other when they express essentially different conditions, so that one of the equations is not a mere transformation of the other. Equations which thus express different conditions of the same Prob- lem, are sometimes called simultaneous equations. Solution of Two Simple Equations Containing Two Unknown Quantities. (121.) From two Equations containing two unknown quantities we may derive a new equation from which one of those quantities shall be eliminated, or made to disappear. The value of the remaining un- known quantity may be found from the new equation ; and this value put for its symbol in one of the given equations, will determine the other unknown quantity. Elimination by Addition or Subtraction. « (122.) The two terms which contain the same letter in two Equa- tions, may be made equal .by multiplying or dividing the equations by proper quantities. That letter will then be eliminated in the su7n, or \5lse in the difference, of the new equations. EXAMPLE. Given the equations 2a; +3?/ =23 ' and 5x—2y=10, to find the values of x and y. Multiplying the first equation by 2, and the second by 3, we have 4a;H-6y=:46, and 15:c— 6^=30. (113. ...4). Adding together the corresponding sides of these equations, we find 19a;=76. (113.. ..2). which gives a;=r4. Putting 4 for x in the first of the two given equations, we obtain 8+3y=23, which gives y=5- 80 SIMPLE EQUATIONS. Observe that if 6?/ had the same sign in the two equations which were added together, this terra would have been eliminated by taking the difference, not the sum, of these equations. Elimination by Substitution. (123.) The value of one of the unknown quantities in an Equation, may be found in terms of all the other quantities in the equation. If this value be then substituted for its representative letter in another equation, that letter will be eliminated. EXAMPL E. Given, as before, 2x+3?/=23, and 5x—2y=z\0^ to find the values of x and y. We will find the value of x in the first equation, as if the value of y were known. By transposition, 2a; = 23 — 3?/ ; dividing both members by the coefficient of x, 23 — 3^ x= -~, We now substitute this value of x for x in the second equa- tion. In doing this, we must multiply this fraction by the coeffi- cient 5 in the first term of that equation. Clearing this equation of fractions, transposing, &c., we shall find y=.5. Putting 5 for y in the first equation, 2x4-15 = 23; which gives a: =4. The values of x and y are thus found to be the same as before. This method of Elimination depends on the evident principle, that equivalent algebraic expressions may be taken, the one for the other ; that is, equal quantities may be substituted for each other. SIMPLE EQUATIONS. f^\ Elimination hy Comparison. (124.) If the value of the same letter be found in each of two Equations, in terms of the other quantities in the equations, that letter will be eliminated by putting one of these values equal to the other. EXAMPLE. Given, as before, 2x-\-3y=23, and 5x—2yz=10, to find the values of x and y. We will find the value of x in each equation, as if the value of y were known. Transposing and dividing in the first equation, . 7 ^ * Transposing and dividing in the second equation, 10 + 2y Putting the first of these values of x equal to the second, which will give y=5. By substituting 5 for y in any of the preceding equations, the value of X will be found to be 4, as in the two preceding solutions. Before applying any of the preceding methods of Elimination, the Equations should generally be cleared effractions, if they contain any ; the necessar}' transpositions must be made ; and similar terms must be added together. Elimination by Addition or Subtraction will generally be found the simplest method, since it is free from fractional expressions, which are likely to occur in the application of the other two methods. Let the student apply each of the three methods to the first ten ol the following Exercises. Equations may be marked, for reference, by the numbers (1), (2) (3), &c., or the capitals (A), (B), (C), &c. Thus (A) 2a;+y=10, would be called equation (A). 82 SIMPLE EQUATIONS. EXERCISES. 1. Given 2x-{-3y=29, and 3x—2y=n, to find the values of a; and y. Ans. x-=.l, and 2/=5. 2. Given 5a;— 3y=9, and 2x-\-5y—\^, to find the values of x and y. Ans. x=:3, and y=2. 3. Given x-\-2yz=l7, and 3a;4-2/=16, to find the values of x and y. Ans. x=3, and y=7. 4. Given 4x+y=34, and 10y—x=:12, to find the values of x andy. J.?^s. x—8,sindy=2. 5. Given 3a;+42/=88, and 6x-\-5y=:128, to find the values of cc and ?/• -4ws. a;=8, and 2/=:16. 6. Given '7x+3y=4:2, and 8y— 2a;— 50, to find the values of x and y. Ans. x=3, and y=7. 7. Given 8y— 3a;=:29, and 6y—Axz=:20, to find the values of x and y. Ans. x=:l, smd y=4:. 8. Given 6x—5y=:39, and 7ic— 32/=54, to find the values of x and y. Ans. x=9, and y=3, 9. Given 12a;— 9?/=3, and 12a;4-16«/=:228, to find the values of a; and y. Ans. x=7, and y=9. 10. Given 5x-\-7y:=201, and 8x—3y=:}37, to find the values of a; and y. Ans. a;=22, and 2/=13. X 1/ 11. Given Ty^- y =99, and 7a; -f -|- =51, to find the values of x and 2/. A?ts. a;=7, and 2/=14. X 1/ X II 12. Given —- + -|- =7, and -— + ^ =8, to find the values of x and y. Ans. a;=:6, and 2/=12. 2a; 4v , «« 5a; 9y ^ , , 13. Given 64 — = -^ , and 77 — = —- , to find the values of o o o 10 X and 2/. Ans. ir=60, and ^=30. a;-|~8 VH"^ 14. Given 21— 6v= — - — , and 23— 5a?= , , to find the values 4 o of a; and y. Ans. a;=4, and y=3. 33;4-4y ^ a; . 6a;— 2?/ ^, y ^ , , 15. Given T =10— — , and — —-^ =14— ^ , to find the o 4 3 6 values of a: and 3/. Ans. x=8, and ?/=4. . . ^ . 2a; , 3y 22 ^ 3x 2y 7 16. Given —- + -^ = — - , and — + -^ =4 -- , to find the values 3 5 5 5 6 10 of a: and y. Ans. x=3, and y=4. „ ^. a:— 2 10— a; y—10 . 2?/+4 2a;+2/ a;-fl3 17. Given -^ _ = :L^_ , and -^^ 8- = ~4~ SIMPLE EQUATIONS. g^ to find the values ofx and y. Ans. a? =7, and y=10 18. Find the values of a: and y in the equations in . ^/t in . 6rc— 35 80 + 3a: ^^, 4a;+3y-8 10x-i-55 = 10y-\ — , and ^^ , =18i -^ . Ans. a:=10, and ?/=15. 19. Find the values of a; and y in the equations y — a , a—x X J — =zc, and y ; — =za. b ^ b Clearing the equations effractions, we have (A) bx—y-\-a-=ihc. and (B) by—a-\-x=bd. By transposition in these two equations, we find (C) bx—y=bc—a, and (D) by-\-x=ibd-\-a. Multiplying equation C by b, in order to eliminate y, (E) b^x—by=b^c-ab. Adding together equations D and E, (122.) b^x-\-x=zb^c-{-bcl—ab-\-a. Dividing both sides of this equation by the co-efficient of a?, Jb^c-\-bd—ab+a ""- P + 1 • The value of y will be found if we multiply equation B by 5, sub- tract equation A, &c. b^d-\-ab—bc-\-a y= — wvT-- X It 20. Given — h - = 1, and bx+cy=de, to find the values of x and y. a a ac — de , de—ab Ans. x= r-, and y= y-. c—b ^ c—b 21. Given 2ca;— 4c=--3 equal to — , (75). Direct and Inverse Ratio. (130.) The direct ratio of the first of two quantities to the second, is the quotient of the first divided by the second ; thus the direct ratio of 3 to 5 is f . The inverse ratio of the first quantity to the second, is the direct ratio of the second to the first ; — in other words, it is the direct ratio of the reciprocals of the two quantities. Thus the inverse ratio of 3 to 5 is |- ; — or it is the ratio of ^ to \, equal to ^-^^, equal to f, (127). Hence inverse is often called reciprocal ratio. The term ratio used alone ahvays means direct ratio. 92 RATIO. Compound Ratio. (131.) A compound ratio is the ratio of the product of two or more antecedents to the product of their consequents ; and is equal to the product of all the simple ratios. The compound ratio of a and b to x and yis — ; xy 1 1 r'l-i . a . h a . h = the product oi the simple ratios — and — , or — and — . X y y X (132.) The ratio of the^rs^ to the last of any numher of quantities, is equal to the product of the ratios of the first to the second, the second to the third, and so on to the last ; that is, it is compounded of all the intervening ratios. For example, take the quantities a, b, x, y. The ratios of the first to the second, the second to the third, &c., are -7- , — , — ; and their product is , — — , which is a : y. o X y . hxy y ^ Duplicate and Triplicate Ratios, (133.) The duplicate ratio of two quantities is the ratio of their squares, and the triplicate ratio is the ratio of their cubes. Thus the duplicate ratio of a to 5 is the ratio of a^ to 5^ j and the triplicate ratio of « to Z> is the ratio of a^ to b^. The subduplicate ratio of quantities is the ratio of their square roots, and the subtriplicate ratio is the ratio of their cube roots. Equimultiples a7id Equisubmultiples. (134.) Equimultiples of two quantities are the products which arise from multiplying the quantities by the same integer, and equi- submultiples are the quotients which arise from dividing the quantities by the same integer. Thus 3a and 3S are equimultiples of a and i, while, conversely, a and b are equisubmultiples of 3a and 35. (135.) Equimultiples, or eqiiisubmultiples, of two quantities have the same ratio as the quantities themselves, (81). PROPORTION. ■ 93 PROPORTION. (136.) Proportion consists in an equality of the ra^zos of two or more antecedents to their respective consequents — but is usually con- fined to four terms. (137.) Four quantities are in Proportion when the ratio of the first to the second is equal to the ratio of the third to the fourth ; that" is, when the first divided hy the second is equal to the third divided by the fourth. Thus the numbers 6, 3, 8, 4 are in proportion, since the ratio ^ equals the ratio \. And the quantities «, 5, x, y are in proportion, when the ratio — equals the ratio — , (129). The first and third terms are the antecedents of the ratios ; the second and fourth are the consequents. The first and fourth are the two extremes ; the second and third are the two means. The fourth term is called a iouxih proportional to the other three taken in order ; thus 4 is a fourth proportional to 6, 3, and 8. (138.) Three quantities are in Proportion when the ratio of the first to the second is equal to the ratio of the second to the third, — ^the second term being called a mean proportional between the other two. Thus the numbers 8, 4, 2 are in proportion, since the ratio |- equals the ratio | ; and 4 is a 7nean proportional between 8 and 2. Direct and Inverse Froj)ortion. (139.) A direct Proportion consists in an equality between two direct ratios, and an inverse or reciprocal Proportion in an equality between a direct and an inverse ratio. Thus the numbers 6, 3, 8, 4 are in direct proportion ; (137). The same numbers in the order 6, 3, 4, 8 are in inverse proportion, since the direct ratio f is equal to the inverse ratio f , (130). The term proportion used alone always means direct proportion. 94 PROPOUTION— VARIATION. Sign of Proportion, (140.) A Proportion is denoted by a double colon (: :), or the sign = between the equal ratios of the proportion. Thus 6 : 3 : : 8 : 4,or 6 : 3 = 8 : 4, or f =| denotes that these numbers are in proportion, and is read 6 is to 3 as 8 is to 4. To denote an inverse Proportion we employ the sign :+: between the two ratios of such proportion. Thus 6 : 39^4 : 8, denotes that 6 is to 3 inversely as 4 is to 8. Inverse Converted Into Direct Proportion, (141.) An inverse is converted into a direct Proportion by inter- changing either antecedent and its consequent ; or by substituting the reciprocals of either antecedent and its consequent. Thus from the inverse proportion 6:3^4:8, we get the direct proportion 3 : 6=4 : 8, by interchanging 6 and 3, or -i : -J=4 : 8, by substituting •!■ and -J. The reason of this is evident from the nature of inver.se ratio, (130). Variation. (142.) Variation is such a dependence of one term or quantity on another, that any new value of one of them will produce a new value of the other, in a constant ratio of increase or diminution. 1. One quantity varies directly as another when their dependence IS such that if one of them be multiplied, the other must be multiplied, by the same quantity. For example, the Interest on money, for a given time and rate per cent., varies directly as the Principal, since the Interest will be dou- bled, or tripled, &c., if the Principal be doubled, or tripled, &c. 2. One quantity varies inversely as another when their dependence is such that if one of them be multiplied, the other must be divided, by the same quantity. For example, the Time in which a given amount of interest will accrue on a given principal, varies inversely as the Rate per cent., since the Time will be doubled, &c. if the Rate be halved, &c. VARIATION. 95 (143.) When one quantity varies inversely as another, the 'product of the two is always the same constant quantity. For as one of the two quantities is multiplied, the other is divided by the same number ; the product of .the two will therefore be multi- plied and divided by \he same number ; hence its value will remain unchmged. Variation — an Abbreviated Proportion. (144.) The two terms of a variation are the two antecedents in a Proportion in which the two consequents are not expressed, but may be understood, to complete the proportion. Thus when we say that the Interest varies as the Principal, for a given time and rate per cent., it is understood, that The Interest on any principal is to the Interest on any other Prin- cipal, for the same time and rate, as the first Principal is to the second. Instead of saying "the Interest varies as the Principal," we may Bay, the Interest is proportional to the Principal ; which is a brief method of expressing a Proportion by means of its antecedents, — the consequents being understood. Sign of Variation, The character ~ placed between two terms, denotes that one of them varies as the other. Thus X'^y, x varies as y, or x is proportional to y. xr^ — denotes that x varies as the reciprocal of y ; or x varies re- ciprocally or inversely as y. X'^ ~ denotes that x varies directly as y, and inversely as z \ z that is, X varies as the quotient of y divided by z. 96 theorems in proportton'. Theorems in Proportion. (145.) A Theorem is a proposition to be demonstrated or proved. — A Corollary is an inference drawn from a preceding proposition oi demonstration. Theorem I. (146.) Two Fractions having a common denominator, are to eacl other as their numerators ; and two fractions having a common nume- rator are to each other inversely as their denominators. First. Let d be the common denominator ; then the ratio ol — to — is a . . and — is the ratio of the numerator a to the numerator c. c Secondly. Let n be the common nimnerator ; then the ratio of n a to n c is n n cn C — — -l- — :::: — z^z a c an a and — is the inverse ratio of the denominator a to the denominatoi a c, (130). Therefore, two fractions having a common denominator, &c. (147.) Corollary. The value of a Fraction varies directly as its numerator, and inversely as its denominator. Theorem II. (148.) In any Proportion, if one antecedent be greater than its con- sequent, the other antecedent will be greater than its consequent; if equal, equal ; and if less, less. Let a '.b '.'. X '. y ; then 4- = —.(137). b y , . J^ow if a be greater than b the first ratio will be greater than a unit, and consequently the second ratio will be • greater than a unit, and therefore x will be greater than y. In like manner if a be equal to b, X will be equal to y, he.. Hence, in any Proportion, if one antecedent, &c. THEOREMS IN PROPORTION. ^ Theorem III. (149.) When four quantities are in Proportion, the product of the two extremes is equal to the product of the two Tneans. Let a \h \\ X \ y\ then is ay^hx. For since the quantities are in proportion, 1-=^. (137). b y \ Clearing the Equation of fractions. ay = bx . (115). Therefore, when four quantities are in Proportion, &c. (150.) Cor. 1. A fourth p7'oportional io three given quantities, ia f* uid by dividing the product of the second and third by the first. bx Thus from the equation ay=.hx, we find y=: — . (151.) Cor. 2. When three quantities are in Proportion, the pro- duv t of the two extremes is equal to the square of the mean. For let a : b : : b : x; then ax=:bb=b^. (152.) Cor, o. A tiican froimrtional between two given quantities, is 4 ^ual to the square root of their 'product. Thus from the equation aa:=Z>2, ^y^ ^^^^ ^_^^-jj^'2'^ Theorem IV. (153.) When the product of two quantities is equal to the product of two other quantities, either pair of factors niay be niade the extremeSy and the other the tneans, of a Proportion. \jet ab=.xy\ then will a : x w y : b. Dividing both sides of the given equation by J, xy Dividing both sides of this last equation by x, j^_ J/ ■ X b Hence a : x : : y : b, (137). In like manner a and h may be taken for the means, and x and y for the extremes. Therefore, when the product of two quantities, &c 7 98 THEOREMS IX PROPORTION. Theorem V. (154.) If three quantities are in Proportion, the first will be to the thh'd as the square of the first to that of the second, or the square of the second to that of the third. Let a '. b :'. b : X', then will a : x \\ a^ : 5^ ; ox a \ x wb"^ : x^. From the given proportion, we find ax^b^, (151). Multiplying both sides of this equation by a, a^x^iab"^. Converting this equation into a Proportion, a:x:\a^ '. b^, (153). And by multiplying both sides of the first equation by a;, we shall, in like manner, find a : x \: b^ : x'^. Therefore, if three quantities are in proportion, &c. Theorem VI. (155.) Four quantities in Proportion are also in proportion by in- version, — that is, when each antecedent and its consequent are inter- changed vdth each other. Let a \b :: X \ y] then isb : a :\ y '. X. From the given proportion we find ay—bx, (149). Making b and x the extremes, and a and y the means, b : a ::y : X, (153). Hence, four quantities in proportion are also in proportion, &c. Theorem VII. (156.) Four quantities in Proportion are also in proportion by alter- nation, — that is, when the two means, or the two extremes, are inter- changed with each other. Let a '.h '.'. X : y\ then IS a '. X :: b : y ] or y : b : : x : a. From the given proportion we find ay=bx, (149). This equation may be converted into the proportion a : X : : b : y, or y : b : : X : a, (153). Therefore, four quantities in proportion, &cc. 99 theorems in proportion. Theorem VIIL (157.) In any Proportion, if the tiuo antecedents, or the two conse- sequents, or an antecedent and its consequent, be multiplied by the same quantity, the products and the remaining terms will be in pro- portion. Let a \h \\ X \ y\ and let n be any numerical quantity ; then will an \h \\ nx : y, &c. From the given proportion we have ay=bx. Multiplying both sides of this equation by n, any=bnx. Converting this equation into a Proportion, an : h :: nx : y; or a : bn :: x : ny ; or a : b : : nx : ny, (153). Therefore, in any proportion, if the two antecedents, &c. (158.) Cor. If the two antecedents, or the two consequents, or an antecedent and its consequent, be divided by the same quantity, the quotients and the remaining terms will be in proportion. For dividing by a quantity is equivalent to multiplying by its re- ciprocal. Theorem IX. (159.) Four quantities in Proportion are also in proportion by com- position, — that is, the sum of the first and second terms is to Xhe first or second, as the sum of the third and fourth is to the third or fourth. Let a \b \\ X \ y then is a-\-b \a\\ x-\-y : x. From the given proportion we have ay—bx. Adding both sides of this equation to ax, ax-\-ay=ax-\-bx. Resolving each member of this equation into its factors, a(x-\ry)=x{a-\-b). Converting this last equation into a Proportion, a-\-b '.awx+y :x, (153). By adding both sides of the first equation to by, it may be proved, in like manner, that a-^^b :b :: x-\-y :y. Therefore, four quantities in proportion are also in proportion, &c. 100 , THEOREMS IN PROPORTION. TlIEOREM X. (160.) Four quantities in Proportion are also in proportion by rlivi sion, — that is, the differe?ice of the first and second terms is to the frst or second, as the diffei'ence of the third and fourth is to the third or fourth. Let a\h \\x\y\ then is a — h \ a :: x — y : x. From the given proportion we find ai/=zbx. Subtracting both sides of this equation from ax, ax — ay=zax — bx. Resolving each member of this equation into its factors, a(x—y)z:^x(a—b). ^ Converting this last equation into a Proportion, a — b '.aw X — y : x. By subtracting both sides of the first equation from by, it may be proved, in like manner, that a — b \b w x — y : y. Hence, four quantities in proportion are also in proportion, &c. The QUE M XL (16L) When any number of quantities are in Proportion, the sum of any two or more of the antecedents is to the sum of their conse- quents, as any one antecedent is to its consequent. Let a \b :: c : d \\ X : y, &c. ; then is «+c : b-\-d : : x : y. From tbe given proportion we shall find ay=bx, and cy=dx, (149). Adding together the corresponding members of these two equations, ay-\-cy=bx-\-dx. Resolving each member of this equation into its factors, [a-i-c)y=(b-\-d)x, Converting this equation into a Proportion, a-\-c : b-\-d \\ x '. y. By adding xy to both sides of the third equation, we shall, in like manner, find that a-\-c-\-x : b-\-d-\-y \\ x \ y. Therefore, when any number of quantities are in proportion, cVc. ^^ ^-'^■"^^ir raEOREMS IN PROPORTION. IQl y Theorem XII. (162. ) If two Proportions have an antecedent and its consequents or the two antecedents, or the two consequents, the same in both, the re- maining terms will be in proportion. Let a \h '.'. X \ y, \ '\ /',' ' ' * '■ ," ; and a \h WW. z\ , ^'' ^ , , , then will x \ y \\w\ z. ' »\ ' ;!''''>'? 'V' , 1 / • \ ^ , , ' ' ',>■,':.:.> For the ratio oixXo y is equal to the ratio of w to z, since each of these ratios is equal to the ratio of a to i ; hence X \ y w 10'. z. The two given proportions have an antecedent and its consequei^ the same in both. If the two antecedents were the same in both, the demonstration would be the same, after inter cha?tging the means; and if the two consequents were the same, after interchanging the ex- tremes, (156). Hence, if two proportions have an antecedent and its consequent, &c. Two or more Proportions having an antecedent and its consequent the same in each, form one continued proportion. Thus, a '.b :: u w, a '.b V. iv. X, and a : b v. y : z, form the continued proportion, a : b '.: u : v w iv '. x '. y : z. Theorem XIII. (163.) The sum of the first and second terms in any Proportion, is to their difference, as the sum of the third and fourth is to their differ- ence. Let a :b '.'. X : y, , then is a-\-b : a—b : : x-\-y : x—y. By Composition and Division, in the given proportion. a-{-b : a : : x-\-y : x\ (159) ; a—h : a \\ x—y : x. (160). These two proportions have the antecedents a and x the same in both ; hence a-\-b : a—b : : x-\-y : x—y, (162). Therefore, the sum of the first and second terms in any proportion, &c. 102 THEOREMS IN PROPORTION. Theorem XIV. (164.) The 'products of the corre$j}onding terms of two or more Proportions, are in proportion. Let a : b \ : X : y, ' "^ "• ..';'• ^^d c : d •.'. IV : z\ then is ac : hd : : xw : yz. '■ ' -'"' ' / ' * '< Fro|n the ^ two given proportions, we have '''"■' ' ay=bx, and cz=:dw. Multiplying together the corresponding members of these equations, ac.yz^=bd .XIV. Converting this equation into a Proportion, ac : bd : : xw : yz, (153). In like manner the demonstration may be extended to three or more proportions. Heuce, the products of the corresponding terms, &c. (165.) Cor. Like poivei'S or roots of proportional quantities, are in proportion. For if <2 : 5 : : a: : ?/, by multiplying each term by itself, we shall have, according to the Theorem, a^ : b^ : : x^ '• y^ ', and multiplying these by the given terms, we shall have a^ : b^ : : x^ : y^, &c. Theorem XV. (166.) For any factors in an antecedent and its consequent, or the two antecedents, or the two consequents, in a Proportion, may be sub- stituted any other quantities which have the same ratio to each other. Let a \b \\ nx \ py, and n '. p '.'. r : s ; then will a : b w rx : sy. From the two given proportions, we find an : bp '.'. nrx : psy, (164). Dividing the aiitecedents in this proportion by n, and the consequents by^, a -.b : : rx '. sy, (158). This proves the first affirmation in the Theorem. By inter- changing the extremes in the two given proportions, and afterwards the wea?2S,*(156), the other two affirmations in the theorem may be demonstrated. Therefore, for any factors in an antecedent and its consequent, &e. general solutions of problems 103 General Solutions of Problems. — Formulas. Applications OF Proportion, &c. (167.) In the general solution of a Problem, all the quantities are represented by letters ; and the unknoivn being thus found in terms of all the known quantities, the result discloses a rule for the numerical computation in any given case of the problem. example. To find two numbers whose sum shall be s, and difference d. Let X represent the greater, and y the less number. From the conditions of the problem, we have x-\-y—s, and x—y^^d. • By adding the second equation to the first ; and also subtracting the second from the first, we have 2x— s-{-d, and 2i/=: s— c?; . s-i-d s—d which give x=: , and y=: -7,— • From these general values of x and y, we learn that the greater of two numbers is equal to -J of (the sum + the difference"), and that the less is equal to \ of (the sum — the difference), of the two numbers ; hence the following ride : (168.) To find two ^lumbers frotn their sum and difference, — Add the difference to the sum, and divide by 2, for the greater of the two numbers ; subtract the diflerence from the sum, and divide by 2, for the less number. For example, if the sum of tM'o numbers be 500, and their differ- ence 146, , . 5004-146 „ „ The greater number is =323 ; /it T , 1 1 . 500 — 146 and the less number is =illi. (169.) An algebraic Formula is an equation between the symhols of certain quantities — resulting from the general solution of a Problem, or the investigation of some general principle. Thus the equations which express the values of x and y in the preceding Example, are for7nulas for finding two numbers from the sum and difference of the numbers. 104 PROBLEMS. PROBLEMS In Proportion^ Percentage, Interest, SfC- (170.) A Proportion occurring in the solution of a Problcw, miy b« converted into an Equation by putting the product of the two extremes equal to the product of the two means, (149). EXAMPLE. To divide $1000 between three persons, in the proportions of 2, 3, and 5 ; that is, so that A's share shall be to B's as 2 to 3, and B's to C's as 3 to 5. Let X represent A's share ; y, B's share ; and z, C's share. Then, by the conditions of the problem, v/e have x-{-y-{-z=zlOOO; - x:tj::2:3; and y : z : : 3 : 5. By converting the two Proportions into Equations, we find 3oc=2y, and 57/=3z. We have now three equations from which to find the values of the three unknown quantities, (125). The solution of the Problem may also be effected with one unknown quantity, by finding fourth proportionals for the shares of B and C. Let X represent A's share ; then 2 :3 :: X :—] and 2 : 5 : : a; : — , (150). 3x 5x Hence B's share is represented by—, and C's by—; and the equation of the problem is 3x 5x ^ ^^^ :?:+ Y + Y^Siooo. Ans. The several shares are $200, $300, $500. The general Problem of which the preceding is a particular case» may be stated thus ; — To divide the sum s between three persons in the proportions of a, b, and c. " The Formulas which would be found for the several shares, are as ^^ . ^^ a+b-\-c' a+b-\-c' a-\'b-\-c' These Formulas translated into arithmetical language, would fur- nish a Proposition, or a Rule, which might be applied to any given case of the general problem. PROBLEMS. lOo It may be here remarked that algebraic Formulas express genera] principles and methods of solution with the utmost distinctness and brevity, — but that it is not always possible to translate them into con- cise and perspicuous phraseology. EXEPwCISES. 1. Divide $950 between two persons so that their shares shall be to each other as 3 to 5. This problem might be solved without employing propo7'tion, by ob- serving that the first share will be | of the second. Ans. $356i; $593J. 2. Find the Formulas for dividing any given sum s between two persons so that the shares shall be to each other as any two numbers a and b. . as . bs Ans. and . a-\-o a-f-o 3. Divide the sum of $3000 between A, B, and C, in the proper* tions of 1, 2, and 3. Ans. $500 ; $1000 ; $1500. 4. Divide the sum of $7600 between three persons, in the propor- tions of -J, ^, and i . Ans. $4000; $2000; $1600. 5. A bankrupt is indebted to A $400, and to B $700. He is able to pay to both $900 ; what sum should each of the two creditors re- ceive ? The $900 should be divided in the proportion of 400 and 700, or, of 4 and 7, (158). Ans. $327y\ ; $572-5-^. 6. Three persons engaged in a speculation towards which they contributed, respectively, $300, $400, and $500. The profit amounted to $550 ; what are the respective shares of profit? Ans. $137^; $183^;^ $229^. 7. A, B, and C in a joint mercantile adventure lost $742. A's part of the capital employed was to B's as 4 to 3, and B's was to C's as 5 to 6 ; what amount of loss should be borne by each ? Ans. $280; $210; $252. 8. Four persons rented a pasture, in which the first kept 8 oxen, the second 6, the third 10, and the fourth 12. The sum paid was $40 ; what amount should have been paid by each person ? Ans. $8f ; $6f ; $nj; $13^ 9. A testator bequeathed his estate, amounting to $7830, to his three children, in such a manner that the share of the first was to that of the second as 2 J to 2, and the share of the second to that of the third as 3J to 3. What were the shares? Ans. $3150; $2520; $2160. 10. A, B, C, and D together have $3000 ; A's part is to B's as P to 3, B and C together have $1500, and C's part is to D's as 3 to 4 "What is the sum possessed by each person ? Ans. $500; $750: $750; $1000. 106 PROBLEMS. 11. Three persons contributed funds in a joint speculation as fol lows : A $200 for 5 months, B $400 for 3 months, and C $500 for 4 months. The profit amounted to $600 ; what are the several shares of profit ? Each Dollar contributed produced a Profit proportional to the Time it was in the business. Each person's share of profit is therefore proportional to his amount of capital X its time ; in other words, the respective shares are to each other in the compound ratio of capital and time, (131), Hence, A's share of profit is to B's as 200 X 5 to 400 X 3 ; and A's is to C's as 200 x 5 to 500 x 4. Dividing the antecedent and consequent by 100, these ratios become 2x5 to 4x3 and 2x5 to 5x4, (158). By still further reductions, on the same principle, the ratios become 5 to 6, and 1 to 2. Ans. $142f ; $17lf ; $285f. 12. Two persons rented a pasture for $43. The first put into it 100 sheep for 15 days, and the second 120 sheep for 9 days; what amount of rent should be paid by each person ? Am. $25 and $18. 13. A, B, and C trade together; A ventures $1000 for 5 months, B $1200 for 4 months, and C $800 for 7 months. The profits of the partnership amount to $2310 ; what share of profit should be assigned to each? Ans. $750; $720; $840. 14. An estate consisting of 1000 acres of land is to be divided be- tween three persons, so that the first share shall be to the second as 2 to 3, and the first to the third as 1 to 2. What are the shares ? An^. 222f ; 333^ ; 444|^, acres. 15. Two men contracted to do a certain work for $5000. In ac- complishing the work, the first employed 100 laborers for 50 days ; and the second 125 laborers for 60 days ; — to what shares of the stipulated sum are the two men respectively entitled ? An^. $2000 ; and $3000. 16. A gentleman bequeathed $18000 to his widow and his three sons, in 'the proportions of 2, 2\, 3, and 3|^, respectively. His widow dying before the division was effected, the whole is to be divided pro portionably among the three sons. What are their several shares ? Arts. $5000; $6000; $7000. 17. Find the Formulas for dividing, between two partners, the profits s of a joint adventure, in which the first had the Capital a for the Time h, and the second the Capital c for the Time d. ahs , cds Ans. — — --and-r- — -. ab-\-cd aO-\-cd PROBLEMS. -^Qy Problems in Percentage. (171.) Percentage is an allowance at a certain rate per hundred ; and this rate is called the Q-ate per cent., from the Latin centum, which means a hundred. The ratio of percentage is the rate per cent. -^ 100, and is there- fore equal to the rate per unit. The hads, of percentage is the sum or number on which an amount of percentage is computed. From these definitions it follows, that (172.) The had^ of percentage X the ratio of percentage produces the amount of percentage ; and, conversely, that the amount — the ratio produces the basis of percentage. \^^ The Student may be required to write the Rules of Percent- age which are indicated by the Formulas among the following prob- lems. • 18. A merchant finds that his capital, which is now $12000, has increased in one year at the rate of 20 per cent. ; what was his capital at the beginning of the year ? Let X represent his capital at the beginning of the year ; then 100 : 120 : : a; : 12000. Ans. $10000. 19. "What is the Formula for finding a sum of money which, in- creased at the rate of r per cent., shall amount to the sum a. lOOd? Ans. — — . 100-fr 20. An agent receives $500 to be laid out in merchandise, after deducting his commission of 1^ per cent, on the amount of the pur- chase; What will be the amount of the purchase ? Ans. $492.61'. 21. A merchant obtains an insurance at 2 per cent, on a stock of goods valued at $7500, which includes this amount and the premium for the insurance. What is the sum insured ? Ans. $7653.06'. 22. What is the Formula for finding a sum of money which, di- minished at the rate of r per cent., shall amount to the sum a ? 100a Ans. — — . 100— r 23. The profits of a manufacturing company this year amount to $3096, which is 3i per cent, less than their profits last year. What was the amount ot profits last year ? Ans. $3200. 24. What must be the percentum of profit at which a quantity of merchandise, bought for $3750, must be sold, that the whole amount of profit shall be $1500 ? Ans. 40 per cent. 108 PROBLEMS. 25. What is the Formula for finding the rate 'per cent, at which the sum s must be increased to produce the sum a ? 100(a— s) s 26. A quantity of silk was purchased for $220, and, on account of its having become damaged, was sold for $176. What was the per- centum of the loss sustained 1 Ans. 20 per cent. 27. What is the Formula for finding the rate per cent, at which the sum s must be diminished to leave the sum a 1 lOO(s-a) Ans. ^^ ■ . ^ s 28. What amount of stock in an Insurance Office, at a discount of 5 per cent,, could be purchased for $3800 ? Ans. $4000. 29. A merchant finds that his capital, which is now $4350, has decreased in one year at the rate of 12^ per cent. What was his capital at the beginning of the year ? Ans. $4971.42'. 30. What amount of stock in a manufacturing establishment, at an advance of 6|- per cent., could be purchased for $1200 ? Ans. $1126.76'. 31. A quantity of damaged cloth was sold for $250, — which was at a loss of 16f per cent. For what sum was the cloth purchased ? Ans. $300. 32. The population of a city increased from 7850 to 11775 inhabit- ants, in one year. What was the percentum of increase during the year ? Ans. 50 per cent. 33. An agent receives $2030 to invest in merchandise — himself to retain a commission of 1^ per cent, on the amount of the purchase. What is the sum to be invested ? Ans. $2000. 34. A merchant wishes to effect an insurance on a stock of goods, amounting to $3573, which shall cover both the value of the goods and the premium of insurance. What is the sum to be insured, al- lowing the rate to be j- per cent ? Ans. $3600. 35. What amount of stock in a Savings Bank, at an advance of 5 per cent., could be purchased for $4200 1 and what amount in another, at a discount of 5 per cent., could be purchased for $1995 ? Ans. $4000, and $2100. PROBLEMS. 109 Problems in Interest^ Sfc. (173.) Interest is the price or premium paid for the use of money, and is reckoned at a certain percentum, annually, on the sum for which it is paid. The Principal is the sum for wliich Interest is paid — the Amount is the sum of the Principal and Interest. From the principles of Percentage, (172), it is evident that, (174.) The Principal X the ratio of percentage produces the inter- est for one year. For the ratio of percentage, in this case, is the inter- est of $1 for one year. [t^ The Student may be required to write the Rules of Interest which are indicated by the Formulas among the following problems. 36. What Principal would amount to $1000 in 5 years, allowing the rate of interest to be 6 per cent ? Let X represent the Principal required ; , 6 G.-K . , _ - then X X — -— or — -— is the Interest for one year ; and by adding 5 years' interest to the Principal, 30a: we have x-\ -— - =$1000. 100 Ans. $769j^. 37. What is the Formula for finding the Principal which, at inter- est at r per cent., would amount to the sum a in t years? \ma /^^^•roo+7r 38. What Principal would amount to $2500, in 10 years, allow- ing the rate of interest to be 7 per cent. 1 Ans. $1470.588'. 39. At what Rate per cent, must $1000 be put on interest, to amount to $1150 in 2 years and 6 months? Ans. 6 per cent. 40. What is the Formula for finding the Rate per cent, of interest at which the sum s would amount to the sum a in t years ? Ans. ^^ -. St 41. In how many years would $6000 amount to $7470, allowing the rate of interest to be 7 per cent? A7is. 3 J years. 110 PROBLEMS. 42. What is the Formula for finding the Time in which the sum s would amount to the sum a, if the interest he at r per cent ? Ans. M^> . sr 43. What Principal would produce as much interest in 3 J years, as $500 would in 4 years, the rate of interest in both cases being 6 percent] Ans. $57 If. 44. At what Rate per cent, of interest would $525 produce the game amount of interest in 5 years, that $700 w^ould produce at 5 per cent, in 3 years ? Ans. 4 per cent. 45. A person who possessed a capital of $70000, put the greater part of it at interest at 5 per cent., and the other part at 4 per cent. The interest on the whole was $3250 per annum ; required the two parts. Ans. $45000, and $25000. 46. Tlie sum of $200 is to be applied in part towards the payment of a debt of $300, and in part to paying the Interest, at 6 per cent., in advance, for 12 months, on the remainder of the debt? What is the Amount of the payment that can be made on the debt ? Let X represent the payment ; then (300— a:) X i%q is the Interest on the remainder of the debt ; and we have therefore the Equation, a;+(300-a:)x 1^-0=200. Am. $193.61'. 47. A is indebted to B $1000, and is able to raise but $600. With this sum A proposes to pay a part of the debt, and the Interest, at 8 per cent., in advance, on his Note at 2 years for the remainder. For what sum should the note be drawn ? Ans. $476.19'. 48. Find the Formulas for dividing the sum s into two parts, one of which is to be applied towards the payment of a debt of n dollars, and the other to paying the interest, in advance, on the remainder of the debt, for t years, at r per cent, per annum. 100s— wr^ , rt(n—s) ^^^- -T7^ » and ~~ -. 100— r^ 100— ri What would be the Rule for finding the amount oi payment that could be made on the debt ? Ill m CHAPTER VII. ahithmetical, harmonical, and geometrical progression. ARITHMETICAL PROGRESSION. (175.) An arithmetical progression is a series of quantities which continually increase or decrease by a common difference. Thus 1, 3, 5, 7, 9, is a Progression in which the quantities increase by the continual addition of the common difierence 2. And 15, 12, 9, 6, 3, is a progression in which the quantities de- crease by the continual subtraction of the common difference 3. The first and last terms of the Progression are called the two ex- tremes, and all the intermediate terms the means. The theory of Arithmetical Progression is contained in the follow- ing propositions. The Last Term. (176.) The last term of an increasing Arithmetical Progression, is equal to the first term + the product of the common difference X the number of terms less one ; and in a decreasing Progression it is equal to the first term — the same product. Let a be the first term, and d the common difference ; then in an increasing progression the series will be, a, a-{-d, a-\-2d, a-\-2d, a-\-^d, &c. ; and in a decreasing progression the series will be, a, a—d, a— 2d, a— 3d, a— Ad, &c. In these series the fifth or last term adzAd, a plus or minus A.d, is the first term a plus or minus 4 times the common difference d. And the proposition is evidently true for any number of terms. (177.) Cor. The common difference of the terms in an Arithmetical Progression^ is equal to the difierence between the two extremes -i- the number of terms less one. 222 PROGRESSION. The Sum of the Tioo Extremes. (178.) The sum of the two extremes in an Arithmetical Progression, is equal to the sum of any two terms equidistant from them, or to twice the middle term when the number of terms is odd. Let a be the first term, and d the common difference ; then in an increasing progression the series will be, «, a+d, a-\-2d, a-\-3d, a-\-4:d, &c. Of these five terms the sum of the frst and the last, is a+(a-\-4:d)=2a+4:d. The sum of the seco^z^Z and the fourth, which are equidistant from the extremes, is {a-{-d)-\-{a-\-3d), also =2a-\-4^d. "We see moreover, that the sum of the two extremes is equal to tioice the middle term, a-\-2d. In like manner the proposition will be found true for any number of terms ; as also when the Progression is a decreasing one. (179.) Cor. An arithmetical mean between two given terms, is equal to half the sum of those terms. For the sum of the two given terms, considered as the two extremes of an Arithmetical Progression, is equal to twice the mean or middle term. The Sum of all the Terms. (180.) The S2im of all the terms of an Arithmetical Progression, is equal to half the sum of the two extremes X the number of terms. To prove this proposition we add the several terms of an Arithme- tical Progression to those of the same progression reversed ; thus a, a+ d, a-\-2d, a-\-3d, a+od a-\-2d a-\- d a 2a+3d' 2a+Sd' 2a+3d' 2a-\-3d' The sum {2a-{-3d)-\-(2a-{-3d) &c. of the two series, is the sum of the two extremes in either series X the number of terms ; hence the sum of either series is equal to half the sum of the two extremes x the number of terms. The demonstration will evidently apply to any number of terms. /* PROGRESSION. jj3 Formulas in Arithmetical Progression. (181.) In an Arithmetical Progression, let a be the first term, c?the common difference, n the number of terms, I the last term, and S the sum of all the terms. Then A . . . . l=:a±Ld{n~\) (176); B . . . . S-=^\n{a^l) (180). The sign + is to be prefixed to d{n—\) when the progression is an increasing one, and — when decreasing. In these two Formulas we have the five quantities, a, d, n,l,s; hence if any three of these quantities be given, the values of the other two may be found from the two Equations, (121). HARMONICAL PROGRESSION. (182.) An HARMONiCAL PROGRESSION is a scrics of quantities such that, of any three consecutive terms, the first : the third :: the differ- ence between the first and second : the difference between the second and third. Thus the numbers 3, 4, 6, 12, are in harmonical progression, since 3 : 6 :: 4—3 : 6—4, and 4 : 12 :: 6-4 : 12-6. An Harmonical Proportion consists of four terms such that the first is to the fourth as the difference between the first and second is to the difference between the third and fourth. Thus a, b, c, d, are in Harmonical Proportion, if a : d :: a — h : c — d\ ox a'.d V. h—a : d—c. The numbers 16, 8, 3, 2, are in Harmonical Proportion, since 16 : 2:: 16-8 : 3-2 . The first and last terms of the Progression or proportion are called the two extremes, and all the intermediate terms the means. i]4 PROGRESSION. An Harmonical converted into an Arithmetical Progression. (183.) The reciprocals of the terms of an Harmopical progression, are in Arithmetical progression. Let a, hy c, be three consecutive terms of a decreasing harmomc^J progression. Then a : c :: a— I : h—c, (182) ; Converting this proportion into an equation, wo have ah—ac=ac—bc. Dividing each term in this equation by abc, and leduoitg +ha sev eral quotients to their lowest terms, we find JL-JL _ J- _ i_ c b ~~ b a * Transposing the first and the last term of this equation a b ~ b c ' We thus find that the difference between the reciprocals of a and h, IS equal to the difference between the reciprocals of b and c ; hence these reciprocals are in Arithmetical Progression, (175). The numbers 3, 4, 6, 12, form an Harmonical Progression, (182) : by taking the reciprocals of the several terms we have the Arithmeti- cal Progression m which the common difference of the terms is ^. (184.) An harmonical mean between two given terms, is equal to twice their product divided by their sum. From the first of the preceding equations, namely, ab—ac^=ac—hcy we shall find b=z ; a+c and b is the harmonical mean between a and c. The harmonical mean between 3 and 6, is 3x6x2 _ 36 _ 3 + 6 ~ "9" ~ • PROGRESSION. 1JL5 GEOMETRICAL PRO GRESSIO'N. (185.) A GEOMETRICAL PROGRESSION is a series of quantities in which each succeeding term has the same ratio to the term which immediately precedes it. Thus 1, 2, 4, 8, 16, is an increasing Progression in which each succeeding term is double the one which immediately precedes it ; that is, the ratio of the progression is 2. And 27, 9, 3, 1, -J, is a decreasing progression in which each suc- ceeding term is one-third of the one which immediately precedes it ; and the ratio of the progression is consequently -J. Hence the successive terms of a Geometrical Progression consist of the first term midtiplied continually into the ratio, that is, multiplied into the successive poivers of the ratio. The theory of Geometrical Progression is contained in the following propositions. The Last Term. (186.) The last term of a Geometrical Progression, is equal to the first term X that poiver of the ratio which is expressed by the number of terms less one. Let a be the first term, and r the ratio of the progression ; then, multiplying a continually into r, the series will be a, ar, ur"^, ar^, ar*, &c. Since the ratio r begins in the second term, with exponent 1, its exponent in the last term will always be one less than the number of terms ; hence the last term consists of the first X into that power of r which is expressed by the number of terms minus 1. (187.) Cor. The last term of a Geometrical Progression -^- the first term, gives that power of the ratio which is expressed by the number of terms less one. Thus ar^ -^a^=:r^ ; the number of terms being^z;^. 116 PROGRESSION. Product of the two Extremes, (188.) The 'product of the two extremes in a Geometrical Progres- sion, is equal to the product of any two terms equidistant from them^ or to the squxire of the middle terrn when the number of terms is odd. Let a be the first term, and r the ratio of the progression ; theYi, multiplying a into the successive powers of r, the series is a, ar, ar^, ar^, ar*, &c. Of these five terms the product of the frst and the last, is axar^=a^r^. The product of the second and the fourth, which are equidistant from the extremes, is arxar^, also =:a^r^. We perceive moreover that the product of the two extremes is equal to the square of the middle term ar^. In like manner the proposition will be found true for any number of terms. (189.) Cor. A geometrical mean, or a mean proportional, between two given terms, is equal to the square root of the product of those terms. For the product of the two given terms, considered as the two ex- tremes of a Geometrical Progression, is equal to the square of the mean or middle term. The Sum of all the Terms. (190.) The sum of all the terms of a Geometrical Progression, is equal to the difference between the frst term and the product of the last term X the ratio, -f- the difference between the ratio and a unit. Let iS represent the sum of the terms, and we shall have S=:a-\-ar-\-ar^-{-ar^-{-ar*, &c. Multiplying both sides of this equation by the ratio r, Srz^ar-{-ar^-^ar^-\-ar^-{-ar^, Subtracting the first equation from the second, we find Sr—S= ar^—a; which gives o= . r— 1 In the numerator of this value of S, observe that ar^ is the last term ar^ of the progression X the ratio r. The demonstration will evidently apply to any number of terms. PROBLEMS. 117 (191.) The sum of an infinite number of terms, in a decreasing Geometrical Progression, is equal to \k\.Q first term divided by the dif- ference between the ratio and a unit. In a decreasing progression the terms continually diminish in a constant ratio ; and if the number of terms be infinite, the last term will be 0. The last term X the ratio will then be 0, and the expres- sion for the sum of the terms, found above, will become 1— r The divisor in this case is 1—r, because the ratio of the progression being a. proper fraction, is less than a unit. Formulas in Geometrical Progression. (192.) In a Geometrical Progression, let a be the first term, r the ratio, n the number of terms, I the last term, and s the sum of all the terms. Then . . . . Z=ar«-i, (186), jy,,,,sJ^, (190). When the progression is a decreasing one, and the number ol terms is infinite, E. . . .S=-^ , (191). 1—r In the Formulas C and D we have the five quantities, a, r, n, I, s; hence if any three of these quantities be given, the values of the other t'lvo may be found from the two Equations, (121). The principles which have been established in this Chapter may be applied to the solution of the following Problems in Progressions. 1. The first term of an increasing Arithmetical progression is 3, the common difference of the terms is 2, and the number of terms 20. What is the last term ? and the sum of all the terras ? Ans. 41, and 440. 2. The first term of a decreasing Arithmetical progression is 100, the common difference of the terms is 3, and the number of terms 34. What is the last term ? and the sum of all the terms ? Ans. 1, and 1717. 118 PROBLEMS. 3. What is the sum of the numbers 1, 2, 3, 4, 5, &c., continued to 1000 terms? Ans. 500500. 4. What is the common difference of the terms in an Arithmetical progression whose first term is 10, last term 150, and number of terms 21? Ans. 7. 5. If the third term of an Arithmetical progression be 40, and the Jlfth term 70, what will the fourth term be ? A?is. 55. 6. If the first term of an Arithmetical progression be 5, and the fifth term 30, what will the second, third, and fourth terms be ? Find the common difference, (177), and thence the three interme- diate terms. Ans. \1\\ Yl\\ 23f. 7. If the fourth term of an Arithmetical progression be 37, and the eighth term 60, what are the intermediate terms ? Ans. 42f ; 48^, 54^. 8. What is the sum of 25 terms of an increasing Arithmetical pro- gression in which the first term is -J, and the common difierence of the terms also -J? (176). " Ans. \^2\. 9. The first term of an increasing Arithmetical progression, is 1, and the number of terms 23. What must be the common difference, that the sum of all the terms may be 100 ? Let X represent the common difference ; then 1 + 22a; is the last term, (176) ; 2 + 22a; and -~- X 23 is the sum of the terms, (180). Hence an Equation may be formed from which the value of x will be found. — Or we might substitute the numbers 1, 23, and 100, for a, n, and s in Formulas A and B, (181), and find the value of d, as one of the tivo unknoivn quantities. Ans. -^. 10. If the first term of a decreasing Arithmetical progression is 100, and the number of terms 21, what must the common difference be, that the sum of the series may be 1260 ? Ans. 4. 1 1 . A and B start together, and travel in the same direction ; A goes 40 miles per day ; B goes 20 miles the first day, and increases his rate of travel f of a mile per day. How far will they be apart at the end of 40 days? Ans. 215 miles. 12. One Hundred stones being placed on the ground in a straight line, at the distance of 2 yards from each other ; how far will a person travel who shall bring them, one by one, to a basket which is placed 2 yards from the first stone ? Ans. 1 1 miles 840 yards. PROBLEMS. iiy 13. Find the third ^erm of an Harmonical progression whose first and second terms are 12 and 15 respectively. If X represent the third terni; we shall have 12 : X :: 15 — 12 : x—l5. Arts. 20. 14. What is the first term of an Harmonical progression whose second and third terms are 30 and 20 respectively ? Ans. 60. 15. What is the fourth term of an Harmonical p?-oportion whose first, second, and third terms are 2, 3, and 8 respectively ? A71S. 16. 16. If the first and third terms of an Harmonical progression be 25 and 40 respectively, what will the second term be ? Ans. 30^|-. 17. The first and fourth terms of an Harmonical progression, are 10 and 20 respectively. What are the two intermediate terms ? This problem may be solved by finding two arWwietical means between -^^ and ^, and then taking the reciprocals of the terms thus found, (183). Ans. 12, and 15. 18. The fifth and eighth terms of an Harmonical progression are 20 and 40 respectively. What are the two intermediate terms ? Ans. 24, and 30. 19. The first term of a Geometrical progression is 2, the ratio of the progression is 3, and the number of terms 4. What is the last term ? and the sum of all the terms ? A7is. 54, and 80. 20. The first term of a Geometrical progression is -J, the ratio of the progression is -J, and the number of terms 4. What is the last term ? and the sum of all the terms ? Ans. -^j, and |^. 21. What is the sum of an infinite number of terms in the Geome- trical progression whose first term is 100, and ratio ^? Ans. 133^. 22. What is the sum of an infinite number of terms in the Geome- trical progression whose first term is 300, and ratio ^1 Ans. 450. 23. If the first and third terms of a Geometrical progression are 8 and 72 respectively, what is the second term 1 The second term is equal to the square root of 8 X 72, (189). Or, considering the third as the last term of the progression, 72 -4- 8 ==9 is the square o^ the ratio, (187) ; then 3 is the ratio of the progression ; and the second term is now readily obtained. • Ans. 24. 24. If the third and fifth terms of a Geometrical progression be 75 and 300 respectively, what will the fourth term be ? Ans. 150 10^ PROBLEMS. 25. If the first and fourth terms of a Geometrical progression are 3 and 24 respectively, what are the two intermediate terms ? Ans. 6 and 12. 26. If the seventh and tenth terms of a Geometrical progression are 6 and 750 respectively, what are the intermediate terms ? A?is. 30 and 150. 27. What is the sum of an infinite number of terms in the series 1. h i> ^c., in which the ratio of the progression is evidently -J-? Ans. 2. 28. If a body move forever at the rate of 2000 feet the first second 1000 the second, 500 the third, and so on, what is the utmost distanct it can reach? Ans. 4000. 29. If 10 yards of cloth be sold at the rate of $1 for the first yard $2 for the second, $4 for the third, and so on, what would be the prict of the last yard ? and what would the whole amount to ? Ans. $512, and $1023. 30. If 13 acres of land were purchased at the rate of $2 for the first acre, $6 for the second, $18 for the third, and so on, what w^ould the last acre amount to? Ans. $1062882. 31. Allowing the interest of a sum of money to be $500 the first year, $400 the second, $320 the third, and so on, forever, what would be the whole amount of interest ? Ans. $2500. 32. Two bodies move at the same time, from the same point, in opposite directions. One goes 2 miles the first hour, 4 the second, 6 the third, and so on ; the other goes 2 miles the first hour, 4 the se- cond, 8 the third, &c. ; how far will they be apart at the end of 12 hours ? Ans. 8346 miles. 33. A and B set out at the same time to meet each other. A tra- vels 3, 4, 5, &c. miles on successive days, and B 3, 4^, 6f, &c. miles on successive days. They meet in 10 days; what is the distance be- tween the two places from which they traveled ? Ans. 414f5-| miles. 121 CHAPTER VIII. PERMUTATIONS AND COMBINATIONS.-INVOLUTION.— BINOMIAL THEOREM.— EVOLUTION. PERMUTATIONS. (193.) Permutations are the different orders of succession in which a given number of things may be taken — either the whole number to- gether, or the whole number taken two and two, or th^'ee and three, &c. Thus the different Permutations of the three letters a, b, and c, when all are taken together, are abc, acb, bac, cab, bca, cba. And the different Permutations of the same letters when taken two and ifi^'o. are ab, ba, ac, ca, be, cb. Number of Permutations. (194.) If ?i represent a given number of things, the number of per^ mutations that can be formed of them, will be equal to n{n — \\n—2){n — 3)(?^— 4), and so on, until the number o{ factors multiplied together is equal to the number of things taken in each permutation. To demonstrate this proposition, — suppose that we have n letters, a, b, c, d, &c., to be subjected to Permutations. If we reserve one of the letters, as a, there will remain n — 1 let- ters ; and by writing a before each of the remaining letters, we have n — 1 permutations of n letters, taken two and two, in which a stands first. In like manner we should find n — 1 permutations of ?^ letters, taken tivo and tivo, in which b stands first ; and so for each of the n letters. Hence we shall have n{n—\) permutations oin letters taken two and two. Suppose now that the n letters are to be taken three and three. By reserving «, and proceeding with n — 1 letters as before, we should find {n — l)(w — 2) permutations oi n — 1 letters taken two and two ; and by writing a before each of these permutations, we have 222 PERMUTATIONS. (n—l){n—2) permutations of n letters, taken three and three^ in which a stands first. We should find the same number of permutations of n letters, taken three and three, in which b stands first ; and so for each of the n letters. Hence we shall have n{'n—l)(n—2) permutations ofn letters taken three and three. Suppose now that the n letters are to be taken four and four. By reserving «, and pursuing the operation in the same manner as before, we should find n{n—l){n—2)(n—3) permutations of n letters taken four and four. Thus the demonstration proceeds ; the number o^ factors multiplied together being found always equal to the number of letters taken in each permutation. As an Example of the application of the principle above demon- strated, — suppose it were required to determine the number of Permu- tations, or different orders of succession, that could be formed in a class composed of six pupils, by taking the whole number in each permu- tation. Since the six pupils are to be taken in each Permutation, the num- ber of factors to be employed is six ; hence the number of permuta- tions is 6x5x4x3x2x1=720. If the whole six were to be subjected to Permutations by taking five at a time, the number of permutations would be 6x5x4x3x2 = 720. If the whole six were to be subjected to Permutations by taking four at a time, the number of permutations would be 6x5x4x3=360. [IF^ It will be observed above, that the number of Permutations will be the same, whether the whole number of things, or one less than the whole number, be taken in each permutation. COMBINATIONS. 123 COMBINATIONS. (195.) Combinations are the different collections whicli may be formed out of a given number of things, by taking the same number in each Collection — without regard to the order of succession. Thus the different Combinations which may be formed out of the three letters a, h, and c, by taking two at a time, are ah, be, ac. Observe that ah and ha are not different comhinations, but different permutations, ef the letters a and h. In Permutations we have regard to the order of succession, and may therefore have two permutations of two things. In Combinations we do not consider the order of succession ; so that the combination of two or more things is the same, in whatever order they are taken. Number of Combinations. (196.) If ^ represent a given number of things, the number of comr binations that can be formed out of them, will be equal to . n(n—])(n—2)(n—3) . — ^-^ , and so on, 1.2.3.4 until the number of factors in the dividend, and also in the divisor, is equal to the number of things taken in each combination. To demonstrate this proposition, we observe that the numerator in the preceding expression, is the number of permutatio?is of n things tsikenfour said four, (194). On the same principle, the denominator 4x3x2x1 is the num- ber of permutations of/cK^r things taken all together. Now since there can be but one combination of 4 things taken all together, the number of permutations of n things taken ^bwr and four is 1.2.3.4 times, that is, 24 times, the number of combinations. Hence the number of Combinations of n things taken /owr and four, is 2'^4 of the number of Permutations ; and will therefore be found by dividing the numerator by the denominator. This mode of demonstration may evidently be applied whatever be the number of things taken in each combination. 124 PROBLEMS. PROBLEMS In Permutations and Combinations. It In how many different ways might a company of 10 personi seat themselves around a table? (194). A7is. 3628800. 2. How many different numbers might be expressed by the 10 nu- meral figures, if 5 figures be used in each number ? A^is. 30f240. 3. In how many different ways may the names of the 12 months of the year be arranged one after another? Ans. 479001600. 4. How many different permutations of 8 men could be formed out of a company consisting of 15 men ? Ans. 259459200. 5. In how many different ways might the seven prismatic colors, red, orange, yellow, green, blue, indigo, and violet, have been arranged in the solar spectrum ? Ans. 5040. 6. How many different combinations of two colors could be formed out of the 7 prismatic colors? (196). Ans. 21. 7. How many different combinations of 5 letters may be formed out of the 26 letters of the Alphabet? Ans. 65780. 8. How many different combinations of 2 elements might be formed out of the 56 elements described in Chemistry? Ans. 1540. 9. In how many different ways might a company of 20 men be arranged, in single file, in a procession ? Ans. 2432902008176640000. 10. A farmer wishes to select a team of 6 horses out of a drove containing 10 horses. How many different choices for the team will he be able to make ? Ans. 210. 11. In how many different ways might the planets Mercury, Venus, the Earth, Mars, Jupiter, Saturn, Uranus, and Neptune succeed one another in the solar system ? Ans. 40320. 12. A company of 20 persons engaged to remain together so long as they might be able to combine in different couples in their eve-, ning walks. What time will be required to fulfil the engagement ? Ans. 200 days. 13. How many different permutations of 7 letters might be formed out of the 26 letters of the Alphabet ? Ans. 3315312000. 14. In an exhibition of a Public School, 5 speakers are to be taken from a class of 15 students. How many different selections of the five might be made ? and in how many different ways might the 5 suc- ceed one another in the delivery of their speeches ? Ans. 3003 and 120. 15. Out of a Company consisting of 100 soldiers Six are to be taken for a particular service. How many different selections of the 6 might be made ? and in how many different ways might the 6 chosen be disposed with regard to the order of succession ? Ans. 1192052400; and 720 INVOLUTION. 125 INVOLUTION. (197.) Involution consists in raising a given quantity to any re- quired poiver. This may always be effected by multiplying the quantity into itself as many times less one as there are units in the ex- ponent of the power. Thus aa is a^, the second power or square of a : aaa is a^, the third power or cube of a ; and so on. Obser-w that one multiplication of a into itself produces the second power of a ; two miultiplications produce the third power, and so on ; also that the number of times the quantity becomes o. factor in raising a Power, is equal to the exponent of the Power. (198.) A higher power of a quantity may also be found by multi- plying together two or more lower powers (of the same quantity) the sum of whose exponents is equal to the exponent of the required power. Thus a^ xd^ produces «* ; a^ xa^ produces a^, &;C. Powers of Unity ^ Monomials, Fractions, SfC. (199.) Every power o^ unity is unity, since any number of Is mul- tiplied together produce only 1 ; thus 1x1x1 &c. =1. (200.) A monomial is raised to any required Power by raising its numerical coefficient to that power, and multiplying the exponents of its other factors by the exponent of the poiver. Thus to find the third power of \ax^, we raise 4 to its third power, which is 4 X 4 X 4=r 64, and multiply the exponents of a and a; by 3 : we thus obtain 64a^a;^. Observe that multiplying the given exponents by the exponent of the required Power, is only a brief method of performing the requisite multiplications in the Involution of Monomials. (201.) A fraction is raised to any required Power, by raising its numerator and denominator, separately, to that power. ^, , , . , - a^ . a^ Thus the third power oi — is ^ . (gg), A mixed quantity may be raised to any required Power by involv- ing the equivalent improper fraction. J2G INVOLUTION. The Sign to be Prefixed to a Power. (202.) Every even power of a quantity is positive ; while every odd power has the same sign as the quantity from which it is derived. Thus if a be positive, all its powers, as aa, aaa, and so on, will evidently be positive ; but if a be negative we shall have —a.—a—a^\ a^,—a = —a^; —a^.—a — a^\ a*.— a=— a^, &c. from which it is plain that all the even powers, as the 2d, 4th, and so on, will be positive, while all the odd powers will be negative. EXERCISES. On the Powers of Monomials. 1. Find the square of Saa;^. 2. Find the cube of —2d^x. 3. Find the square of — Aax^, 4. Find the cube of 3a^x. 6. Find the square of a6 2 ^3^ 6. Find the cube of —a'^x^y. 7. Find the square of -Jaaj^. 8. Find the cube oi ^ay^ . 9. Find the square of — f«6". 10. Find the cube of — fax^. 11. Find the 4th power of 2a". 12. Find the 4th power of —^a^. 13. Find the 4th power of —3x^. 14. Find the 5th power of ■J-a;^. 15. Find the 6th power of —ax^, 16. Find the 6th power of2y^. 17. Find the 7th power of —a^y^, 18. Find the 7th power of -Jy. 19. Find the 8th power of a^b"'. 20. Find the 8th power of —^x^. Ans. 9a^x*. A71S. - -Sa^x^. Ans. IQa^x^. Ans. 27a^x^, Ans. aH^c\ Ans. — a^x^y^. Ans. ia^x*. Ans. ia^yK Ans. iaH^\ Ans. — ^^x^. Ans . 16a^\ Ans. Y6«®. Ans. 81a;8. Ans. A^^^ Ans. a^ajia. Ans. 642/12. Ans. — a^^tf. Ans. -ife/^ Ans. a^e^sn^ Ans. ^k^^^ binomal theorem. 227 Powers of Polynomials^ (203.) The Powers of a binomial, or of any 'polynomial, may be obtained by successive multiplications of the quantity into itself, (197). Thus the 2d power or square oia-\-h, is (a+^)(a-f^>)=:a2_j_2a5+i2. And the 3d power or cube of «4-i, is («+6)(a+^>)(a+^')=a^ + 3a26+3a&2_|_j3^ The Involution of Binomials, however, and thence of Polynomials, is greatly facilitated by the application of Newton's Binomial Theorem. (204.) This Theorem explains a general method of developing a Binomial according to any exponent with which the Binomial may be afiected. In developing (a+i)^ by the Binomial Theorem we should obtain the 2^ power or square of (a+i). In developing [a-^-h^ we should obtain the square root of (a-j-i) ; and so for other exponents. Development of {a±.bY ; • n representing any exponent. 1. The Exponent of . By applying the principles of the Binomial Theorem, we obtain a^+3a^.2b+3a(2by-\-(2b)^ ; which may be developed into Ans, a^-^6a^b-\-12ab^-\-Sb^. 4. Find the square of 3a+y- Ans. 9a^-\-Qay-{-y^. 5. Find the cuhe of 2a— 3a;. A7is. 8a^—36a^x+54:ax^—27x^. 6. Find the square of a+ac — y, By operating on {x—y) as if it were a monomial, and applying the Binomial Theorem to a-\-{x—y), we obtain a^-^2a(x-y)+{x-tj)^; which may be developed into Ans. a^-\-2ax—2ay-\-x^—2xy+y^. 7. Find the square of a— 5+2/. J.?^s. a^—2ab-\-2ay-\-b^—2by-\-y^. 8. Find the square of a— a?— y. ^ws. a^—2ax-{-x^—2ay-\-2ocy-{-y^. 9. Find the square of a-{-b~2x. Ans. a^ + 2ab-\-b^ —4:ax—4bx-{-4:X^ . 10. Find the square of a^ + a;^- This square, according to the Binomial Theorem, may be indicated thus : (a2)24.2a2a?3 + (a;3)2; which may be developed into Ans. a^-\-2a'^x^-\-x^. 11. Find the cube of a2_y 3^ ji^^ a^ — 3a^y^ + 3a^y^—y^. 12. Find the cube of ]+2a;2. Ans. l-{-ex^ + 12x^ + 6x^. 13 Find the fourth power of a4-^- Ans. a^-\-4ta^x+6a^x^-{-4ax^-\-x^ 14. Find the fifth power ofa—y. Am. a^—5a^y-\-10a^y^ — l0a'^y^-\-5ay*'{-y^. EVOLUTION. J3^ EVOLUTION. (206.) Evolution consists in extracting any required root of a given quantity, regarded as the corresponding power of the root to be found. Extracting the square root consists in finding a quantity whose square is equal to a given quantity ; extracting the cube root consists in finding a quantity whose cube is equal to a given quantity ; and so Roots of Unity ^ Monomials^ Fractions, Sfc. (207.) Every root oi unity is unity, since the square, or cube, &c. of 1 is 1 ; thus 1x1 = 1; 1x1x1 = 1; and so on. In general terms, any power or root of a unit is a unit. (208.) Any required Root of a monomial will be found by extract- ing the root of its numerical coefficient, and dividing the exponents of its literal factors by the integer corresponding to the root. Thus the square root of 25 a^x is 5aa;^, found by extracting the square root of 25, and dividing the exponents of a and x by 2. And the cube root of 21a^x^ is '^a'^xJ^, found by extracting the cube root of 27, and dividing the exponents of a and x by 3. The correctness of this method will appear from considering that the Extracting of a Root is the reverse of raising the corresponding Power, (200). Hence also the propriety of denoting roots by fractional expo- nents. a^ denotes the square root of a, because a^xa^=a. (209.) Any required Root of a fraction will be found by extracting the root of its numerator and denominator, separately. Thus the square root of q-o—j is - — r » found by extracting the square root of the numerator, and the square root of the denominator. A Root of a mixed quantity would be found by extracting the root of the equivalent improper fraction. 132 EVOLUTION. (210.) A Root whose exponent is r£Solvable into two factors, may be found by extractfllg in succession the roots denoted by those factors. The 4th root may be found by extracting the square root of the square root, — the exponent \ which denotes the 4th root, being ^ X -J-. Thus the 4th root of a:* is the square root of ic^ ; the 4th root of 81 is the square root of 9 ; and the 4th root of 10000 is the square root of 100. The 6th root may be found by extracting the square root of the cuhe root, or the cube root of the square root, — \ being equal to -J- X -J. The correctness of this method of extracting Roots, is evident from considering, that, by raising in succession the powers denoted by two or more factors, we shall obtain the power denoted by the product of those factors. Thus the square of the cuhe of a is {a'^Y^=za^, (200), and there- fore, conversely, the square root of the cube root is the sixth root Bx)ots of Powers or Powers of Roots. (211.) The Numerator of a fractional exponent denotes a power of the quantity affected, and the Denominator a root of that power. Thus CL^ denotes the square root of the first power of a, or simply the square root of a. In like manner a^ denotes the cule root of the square of a ; c^ denotes the 4th root of the cube of a ; and so on. But, observe that (212.) A Root of any power of a quantity, is equal to the same power of the same root of the quantity. To illustrate this principle it may be shown that the cuhe root of the 6th power of a, is equal to the 6th power of the cube root of a. The cuhe root of a^ is a- since a"^ .a^ .a^ =za^ ; and the 6th power oi a^ is also a^, since the cube of a^ is a, and the square of this cube. — which gives the 6th power of a* — is a^, (200). To give an application of this principle to numbers, — the cube root of the square of 8, is 4, and the square of the cube root of 8 is 4 EVOLUTION. • 133 Equivalent Exponents. (213.) Two Exponents which are numerically equivalent, are also equivalent exponentially ; and may therefore be substituted, the one for the other. 2 1 Thus a* z=a^ ; the fourth root of a^ is equivalent to the square root of «. The 4th root of a^ is a^, since a^ raised to the 4th power produces a"^. 3. 1 4. 2. A In like manner x^—x^\ y^z=zy'^ ; a^z^a^ ; &c. To give an example in numbers, — the fou7'th root of the square of 9, that is, of 81, is 3, and the square root of 9 is also 3. On this principle a fractional exponent may always be taken in its lowest terms. The Sign to be Prefixed to a Root. (214.) Every odd Root of a quantity has the same sign as the quantity itself. For the quantity itself is an odd power with reference to such root^ (206), and an odd power has the same sign as the quantity of which it is a power, (202). (215.) Every even Hoot of a positive quantity is ambigtwus ; that is, the quantity itself does not determine whether the root is positive or negative. This follows from the principle that every even power of a nega- tive, as well as of a positive quantity, is positive, (202). The square of — a, as well as of -f-a, is +«^ 5 hence the square root of a^ is -[-« or else ~a ; and when it is not known whether a^ was derived from a or —a, it is uncertain which of these two is the square root. This uncertainty as to the sign of the root, is expressed by the am- biguous sign ± plus or minus ; thus the square root of a^ is ±a. For a like reason the fourth root of a* is ±a. The square root of 9 is d=3 ; the fourth root of 16 is ±2 ; &c. (216.) An even Hoot of a negative quantity is impossible, since there is no quantity which can be raised to an eve7i negative power » (202). ]0^ EVOLUTION. Thus no quantity multiplied into itself will produce — a^ ; this quantity therefore has no square root. For a like reason —a* has no square root or fourth root. In like manner —9 has no square root ; —16 has no square root or fourth root ; —25a^ has no square root, &c. EXE RC I s E s On the Roots of Monomials. 1. Find the square root of 4^a^x^. Ans. dz2ax^, 2. Find the cuhe root of 8a ^2/^* Ans. 2ay^. 3. Find the square root of 9a*x. Ans, ±3a^x^, 4. Find the cube root of —21y'^. Ans, —oy^, 5. Find the square root of 16a;* Ans. ±Ax^. 6. Find the cube root of a ^rr^/^. Ans. ax^y^. 3 7. Find the square root of 25a'. Ans. ±6a'^. 8. Find the cube root of —643/^, Ans. —^y^. 9. Find the square root of 36a;^ . Ans. ±6a;'. 1. « 10. Find the cube root of 8<2^*. Ans. 2a^y^. 11. Find the square root of :|a'*?/2, Ans. ±:^^y. 12. Find the cube root of —Sa^x^"*. Ans. —2ax'^. 13. Find the square root of |a;2/*. Ans. zt^^y^. 14. Find the cube root of —^ax^. Ans. —\a^x. 15. Find the fourth root of \^a^y^ . Ans. ±2ay^. 16. Find the cube root of —^fX^, Ans. — f a;^. 17. Find the fourth root of j3^fl5y 2 Ans. ±la^y'^. 18. Find the fifth root of —ai0a;2y. Ans. —a^x^y^. 19. Find the sixth root of (2®a?2y. Ans. ±:a^x%j^. 20. Find the sixth root of ba^tj^''. Ans. ±.h^a^y^' evolution. 135 Roots op Polynomials. (217.) The method of extracting any required Root of a Polynomial, may be discovered from the manner in which the corresponding power of a polynomial is formed. This subject will be elucidated under the appropriate Rules. RULE XVIII. (218.) To Extract the Square Root of a Polynomial, 1 . Arrange the Polynomial according to the powers of one of its letters, and take the square root of the left hand term, for the first term of the root. 2. Subtract the square of the root thus found from the given Poly- nomial ; divide the remainder by twice the root already found, and annex the quotient to both the root and the divisor. 3. Multiply the divisor thus formed by the last term in the root; subtract the product from the dividend ; divide the remainder by twice the root now found ; and so on, as before. EXAMPLE. To extract the Square Root of a^-\'2ab-\-h^. The required Root, we already know, is a +5, since the square of this binomial is the given trinomial : — our object is to show that the root a-\-b would be found by the foregoing Rule. a^+2ah-\-h^(a-{-b 2a-{-b) 2ab+h^ 2ab-hb^ The first term a of the root, is the square root of a^, the left hand term of the given polynomial. Subtracting the square of a, we have the remainder 2ab-\-b^. We have now to discover a divisor of this remainder, which will give, for a quotient, the next term h of the root. 2a, that is, twice the root already found, divided into 2ab, gives b ; and b annexed to 2a makes the divisor 2a -{-b. This divisor multiplied by b, produces 2ab-{-b^f — which completes the operation. The Rule is framed in accordance with the process thus discovered. 136 EVOLUTION. EXERCISES On the Square Root of Polynomials. 1. Find the square root of the polynomial a*— 4a3-f-6a2_4^_|_l. ^^5 a^—2a-{-l. 2. Find the square root of the polynomial a^-\-4:a^x+6a^x^-\-4:ax^-\-x*. Ans. a^-^2ax+x^. 3. Find the square root of the polynomial 1— 6y+13y2_i2i/3+42/*. Ans. l—2y-\-2y^. 4. Find the square root of the polynomial 4a;6_4a;4 + i2a;3+a?2_6a;+9. Ans. 2x^-x+^. 5. Find the square root of the polynomial 4:a*-\-12a^x-\-13a^x^ + 6az^+x*. Ans. 2a^-{-3ax+x^. 6. Find the square root of the polynomial I_6a+13a2_i2«3_|.4^4^ j^^s. 1— 3a+2«2. 7. Find the square root of the polynomial x^-\-2ax^-\-3a^x^-i-2a^x-^a*. Ans. x^+ax-{-a^. 8. Find the square root of the polynomial 9a* — 12a^y-\-28a^y^ — iea2j^-i-16y^. Ans. Za'^—2ay-\-^y^. 9. Find the square root of the polynomial 25-f34a;2 + 12a?3 + 20a;4-9a;*. Ans. 5+2a?+3a;2. 10. Find the square root of the polynomial a*+4a32/— 12ay+4«22^2_|_9_6^2^ ^^5^ a2^2ay—Z. 11. Find the square root of the polynomial 4+24a;2 — 16a;+4a;* — 16a;3. Ans. 2—^x+2x^. 12. Find the square root of the polynomial l—2y-\-ly^—2y^-\-6y*'-\-\2y^+^y^. Ans. l—y-^3y^ + 2y^. evolution. 137 Square Root of Numbers. The preceding Rule is substantially the same as the Arithmetical Rule for extracting the square root of Numbers. To show its applica- tion, however, to the latter purpose, we must premise the following principles. (219.) If a Number be separated into periods of two figures each, — from right to left, — these periods will correspond, respectively, to the units, tens, hundreds, &c., in the Square Root of the number. For since the square o^ten is 100, the square of the tens figure in the root will leave ttco vacant places in the right of the given number ; these two places must therefore correspond to the units in the root. And since the square of a hundred is 10000, the square of the hundreds in the root leaves four vacant places in the right of the number ; and the first two corresponding to the unitSf the next tivo must correspond to the tens in the root. In like manner it is shown that the third period of two figures corresponds to the hundreds in the square root of the given number ; and so on. (220.) If a Number be divided into any two parts, the Square of the number will be equal to the square of the first part -f twice the first X the second + the square of the second part. For, a representing the first part of the number, and h the second, a-{-b will be equal to the number ; and (a^bY=a^-{-2ab+b^. "With these principles established we may proceed to the following EXAMPLE s . 1. To extract the Square Root of 529 5'29 ( 23 4 43) 129 129 The first period 29 corresponds to the units figure, and the 5 to the tens figure in the root, (219). The greatest integral square root of the 5 is 2 tens, the square of which is 4 hundreds, and thi$ subtracted leaves 129. 138 EVOLUTION. Since the root of the given number consists of two parts, te9is and U7iits, and the square of the 2 tens has been subtracted, the remainder 129 = ttaice 2 tens X the units + the units'^, (220). Doubling the root, 2 tens, already found, we take the product, 4 tens, for a divisor. The in 4 tens, or 40, may be omitted in finding the next figure in the root, if we omit the 9 in the corresponding place in the 129. We therefore say 4 in 12, 3 times. This 3 annexed to the 4 makes the 4 now become 4 tens. Then 43, or 4 tens + 3 units, X 3 units, produces the remainder 129 ; and thus completes the operation. 2. To extract the Square Root of 36 84 49 36'84'49 ( 607 36 1207 ) 84 49 84 49 Pointing off periods of two figures, from right to left, we find that the root will contain units, tens, and hundreds, (219). The square root of the left hand period 36, is 6, the square of which subtracted leaves no remainder in that period. We take the next period 84 to find the next figure in the root. Doubling the root 6, for a divisor, and omitting the 4, we say 12 in 8, time. Including the next period 49, and doubling the root 60 now foimd, for a divisor, we say 120 in 844, 7 times. Annexing 7 to 120, the divisor becomes 1207, which multiplied by 7 produces 8449 ; and thus the operation is completed. Square Root of Decimals. (221.) In extracting the Square Root of a Decimal Fraction, the periods must be taken from the decimal point towards the right ; and a must be annexed, if necessary, to complete the last period. The last period must he complete, because, by the principles of decimal multiplication, the square of a decimal Fraction must contain tivice as many decimal figures as are in the root. The number of decimal figures to be made in the root, is therefore the same as the number of decimal periods. When an exact root cannot he found, decimal periods of 00 each may be annexed, and the root continued in decimals to any required exactness. EVOLUTION. 139 EXE RCISE S On the Square Root of Numbers, 1. Find the square root of 11236, and of §ff. Am, 106; and 1} 2. Find the square root of 38809, and of ^f . Ans. 197; and f f 3. Find the square root of 75076, and of fffi-. Ans. 274; and ||. 4. Find the square root of 22801, and of f||f. Ans. 151 ; and |^. 5. Find the square root of .582 169, and of 127i|. Reduce the |^f to an eqivalent decimal, before proceeding with the evolution. Ans. .763; and 11.292'. 6. Find the square root of 475.125, and of 3461^. Ans. 21.797' ; and 18.616'. 7. Find the square root of 37.4780, and of 470^. Ans. 6.121'; and 21.684'. 8. Find the square root 839.103, and of 500^. Ans. 28.967'; and 22.366 . RULE XIX. (222.) To Extract the Cube Root of a Polynomial, 1. Arrange the Polynomial according to the powers of one of its letters, and take the cube root of the left hand term, for the first term of the root. 2. Subtract the cube of the root thus found from the given Polyno- mial, and divide the remainder by 3 times the square of the root al- ready found, and annex the quotient to the root. 3. Complete the divisor by adding to it 3 times the product of the last term in the root multiplied into the preceding part of the root, and also the square of the last term. 4. Multiply the divisor thus formed by the last term in the root ; subtract the product from the dividend ; divide the remainder by 3 times the square of the root now found ; and so on, as before. 340 EVOLUTION. EXAMPLE. To extract the Cube Root of a3 + 3a25-|-3aS2_|.j3, The required Root, we already know, is a-{-b, (203), and our ob- ject is to show that this root would be found by the Rule. a^ Sa^+Sab+b^ ) 3a^b+3ab^+b^ 3a^b+3ab^-{-b^ The first term a of the root, is the cube root of a^, the left hand term of the given polynomial. Subtracting a^, we have the remainder 3a^b-\-3ab^-{-b^. We have now to find a divisor of this remainder, which will give, for a quotient, the next term b of the root. 3a^ , that is, 3 ti7nes the square of the root already found, divided into 3a'^b, gives b. Adding 3ah, and also b"^^ to 3a'^, the completed divisor is 3a2 + 3«6+i2^ _ This divisor multiplied by b, produces the last dividend, and thus shows that the operation is completed. EXERCISES On the Cube Root of Polynomials. 1. Find the cube root of the polynomial 2. Find the cube root of the polynomial l — Qx+2lx^—Ux^-{-63x*—54:X^ + 27x^. Ans. 1— 2a;4-3a:2. 3. Find the cube root of the polynomial 8 — 12a;4-30x2— 25a;3 + 30a;* — 12a:5 + 8a:6. Ans. 2-ic+2x^2. 4. Find the cube root of the polynomial 8-{-36y^ + 24:y+32y^-\-6y^+y^ + 18y^. Ans. 2-{-2y-\-y^. 5. Find the cube root of the polynomial 9a*—3a^ — 13a^-\-a^-l2a+S+iSa^. Ans. a^—a-\-2. 6. Find the cube root of the polynomial 2lx'^—Ux^ — 5^x-{-63x^—e>x^+x^ + 27. Ans. a:2_2a;+3. 7. Find the cube root of the polynomial evolution. 141 Cube Root of Numbers. The method of extracting the Cube Root of Numbers involves the following principles. (223.) If a Number be separated into 'periods of three figures each, — from right to letl, — these periods will correspond, respectively, to the units, tens, hundreds^ Sec, in the Cube Root of the number. For since the cube of ten is 1000, the cube of the tens figure in the root, will leave three vacant places in the right of the given num- ber ; these three places must therefore correspond to the units in the root. And since the cube of a hundred is 1000000, the cube of the hun- dreds in the root leaves six vacant places in the right of the number ; and the first three corresponding to the units, the next three must cor- respond to the tens in the root. In like manner it is shown that- the third period of three figures corresponds to the hundreds in the cube root of the given number, and so on. (224.) If a Number be divided into any two parts, the Cube of the number will be equal to the cube of tJie first part + 3 times the square of the first X the second -f- 3 times the first X the square of the second -f- the cube of the second. For a representing the first part of the number, and b the second^ a-\-b will be equal to the number ; and (a+b)'^=za'^ + 3a^b-{-3ab^ + b^. As an example of the application of this principle to numbers, let 275 be divided into the two parts 200 and 75 ; then (2004-75)3=2003-1-3x2002x75 + 3x200x752 + 753. If the values of the several terms on the right be computed, and added together, we shall obtain the cube of 275. We might now proceed to extract the Cube Root of Numbers in a manner corresponding to the Rule already given for the cube root of Polynomials ; but the method by the following Rule is more direct and simple. 142 EVOLUTION. RULE XX. (225.) To Extract tlie Cube Root of a Number. 1. Separate the number into periods of three figures each, from right to left, observing that the last period will sometimes have but one or two figures. 2. Take the greatest integral cube root of the left hand period, for the first figure of the root required, — subtract its cube from said period, — and to the remainder affix the next period for a dividend. 3. Take 3 times the square of the root already found, for an in- complete divisor ; divide it into the dividend exclusive of its two right liand figures, and annex the quotient to the root. 4. Complete the divisor by annexing to it 00, and adding the pro- duct which arises from annexing the last figure in the root to 3 times the other part of the root, and multiplying the result hij the last figure. 5. Multiply the completed divisor by the last figure in the root ; subtract the product from the dividend ; and to the remainder affix the next period for a new dividend. 6. Find the next incomplete divisor by adding to the last complete divisor the pi'oduct which completed it, and the square of the last figure in the root ; divide, and complete the divisor, as before ; and so on. In applying this Rule it will be convenient to have the following Table of Roots and Cubes, Roots, 1, 2, 3, 4 5, 6 7, 8, 9. Cubes, 1, 8, ... 27, ... 64, ."^ 125, . . 216, . . 343, . . 512, . . 729. We will first apply, and then demonstrate this Rule. EXAMPLE. To extract the Cube Root of 95 44 39 93. 48 4800 + 625 = 54 25 5425+ 625+ 25 = 60 75 607500+ 9499= 6169 99 95'44 3'9 93 ( 457 64 3144 3 27 12 5 43 1 8 993 43 1 8 993 EVOLUTION 143 The greatest integral cube root of the left hand period 95, is 4, the cube of which subtracted leaves the remainder 3 1 . Affixing the next period, we obtain the dividend 31443. Three times the square of 4, the root already found, is 48. We divide 48 into. 314, excluding the 43. The quotient would appear to be 6 ; but the divisor being as yet too small, we take 5 for the quo- tient. To complete the divisor, we annex 00 to it, and add 625 ; — the 625 being obtained by annexing the 5 in the root to 3 times the 4, and multiplying the result 125 by 5, Multiplying the completed divisor 5425 by 5, subtracting, and af- fixing the next period to the remainder, we find the dividend 43 18 99 3. For the next incomplete divisor, we add to the last complete divi- sor the product 625 which completed it, and the square 25 of the 5 in the root. Dividing by 6075, the quotient is 7. To complete this divisor, we annex 00, and add 9499, — this last number being obtained by annexing the 7 to 3 times the 45, and mul- tiplying the result 1357 by 7. The divisor now found multiplied by 7 produces the last dividend. Demonstration of the Rule for the Cube Root of Numbers. To facilitate the Demonstration, we will take a number containing but two periods, — whose root will therefore consist of tens and unitSf (223). 48 4800 + 625 =5425 91'125.( 45 64 27125 27125 The 4 in the root, found according to the Rule, we know, is 4 te?i!i, (223) ; and the second figure in the root will be units. Let t represent the tens, and u the units of the root, then the given numher rzz-t"^ -\-'^t'^u-{-'itu'^ -^-u"^ , (224). Hence the cube 64, — which is 64 thousand,— ^i the 4 tens, sub- tracted, leaves 27125 = 3^2^^_^3^^2 4_^3 By the Hule for the cube root of a Polynomial, the incomplete divi- sor of this remainder — to be used for finding the next figure or term u of the root, is 3^2 _ 3 ^imes (4 ^ms)^— 4800, J44 EVOLUTION. In dividing we omit the 00, and at the same time exclude the 2% in the corresponding places of the dividend 27125. By the Rule just referred to, the quantity to be added to complete the divisor St^, is 3tu-{-u^ = 3x^ tens x 54-25 = 600 + 25 = 625. But this quantity 3tu-\-u^ =:{3t-\-u)u ; that is, the last figure ii the root annexed to 3 times the other part of the root, and the result multiplied by the last figure. Hence the divisor may be completed accordiiig to the last Rule. Suppose now that the given number contains three periods, and iti root, consequently, three figures. Regarding the two figures already found as constituting the first part of the root, the next incoinplete divisor, by Rule XIX, would be 3 times {t+uy=3t^-{'e>tu+3u^ ; which is equal to 3t^ ■\-3tu-\-u^ -^{3tu-\-u^)-\'U^ . The three left hand terms of this last expression make up the last C07}iplete divisor; the next two terms, (3tu-\-u^), or (3t-]-u)u, are the product which was added to 3t^ to complete that divisor; and the re- maining term u^ is the square of the last figure in the root. Hence the second incomplete divisor may be found according to the last Rule. In like manner the Rule may be shown to be applicable, whatever be the number of figures in the required root. Cube Root of Decimals. (226.) In extracting the Cube Root of a Decimal Fraction, the pe- riods must be taken from the decimal point towards the right, and or 00 must be annexed, if necessary, to complete the last period. The last period must be cojnplete, because, by the principles of de- cimal multiplication, the cube of a decimal Fraction must contain 3 times as many decimal figures as are in the root. The number of decimal figures to be made in the root, is therefore the same as the number of decimal periods. When an exact root cannot be found, decimal periods of 000 each may be annexed, and the root continued in decimals to any required exactness. BVOLUTIOPC. 145 EXERCISE S — On the Cube Root of Numbers. 1. Find the cube root of 262144, and of J^Yy- Ans. 64 ; and |J. 2. Find the cube root of 2406104, and of-f^%^, Ans. 134; and ||. 3. Find the cube root of 22906304, and of ^Yo^- A?ts. 284 ; and |-J 4. Find the cube root of 479.2735, and of 8377y233_ . Ans. 7!825' ; and 20.309'. 5. Find the cube root of 5371.3745, and of 3059^. A?ts. 17.513'; and 14.51' 6. Find the cube root of 403.73331, and of .71200,^. Ans. 7.390' ; and .892'. 7. Find the cube root of 4370.666, and of ffj-f. Ans. 16.34'; and i|. «. Find the cube root of 20796875, and of 3511l-Oi. Ans. 275 ; and 15.20. 9. Find the cube root of .202262003, and of fj|f. A?ts. .587; and f J. 10. Find the cube root of 103823, and of 2460f. Ans. 47 ; and 13.5. Extraction of the nth. E,oot. (227.) Any Root whatever of a Polynomial might be extracted, — by taking the root of its left hand term, — with this root forming an in- complete divisor, — with the quotient term, and the root already found, completing the divisor, — and so on, in a manner depending on the order of the root to be extracted. But this method, which is preferable for the squai'e and cube, be- comes too complicated when applied to the higher roots. By dispensing with the completed divisors, the operation may be simplified, and the process of Evolution generalized, as follows. 10 J4g EVOLUTION. - RULE XXI. (228.) To Extract any Root of a Polynomial. 1 TJunderstanding the order of the root to be denoted by n, — ar- range the Polynomial according to the powers of one of its letters, and take the ^th root of its left hand term, lor the first term of the root. 2. Subtract the nih. 'power of the root found from the given Poly- nomial ; and divide the remainder by n times the {n—\) power of this root, for the second term of the root. 3. Subtract the wth power of the root now found from the given Po- lynomial, and, using the same divisor as before, proceed in the same manner till the wth power of the root becomes equal to the given Poly- nomial. This Rule may also be applied to Numbers, by taking n figures in each period, from right to left, for integers, and from the decimal point towards the right, for decimals ; and there will be less liability to error in finding the quotient figure, if new divisors be found for the second and subsequent remainders. EXAMPLE. To extract the 4th root of 30 49 800 625. 30'4980'0625 ( 235 2*= 16 4x2^ri:32 ) 144980 .... Exclude 980 in dividing. 23 ^= 279841 4 X 233 =48668 ) 251390 6"25 . . Exclude 625 in dividing. 235*= 3049800625 The preceding root might also be found by two extractions of the square root, (210) ; thus The square root of 3049800625 is 55225 ; and the square root of the latter number is 235. 147 i i CHAPTER IX. IRRATIONAL OR SURD QUANTITIES.— IMAGINARY QUANTITIES Perfect and Imperfect Powers. (229.) A Perfect Power, of any degree, is a quantity which has an exact root of the same degree ; — otherwise, the quantity is called an Imperfect Power. Thus 4 and 9a^ are perfect squares, having the exact square roots 2 and 3a ; while 2a and 8x^ are imperfect squares, since they have no exact square roots. In like manner 8 and 27a' are perfect cubes ; while 9 and 25ac2 are imperfect cubes, since they have no exact cube roots. The Polynomial a^-i-2ab-\-h^ is a perfect square, whose root is a-\-b; (218); and a^ — 3a^xi-3ax^—x^ is a perfect cube, whose root is a — x. Irrational or Surd Quantities. (230.) A Rational quaniity is one which can be accurately express- ed without any 2>2c?zc«^cc^ roo^; as 2, 3a, or |a;. An Irrational or Surd quantity is one which can be accurately expressed only under the form of a root, — being the indicated root oi an imperfect power. Thus 2^ is an irrational quantity, since the exact square root of 2 cannot be determined. Also a^ and x^ are irrational quantities. By the common Rule for the square root of numbers, we should find the square root of 2=1.414213' ; but other decimal figures would succeed without end. The term irrational when applied to a quantity, implies that such quantity has no determinable ratio to unity. Irrational or Surd quantities — being expressed under the form of roots — are also called Radical quantities, from the Latin Radix, a root. 148 IRRATIONAL OR SURD QUANTITIES. « Radical Sign. (231.) The radical sign ^ prefixed to a quantity, denotes tha square root of the quantity ; and is therefore equivalent to the exj^o^ nent -J-. Thus V a is the square root of a ; equivalent to a^. With the i?idex 3 affixed to it, this sign denotes the cube root . and is then equivalent to the exponent -J-. With the index 4, it de notes the 4th root ; and so on. Thus y a is equivalent to a^ ; and \/a to a*. The radical sign may always be thus superseded by a fractioncC exponent; and this should be done whenever any obscurity wouU arise in calculation from using this sign. For example, in extracting the square root of y «, we substitute the exponent -J, and find the required root to be a^'^^^c^\ (208). A quantity preceding the -y/, without + or — in-terposed, is a co- effidera or multiplier of the Surd ; and when no coefficient is express- ed, a unit is understood. Thus 5 Va is 5 times the square root of a ; and yax is the same as iVax. Similar and Dissimilar Surds. (232.) Similar Surds are such as express the sa7ne root of the same quantity ; otherwise, the Surds are dissimilar. Thus Vs and 5^/3 are similar ; while ^3 and 5V3 are dissimi lar. So (a-^-hy and Sya-i-b are similar Surds. A Rational Quantity under the Form of a Surd. . (233.) A rational quantity may be expressed under the form of the square root, or cube root, &c., by placing the corresponding power of the quantity under the exponent or sign of the root. To express Sa under the form of the square root, we plar^e the square of 3a, which is Oa^^ under the exponent or sign of the square root, thus 3a=(9a2)*or VQ^. irrational or surd quantities. i49 Transformation or Reduction of Surds. Certain Transformations of radical quantities are sometimes neces- sary, to adapt them to the purposes of calculation. These depend chiefly on the two following principles. (234.) When two or more Factors have the same exponent, that exponent may be transferred to the product of those factors ; and, Conversely, the exponent of a Product may he transferred to each of the factors which compose that product. Thus a^x^ is equal to (aic)^ ; the product of the squares of a and X, is equal to the square of the product of a and x. And a^-jjl jg equal to {axf' \ the product of the square roots of o and X, is equal to the square root of the product of a and x. These principles result from the methods of finding the powers and roots of quantities ; thus by squaring ax, — and extracting the square root of ax, — we have (a.'c)2=a2a;2^ (200); and (aa:)*= A*, (208). These methods of illustration may be applied to any other power or root as well as to the square and square root. Thus the product of the cuhe roots of two or more factors is equal to the cube root of the product of those factors ; and so for any power or root. To give examples in numbers ; — V^X V25=\/l00; or 2x5 = 10. V8x-v^27r='v^216; or 2x3 = 6. (235.) The Exponent of a quantity may be changed to any equiva- lent expression, without altering the value of the power or root de- noted. This proposition is substantially the same as one already demon- Btrated, (213). 1 2_ 3^ 4. As examples, a^=a*=a^=aS; 1 2 1 4^=44 = 16^=2. The following are the principal Transformations required in the calculation of Irrational or Surd quantities. 150 IRRATIONAL OR SURD QUANTITIES. Product of a Rational and an Irrational Factor. (236.) A rational multiplier or coeffudent may be put under the form of the Surd annexed, and the product of the two radicals be then affected with the same fractional exponent or radical sign. T\ms aV^^ — Vci^V^ — V'^a?, hy putting the coefficient a un- der the form of the square root, (233), and transferring the sign -y/, which is equivalent to the expo7ient ^, from the two factors to the pro- duct 3^2, (234). In like manner 2V'5 = '/4x 1/5=^20. As an example of the utility of this Transformation, — suppose it were required to find an approximate value of 2^/5. By the common method of extracting the square root, we might find an approximate value of y 5, and such value multiplied by 2 would be an approximate value of 2y 5. It is evident, however, that this process would double, in the pro duct, the deficiency in the value found of Vs ; and the error thus arising will be greater as the coefficient of the Surd is greater. By taking y 20, instead of 2y 5, the approximate value will be found independently of the preceding source of error. Surds Pceduced to Simpler Forms. (237.) A Surd is simpilified by resolving the quantity under the -y/, or exponent, into two factors — one of which is a perfect power of like degree — and putting the extracted root of this factor as a coeffi- cient to the indicated root of the other. Thus V50 = V'25x 1/2 =5^/2; or, 50^=: 25* X 2^=5x2*, (234). In like manner, V^a^ -\-Qa^ = ^/^Vl-\-2a = 2a^/l + 2a. In these examples the given Surds are square roots, and the factors 25 and 4^^ are accordingly perfect squares. As an example of the cube root, we have •^54 = ^27 X \/2=:3v^2 ; 27 being a perfect cube. When the given Surd is preceded by a Coefficient, this coefficient must be multiplied into the one found by the process of simplifying. Thus 3V^32a3=:3A/i6a2y^2a=:3x4a-/2a=:12al/2a. IRRATIONAL OR SURD QUANTITIES. j^j If, in the preceding process, the square or cube factor be the great- est square or cube thus entering into the composition of the quantity under the -y/, the Surd will be reduced to its si7nplest form. Thus in each of the examples above given, the Surd is reduced to its simplest form. The second example would also give which is not the simplest form, since a^ ig gtiH under the ^ . (238.) A Fractional Surd may be reduced to an integral surd, by multiplying both its terms, if necessary, to make the denominator a perfect poiver, and then resolving into factors, and proceeding as be- fore, (237). For example, ^J - = yj - = ^J^ ^6= \ -/G ; "We multiply both terms of the first Fraction by 2, to make the de- nominator 16 a perfect square. The reciprocal of this denominator is the square factor, and the numerator 6 is the other factor. Surds which are apparently dissimilar, often become similar when reduced to their simplest forms. They are thus prepared for Addition or Subtraction, as will be seen hereafter. Surds of Different Roots reduced to the Same Root. (239.) Two or more Surds of different roots, may be reduced to equivalent ones of the same root, — by reducing their fractional expo- nents to a common denominator, — raising each quantity under the -/ to the power denoted by the numerator of its new exponent, — and ta- king the root denoted by the common denominator. Thus to reduce y'S and y 4 to the same root. Reducing the exponents -J- and -J to a common denominator, we find y/5z=fi, and 1/^=4^, (235). Cuhing the 5, and squaring the 4, according to the numerators of their new exponents, (211), w^e find /5 = 125%and-^4 = 16^. The square root and the cuhe root have thus been both reduced to the sixth root. — This kind of reduction is sometimes necessary in the Multiplication and Division of Surds, — as will be seen hereafter. 152 EXERCISES. EXERCISES On the Transformation or Reduction of Surds. 1. Find the Product of 3 ■v/'ia. (236). Ans. V^- 2. Find the Product of 2 v^. Ans. \f2\x. 3. Find the Product of 4Va6. -^^s. ^/\^ab. 4. Find the Product of a y^. Ans. yEoFx. 5. Find the Product of 3a;V7. -Ans. -yj'^x^ 6. Find the Product of ax^lO. Ans, ^lOa^x^. 7. Find the Product of 7 i/o+T. • -^ns. ^49^+49. 8 Find the Product of 2v^l— a;. Ans. ^S—Sx. 9. Find the Product of {a-{-b)-\/2. Ans. '\/2(a-\-by. 10. FindtheProductof (a— a:)V3. Ans. l/S{a—xy' 11. Find the Product of (/4 to Surds of the same root. (239). By reducing the exponents of the Surd factor to the common deno- minator, we find 2^=26 = 86; and4»=46 = 166. Ans. 5V8, and 3^16. Prefixing the rational co-efficie?its, we have 5 V2=5 Vs, and 3 V^^S Vl6. 37. Reduce 2-^/5 and S-y/S to Surds of the same root. Ans. 2^25, and 5 V 27. 38. Reduce a-/^ and x-\/2 to Surds of the same root. Ans. ay 125, and x^^A. 39. Reduce lOVlO and 2y^'^x to Surds of the same root. Ans. 10^/100, and 2t/3x. iO. Reduce 7 V^^ and 2y'y/xy^ to Surds of the same root- Ans. l^^Qly^, and 2y^^/x^'^ 41. Reduce a^a-y^s. j^j^^j ? -yja^x to Surds of the same root. ^ws. a2a,:*/3^ and ^\/'^x^. 42. Reduce ;— V 2 and 5y\/ j— to Surds of the same root. 2x^ -^ ^ 2a; ^«s. ^ V4, and 5y\/^ 43. Reduce 2a; -/a and^-\/2 to Surds of the same root. 4/2 ,1 4/1 ^ns. 2^/0 , and-^Va* 44. Reduce f \/lO and 31^- to Surds of the same root. a^ Ans. iVlOO. and ^l\/-3' 45. Reduce 7^1/— and^V ^ to Surds of the same root. 2yy a Zxy y2 Am. ^ 1 2/i-,and ~ 1 2/j_. 2y V^4' 2aJ V7/0 ADDITION AJID SUBTRACTION OF SURDS. jgg ADDITION AKD SUBTRACTION OF SURDS. (240.) 1. The Sum, or Difference, of similar surds is obtained by prefixing the sum, or difference, of their coefficients as a coefficient to the common radical factor. 2. Dissimilar surds can be added together, or subtracted the one from the other, only by the proper sign ; but Surds apparently dis- similar often become similar when reduced to their simplest forms, (237). E XAMPLE. To Add together 5 -/BOa and 3 '/125a. Reducing the surds to their simplest forms, we find 5VS0a=5-/l6-v/5a=20 V5a ; and Z^/\25a—'i^25^/^a=zl6^'Ea. The two given Surds have thus become similar, (232j_ Adding together the coefficients 20 and 15, we find the Sum 'i5^/6a. The Difference of the two given Surds, is (2^ — l5)^/5a=:6^/5a. The Addition or Subtraction of similar Surds is evidently nothing more than the addition or subtraction of similar monomials ; thus 20 times v^ 4-15 times -y/^a is 35 times V^/ just as 20a-\-l5a is 35a. EXERCISES On the Addition and Subtraction of Surds. 1. Find the Sum of 3^/27 and 2V48. Ans. 17-/3. 2. Find the Difference between ^50 and -/72. Ans. -v/2. 3. Find the Sum of 7^28 and G-v/GS. Ans. 32'/7. 4. Find the Difference between 21^2 and 5^/18. Ans. 6V2. 5. Find the Sum of y^lSO and V'405. Ans. 15\/5. IgU ADDITION AND SUBTRACTION OP SURDS. 6. Find the Difference between -x/lS and 2^/60. Ans. 7-\^2. 7. Find the Sum of VT^a and V^7a^. Ans. 5aV^' 8. Find the Difference between 3-/24a;2 and -\/54^. Ans. Src/G 9. Find the Sum of 4'/3a and -\/48a. ^;zs. S-y/Sa. 10. Find the Difference between V^a^ and ■\/9a^. Ans. a^a. 11. Find the Sum of 3 v^4 and yy^I Ans. 10v^4 12. Find the Difference between 9^/200 and '/288. J.m. 78'/2. 13 Find the Sum of 4^54 and 2v^250. Ans. 22|/2. 14. Find the Difference between 5^/9^' and 3V^- -^^s. 12a;V^. 15. Find the Sum oi^Vl^a and Y^- * -^.ns. 7 ^/2a. 16. Find the Difference between S-J^IO and 5{/l0- ^ns. 2^/10 17. Find the Sum of 5^98^ and lO-v/sS- -^.ns. ASV^x. 18. Find the Difference between 3a^5 and a\^5. Ans. 2a \/ 5. 19. Find the Sum o£ aV^ and 36^^^. Ans. (a+35)^^. 20. Find the Difference between 5\^5 and 2a V^. Ans. (5— 2a)-\/*^ 21. Find the Sum of (a 4- 1)2 and V4a+4. ^ws. 3 (a+l)2 22. Find the Difference between -y/l+a? and 3 (l+a;)2. ilws. 2(l+a:)2 23. Find the Sura of 2 {a—x)i and V9a— 9a:). 24. Find the Difference between l/2-\-y and 4 (2/+2)3. ^?zs. 3 (24-y)^ 2f. Find the Difference between 4 (l+x^)^ and 4 v^^+T. 4 -i ^ws. 4(l + a:2I_ic2-f 1^). MULTIPLICATION AND DIVISION OF SURDS. 157 MULTIPLICATION AND DIVISION OF SURDS. (241.) 1. The Product, or (Quotient, of two Surds of the same root^ is obtained by prefixing the product, or quotient, of their coefficients as a coefficient to the product, or quotient, of the radical factors, — the latter being affected with the sanie fractional exponent or radical 2. Surds of different roots may be reduced to equivalent ones of the same root, (239), and then multiplied, or divided, as above. But 3. A7iy two roots of the same quantitij may be multiplied into each other, by adding together their fractional exponents ; or divided, the one into the other, by subtracting the exponent of the divisor irom that of the dividend. EXAMPL E. To find the Product of 2^10 x 3 ^/2. Since it is immaterial in what order the four factors are taken, we may take them in the order, 2x3V'lOX'/2;whichgivestheProduct6'/20, (234), =12/5, (237J The Qwoi^e?^^ of 2'v/l0-^3'v/2 is f -y/S, since this quotient multi- plied by the divisor produces the dividend. EXERCISES On the Multiplication and Division of Surds, 1. Find the Product of 5 V'S X 3^5. Ans. 30/10. 2. Find the auotient of 6'/54-f-3/2. Ans. 6/3. 3. Find the Product of /lOS x 2/6. Ans. 36/2. 4. Find the auotient of 2/96^/54. Ans. 2f. 5. Find the Product of 3 /5aa?-f- 4 /20a. Ans. 120a /a;. 6. Find the auotient of 4/r2a^2/6. Ans. 2/2a. 7. Find the Product of /Sao: X 3 /ai. Ans. 3aa;/3 8. Find the auotient of 6/12^-^3/4. Ans. 2a:/3. 9. Find the Product of /l8 x 5/4. Ans. 10/9. 10. Find the auotient of 43/72-^-2^18. Ans. 2^ 4. 158 * MULTIPLICATION AND DIVISION OF SURDS. 1 1 . Find the Product of 5 -/a x 3 ^a. By reducing the surds to the same root, we obtain 5Va^ and 3 1/^2, (239). These are to be multiplied together as before. Arts. IS-v^a*. But the given roots of the same quantity a, may also be multiplied into each other, by adding together their fractional exponents \ and \. Thus, 5a^x3«■3■=15a6=rl5V«^ as before. Either of these two methods may be applied to roots of the same quantity. The first only is applicable to different roots of different quantities. 12. Find the Product of 3^/^ X 2^/x. Ans. 6 V^- 13. Find the auotient of 2^3 ^5^/3. Ans. ^^/Z. 14. Find the Product of 4^/3 X 2>/2. Ans. 8^108 15. Find the Quotient of 8 -v/a^ 4 -y/tt. Ans. 2 -y/a. 16. Find the Product of v^ X 3 Vaa?. Ans. 'i^a^x^, 17. Find the Quotient of 4-/^^ -^ 2 y'a?. Ans. 2x\Jx. 18. Find the Product of -J-v/a X -J- VlO- -A-ns. \y/2^^. 19. Find the Quotient of a^-v/'J-^tt -/I". Ans. a^ Vi- 20. Find the Product of (3 + 2/5) X (2— -y/S). In cases of this kind, in which a Surd is connected with another quantity by the sign + or — ,^the Multiplication, or Division, must be performed as on polynomials. 3 + 2/5 2- -v/5 6+4/5 -3/5-10 6+ /5 — 10 = /5— 4. Each term of the multiplicand is multiplied by each term of the Qiultiplier, and the partial products are then added together. Observe that 2/5x — /5 = — 2/25= — 10. 21. Find the Product of (2 + 3/2) x (1 + 5/2). Ans. 13/2+32. RAnONALIZATION OF SURD DIVISORS. 15$ 22. Find the Product of (4- -/S) X(2+3'v/3). Ans. 10V3 — 1. 23. Find the Product of (5+2^6) x (1 + 2V6). Ans. 12V64-29. 24. Find the Product of {l-4'/7) x(3-3'/7). Ans. 87-15V7. 25. Find the auotient of ('/20 + '/l2-^(-v/5+ -/S). Ans. 2. RATIOK ALIZATIOIf OF SURD DIVISORS. (242 ) In compiiting an approximate value of an irrational nume- rical expression, it is expedient that a surd divisor or denominator be made rational. For example, suppose we wish to compute an approximate value of --— , 2 divided by the square root of 3. ■v/3 If we extract the square root of 3, for a divisor, a regard to accu- racy will require that the root be continued to several figures, and hence will arise the inconvenience of dividing by a large number. By multiplying both terms by the denominator, we have = = , in which the divisor is rational. -v/S y9 3 The value will therefore be found by taking -J of the square root of 12 ; and by this method the computation is much simplified. In pursuance of the object at present in view, it is necessary To find a Multiplier of a given Surd which will cause the Product to be Rational. (243.) 1. A monomial Surd will produce a rational quantity by being multiplied into itself with its exponent subtracted from a unit. Thus a^ multiplied by a^~^, or a^xa^=a, (241 . . .3). 2. A binomial in which one or both terms contain an irrational square root, will produce a rational quantity by being multiplied into itself with a sign changed. Thus (/ 3+/ 2) x(/ 3-/ 2) = 3-2 = l. The product in this case is readily found on the principle, that the Product of the sum and difference of two quantities is equal to the dif ference of the squares of the two quantities. 150 RATIONALIZATION OF SURD DIVISORS. 3. A trinomial containing irrational sqimre roots will produce a binomial Surd by being multiplied into itself with a sign cha^iged ; and this binomial may be rationalized as above. These principles provide for the most useful cases of the subject under consideration. — In applying them to the rationalization of surd denominators, both terms of the given Fraction must be multiplied by the same quantity, (81). EXERCISES On the Rationalization of Surd Denominators. 1. Reduce — 5 to a Fraction having a rational denominator. . 2V9+3 2. Reduce / to a Fraction having a rational denominator. Ans. . 2a ... ^ 3 Reduce to a Fraction having a rational denominator. 2— V? 4. Reduce t° ^ Fraction having a rational denominator. '\/a — '\/x A71S. — — - . a—x 10 ... 5. Reduce 7— to a Fraction having a rational denominator. a+VlO 4 /g C. Reduce — ^ to a Fraction having a rational denominator 2 8 7. Reduce — /q 1 /04,i ^^ ^ Fraction having a rational denominator. Ans, 4+2^2-2/6. INVOLUTION AND EVOLUTION OF SURDS. jgj INVOLUTION AND EVOLUTION OF SURDS. (244. ) The Powers and Roots of irrational quantities are obtained, or i7idicated, according to the general principles of Involution and Evolution which have been established in the preceding Chapter. We present here how^ever a particular case of the Square Root of Binomial Surds. (245.) A Numerical Binomial of the form a±'\/b admits of a square root in a rational and an irrational term, or two irrational terms, whenever a^ —b is Vi, perfect square. To determine the method to be pursued in this case of evolution, we must find Formulas for the Square Root of fl±/ b. The square of the siwi of any two quantities, is equal to the sum of their squares + tivice tlteir product^ (59). The binomial a-\--^^ b may therefore represent the square of the sum of a rational and an ir- rational numerical term, or of two irrational terms, in the square root; a representing the sum of the squares of the two terms, — which sum will necessarily be rational, — and -y/ b representing twice the prodmt of the two terms. In like manner a — y/b may represent the square of the difTerenc^ of a rational and an irrational numerical term, or of two irrational nu- merical terms, in the square root, (60). If therefore we take % and y to represent the two terms of the square root of a±Y/ b, we shall have, (1) x'^y^^a-, (2) X ■\-y =^/a+y/b^ (3) X —y —^/a—^b\ Multiplying together equations (2) and (3), we have, (4) x^-y^ = ^/'^(^^, (58.) Adding together the (4) and (1), and also subtracting the (4) from the (1), and dividing by 2, we shall find, . 11 162 INVOLUTION AND EVOLUTION OF SURDS. (5) (6) y^= 2 Extracting the square root of each of these equations, ._A/«+Va; we find a?= V ^ ' *" "^ and y= V -T, 2 Substituting these values of x and y in equations (2) and (3),— and interchanging the first and second members, — we have, (A) V„-+73.\/^±^* + \AI^*; . (B) Va-=7^=V'^i^-\/"-=^- These are the Formulas required. The right hand member of each will contain at most but two irrational terms, when {a^—h) is a perfect square^ that is, has an exact square root. EXAMPLE. To find the Square Hoot of 6 + 2/ 5, or 6+ -v/20, (236.) Substituting 6 for a, and ^/2^ for -yjh, in Formula (A), And since •v/36— 20 = v'l6=:4, the second member reduces to Y f- Y = ^J^ + 1, the Root required. The root 'yJ^-\-\ may be verified by squaring it. Thus (V5+l)2=:5+2V54-lrT64-2V5. Formula (B) would give the square root of 6 ^2^/5. INVOLUTION AND EVOLUTION OF SURDS. 168 EXERCISES On the Powers and Roots of Surds. It will be readily perceived that by cancelling the exponent or sign of a root, that root is raised to the corresponding power. Thus the square of y'a is a ; the cube of -y/a is a, &c. 1. Find the Square, and also the square root, of sV^- The square of 5 X 4^ is 25x4^=25^16, (200). Keducing this result to its simplest form, we have 25 Vl6=:25 ^8 ^2=^50 ^2, (237). The square root of 5 X 43" is 5^x46, or ^6 Va, (208). Multiplying together the two factors of this root, we have ^5V4 = Vl25V4=: V5OO, (241 ... 2). 2. Find the Square, and also the square root, of 9 ^/Z. Ans. 81 V^Tand 3 ^/3. 3. Find the Cube, and also the cube root, of 3'v^2. Ans. 54 V^Tand V18. 4. Find the Square, and also the square root, of 2 -y/S. Ans. 4 V^, and «/24. 5. Find the Cube, and also the cube root, of 3a -y/o. Airs. 81a3 y^sT and ^/zfd^. G. Find the Square, and also the square root, of x^ 'v/5. Ans. x^ -^25, and x V^T 7. Find the Square, and also the square root, of 3 -y/a+l. Ans. 9(a+l), andy9(a+l). 8. Find the Square, and also the square root, of 4 y 1 — x. Ans. 16 y"(l-a:)2, and 2 ^/T-^ 9. Find the Cube, and also the cube root, of 2 -y/a'^—x. Ans. 8 V(a2_a;)3, and \/4.{a^-xy 10. Find the Square, and also the square root, of o^2—a. Ans. ^i2-a\. andy9(2-a). jg4 EXERCISES. 11. Find the Square of 24-^3, and of 2 — -/S. These squares may be found by multiplying each binomial into itself, or, more readily, by applying the propositions relating to the squares of the simi and the difference of two quantities. Arts. 74-4i/3, and 7-4V3. 12. Find the square of 3 + 2^5, and of 5-3-/^- A71S. 12v^5 + 29, and 43-30 V2.' 13. Find the Square of V5-^2Va, and of V^ — 5^/x. Ans. 5 + 2 ^/a, and 3 — 5 ^x. 14. Find the Square of V2-|-3\/5, and of ^/a—2x^/x. ^;zs. 6'y/i0-f47, anda— 4-v/aa;+4ar. 15. Find the Cube of V l + ov^4, and of Va-'^V^- Ans. l+5'v/4, and «— S-y/^. 16. Find the Square of '/3+ai/3, and of -v/3— 3\/«- Ans. 3 + 6«+3a2, and 3 — 6y'3^_j_9^, 17. Find the Cube of 2+2-/^, and of 2— a'/2. Ans. 84-24«+(24 + 8a)'/^and S-\-l2a-{l2a^2a'^)^/2. 18. Find the Cube of (^^ 4.^2)3 and of (y^l- Vy)* Ans. V«+ V'2, and V^T^V^ 19. Find the Square root of a+2^/ax-]-x, ( 218 ). Ans. V«+ V^' 20. Find the Square root of a-f 2-\/a+l, and of 34-2a'/3+«2 Ans. -\/«-f 1, and V^ +^» 21. Find the Square root of a;— 2'v/ar+l, and oi^ 5—2x-\/5-{-x^. Ans. \/x—l, and -x/Q—x. 22. Find the Square root of 23 + 8'/7, ( 245 ). Formula A. Ans. 4+ '/7. 23. Find the Squaie root of 19^ 8'/3, and of 7— 2V'lO- A?ts. 4+^/3, and ■\/^ — -\/2. IMAGINARY QUANTITIES. J^gg IMAGINARY QUANTITIES. (246.) An even root of a negative quantity is impossible, and the symbol of such a root is therefore called an imaginarij or impossible quantity, in contradistinction to real quantities. Thus y/— 4, the square root of —4, is imaginary, since there is no quantity whose square is — 4 ; and for a like reason -y/^lG is ima- ginary. An Imaginary Quantity results in calculation from some impossi- bility in the conditions of a Problem, and may therefore be Regarded as a sign of such impossibility. As an instance of this, suppose it were required To find a number whose square subtracted from 5 shall leave 9. If X represent the required number, the Equation of the problem will be From this equation, ac^^i— 4 ; and hence a;=zy^— 4. The value of x being imaginary or impossible, shows that the problem is imjjossible ; that is, that there is no number whose square subtracted from 5 will leave 9. * Imaginary Quantities may become the subjects of calculation like real quantities. Thus we may wish to verify an imaginary value of the unknown quantity in an Equation, by subjecting this value to all the operations which are performed on its symbol in the equation. Calculus of Imaginary Quantities. (247.) All the principles which have been established for the cal- culus of radical quantities are applicable to imaginary quantities, ex- cept that the ambiguous sign ± does not precede the product of two imaginary quantities, as is the case, in general, with even roots of real quantities. The sign which affects the Product of two imaginaries may always be determined by means of imaginary and real factors into which such quantities may be resolved. 166 IMAGINARY (QUANTITIES. (248.) An Imaginary (Quantity may always be resolved into the like root of ( — 1) multiplied into the like root o^ 2i positive quantity which is equal to the negative quantity in the given imaginary. Thus —4 being equal to— 1x4, we have -y/— 4 = ^/ — 1 X-y/"^ > and —a beingequalto— Ixa, wehave -/— « = -/ — ! X\/«;(234-) By means of this transformation it may be shown, that (249.) The Product of two imaginary square roots is the negative square root of the product of the two quantities under the -y/, if the given roots are preceded by like sig?is, -f or — ; otherwise, it is the positive square root of that product. For example, ^J —a and, -yf—a On general principles of calculation, the Product of the square roots of —a and —h, is equal to the square root of the product ah, which would be +^/ah, or else —^/ab, (215). But the ambiguity in the sign to be prefixed to an even root in general, is removed when we know the factors which entered into the ccmiposition of the quantity whose root is considered. When a^, lor example, is known to have been derived from axa, the square root of a^ is a, and not —a. To determine the sign of ^/ab in the first example, we have .yfZTa . yf-b^^fZTx . ^a . -/^l . -y/^, (248) = i/^ . -/^ . ^/a^/b =:('V/— l)^V^=~lV^^» 0^ —yfab. In the second example, we have yTTa . -yfZrh^yfZTi . ^a . -y^l . ^/b, (248), =:V^1 . —^T-^ . ^/ay/b =:_(yCri)2y^__(_l)y^=4_y^. In this example the square (y^l)^ has the sign — before it, be cause it results from multiplying -v/— 1 by — -y/ — 1, (42) ; an6 — ( — 1) becomes -f I hy changing the sign in subtracting. 167 CHAPTER X. QUADRATIC AND OTHER ECiUATIONS. (250.) A GluADRATic Equation, or an equation of the second de- gree, is one in which the highest power of the unknown quantity is its square, or second power, as 3a;2 = 12; or 3a;2+4a;=20. A Cubic Equation, or an equation of the third degree, is one in which the highest power of the unknown quantity is its cube, or third power ; and in like manner is defined a Biquadratic Equation, or an equation of the fourth degree, and so on. An Equation containing two or more unknown quantities is of the degree which corresponds to the greatest number of unknown factors in any of its terms. Thus 3xy-\-y=a is an Equation of the second degree, its first term containing the unknown factors xij. And x^y—y=ib is an equation of the third degree, its first term containing three unknown factors xxy. Pure and Affected Equations, (251.) A Pure Equation is an equation which contains but one power of the unknown quantity ; and is a Simple Equation, a Pure Quadratic, or a Pure Cubic, &c., according to its degree. Thus 3a;2 is a pure quadratic ; a;^ = 64 is a pure cubic. An Affected Equation is one which contains different powers of the unknown quantity. When these powers are in regular ascension, beginning with the first power, the Equation is also called a complete equation. Thus a;2-f-3a:=10 is an affected, and also a complete quadratic ^ x^ — 2x'^=zl6, or rc^— 3xnzll0, is an affected cubic ; and a;^ + 3a;2-}- 4a: =2 8 is a complete cubic. 169 GENERAL PROPERTIES OF EQUATIONS. Roots of Equations. (252.) A. Root of an Equation is a value of the unknown quantit} in the equation. It will presently be shown that an equation of the 2d degree has two roots, of the 3d degree three roots, and so on. ^ In the Simple Equation 3a?=15, the value of the unknown quantity a; is 5 ; then 5 is the root of the equation. It is evident that in a Simple Equation there can be but 07ie valu« of the unknown quantity that will satisfy the equation. An equation of the 1st denfree has therefore but one root. General Properties of Equations. 1. Divisors of an Equation, (253.) If a be a root of an Equation of any degree, containing but one unknown quantity, x ; the equation — with all its terms transposed to one side — will be divisible by x — a. Let a be a root of the Equation x^-\-mxz=is\ then will the equation x^ -{-mx—s^z^) be divisible by x -a For let r be the remainder, if any, after the quotient q has beea obtained ; then will x'^-\-mx—Sz=i{x—a^q-\-r^::^^ ; the dividend being equal to the remainder added to the product of th$ divisor and quotient. But a being a value oix, (252), we have x—a — ^ ; then the {x—d) X 9 is 0, (43) ; and consequently r=0 ; that is, the division will leave 710 remainder. The preceding demonstration is applicable to an equation of the third, fourth, or any higher degree. (254.) Conversely, If an Equation of any degree, containing but one unknown quantity, x, — with all its terms transposed to one side — be divisible by x — a ; then a will be a root of the equation. This is evident from considering that the divisibility of the Equation, as shown above, depends on the condition that a;— a=:0, or that a is a value of X. GENERAL PROPEilTIES OF EQUATIONS. 1 59 2. Number of Roots of an Equation. (255.) Every Equation containing but one unknown quantity, has just as many t'oots as there are units in the exponent of the highest power of the unknown quantity in the equation. Let a represent a root of the cubic Equation X^-{-77lX^ -\-7tXz:^S. \ Transposing s to the first side of the equation, we have x^-\-mx^-\-nx—s = 0. Dividing this equation by x—a, (253), we shall obtain an equatior of the second degree, which may be represented by x^+px—q=0. Let 5 be a root of this equation. Dividing the equation by x--b we shall obtain an equation of the first degree, represented by X — U=zO. The binomials x—a, x—b, and x—u, which represent the two di visors and the last quotient, are the factors of the dividend. x^ -\-7nx^ -\-nx-—s. The original equation is thus resolved, representatively, into {x—a){x—b)(x—u) = 0. Since this equation is divisible by each of these three binomial fac tors, it follows that a, b, and u are th?'ee roots of the equation, (254). And since the given equation — being of the 3d degree — cannot be resolved into more than three binomial factors, each containing the first power of x, it cannot have more than three roots. The same method of demonstration will show the proposition to be true for an Equation of any other degree. The several roots of an Equation are not necessarily unequal, though such will usually be found to be the case. The preceding Pro- position shows that an Equation may be resolved into binomial factors, each containing a root of the equation. Two or more of these roots may be equal to each other. In the Equation a;^— 5a?+6=^(a: — 2)(a: — 3) = 0, the two roots are 2 and 3. B In the Equation a;^— 7x2 + 160:— 12=r(a;—2)(ir—2)(a;—3)=i:0 ^ft^ the three roots are 2, 2, and 3. ]7q solution of pure equations. Solution of Pure Equations of the Second and Higher Degrees. The Equations belonging to this class are those which, in their simplest forms, contain but one power of the unknown quantity. RULE XXII. (256.) '^or the Solution of a Pure Equation. 1. Reduce the Equation to the form x^=iS', in which x^ must be positive. 2. Extract that root of both sides of the equation which corresponds to the power of the unknown quantity EXAMPL E. To find the value of x in the Equation — +5 = 2iB2_-23. 4 Clearing the equation of its Fraction, we have ic2 + 20=:8x2— 92. Transposing, and adding similar terms, we find -7a;2r=-112. Dividing both sides of this equation by the coefficient —7, Extracting the square root of both sides, x=±:4., (215). The two values of x are thus found to be 4 and —4, (255), either of which will satisfy the given Equation. It is evident that, in a Pure 'Equsiiion oi' the second degree, the two values of the unknown quantity will always be equal, with 0071- trary signs. The unknown quantity may enter an Equation in a surd expres- sion, which it will be necessary to rationalize in the solution of the equation. Thus in the equation ■y/x-\-^l-\-x=a, it would be necessary to rationalize x, that is, to clear it of the radical sign before the value of X could be determined. The following observations will assist the student in the RATIONALIZATION OF SURD QUANTITIES IN AN EQUATION. l*Jl Rationalization of Surd Quantities in an Equation. (257,) A Surd quantity in an Equation will be rationalized by transposing all the other terms to the other side of the equation, and raising both sides to the power corresponding to the indicated loot. To rationalize x in the Equation V^c+\—a=zh. By transposition, V x-\-\^=a-\-b Squaring both sides, x-\-\=a^ -\-2ah-\-h^ . (258.) Two Surds in an Equation may be rationalized by successive involutions, — in the first of which it will generally be expedient to make one of the Surds stand alone on one side of the equation. To rationalize x in the Equation '\/x-\-Vl'-{-x=a, By transposition, -y/x =a— VT^- x. Squaring both sides, x=a^— 2a Vl-{-x +14-^' By transposition , 2a Vl-{- x=a^ + l. Squaring both sides, 4a^(l-]-x)=a^-\-2a^-{-l. (259.) When an Equation contains a Fraction whose terms are both irrational, it will sometimes be expedient to rationalize its de- nominator before clearing the equation of the fraction. To rationalize x in the Equation ■\/oo _ Multiplying both terms of the Fraction by l — 'x/x, (243... 2), ya;— 1-x ■.a. Clearing this equation of its Fraction, and transposing, we have ^x=za—ax-{-x. The Surd in this equation will be rationalized by squaring hath sides, as in the preceding examples. 172 EXERCISES. By the preceding methods we may generally rationalize one or more Surds contaming the unknown quantity in an Equation. Other expedients, however, such as the extraction of roots in possible cases, in the course of the operation, will sometimes be requisite ; but thes<* must be left to the care and skill of the computer. EXERCISES ^ On Rationalization, and Pure Equations, 1. Find the value of ic in the equation 24 — V2a;2+.9=rl5. Ans. x=±^ 2. Find the value oix in the equation 13 — V3a:2 + 16=:5. Ans. a:=±4 3. Find the value of x in the equation 35-1- -^^115— 40. Ans. x—l2>^ 4. Find the value of x in the equation \-[-2^x—■^/4.x-\-2l. Am. x=25 6. Find the value of x in the equation Vx—32 = V^—YV^^' 'Ans. x=50 6. Find the value of x in the equation 3 + '/H-4x -/a:— 4 = 10. A71S. Xz=ztV^5 7. Find the value of a; in the equation a+V^— 3x Va;-h3=4a. Ans. x=±3\/a^-{-l. 8. Find the value of a; in the equation 9. Find the value of a; in the equation 4v^6ar— 9 -y/Ga;— 2 A^Qx+Q V6^+2 Ans. x=6, 10. Find the value of a: in the equation /—- 7=^=6. Ans. x=-^.~~. solution op complete equations. . j73 Solution of Complete Equations of the Second Degree. These Equations in their simplest forms contain no other power of the unknown quantity than its square and first power. The value of the unknown quantity will be found by the following RULE XXIII. (260.) For the Solution of a Complete Equation of the Second Degree. 1. Reduce the Equation to the form x^-\-bxz^s\ in which x^ must be 'positive. 2. Add the square of half the coefficient of a;, in the second term, to each side of the equation : — the first side will then be a perfect square. 3. Extract the square root of each side, and the result will be a Simple Equation, — from which the value of x may readily be found. EXAMPLE. To find the value of x in the equation 3a: — 5 ^ 5x — 5 ^;^+ -F- =^^- ^=3 • Clearing the equation of its fractions, transposing, and adding sim- ilar terms, we shall find —3x^-i-Ux=—5. Dividing both sides of this equation by —3, 2 14a: 5 3 -3- Adding the square of half the coefficient of x, in the second term, to each side, — which is called completing the square^ 3 14a; 49 _ 5 49 _ 64 ^ 3+9 ~3"'""9"-"9'' Extracting the square root of each side, 7 8 3 3 7 8 7 ft 1 Whence x=-+--=5; or 0:= - - - z=- -. (255). 174 SOLUTION OF COMPLETE EQUATIONS. Either of these two values of a?, 5 or — ^, will satisfy the given equation; and each of the binomials x—5 and x-\-^ will divide the equation ' x^-^-|=0. (253). By performing the division it will be found that the left hand sid^ of this equation, is the product of the two binomials ; that is, On the preceding Rule we remark, 1. The method of completing the square in the first member, re- sults from the composition of the square of a Binomial. Thus the square of the binomial a+6 is a'^ ■\-2ha-\-h'^ , in which h is half the coefficient of a in the second term. 2. If in reducing the equation to the required form, the first term should become —x^, the signs of all the terms in the equation must be changed (117), before completing the square, — otherwise the root of the first side would be imaginary ^ (246). 3. The square root of the first side of the equation — after the com- pletion of the square— will always be the square root of the ^rs^ term, -f- or — half the coefficients of x in the second term, according as the second term is + or — . Hence, without completing the square on the first side, we may shorten the solution, by observing that 4. In an equation of the form x^ -^bx=s, the value of iris half the co-efficient of a; in the second term, taken with a contrary sign, -4- the square root of (the second member of the equation -j- the square of said half co-efficient). 1 ^^ 5 Thus from the equation x^ b~^=^3" ' ^^ immediately find o o 3— V 3^9 ~3— V 9 "~3— EXERCISES. / jy^ EXE R CISE S On Quadratic Equations with One Unknown Qiuantity, Wh' Find the value of a; in the equation ic2 — 15=45— 4a:. Ans. x=& ox —10, 2. Find the value of x in the equation a;2-|-10=:65 + 6a:. Ans. x = l\ or —5. 3. Find the value of a; in the equation 2a;2 + 8a:~30 = 60. Ans. x=6 or -9. 4. Find the value of x in the equation 3a;2-3a;+9 = 8j-. Ans. rr^^ or f . 5. Find the value of in the equation 5a:2+4a;-90 = 114. J.ws. a;=6 or ~6|. 6. Find the value of a; in the equation \x^—lx-\-2 = ^. Ans. x=4. ox —'^^. 7. Find the value of a; in the equation 2x2+12a:+36=356. -4«s. a;=10 or —16. 8. Find the value of a; in the equation 4a;— 46= . Ans. a;=12 or — f. X * 9. Find the value of a; in the equation 8a;2-f 6 = 7a;+l71. Ans. x=5 or -4^. 10. Find the value of a; in the equation 5a;— 23 = . Ans. x=z5 ox —1. X 11. Find the value of <«; in the equation 3a;2 4-6 = 3a;+5i. Ans. x=i ox ^. 12. Find the value of a; in the equation Y - I +20|=42f. Ans. x=7 or -6^. 13. Find the value of a; in the equation 8-x x-2 , 2a;-ll ^ ^ ^, -2- = -Q- + ^ZF • ^^'- ^=^ °' ^*- 14. Find the value of a; in the equation 7/g 8 a;4-4 = 13 . -4?zs. a;=r4 or — 2. X 15 Find the value of a: in the equation a;— 3a 9(b—a) ^ „, „, „ — ^— = -^ -. Ans. x=3b, or 3{a—b). 176 EXERCISES. Another Method of Solving Quadratic Equations. (261.) A Binomial of the form ax^-{-bx will be made z. perfect sqtcare by multiplying it by 4 times the coefficient of x^, and adding the square of the given coefficient of x. Thus (ax^-{-hx)^a+h^=4.a^x^+^abx-\-h^={2ax-{-hY, (59). To apply this principle to solution of the Equation 3a:2 — 2a:=r65. Multiplying both sides of the equation by 4 times the coefficient 3, and adding the square of the coefficient 2 to both sides, we have 36a;2_24a;+4=780 + 4=:784. Extracting the square root of both sides, we find ± 28 + 2 6a?— 2= ±28; which gives x=i — — - =5 or —4^. The square root of the first side of the Equation prepared as above, will be X multiplied into twice the given coefficient of a;^, + or — the given coefficient of x in the second term, according as this term is 4- or — , This is evident from the preceding illustration, (261.) From these principles we derive RULE XXIV. (262.) To Reduce an Equation of the form ax^+bx^s to a Si?nple Equation. 1. Double the coefficient of a;^, and divide the first member by x. 2. Multiply the second member by 4 times the coefficient of a; 2, add the square of the given coefficient of x, and extract the square root of the sum. Applying this Rule to the numerical equation 3x^-5x=z50, we immediately obtain 6x —5 = ± ^50 xl2-\-26=± -/625. When the coefficient of x^ is unity, the multiplier of the second member will be simply 4 ; and when the coefficient of x is unity, the square to be added after multiplying the second member is 1. This method of solving a (Quadratic is preferable to the one first given, whenever the coefficient of x would give rise to a. fraction in dividing the Equation by the coefficient of x''^, or in completing the square, according to that method. « EXERCISES. 177 16. Find the value o{ x in the equation 3a;2 + 2a;— 9=76. Ans, oc=5, or — 5J. 17. Find the value of a; in the equation 2x^ — Ux-{-2 = 18. Ans. x=8, or —1. 18. Find the value of ic in the equation x^ — l2x+50=0. Ans. a:=6±V^^^T4^ 19. Find the value of a: in the equation |a;2_i2^=|a:+15f Ans. x=8, or —7. 20. Find the value of a; in the equation • 3x^-l-=z5i-^2x. Ans. x=i±:i^l9. 21. Find the value of x in the equation 5x x^ — +i=rO. Ans. x—^, or -J. 22. Find the value of x in the equation 7.T — 8 a?4-4=rl3 ^ . Ans. xz:z^, ox —2. X 23. Find the value of a; in the equation x^ 4a? -w- + -r -341=0. Ans. a:=9, or -112 o D 24. Find the value of x in the equation 0:2 - I = - - ^. ^^s. oc=^±^\/ 97. 25. Find the value of ic in the equation 10 14— 2a; ^^ 5 — =2f . Ans. x=3, or 1{?. X x^ ' *^ 26. Find the value of x in the equation — — +3:c+3 = 13. Ans. x—6±^—4c. 27. Find the value of x in the equation 17x llx x^-\ = 4. Ans, a:=— t, or —8. 4 4 ^ 28. Find the value of x in the equation, 3a:-3 3a;-6 . 5^ — =2a;H — , Ans. a?=4, or —1. a: — 3 2 29. Find the value of a; in the equation a;+4 1-x 4a?-f-7 . — = — 1+ — - — .. Ans. a;=21, or 5 3 a;— 3 9 30. Find the value of a; in the equation x-{-a ox ^ - , . , —0. Ans. x=±a'x/—-^, * X X — a ^ 12 J78 EXERCISES. Equations in which the Unknown Quantity is contained in a Surd Expression, 31. Find the value of a in the equation 5 + Va^^+36 = 15. By transposition, y^M- 36 = 15—5 = 10. Squaring both sides, a;3 + 36 = 100; (257). from which, x^ = 64. We have now a pure cubic equation, in which x has necessarily three values or roots, (255). Extracting the cube root of each side of the last equation, we find 07=4, which is one value of a. To find the other two values of x, we must reduce the cubic equa- tion to a quadratic by division, (253). Dividing each side of the equation a;' —64=0 by a;— 4, we find a;2_f_4a:4-l6=0, or x^-\-4.x=z — lQ ; hence x=:—2±V—^^' We have thus found x=A, or — 2+\/ — 12, or — 2— -v/ — 12 ; the first value being real, the other two i7naginary. Each of these imaginary values of a;, as well as the real value 4, will satisfy the equation a;^ = 64. Thus X being equal to —2+ V— 12, we have, (247), (249), ir2=(-2+/^12)^=4-4v^-12-12=-8-4v^^Il2. and a;3=(-8-4-v/-12)(-2 + -v/^T2) = 16+48 = 64, In like manner the other imaginary value of x may be verified. 32. Find the value of x in the equation 6 + V3a;+4 = ll. Ans. x=7, 33. Find the value of x in the equation 24 — •v/2a;2 4-9 = 15. Ans. x=±6. 34. Find the value of x in the equation 20-1/53+40=4. Ans, x=6, or - + — 3+-v/^27, or —3—/ — 27. EXERCISES. 179 35. Find the value of a; in the equation From this equation we shall find a;=z25 or 16. The value 25 will not satisfy the original equation if the square root of x be restricted to its pesitive value ; but this root is ± ^x, and 25 satisfies the equation for the negative root. Two values of x may be found in each of the next three Exercises ; but only that value is given in the Ans. which corresponds to positive roots in the given Equation. 36. Find the value of a; in the equation 2Vic+Va;+9 = 13. Ans. x=zl6, 37. Find the value of x in the equation 38. Find the value of « in the equation 4:Vx-\-l6=z7Vx-i-l6—x — G, Ans. x=9. 39. Find the value of x in the equation ^i- =Va-\-V2x^-{-x^, Ans, a:=12, or 4. 40. Find the value of a; in the equation By rationalizing the denominator, (243... 2), we shall find Extracting the square root of each side of this equation, we have x—2= ■ ^: -. Ans. x=5, or 3. o 41. Find the value of a; in the equation x2-Gx+9= ^±V(^!llli) ji^s, x=5, or 4. x—^{x^ — l6) 42. Find the value of x in the equation ^x^ + 37 X(x^ +37)^=6^. Ans. x=3, or-f ±1 V^. 180 EXERCISES. Equations of a Quadratic Form with reference to a Power or Root of the Unknown Quantity. (263.) Any Equation containing the unknown quantity x in "but two terms — with its exponent in one double its exponent in the other — is a Quadratic with reference to the lower jpoiver of x ; and the value of such power may be found accordingly. To find the value of a? in the equation The \ii^ex fractional power a;- is the square of the lower x^ ; and the equation is therefore quadratic with reference to x^. Completing the square, we have a;i_f-4a;4_|_4_2] +4 = 25. Extracting the square root of each side, a;4_f-2=±5; from which a;* =3, or —7. By raising each of these values to the A^th power, we find a;=81, or 2401. The first of these two values of x is easily verified. In verifying the value 2401 it must be observed that its 4th root is —7, and that 4a;^ is therefore —28. When a; is in a fractional power, in the following Exercises, only that value will be given in the Ans. which satisfies the given form of the Equation. — Imaginary values of x are also omitted. 43. Find the value of x in the equation a;4-2a;2 + 6 = 230. Arts. «=±4. 44. Find the value of x in the equation a;6 + 20a73_l0 = 59. Ans. a;=y3, 45. Find the value of x in the equation 2a;4-r2;2 + 20 = 23. Ans. x=^^Q>. 46. Find the value of x in the equation ^x^'—5x^= — l^. , Ans. x-Z^. 47. Find the value of x in the equation 6o;3"— 5a;3"+ll84=0. Ans. x-B. EXERCISES 181 48. Find the value of x in the equation In this equation, we must regard the binomial (a: +12) as the un- known quantity ; and to simplify the operation we may represent this binomial by y. Then y^-\-yi=Q which gives 2/* = 2 or — 3. By restoring the binomial value of y, we have (ir+12)i=:2, or -3. Ans. a:z=4. 49. Find the value of a; in the equation (2a:+6)*-6 = -(2a;+6)^. • Ans. x=5. 50. Find the value of a; in the equation a;2-f-ll4--v/a;2-fn=42. Ans. x=±:5. 51. Find the value of x in the equation 52. Find the value of x in the equation This equation may be reduced to the form of a qicadratic, thus. — The first two terms of the square root of the first side will be found to be 2x'^ — X ; and the remainder will be — x"^ -\- - , which may be put under the form — ^(2x^—0:). Now the square of the root found, and the remainder, are together equivalent to the first side of the equation ; hence we have (2a;2-^)2_^(2:c2-a;)r=33. Ans. a;=:2, or-1^. A Biquadratic equation may be reduced to the form ,of a Quad- ratic, as above, whenever the remainder — after having found the first two terms of the square root of the first side — can be resolved into two factors, one of which is the same as tlie part of the root thus found. 182 EXERCISES. 53. Find the value of x in the equation a:3— 8a;2-f-19a:=:12. Transposing the 12, and multiplying by x, we have ^4_8^3_|_i9^2_i2a; = 0. We have now a Biquadratic equation which may be reduced to the form of a Quadratic by the method just exemplified. We fi.lid the first two terms of the square root of the first side of the Equation, by the common Rule. ^X^—^X /—8a;3 + 19:^2 /-8:c3 + 16ic2 3.'?;2 — 1207. The remainder, 3a;2_12a7, may be resolved into 3(a;2_4a:)^ in which the binomial factor is the same as the part of the square root above found. Then (a:2_4a')2 + 3 (a;2— 4a;)=:0. Ans. x—\, 3, or 1. 54. Find the value of x in the equation a;4_2a;3 + a;^5112. Ans. a: =9. K>^. Find the value of x in the equation x^-\-2x:^-lx'^-'6x=-\2. Am. a;=:2, — 3, l,or — 2 k)^. Find the value of a; in the equation ^4_iOx3-|-3oa:2— 50:e + 24i=:0. Am. a:i=l, 2, 3; or 4. 57. Find the value of a; in the equation a;4 _l2a;3 + 44:^2 _48a:_z 9009. AjIIS. a:rz:13. 50. Find the value of x in the equation Am. x=2a±\/Qa^ ±-/s+ 16a*. PROBLEMS. |gjt PROBLEMS In Pure Equations and Affected Quadratics containing but One Unknown Quantity. 1. Find two numbers such that their product shall be 750, and ^he quotient of the greater divided by the less, 3;J-. Let X represent the greater of the two numbers ; 750 then NAiIl represent the less ; and the Equation will be 750 ^2 X-. or -— - =34^. X 750 ^ From this equation we shall find a;=:50, or —50. Each of these values Mall satisfy the Equation of the problem ; but only the positive one can be taken to answer the conditions of the problem itself, in which the required numbers are understood to be positive, as in the problems of common Arithmetic. Ans. 50, and 15. 2. Find a number such that if \ and -J of it be multiplied together, and the product divided by 3, the quotient will be 298f . Ans. 224. 3. A mercer bought a piece of silk for £16 4s. ; and the number of shillings that he paid per yard, was to the number of yards, as 4 to 9. How many yards did he buy ? and what was the price per yard ? Let X represent the number of shillings he paid per yard ; 9a; then 4 : 9 :: a? : — - , the number of yards. 4 But without forming a Proportion, the number of yards is readily known to be |- of the price per yard. Ans. 27 yards, at 12^. per yard. 4. Find two numbers which shall be to each other as 2 to 3, and the sum of whose squares shall be 208. Ans. 8 and 12. 5. A person bought a quantity of cloth for $120 ; and if he had bought 6 yards more for the same sum, the price per yard would have been $1 less. What was the number of yards? and the price per yard ? Ans. 24 yards, at $5 per yard. 184 PROBLEMS. 6. Divide the number 20 into two such parts that the squares of these parts may be in the proportion of 4 to 9. Ans. 8, and 12. 7. A merchant bought a quantity of flour for $100, which he sold again at $5J per barrel, and in so doing gained as much as each bar- rel cost him. What was the number of barrels? Ans. 20. 8. Divide the number 800 into two such parts that the less divided by the greater, may be to the greater divided by the less, as 9 to 25. Let X represent the less number : — we shall then have the Pro- portion ■ 800— a; X which will be converted into an Equation by putting the product of the two extremes equal to the product of the two means. Ans. 300, and 500 9. Two fields which differ in quantity by 10 acres, were each sold for $2800, and one of them was valued at $5 an acre more than the other. What was the number of acres in each ? Ans. 70, and 80. 10. A and B started together on a journey of 150 miles. A tra- veled 3 miles an hour more than B, and completed the journey 8^ hours before him. At what rate did each travel per hour? A71S. 9, and 6 miles. 11. A man traveled 105 miles, and then found that if he had gone 2 miles less per hour, he would have been 6 hours longer on his jour- ney. At what rate did he travel per hour ? Ans. 7 miles. 12. A person has two pieces of silk which together contain 14 yards. Each piece is worth as many shillings per yard as there are yards in the piece, and their whole values are in the proportion of 9 to 16 ; how many yards are there in each piece ? A?is. 6, and 8 yards. 13. A merchant sold a piece of linen for $39, and in so doing gained as much per cent, as it cost him. What was the cost of the Unen? Ans. $30. 14. A grazier bought as many sheep as cost him $100. After re- seiving 5 of the number, he sold the remainder for $135, and gained $1 a head on them : how many sheep did he buy ? Ans. 50. 15. Find two numbers which shall be in the proportion of 7 to 9, and have the difference of their squares equal to 128. Ans. 14, and 18. 16. An officer would arrange 2400 men in a solid body, so that each rank may exceed each file by 43 men. How many must be placed in rank and file ? A?is. 75, and 32. PROBLEMS. I §5 17. Two partners gained £18 by trade. A's money was employed in the business 12 months ; and B's, which was £30, 16 months, A received for his capital and gain £26 ; what was the amount of his capital 1 Let X represent A's capital; then 26 — x will be his gain ; and since the gain is in the coiiipound ratio of the capital and the time it was employed, we have 12a:+16x30 : 12a;:: 18 : 26— a;. The first ratio in this proportion may be simplified by dividing the antecedent and the consequent by 12, (158). Ans. £20, 18. A detachment from an army was marching in regular column, with 5 men more in depth than in front ; but upon the enemy's coming in sight, the front was increased by 845 men ; and by this movement the detachment was drawn up in five lines. What was the number of men? Ans. 4550. 19. A company at a tavern had £8 \5s. to pay, but before their bill was settled, two of them went away, when those who remained had 10s. apiece more to pay than before. How many were there in the company at first? Ans. 7. 20. Some gentlemen made an excursion, and each one took the same sum. Each gentleman had as many servants as there were gen- tlemen, and the number of dollars which each had was double the whole number of servants ; also the whole sum taken with them was $3456. What was the number of gentlemen? Ans. 12. 21. Divide the number 20 into two such parts, that the product of the whole number and one of the parts shall be equal to the square of the other. Ans. 10/ 5 — 10, and 30 — 10^5. 22. A laborer dug two trenches, one of which was 6 yards longer than the other, for £17 16*., and the digging of each cost as many shillings per yard as there were yards in its length. What was the length of each? Ans. 10, and 16 yards. 23. There are two numbers whose product is 120, and if 2 be added to the less, and 3 subtracted from the greater, the product of the sum and the remainder Avill also be 120. What are the two num- bers? Ans. 8 and 15. 24. Two persons lay out some money on speculation. A disposes of his bargain for £11, and gains as much per cent, as B lays out; B's gain is £36, and it appears that A gains 4 times as much per cent, as B. What sum did each lay out? Ans. A £5, B £120. 25. A set out from C towards D, and traveled 7 miles an hour. After he had gone 32 miles, B set out from D towards C, and went each hour ^ of the whole distance ; and after he had traveled as many hours as he went miles in one hour, he met A. Required the distance between the two places. Ans 152, or 76 miles. 2gg SOLUTION OF TWO EQUATIONS. Solution of Two Equations — One or Both of the Second or a Higher Degree — Containing Two Unknown Quantities. (264) For the solution of two Equations, containing two unknown quantities, the method which naturally occurs is, 1. By elimination between the given equations to derive a new equation containing but one of the unknown quantities, and thence to find the value of that quantity. 2. By substituting this quantity for its symbol in one of the equa- tions containing the other unknown quantity, to determine thence th© value of that quantity. There are, however, some facilitating expedients to be applied, in certain cases, to equations of the second and higher degrees : these will be exemplified as we proceed. But the solution of two Equations — one or both of the second or a higher degree — containing two unknown quantities, may be impossible by the method of quadratics, — from the impossibility of deriving from them a new equation containing but one unknown quantity, which will admit of a quadratic solution EXAMPLES AND EXERCISES. 1 . Find the values of x and y in the equations 2a?+2/=10, and 20:2— 3^3/+ 32/2 =54. From the first equation, we have Then 2.- ^(^00-^0^+^^^)- and .y= i^^Z^. 4 "^2 By substituting these values in the second equation, we find 2(100— .203/+?/^). 10?/-?/ 4 ^ + 32/2=54. The value of y may be found from this equation ; and by substi- tuting the value of y for y in the first equation, the value of x may readily be determined. Observe that the left hand fraction in the last equation may be re- duced to lower terms; and the solution of the equation be thus some- what simplified. Am. x=z3, or -Q-; ?/=4, or —J. EXAMPLES AND EXERCISES. |§y 2. Find the values of a: and y in the equations iC+?/=9, and x^^-\-y^=z^5. Ans. 7/=rdbl, or ±~^* y 6 yG SOLUTION OF EQ,UATIONS. 5^39 13. Find the values of x and y in the equations 4:X^—2xy=l2, and 2y^-\-3xij=8. Ans. x=dz2,0Y ^3^-^; 2/=zztl, ox ±8^^. 14. Find the values of a; and y in the equations 3y^—x^—39, and 0^2+43:?/ ==256— 47/2. Ans. a;=d=6, or ±102; 2/=±5, or+59. Solutions by Means of Two Auxiliary Unknown Quantities. (266.) When the unknown quantities are similarly involved m each of the two Equations, the solution will sometimes be facilitated by substituting for the two unknown quantities the sum and difference of two other unknoivn quantities. 15. Find the values of x and y in the equations 2 2 x^y=\2, and ^+?L=18. y X If we assume x equal to v-\-z, and y equal to v — 2^, we shall have x-\-y=.2vz=i\2 ; and hence 2; = 6. Then x — Qi-^z, and y=6 — z. Substituting these values of x and y in the second equation, we have (5±^+(5z:^:.i8. 6—2; 6+2; Clearing this equation of its fractions, (64-:s)^ + (6— 2)3 = 18(36— 2^2), By developing both sides of this last equation, and proceeding with the solution in the usual manner, we shall find 2;=: ±2. Having now found the values of both v and z, the values of x and y are easily obtained. An?,. a:=::8or4; 3/=:4or8. 16. Find the values of a: and y in the equations x-\-y—\^, and a:3 + 7/3 = 280. Ans. x—\ or 6 ; y—^ or 4. 17. Find the values of x and y in the equations a;+?/i=ll, and 0:4+7/^=2657. ^ .Ans. ir=4 or 7, 2/=7 or 4. 18. Find the values of ic and y in the equations a?4-2/=10, and a;^ +7/^ = 17,050. J.rtS. a!;=3 or 7, ^=7 or 3 # ^90 MISCELLANEOUS SOLUTIONS AND EXERCISES. Miscella7ieous Solutions and Exercises, 19. Find the values of a and y in the equations x7j=6, and x^ -{-x=zl8—y^ —y. From the second equation, x^-Jry^+x+y=lS. Adding twice the first, This last equation may be put under the form {^+y)^ + (x+y)=30; which is quadratic with reference to x-\-y, and from which we may therefore find the value of x-{-y. Ans. x=z2,ov3; or— 3=f-v/3; y=z3,0Y2; — or 3dt -/S- 20. Find the values of a: and y in the equations x-{-y=:Q, and x^y^-j-4:xy=96. Ans. x=2, or 4 ; or 3=^/21 2/=4, or 2 ; or 3±'/21 21. Find the values of ao and y in the equations .3 3. 1. .1 x^y^=z2y^, and 8x^ — y^=:\4:. ? 2 ,1 1 2 Dividing the first equation by y^, we find x'-^=z2y^, or t/2— ^a^ and by substituting this value of ^/^ in the second equation, that equa- tion will become quadratic with reference to x^. Ans. a?i=2744, or 8 ; yz=960i, or 4. 22. Find the values of x and y in the equations x^y—xy=6, and x^y—y=2l. Dividing the first equation by the second, we have x^y -xy 6 x^y y~~21 This equation will be simplified by reducinof each of its two frac- tional members to its loivest terms. " \ A71S. a; =2, or 2 ; y=3, or — 24. 23. Find the values of x and y in the equations a;^-fa;7/2=39, and x^y+y^=z2^^. Ans. x — o] y—2. MISCELLANEOUS SOLUTIONS AND EXERCISES. 19| 24. Find the values of x and y in the equations Dividing each side of the second equation by the corresponding side of the first, we shall find, x^-}-xy-\-y^ =79. Squaring Xhe fii&t equation, and subtracting, we have 3xy=(j3 A?is x = 7, or —3 ; y=o, or —7 25. Find the values of x and y in the equations x^ — y^=z5, and x^ — 7/^ = 65. Ans. x=zt3, ov y=±: o 26. Find the values of a; and y in the equations x-i-y=:60, and 2{x^ ■i-y^) = 5xy. The second equation may be put under the form x^ -\-y^ —2^ocy=(S , and the solution will be facilitated by subtracting this from the squaie of the first equation. Ans. a;=40, or 20 ; y=20, or 40. 27. Find the values otx and y in the equations a>w=z8, and \-^ = 9, y ^ Multiplying the two equations together, we find x^ -{-y^ =72 ; and, multiplying this equation by x^, we have x^ + x^ y^ = 7 2x^ . From the first equation, x^y^ = 8^=5l2 ; and if this number be substituted in the preceding equation wc shall have a quadratic with reference to x^. Ans. 07=4, oi* 2; y=2, or 4. 28. Find the valaes of a? and y in the equations xy=25, and x^-\-y^ =z\Qxy. The second equation will be reduced to the same form as the second in the preceding example, by dividing it by xy. Ans. x=5 . y=.5, 39. Find the values of x and y in the equations x^—y^—{x-\-y) = S, and {x—yy{x-\-y)^32. Dividing each equation by ic -[-?/, we have X — V— 1= . and {x—yY= — ■ — . ^ x-^y ^ ^' x-\-y Transposing — 1, and squaring, we obtain (--^)^ = (r+-y)+l)'=^; from which will result a quadratic with reference to x-\-y. Ans. x=:5\ y=S. 192 MISCELLANEOUS SOLUTIONS AND EXERCISES. 30. Find the values o^ x and y in the equations x-{-\^xy-\-y=T, smd x^ -\-xi/-{-y^ =z2l. Dividing the second equation by the first, we have x—Vocy+y=^^.^ We shall now obtain two equations of simpler forms by adding th( third equation to the first, and subtracting it from the first, A?is. x=zl, or 4 ; ?/=4, or 1 31. Find the values of a; and y in the equations x-i-y . — — 2xy 12 ——=Vocy-\-^, and V^y=^--_^+-j' From the first equation, by transposition, Vooy=—^ 4. Clearing this equation of its fraction, and squaring, 4:xy=:{x+y)^—^6 {x-{-y)-\-64:. By equating the two values of -y/xy, from the second and third equations, 2xy \2_jx-\-y ^-\^j~6'^~~2 ' 2xy _x-{-y 32 ^^ x+y~~2 y* Clearing this equation of its fractions, 64 ^xy={x-\-yy — ^{x-\-y)' We shall now obtain a simple equation by subtracting this last equation from the fourth equation. Ans. a: = 2, or 18 ; y=z 18, or 2. 32. Find the values of ic and y from the equation and proportion xy"^ — a:z=:3 ; x'^y^ — x"^ : aj2-j-a;2?/2-i-a;2^4 . . 5:7. In any proportion the difference of the first and second terms 2S to the first, as the difference of the second and third zs to the second, (160). Hence from the given proportion, we shall find 2x2+a:2y2 : x^y^—x^ . : 2 : 5. Dividing the first antecedent and consequent by a;^, (158) 2+2/2 :y^-i : : 2:5. 6 / — 3 Ans. a;z=l, or — - ; y=2, or. / — . MISCELLANEOUS SOLUTIONS ANt EXERCISES. ^93 33. Find the values of a? and y in the equations x^-\-xy+y^ = 13, and x'^ + x^y^ +y^ =91. Dividing the second equation by the first, x''^—xy-\-y^=i7. Adding the third equation to the first, and dividing the result by 2 ; and also subtracting the third from the first, and 2xy=Q. The solution now proceeds by adding the latter of these two equa- tions to the former, and also subtracting the latter from the former, and extracting the square roots of the resulting equations. Ans. x=±:3, 7/=±l. 34. Find the values of x and y in the equations x^y=x^y''^—x^, and x^y^ -{-x^ =x^y^ —x^ . Dividing each equation by x^y we have y=y^ — l, and y^ -\-\=zxy^—x. The value of y is to be found from the first of these equations, and substituted in the second. Ans. x=^^5 \ y=^±^^6. 35. Find the values of a; and y in the equations xy:=x^ — 7/2, and x^ -\-y^ ^zx"^ — y^. If we assume y to be equal to xv, and substitute this product for y in the two equations, (264,) we shall have x'^v^^x'^ — x^v^y and x'^-\-x^v^=zx^ — x^v^. These equations may be solved in the same manner as those in the preceding example. Or the value of a; may be found, in terms of y, from the first of the given equations, and substituted in the second. Then for the second equation, Ans. 2;=i(5±y5); y^±l^6. 194 MISCELLANEOUS SOLUTIONS AND EXERCISES. 36. Find the values of x and y in the equations Subtracting the first equation from the second, 2x^y+2xy^ = n^Q. Adding this to the second equation, we obtain By extracting the cube t'oot of this equation, we shall fin« the Value of a; -f?/, which maybe substituted fora:-f-y in the third equition. Ans. a;=rll; y=zd. 37. Find the values of x and y in the equations a??/=:320, and a;^— 2/^ = 61 (a;— 2/) 3. Dividing the second equation by x — y, we have x^+xy-\-y^=z^l (x—y) 2. By converting this equation into a. proportion, (153,) x'^-\-xy-\-y'^ : [x—yY : : 61 : 1. The solution now proceeds by developing the terra {x—yY, — sub- tracting each consequent from its antecedent, and forming a proportion oi Xhs consequents smd. remainders, &c, (160). Ans. a;=20 : ^=16. 38. Find the values of x, y, and z, in the equations, x^+y^-{-xy=37 ; x^-{-z^-\-xz=4:9 ; y^-{-z^+yz=61. Subtracting the first equation from the second, and decomposing, (z-y){z-]-y)-\-(z—y)x, or {z-\-y+x){z—ij) = 12 ; 12 from which z-^ii-\-x:=: * z-y Proceeding in like manner with the second and third equations, 12 we shall find y-\-x-^z^= * y—x Hence the right hand members of the last two equations are equal to each other, (113 1); and since the numerators are the same, we have z — y^^y — X, from which 2y=zx-\-z. By substituting 2y for x+z in the sixth equation, we shall find 7/2— ya;=:4. The value of a; from this equation, is to be substituted in the first equation. Aiis. x=o; 7/=:4 ; zz=z^. PROBLEMS. 295 PROBLEMS In Quadratic Equations of One or More Unknown Quantities. 1. The area of a rectangular lot of ground is 384 square rods, and itB lerigth is to its breadth as 3 is to 2. Required the length and breadtli of the lot. ^ The areay in square measure, of a rectangle, is expressed by the product of the number of linear units in its length x the number of linear units in its breadth. The length and breadth must bo taken in the same denomination in multiplying : the area will be found in the corresponding denomina- tii)n of square measure. Let X represent the length, and y the breadth of the lot ; then by the conditions of the problem, a??/=384, and a; : 2/ : : 3 : 2. 2x Or, if X represent the length, - will represent the breadth, and we shall then have 3 Ans. 24, and 16 rods. 2. The length of a rectangular garden exceeds its breadth by 6 rods, and its area is 216 square rods. What are the length and breadth of the garden ? Ans. 18, and 12 rods. 3. Find two numbers whose sum shall be 24, and whose product shall be equal to 35 times their difference. Ans. 14 and 10. 4. Divide a line 20 inches in length into two such parts that the rectangle or product of the whole line and one of the parts shall be equal to the square of the other part. Ans. lOVS— 10, and 30 — lO-y/^. 5. Find the dimensions of a rectangular field, so that its length shall be equal to twice its breadth, and its area 800 square rods. Ans. 40, and 20 rods. 6. The sum of the two digits of a certain number is 10, and if their product be increased by 40, the digits will be reversed. What is the number ? ''A71S. 46. 196 PROBLEMS. 7. The sum of two fractions is 1-^, and the sum of their reciprocals is 3^ ; what are the two fractions 1 Ans. -J- and |. 8. The perimeter, or sum of the four sides of a rectangle, is 112 rods, and its area is 720 square rods. What are the length and breadth of the rectangle? Ans. 36, and 20 rods. 9. Divide the number 60 into two such parts that their product shall be to the sum of their squares as 2 to 5. Ans. 20 and 10. 10. A merchant bought a piece of cloth for $120, and after cutting off 4 yards, sold the remainder for what the whole cost him — by which he made $1 a yard on what he sold. How many yards did the piece contain? Ans. 24. 11. Divide the number 100 into two such parts that the difference of their square roots shall be 2. Ans. 64, and 36. 12. A garden which is 20 rods square is surrounded by a walk whose area is equal to -| of the area of the garden itself What is the breadth of the walk ? ^^s, Sy^S— 10 rods. 13. The sum of the squares of two numbers is 325, and the dif- ference of their squares is 125. What are the numbers ? Ans. 15 and 10. 14. The area of a rectangular court-yard is 875 square rods, and if its length and breadth were each increased by 5 rods, its area would then be 1200 square rods. What are the dimensions of the yard ? Ans. 35 and 25 rods. 15. The difference of two numbers is 4, and the difference of their cubes is 448. What are the two numbers ? Ans. 8 and 4. 16. A grocer sold 80 pounds of mace and 100 pounds of cloves for £65, and finds that he has sold 60 more of cloves for £20 than of mace for £10. What was the price of each per pound. Ans. 10s. and 5s. 17. The fore- wheel of a carriage makes 6 revolutions more than the hind wheel in going 120 yards ; but if the circumference of each be increased 1 yard, it will make only four revolutions more in going the same distance. What is the circumference of each wheel? Ans. 4, and 5 yards. 18. Find four numbers in arithmetical progression, such, that the product of the two extremes shall be 45, and the product of the two means 11. L.et a? be the first term, and y the common difference of the terms ; then the numbers will be X, x-\-7j, x-\-2]j, x-\-'3>y; (175). and by the conditions of the problem we shall have a;2-|-3a^2/=45; and a;2-f-3a:2/+2?/2=77. Ans. 3,1, 11, and 15. PROBLEMS, icj-y 19. A farmer has a field 16 rods long and 12 rods wide, which he wishes to enlarge so that it may contain just twice as J^iuch area, M'ithout altering the proportion of the sides. What will be the dimen- sions of the field when thus enlarged? ^^5 16-v/2 ; and 12'/2. 20. Find three numbers, such, that the difference of the first and second shall be two less than the difference of the second and third, their sum 17, and the sum of their squares 115. The solution will be facilitated by assuming x to represent the second number, and y the diflTerence of the first and second. Ans. 3, 5, and 9. 21. There are two square gardens which together contain 1025 square rods, and a side of the one exceeds a side of the other by 5 rods. What are the sides of the two gardens 1 Ans. 20, and 25 rods. 22. Find two numbers, such, that their sum, their product, and the difference of their squares shall all be equal to one another. Take x-{-y to represent the greater, and x — y the less number. Ans. I =h -v/f, and -\ ± -/i • 23. A merchant received $12 for a quantity of linen, and an equal sum, at 50 cents less per yard, for a quantity of calico, which exceeded the quantity of linen by 32 yards. What was the quantity of each ? Ans. 16, and 48 yards. 24. Find two numbers whose sum multiplied by the greater shah be equal to 192, and whose difference multiplied by the less shall be equal to 32. The solution will be facilitated by taking x to represent one of the required numbers, and ir?/ the other. Ans. 12, and 4. 25. Three merchants gained $1444 ; of which their respective shares were such that B's, added to the square root of A's, made $920: but if added to the square root of G's it made $912. What was the share of each ? A?is. $400, $900, and $144. 26. The sum of three numbers in harmonical progression is ]3> and the product of the two extremes is 18. What are the numbers ? If X and y represent the two extremes, the mean term will be — -^ (184) x+y ^ ^\ Ans. 6, 4 and 3. 27. There is a rectangular field whose length is to its breadth as 4 to 3. A part of this field, which is equal to ^ of the whole, being in meadow, there remain for ploughing 1296 square rods. What are the dimensions of the field ? Ans. 48. and 36 rods. 198 PROBLEMS. 28. The sum of three numbers in geometrical progression is 21, and the sum of their squares is 189. What are the numbers ? If a; and ?/ represent the two extremes, tho-mean term will be Vi^, (189). Ans. 3, 6, and 12. 29. A and B set out from two places which are distant 110 miles, and traveled towards each other. A went five miles an hour ; and the number of hours in which they met was greater, by four, than the number of miles B went per hour. What was B's rate of traveling ? Ans. 6 miles per hour. 30. The arithmetical mean between two numbers exceeds the geo- metrical mean by 13, and the geometrical mean exceeds the harmoui- cal mean by 12. What are the numbers ? Ans. 234 and 104. ^ 31. Three merchants made a joint stock, by \fhich they gained a sura less than that stock by $80. A's share of the gain was $60, and his contribution to the stock was $17 more than B's; also B and C together contributed $325. How much did each contribute ? Ans. $75, $58, and $267. 32. Of three numbers in Geometrical Progression the greatest ex- ceeds the least by 15, and the difference of the squares of the greatest and the least is to the sum of the squares of the three numbers as 5 to 7. What are the numbers ? Assume x to represent the first term, and y the ratio of the pro- S^^^^^on. ^^^3 5^ 10^ ^„^ 20. 33. Two persons set out from difierent places, and traveled towards each other. On meeting, it appeared that A had traveled 24 miles more than B, and that A could have gone B's journey in 8 days, while B would have been 18 days in performing A's journey. What distance was traveled by each] A?ts. 72, and 48 miles. 34. The joint stock of two partners was $416. A's money was in the business 9 months, and B's 6 months. When they shared stock and gain, the first received $228, and the second $252 ; what was each man's amount of stock ? Ans. A's $192, B's $224. 35. The sum of $700 was divided among four persons, A, B, C, and D, whose shares were in Geometrical Progression ; and the dif- ference between the greatest and the least was to the difference be- tween the two means as 37 to 12. What were the several shares ? Ans. $108, $144, $192, and $256. CUBIC AND HIGHER EQUATIONS. J 99 Solution of Affected Cubic and Higher Equations. Various methods have been devised for the solution of Affected Equations of the third and higher degrees. Some of these methods are very prolix, — vi^hile others are of limited application ; we shall ex- plain those which are the most useful in a practical point of view, without attempting a full exposition of this subject. Under the head of General Properties of Equations, we have al- ready noticed the divisors, (253), and the number of roots, (255) of equations ; we here present the General Law of the Coefficients of Equations. (267.) When the terms of an Equation containing but one unknown quantity x, are all arranged, according to the descending powers o^ x, in the tirst member — with the known or absolute term for the last term — and the coefficient of the first term is unity ; then, 1 . The coefficient of the second term is equal to the sum of all the roots of the equation, with their signs changed. 2. The co-efficient of the third term is equal to the sum of the jn'o- ducts of all the roots combined two and tico, with their signs changed, &c. 3. The known or absolute term is equal to the product of all the roots, with their signs changed. To demonstrate these principles with reference to a Cubic Equa- tion, let the three roots be denoted by fij, h, and — c ; then x—a, x—h, and a;+care the divisors of the equation, and the equation may fee ac- cordingly resolved into {x~a) (x-b) (x+c) =0, (253). By performing the multiplication which is here indicated, and de- composing the terms containing the like powers of x in the product, we find x^ + {c—a—b)x^-i-{ab—ac—bc)x-{'abc=zO. In this equation the coefficient of x^ is the sum of the roots a, b, and — c, with their signs changed ; the coefficient of x is the sum of the products of the roots combined tioo and two, with their signs changed ; and the known or absolute term abc is the product of all the roots, with their signs changed. The same principles may be demonstrated, in like manner, in refer- ence to an Equation of the second, or of any of the higher degrees. 200 CUBIC AND HIGHER EQUATIONS. An application of these principles may be made to the equation x^-{-3x — 10=:(a;— 2) (z-{-5) = 0, whose roots are 2, and —5 ; or to a;3_192'4-30 = (x— 2) {x+5) (a;— 3) = 0, whose roots are 2, —5, & 3 In the second equation it will be observed that the second term containing x^, is wanting, since its coefficient, that is, the sum of th< roots 2, —5, and 3, is ; and that 19a? therefore corresponds to the third term, (267. ..2). Determination of the Integral Roots of Equations. (268.) If an equation containing but one unknown quantity a:, witl all its terms transposed to one side, be divisible by x-\- or —any num her, that number, with a contrary sign, will be a root of the equation (254). The trial numbers to be used in this division, are the factors or dh visors of the known term of the equation, since that term is equal to the product of all the roots of the equation, (267... 3). When an equation has any integral roots, such roots may be readily determined by an application of these principles. EXAMPLE. To find the values of x in the equation a?3 + 3a;2_4a: = 12. The divisors of the known term 12 are ], 2, 3, 4, 6, and 12; and it will be found, on trial, that the equation a:34-3a;2— 4a: — 12 = 0, is divisible by a: — 2, aj-f-2, and a;-|-3 ; hence the values of x, or roots of the equation, are 2, —2, and —3. After any one of the three roots has been determined, the two re- maining ones may be obtained directly from the quadratic equation which results from dividing the given equation by x-\- or — the root already found. a:-2)a;3 + 3a;2-4a;-12( x^ + 5x^Q>. By dividing the given equation by a:— 2, we thus find a:2 + 5:r+6 = 0, or a;2 + 5a;r= — 6, which gives x=i — 2 or — 3 By this method the last two roots are found the same as before. CUBIC AND HIGHER EQ,UATIONS. 201 Solution of Equations by Approximation. (269.) The following method of solution may be applied to an Equation of any degree — even to one in which the unknown quantity is left, without rationalization, in a surd expression. 1 . By trial find two numbers — differing by a unit or less — which being substituted for x in the given Equation, will produce results, the one less and the other greater than the known tenn of the equa- tion ; then, The difference between the two results, Is to the difference between the two assmined numbers, As the difference between either result and the known term, Ts to the correction, nearly, required in the corresponding as- sumed number. 2. Take the corrected root thus obtained for one of two numbers to be substituted for x, and find, and apply, a correction as before. We shall thus obtain a nearer value of the unknown quantity ; and the approximation may be carried, in like manner, to any required exactness. EXAMPLE. To find an approximate value of x in the equation. a;3 + a;2-|-a;=100. First, It will be found that x is more than 4, and less than 5. Sub stituting these numbers for x, we have 64 x^ 125 16 ^2 25 4 X 5 84 155 The difference between the two results is 155—84 = 71 ; and the difference between the less result and the known term 100 is 16. Then 71 : 1 : : 16 : the correction . 225. This correction, added to the less assumed number, gives 4.225 for an approximate value of x. Secondly, By substituting 4.2 and 4.3 for x, we have 74.088 x^ 79.507 17.64 a;2 18.49 4.2 X 4.3 95.928 102.297 202 CUBIC AND HIGHER ECtUAT|pN&. Forming a proportion between the difference of these two results, and the difference between the greater result and the known term 100, 6.369 : ,1 :: 2.297 : the correction .0^^. By subtracting this correction from the greater assumed number 4 3, we have 4.264 for a nearer value of a;. For the next approximation we should take 4.264 and 4,265 to to be substituted for x in the given Equation. The value of x would then be found to be 4.2644299 very nearly. In the Proportion for finding the correction, it is best to- employ the less error in the results of the substitution. Thus in the first substitution, in this Example, thd error in the less result 84 is (100 — 84) = 16, and this being less than the error m the 155, we employ 16 in the first proportion. But in the second substitution, the error in the greater result 102.297 is less than the error in the 95.928, and we accordingly use (102.297 — 100) = 2.297 in the second proportion. Each approximative solution will generally double the number of true figures in the root. Thus in the preceding Example we found by trial that 4 is the first figure in the root, and the first solution gives 4.2 for the first two correct figures ; the next solution gives 4.264 ; and the number of figures will again be doubled by a third solution. This property determines the number of figures which need be found in the successive corrections of the assumed numbers. To find the other Roots of the given Equatioji, we would divide- a3+a;2+a?— 100 = by a;— 4,2644, &c. = 0, (253). We should thus obtain a quadratic equation, from which the o.ther two values of x might be determined, according to the usual method. H^ When all the Roots of an Equation have been found, we may verify them by the property that, with their signs changed, their sum must be equal to the coefficient of the second term of the equation, (267. ..1). Thus the sum of the three roots of the equation in the preceding Example, with their signs changed, would be U7iity, which is the co- eflficient of the second term x^. CUBIC AND HIGHER EQUATIONS. 203 EXERCISES . On Affected Cubic and Biquadratic Equatitms. 1. Find the values of a? in the equation x^-Qx^ + llxzzz^, (268). Ans. x=:l, 2, or 3. 2. Find the values of x in the equation a;3_9^2_|_26a;=24. Ans. x=z2, 3, or 4. 3. Find the values oi x in the equation a;3-3aj2— 6a:=:-8/ Ans. a; = l, 4, or —2. 4 Find the values of x in the equation 2a;3_6a;2_8a;=-24. Ans. x=2, 3, or —2. 5. Find the values of x in the equation x* + 2x^ — 13x^ — Ux-{-24: = 0. Ans. 1, —2, 3, or —4. 6. Find an approximate value of x in the equation x^ + 10a;2 + 5a;=260, (269). Ans. x=4:.lH\ 7. Find an approximate value of x in the equation x^-15x^ + 63x=50. Ans. a;= 1.028.* 8. Find an approximate value of x in the equation x^-l7x^-\-54.x=:350. Ans. a;= 14.95'. 9. Find an approximate value of x in the equation a;4_3a;2__75a,_ 10000. Ans. a!=10.23\ 10. Find an approximate value of a: in the equation 2a;* — 16a;3+40a:2_30a;+l=0. Ans. a;= 1.284'. 11. Find an approximate value of « in the equation (ia;2_i5)2_|.^^^_90. Ans. x=l0.52\ 204 general method of elimination. Elimination by the Method of Common Divisor. (269.) The following is a general method of Elimination, and, for Equations of the higher degrees, it will sometimes be found preferable to any other. Transpose all the terms of the two Equations to one side ; then di- vide one into the other, and the reinainder into the divisor, and so on, as in finding the Greatest Common Measure, (66,) until one of the two unknown quantities is eliminated from the remainder; and put this remainder =0. To eliminate x from the equations a;2 + a:?/=10, and a:?/ + 2?/2 = 24. xy+2y^—24: X x'^-\-xy—lQ \ x^y+2xy^24:—x(y J x^y-^xy^ — lOy x{y^—2i)+10y xy^—2^x-\-10y, or x{y^ —2A) -{-}0y. x^-\-xy—10 3/2-24 ) x^(y^—24:)-\-x{7j^—24:y) — 10y^-{-24.0(x-^y / x^{y^-2A)-^10xy ^ x(y^ —24:y) — 10xy—10y^~^~2i0 %3-24?/) + 10?/2 — 10xy—20y^-\-2i0 x{y^—24:)-{-l0y _^ a:^+2y2_24 )%3_24y)-|_i07/2(^2_24 /a:(y3—24?/)4-23/*— 487/2—242/2 + 576 -23/4 + 82?/2-576 = 0. In the remainder — lOo:?/— 20i/2 + 240, we cancel the factor 10, and change the signs, for the next divisor. In dividing into this di- visor, we take the binomial y^—24: for the quotient, and multiply the divisor by this binomial. The first remainder is equal to 0, because the divisor and dividend arc each equal to ; and it follows hence that each subsequent re- mainder is eqtcal to 0. The operation will he much Tnore simple if we divide the first equation hy the second : the result will be the same. 205 CHAPTEU XI. general description of problems. Miscellaneous Problems. 1. Determinate Problems. (270. ) A Determinate Problem is one in which the given condi- tions determine the values of the unknown or required quantities. A Determinate Problem is represented by as many independent equations as there are difierent conditions to be expressed, or unknown quantities to be determined, (120.) All the Problems which have hitherto been proposed in this work, are determinate ; and no example of this kind need be here given. 2. Indeterminate Problems. (271 .) An Indeterminate Problem is one in which the given condi- tions do not determine the values of the required quantities, — admitting either of an unlimited number of values to those quantities, or else of a variety of values, within certain limits. An Intermediate Problem is represented either by a less number of independent Equations than there are unknown quantities to be deter- mined, or by an identical equation. We give an example of each of these forms of indeterminateness. E XAM ple I. To find three numbers such that the first shall be 5 less than the second, and the sum of the second and third shall be 12. This Problem contains but two conditions ; and if we represent the three required numbers by x, y, and 0, we shall have only the two Equations By subtracting the first equation from the second, we have 206 GENERAL DESCRIPTION OF PROBLEMS. This equation will admit of an unlimited number of values of x and z ; for we may assume any value whatever for one of the letters, as ar, and determine thence the corresponding value of z. Thus if x=\, z=z6l: if x=l, 2 = 6^; if a;=J, z=z6^, &c.; and from the values of x or z, we might obtain the corresponding values of y from one of the given equations. If, however, the required numbers were limited to integral values. the third equation would be satisfied only by x=l, 2, 3, 4, 5, or 6, and z = 6, 5, 4, 3, 2, or 1. In the first equation y=5-\-x, which would give y=z6, 7, 8, 9, 10, or 11. ExampleII. To find a number such, that |- of it, diminished by ^ of it, and by 5, shall be equal to ^^ of the excess of 5 times the number above 60. The equation of this problem will be 3a; x _5x—60 T~S~'^~ 12 . Clearing the equation of its fractions, 9x—4:X—60=5x—60; or 5x—60=z5x—60. This last is an identical Equation, './hich will be satisfied by attri- buting to X any numerical value whatever. The problem is therefore entirely indeterminate. We may obtain an expression for the value of x from the last equation. Thus, by transposition, 52:— 5a;— 60 — 60. By adding similar terms, and retaining ic as a symbol in the first member, we have 0.^=0; which gives x=^. Hence ■§• is a symbol of an indeterminate quantity. The same thing will appear from considering that the quotient of 0-f-O is any quantity whatever; inasmuch as ih.Q divisor Ox any quantity will produce the dividend 0, (43). GENERAL DESCHIPTION OF PROBLEMS. 207 3. Impossible Problems. (272.) An Impossible Problem is one in which there is some condi- tion, expressed or implied, which camiot be fulfilled. - An Impossible Problem is represented by a greater number of independent equations than there are unknown quantities to be deter- mined ; or by an equation in which the value of the unknown quantity is negative— zero — infinite — or imaginary. We subjoin an example of each of these forms of impossibility Example I. To find two numbers whose sum shall be 10, difference 2, and pro- duct 20 Representing the two numbers by x and y, we shall have a:+?/=10; a;— ?/=2 ; xy=20. From the first and second equations the values of x and y will be found to be a: = 6, and y=4. The third equation cannot, therefore, bo fulfilled ; that is, the problem is impossible. If the third equation were xy=z2^, the problem would be possible, but this would not be an independent equation, since it may be derived from the other two. Thus, squaring the first and second equations, and subtracting, we find 4a;2/=96, or a;?/=24. Example II. To find a number which, added to 17 and to 53, will make the first sum equal to \ of the second. If X represent the number, the equation will be 17^ 53+a; 17 + 0:=---. From this equation we shall find x=z — 5. This number, added to 17 and 53, gives 12 and 48, and 12=^ of 48. The problem is impossible in an arithmetical sense, according to wliich addition always implies augmentation ; and it is in this sense only that the problem would be considered. To make it arithmetically consistent, it should be stated thus : To find a number which, subtracted from 17 and from 53, will make the first remainder equal to \ of the second. 20g GENERAL DESCRIPTION OF PROBLEMS. Example III . To find a number such, that if 82 be increased by 3 times that number, ^ of the sum will be equal to ISf. If z represent the number, the equation will be Clearing the equation of its fractions, we find 82+3a:=82; which gives 3a?=:82— 82=0 ; and oc=q = 0, (50). Hence no number can be found that will fulfil the conditions of the problem ; that is, the problem is impossible. The result shows that i of 82 itself is equal to 13|. Example IY. To find a number such, that the sum of ■} of it and f of it, dimi- nished by 2, shall be equal to -^^ of it increased by 3. If X represent the number, the equation will be X 2x ^^ \lx ^ From this equation we shall find 3a;+8a;— 24==lla;+36; or 11a;— lla; = 0a;=60 ; and x=:^^ — CD, infinity ; (50). The result shows that it would require a number infinitely great y to fulfil the conditions of the problem. The problem is therefore im- possible, E X A M P L E V. To divide the number 24 into two such parts, that their product shall be 150. If a; represent one of the two parts, 24— a; will represent the other, and the equation will be 2Ax—x^ = 150 or a^2_24^— — 150, (117); which gives a:=:12± ^144 — 150, =:rl2±y-6. In this value of a:, the part -xZ—Q is i?7iagina7'7/, that is, it is an impossible quantity, (246) ; hence the problem is impossible. GENERAL DESCRIPTION OP PROBLEMS. • gQg That the preceding problem is impossible, will also appear from the following general proposition ; viz : (273.) The square of one half oi any quantity is greater than the product of any two unequal purU of the quantity. Tills proposition may be thus demonstrated. Let s represent any number, and c? the difference of any two parts of that number ; then = the greater part ; The product of these two parts is 4~ This product will vary directly as its numerator s'^—d?, (147) ; and will therefore be the greatest possible when <^=0. In that case the product becomes — , which is the square of—, half the pro- posed number. The student may apply this proposition to any number taken at pleasure. It will be found that the product of the two parts is greater as the parts are more nearly equal to each other ; and greatest when they are the halves of the assumed number. To one or another of the preceding classes every problem may be referred ; that is, every problem is, by its conditions, either determin- ate, indeterminate, or impossible. And the following principles have been established with respect to the Signification of the Different Forms under which the Valtce of the Unknown Qita7itity may be found in an Equation. (274.) 1 . Positive values of the unknown or required quantities, ful- fil the conditions of problems in the sense in which they are proposed. 2. A value of the unknown quantity of the form ■§-, shows that the problem from which the equation was derived is indeterminate. 3. A negative value of the unknown quantity, in an equation of the first degree, indicates an impossibility in the problem, produced by taking this quantity additively, instead of subtr actively, or vice versa. 4. When the value of the unknown quantity in an equation is zero, infinite, or imaginary, the problem from which the equation was de- rived is impossible. 14 210 MISCELLANEOUS PROBLEMS. MISCELLANEOUS PROBLEMS- 1. A, B, and C together have $2000. B has $100 less than twice as much as A, and C $400 less than twice as much as the other two together, What sum has each? Ans. A, $300 ; B, $500 ; C, $1200. 2. A gentleman has three plantations. The first contains 250 acres, the second as much as the first and -^ of the third, and the third as much as the first and second. What is the whole number of acres ? Ans. 1200 acres. 3. A company of workmen had been employed on a piece of work for 24 days, and had half finished it, when, by calling in the assistance of 16 more men, the remaining half was completed in 16 days. What was the original number of men ? Ans. 32 men. 4. A's money was equal to f of B's. A paid away $50 less than |- of his, and B $50 more than f of his, when it was found that the latter had remaining only ^ as much as the former. What sum had each at first? Ans. A, $300 ; B, $400. 5. A person wishing to enclose a piece of ground with palisades, found, that if he set them one foot asunder, he would not have enough by 150, but if he set them one yard asunder, he would have too many by 70. What was the number of his palisades ? A71S. 150 palisades. 6. From two tracts of land of equal size, were sold quantities in the prbportion of 3 to 5. If 150 acres less had been sold from the one which is now the smaller of the two, only |- as much would have been taken from it as from the other ; how many acres were sold from each? » Ans. 150, and 250 acres. 7. A and B had adjoining farms, which, in quantity, were in the ratio of 4 to 5. A sold to B 50 acres, and afterwards purchased from B one-third of his entire tract, when it was found that the original ratio of their quantities of land had been reversed. How many acres had each at first? Ans. A, 200 ; B, 250 acres. 8. A waterman can row down the middle of the stream, on a certain river, 5 miles in J of an hour ; but it takes him 1 ^ hours to return, though he keeps along shore, where the current is but half as strong as in the middle. What is the velocity of the middle of the stream ? Ans. 2-J miles per hour. 9. A farmer has three flocks of sheep, whose numbers are in the proportion of 2, 3, and 5. If he sell 20 from each flock, the whole number will be diminished in the proportion of 4 to 3 ; how many has he in each flock? Ans. 48, 72, and 120 sheep MISCELLANEOUS PROBLEMS. 2il 10. The sum of $1170 is to be divided between three persons, A, B, and C, in proportion to their ages. Now A's age is to B's as 1 to Ij, and to C's as 1 to 2 ; what are the respective shares? Ans. $270 ; $360 ; $o40. 11. A hare is 50 leaps before a greyhound, and takes 4 leaps to the greyhound's 3 ; but 2 of the greyhound's leaps are as much as 3 of the hare's. How many leaps must the greyhound take to catch the hare ? Ans. 300. 12. A vintner has two casks of wine, the contents of which are in the proportion of 5 to 6, and if ^ of the quantity in the second were to be drawn off, the contents of the two casks would be equal. How ma/iy gallons are there in each ? Ans. This problem is indeterminate ; how is its indeterminate- ness indicated ? 13. A person looking at his watch, and being asked what o'clock it was, replied that it was between eight and nine, and that the hour and minute hands were exactly together. What was the time ? Ans. 43m. SSy'^yS. past ei^ht. 14. A criminal having escaped from prison, traveled 10 hours be- fore his escape was known. He was then pursued, and gained upon 3 miles an hour. When his pursuers had been 8 hours on the way they met an express going at the same rate as themselves, who had met the criminal 2 hours and 24 minutes before. In what time from the com- mencement of the pursuit will the criminal be overtaken ? Ans. 20 hours. 15. A regiment of militia containing 875 men is to be raised from three counties. A, B, and C. The quotas of A and B are in the pro- portion of 2 and 3, and of B and C in the proportion of 4 to 5. What is the number to be raised by each ? Ans. 200, 300, and 375 men. 16. If 19 pounds of gold, in air, weighs 18 pounds in water; 10 pounds of silver, in air, weighs 9 in water; and amass of 106 pounds, composed of gold and silver, weighs 99 pounds in water ; what are the respective quantities of gold and silver in the mass ? Ans. 76, and 30 pounds. 17. A farmer having mixed a certain quantity of corn and oats, found that if he had taken 6 bushels more of each, there would have been 7 bushels of corn to 6 of oats ; but if he had taken 6 bushels less of each, there would have been 6 bushels of corn to 5 of oats. How many bushels of each were mixed? Ans. 78, and 66 bushels. 18. Two persons, A and B, can perform a piece of work in 16 days. They work together for 4 days, when A being called off, B is left to fniish it, which he does in 36 days more. In what time could each do it separately ? Ans. 24, and 48 days. 212 MISCELLANEOUS PROBLEMS. 19. A merchant has two casks containing unequal quantities of wine. Wishing to have the same quantity in each, he pours from the first into the second as much as the second contained at first ; then he pours from the second into the first as much as was left in the first ; and then again from the first into the second as much as was left in the second, when there are found to be 16 gallons in each cask. How many- gallons did each cask contain at first? Ans. 22, and 10 gallons. 20. A fisherman being asked how many fish he had caught, re- plied. If 5 be added to one-third of the number that I caught yesterday, it will make half the number I have caught to-day ; or if 5 be sub- tracted from three times this half, it will leave the number I caught yesterday. How many were caught each day ? Ans. This problem is impossible ; how is its impossibility indicated ? 21. A laborer engaged for n days, on condition that he should re- ceive p pence for each day that he worked, and forfeit q pence for each day that he idled. At the end of the time he received s pence ; how many days did he work ? and how many was he idle ? Ans. Worked— ; was idle— days. p+q p+q 22. A, B, and C engage in a joint speculation. A invests $2000 for 5 months, B $2400 for 4 months, and C $1600 for 7 months, The profits amount to $4620 ; what is each man's share of profit ? Ans. $1500, $1440, ^1680. 23. A farmer wishes to mix rye worth 40 cents a bushel, and oats worth 26|- cents a bushel, in such quantities as to produce 100 bushels which shall be worth 30 cents a bushel. What quantity of each must be taken* A?is. 25, and 75 bushels. 24. Three persons engage in a joint mercantile adventure, in which the first has the capital a for the time b, the second the capital c for the time d, and the third the capital e for the time f. Their profits amount to s ; what is each partner's share of profit ? abs cds efs ab-\-cd-\-ef' ab-{-cd-{-ef' ab-{-ed-{-ef' 25. A church which cost $40,000 is insured, annually, at IJ per cent., for such an amount, that, in case of its being destroyed by fire, the Insurance Company shall be liable for the cost of the edifice, and the premium of insurance. What is the sum insured ? Am. 40609.13'. 26. Four towns are situated in the order of the first four letters of the alphabet. The distance from A to D is 34 miles ; the distance from A to B is to the distance from C to D as 2 to 3 ; and J of the distance from A to B added to half the distance from C to D, is 3 times the distance from B to C. What are the respective distances ? A71S 12 4, and 18 miles. MISCELLANEOUS PROBLEMS. 213 27. A commission merchant receives the sum of s dollars to invest in merchandise — himself to retain a commission of r per cent, on the amount of the purchase. What is the sum to be invested ? 100s Ans. . 100-fr 28. A sum of money was equally divided among a.number of per- sons, by first giving to A SlOO and i of the remainder, then to B $200 and i of the remainder, then to C |!300 and ^ of the remainder ; and so on. What was the sum divided ? and the number of persons ? Ans. $2500, and 5 persons. 29. The sum of $500 is to be applied in part towards the payment of a debt of $900, now due, and in part to paying the interest, at 7 per cent., in advance, on the remainder of the debt, on which a credit of 12 months is to be allowed. What is tlie amount of payment that can be made on the debt? Ans. 469.89' 30. A besieged garrison had such a quantity of bread as would, if distributed to each man at 10 ounces a day, last 6 weeks ; but having lost 1200 men in a sally, the governor was enabled to increase the allow- ance to 12 ounces per day. What was the original number of men? Ans. 7200 men. 31. A is indebted to B the sum of s dollars, and is able to raise but a dollars. With this latter sum A proposes to pay a part of the debt, and the interest, at r per cent., in advance, on his note at n years, for the remainder. For what sum should the note be drawn. ^ ■ 100 (s-a) Ans. r^^ 100— m • 32. The crew of a ship consisted of her complement of sailors and a number of soldiers. Now there were 22 sailors to every three guns, and 10 over. Also the whole number of men was 5 times the number of soldiers and guns together. But after an engagement, in which the slain were ^ of the survivors, there wanted 5 of being 13 men to every 2 guns. Required the number of guns, soldiers, and sailors. Ans. 90 guns, 55 soldiers, 670 sailors. 33. Two sums of money, amounting together to $600, were put at interest — the smaller at 2 per cent, more than the other. The interest of the larger sum was afterwards increased, and that of the smaller di- minished, 1 per cent. By this the interest of the whole was augmented one-tirentieth. But if the interest of the greater sum had been so in- creased, without any diminution of the other, the interest of the whole would have been increased one-tenth. What were the two sums ? and the two rates of interest ? Ans. $400 at 6 per cent. ; $200 at 8 per cent. 34. Two persons purchase 300 acres of land at $2 per acre, each one paying $300 ; but the first takes the more fertile portion of the 214 MISCELLANEOUS PROBLEMS. tract, at 25 cents above the mean price per acre, and the second the remainder at 25 cents below the mean price per acre. How many acres has each ? Ans. This problem is impossible ; how is its impossibility indicated ? 35. The area of a field is 432 square rods, and the sum of its length and breadth is equal to twice their difference. Required the length and breadth of the field. Ans. 36, and 12 rods. 36. A and B lay out some money on speculation. A disposes of his interest in the business for <£ll, and gains as much per cent, as B lays out ; B gains X36, and it appears that A gains 4 times as much percent, as B. "What was the capital of each? Ans. £5, and £120. 37. A garden which is 12 rods in length, and 8 rods in breadth, is surrounded by a walk whose area is equal to ^ of the area of the garden itself. Required the breadth of the walk. Ans. 54-'\/29 38. A and B hired a pasture, into which A put 4 horses, and B aa many as cost him 18 shillings a week. Afterwards B put in two addi- tional horses, and found that he must pay 20 shillings a week. At what rate was the pasture hired 1 Ans. 30s. per week. 39. A gentleman bought a rectangular piece of ground, at $10 for every rod in its perimeter. If the same area had been in the form of a square, and had been purchased in the same way, it would have cost $20 less ; and a square piece of the same perimeter would have con- tained 12J square rods more. What were the length and breadth of the lot? Ans. 16, and 9 rods. 40. A person being asked the ages of himself and his wife, replied, that the product of their ages added to the square of his age, would make 1560, but added to the square of hers would make 1144. What were their ages ? Ans. 30, and 22. 41. A and B purchased a farm containing 900 acres for which they paid $900 each. On dividing the land, it was agreed that A should have his choice of situation, and pay 45 cents per acre more than B. How many acres should each have taken ? and at what price per acre ? Alts. A 400 acres at $2.25; B 500 acres at $1.80. 42. A capital of $13,000 was divided into two parts, which were put at interest in such a manner that the income was the same from each. If the first part had been at the same rate of interest as the second, it would have produced an income of $360 ; and if the second part had been at the same rate as the first, it would have produced an income of $490. What were the two rates of interest? Ans. 7 and 6 per cent. 43. A departs from London towards Lincoln at the same time at which B leaves Lincoln for London. When they met, A had traveled 20 miles more than B, having gone as far in 6| days as B had in all the time; and it appeared that B would not rcacli London under 15 MISCELLANEOUS PROBLEMS 21 S days. What is the distance between the two places ? and how far had each man traveled ? Ans. Distance, 100 miles ; A had gone 60, B 40 miles. 44. The product of the two dimensions of a rectangular piece of land, subtracted from the square of the greater dimension, leaves 300 square rods, and subtracted from the square of the less, leaves 200 square rods. What are the dimensions of the piece ? Ans. This problem is impossible ; how is its impossibility indicated ? and in what does this impossibility consist % 45. A person being asked the ages of his two children, replied, that the difference of their ages was 3 years, and the product multiplied by the sum of their ages was 308. Wkat were their ages ? This problem will result in an Affected Cubic Equation. Ans. 1 and 4 years. 46. A gentleman who had a square lot of ground, reserved 10 square rods out of it, and sold the remainder for $432, which was as many dollars per square rod as there were rods in a side of the whole square 1 What was the length of its sides ? Ans. 8 rods. 47. A and B set out together from the same place, and travel in the same direction. A goes the first day 28 miles, the second 26, and so on, in arithmetical progression ; while B goes uniformly 20 miles per day. In how many days will the two be together again ? Ans. 9 days. 48. A farmer wishes to build a crib whose capacity shall be 1620 cubic feet, and whose length, breadth, and height shall be in an arith- metical progression decreasing by the common difference 3. What must be the dimensions of the crib? It may be well to remind the student here, that cubic measure^ or measure of capacity, is found by multiplying Xogeih.ex length, breadth, and height or depth. Ans, 15, 12, and 9 feet. 49. One traveler sets out to go from A to B, atthesame time at which another sets out from B to A. They both travel uniformly, and at such rates, that the former, 4 hours after their meeting, arrives at B, and the latter at A, in 9 hours after. In how many hours did each one per- form the journey ? Ans. 10, and 15 hours. 50. A lady on being asked the ages of her three little boys, answered that they were in harmonical progression ; and the sum of their ages was 22 years ; and that if the ages of the two elder were each in- creased by -J of itself, the three would then be in geometrical progres- sion. What were the respective ages? A71S. 4, 6, and 12 years. 51. A person wishes to construct two cubical reservoirs which shall differ in their linear dimensions by 4 feet, and which shall together contain 5824 cubic feet. What must be the dimensions of the two reservoirs? Ans. 12, and 16 feet. 216 MISCELLANEOUS PROBLEMS. 52. Two partners, A and B, divided their gain, which was when B's share was found to be $20. A's capital was in trade 4 months ; and if the number 50 be divided by A's capital, the quotient will be the number of months that B's capital, which was $100, con- tinued in trade. What was A's capital ? and the time B's was in trade ? Ans. A's Capital $50 ; B's 1 month in trade. 5o. Let there be a square whose side is 110 inches ; it is required to assign the length and breadth of a rectangle whose perimeter shall be greater than that of the square by 4 inches, but whose area shall be less than the area of the square by 4 square inches. Ans. 126, and 96 inches. 54. There is a number cons^ting of three digits which increase from left to right by the common difference 2 ; and the product of the three digits is 105. Required the number. Ans. 357. 55. A person bought 2 pieces of cloth for $63. For the first piece he paid as many dollars per yard as there were yards in both pieces, and for the second as many dollars per yard as there were yards in the first more than in the second ; also the first piece cost six times as much as the second. What was the number of yards in each piece ? Atts. 6, and 3 yards. 56. There is a number consisting of 4 digits which decrease from left to right by the common difierence 2 ; and the product of the four digits is 945. Required the number. Ans. 9753. 57. A gentleman purchased two square lots of ground for $300 ; each of them cost as many cents per square rod as there were rods in a side of the other, and the greater contained 500 square rods more than the less. What was the cost of each lot ? A?is. $180, and $120. 58. A merchant bought a number of bales of cloth. The number of pieces in each bale was 10 more than the number of bales, and the number of yards in each piece was 5 more than the number of pieces in each bale ; and the whole quantity was 1500 yards. W^hat was the number of bales ? Ans. 5 bales. 59. A person dies, leaving children, and a fortune of $46800, which, by his will, is to be divided equally amongst them. Immedi ately after the death of the father, two of the children also die, in consequence of which each surviving one receives 81950 more than he was entitled to by the will. How many children did the father leave ? Ans. 8 children. 60. A coach set out from Cambridge for London with 4 more out- side than inside passengers. Seven outside passengers went at 2 shillings less than 4 inside ones, and the fare of the whole amounted to £9. At the end of half the journey, 3 more outside and one more inside passenger were taken up, in consequence of which the fare of MISCELLANEOUS? l'R0BLEM3. 217 the whole was increased in the proportion of 17 to 15. Required the number of passengers at first, and the fare of each. Ans. 5 inside, and 9 outside passengers ; fares 18 and 10 shillings. 61. In a purse which contains 24 coins of silver and copper, each silver coin is worth as many pence as there are copper coins, each copper coin is worth as many pence as there are silver coins, and the whole is worth 1 8 shillings. How many were there of each kind of coins? Ans. 6, and 18. 62. A and B travelled on the same road, and at the same rate, from Huntingdon to London. At the 50th mile stone from London, A overtook a drove of geese, which were proceeding at the rate of 3 miles in 2 hours ; and 2 hours afterwards met a wagon which was moving at the rate of 9 miles in 4 hours. B overtook the same drove of geese at the 45th mile stone, and met the same wagon 40 minutes beibre he came to the 31st mile stone. Where was B when A reached London ? Ans. 25 miles from London. NEW AKD OIPROVED NATIONAL SCHOOL BOOKS, PUBLISHED BY PRATT, WOODFORD & CO., No. 4 COURTLANDT-STKEET, N. Y. P., "W. & Co. would respectfully call the attention of all interested in the 'subject of education to the following works published by them, aa text-books, in nearly every branch of study ; all of which are prepared by practical teachera of high reputation, and many of them are in use in almost every State of the Union- They have stood the test of the school- room, and received the sanction and approval of many of the best educators in the country from whom numerous testimonials and recom- mendations are in our possession. BULLIONS' SERIES OF GRAMMARS AND ELEMEN- TARY CLASSICS. This series consists of the following works, viz: I.— PRACTICAL LESSONS IN ENGLISH GRAJVIMAR. This little book contains a brief synopsis of the leading pnuciplos ol Ei\glish Grammar, every part of which is illustrated by a great variety of exercises, of the simplest character, adapted to the capacity of pupiht at an early age. — ^New edition, revised and improved. • II^THE PRmCIPLES OF ENftLISH GRAJ^BIAR. This Work is intended ao a school Grammar, for the use ol classei pvirsuinj this branch of study in the common schools, or of the junioi dasdes in academics. It embraeea all that is important on the subject., expressed with accuracy, brevity, and simplicity, and is peculiarly adapt- ed to the purposes of instruction in public achools. m.— THE ANALYTICAL AI7D PRACTICAL EI^T GLISH GRAMMAR. This -work, designed for the more advanced classes in schools and academies, is prepared on a more extended plan than the preceding, though not essentially different from it The arrangement (except in Byntax), the definitions and rules, are the same, but with much greater fulness in the illustrations and exercises, intended to lead the student into a thorough and critical acquaintance with the structure and use of the English Language. IV.— EXERCISES m ANALYSIS AND PARSING. This little work consists of selections in prose and poetry from stand ard writei*s, so arranged as to furnish a convenient and progressive course of Exercises in Analysis and Parsing, in every variety of style, with such occasional references to the grammare as are deemed necessary to explain peculiar or difficult constructions. To tliis is prefixed directionn for the analysis of sentences and models both of analysis ^nd parsing. v.— THE PRD^CIPLES OF LATIN GRAMMAR. This work is upon the foundation of Adam's Latin Grammar, so Jong and favorably known as a text-book, and combines with all that is excel- lent in that work many important corrections and improvements suggest- ed by subsequent writers, or the results of the author's own reflection atti observation, during many yeai-s, as a classical teacher. VL-JACOBS' LATIN READER. . This work forms a sequel to the Grammar, and an introduction to the etudy of Latin classic authors. It begins with a series of simple and plain sentences mostly selected from classic writera, to exemplify anf» illustrate the leading constructions of the language, followed by Reivling Lessons, of pure and simple Latin, chiefly narrative, by which the pupil, while he becomes familiar with the construction of the language, is also made acquainted with many of the most prominent characters and mytho- logical fables of antiquity, as well as with the leading events of Roman history. Throughout the work, references are constantly made, at the foot of the page, to the Grammar and Introduction, when necessary to explain the construction or assist the pupil in his preparations. VII.— FIRST LESSOJ^S IN GREEK. Thus work 13 intended chiefly for those wlio begin the study of Gree^i at any eni-ly aj^e ; and for this reason contains only the outlines of Gram- mar, expi eased in as clear and simple a manner aa[|i|ssible. It is com- plete in itself, being a Grammar, Exercises, Reading^ook, and Lexicon, bU iu one ; so that the pupil, while studying this, needs no other book on the subject. The knowledge acquired by the study of this work will be an important preparation to the young student for commencing tl\e tftudy of Greek Grammar with ease and advantage. VIII.— THE PRINCIPLES OF GREEK GRAMMAR. This work is intended to be a comprehensive manual of Greek Gram- mar, adapted to the use of the younger, as well as of the more advanced students, in schools and colleges. Both in Etymology and Syntax, the leading principles of Greek Grammar are exhibited in definitions and rules, as few and as brief as possible, in order to be easily committed to Tiemory, and so comprehensive as to be of general and easy application This work is tiow more extensively used than any other of the kind in the country. DC.— GREEK READER. Tliis work, like the Latin Reader, is properly a sequel to the Greek Grammar, and an introduction to the study of the Greek classic authors. It seeks to accomplish its object in the same way as the Latin Reader. (See above, No. VI.) With these are connected SPENCER'S LATIN LESSONS, with exercises in parsing, introduc- tory to Bullions' Latin Grammar. In this series of books, the three Grammars, English, Latin, and Greek, are all on the same plan. Tlie general arrangement, definitions, rules, etc., are the same, and expressed in tlie same language, as nearly as the nature of the case would admit. To those who study Latin and Greek, much time and labor, it is believed, will be saved by this method, both to teacher and pupil ; the analogy and peculiarities of the different languages being kept in view, will show what is common to all, or pecu- liar to each ; the confusion and difficulty minecessarily occasioned by the use of. elementary works, diftering widely from each other in language and structure, will be avoided ; and the progress of the student rendered much more rapid, easy, and satisfactory. No series of Grammars having this object in view, has heretofore been prepared, and the advantages which they offer cannot be obtained in an equal degree by the study of any other Grammars now in use. They form a complete course of elementary books, in which the substance ol the latest and best Grammars in each language has been compressed into a volume of convenient size, beautifully printed on superior paper, neatly and strongly bound, and are put at the lowest prices at whicli they can be afforded. The elementary works, intended to follow the Gi-ammars, namely, tha Latin Reader, and the Greek Reader, Are also on the same plan — are pre- pared with special reflirences to these works, and contain a course of elementary instiuction so unique and simple, as to furnish great facilities U> tl*e student in tliese languages. 4 I BULLIONS' SERIES OF LATIN CLASSICS. This series contains the following works, to which others, in course 0/ preparation, will soon be added, viz : I.— CAESAR'S COMMENTARIES ON THE GALLIC WAR. In this work, the plan of the Latin Reader is carried on throughout. The same introduction on the Latin idioms ia prefixed for convenience of reference, and the same mode of reference to the grammar and introduc- tion is continued. The Notes are neither too meagre nor too voluminous ; they are intended not to do the work of the student for him, but to direct and assist him in doing it himseK. It is embelUshed with a beauti- ful map of Gaul, and several wood-cuts representing the engines of war nsed by the Romans. IL— CICERO'S SELECT ORATIONS, With notes, critical and explanatory; adapted to Bullions' Latin Grammar, and also to the Grammar of Andrews and Stoddard- Tliia selection contains the four orations against Catiline. — The oration for the Poet Archias, — for Marcellus, — for Q. Ligarius, — for king Deiotarus, — for the Manilian law, — and for Milo. The notes are more extended than those in Caesar's Commentaries, especially in historical and archasological notices, necessary to explain the allusions to persons and events in which the orations abound, a knowledge of wliich is indispensable to a proper understanding of the subject, and to enable the student to kefep in view the train of argument pursued. — In other respects, the proper medium between too much, and too Uttle assistance has been studied, and constant reference made to the Grammar, for the explanation of imcommon ax difficult constructions. m.—SALLUST'S CATILINT3 AND JUGURTHA, On the same plan. Published also by the same — THE WORKS OF VIRGIL, with copious notes,