Cl.$A£v,.£.l Bk. AVv' E £ o ~ - . THE ETHEL CARR PEACOCK MEMORIAL COLLECTION Matris amori monumentum TRINITY COLLEGE LIBRARY DURHAM, N. C. 1903 Gift of Dr. and Mrs. Dred Peacock Digitized by the Internet Archive in 2016 with funding from Duke University Libraries https://archive.org/details/elementarypracti01dodd L ELEMENTARY AND PRACTICAL ALGEBRA: IN WHICH HAVE 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. FOURTEENTH EDITION. NEW YORK: PRATT, OAKLEY & COMPANY, 21 MURRAY STREET. ISfiO. Entered, according to Act of Congress, in tne year - 1852, By JAMES B. DODD, In the Clerk’s Office of the District Court of Kentucky. I. P. JONES i. CO., STEREOTYPEaS. I S3 William street S'12,.02. 0U5°, P I PREFACE. The following work is designed to furnish, a practicable course of Algebra for the younger classes of students, without the omission of any thing important to a thorough education in the subjects which it embraces. It aims at the most methodical arrangement , the clearest expositions , the best elementary exercises , and the most varied and useful applications : — in all these respects presenting some new features , which have been adopted as improvements in the method of teaching this science. All that is appropriate to an Algebraic treatise in a general course of mathematical studies, or necessary in preparation for the higher Avorks of the course, has been introduced, with the exception of a few subjects Avhich are more exclusively pre- liminary to the Differential and Integral Calculus. These, with whatever else may be considered useful in a larger work, will shortly be added to the present treatise. Between this work and the 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. 234^7 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 L II, III, an Integral Surd, (238). — Surds of Different Roots reduced to the Same Root, (239). — How to find the Sum or Difference of Similar Surds — of Dissimilar Surds, (240). — Howto find the Product or Quotient of Surds of the same root — of different Roots — of any two Roots of the Same Quantity. (241). — Expediency of rationalizing a Surd Divisor or Denominator, (242). — How a Monomial Surd may be made to produce a Rational Quantity — a binomial Surd — a trinomial Surd, (243). . — Of the Powers and Roots of Irrational Quantities, (244). — Of the Square Root of a Numerical Binomial of the form a±y/b (245). — Of Imaginary Quantities — From what an Imaginary Quantity results. (246). — Of the Cal- culus of Imaginary Quantities — By what means the Sign affecting the Pro- duct of two Imaginaries may be determined, (247). — Resolution of an Imaginary Quantity, (248). — Product of two Imaginary Square Roots, 1 2 4 9 ) ANALYSIS OF CONTENTS. £111 CHAPTER X. Quadratic and Other Equatations. — 167 . .. 204. A Quadratic Equation — A Cubic Equation — A Biquadratic Equation, (250). — A Pure Equation — An Affected and a Complete Equation, (251). — A Root of an Equation, (252). — Principle for determining a Divisor of an Equation, (253). — Principle for determining a Root of an Equation, (254). Number of Roots of an Equation — Whether the several Roots of an Eqau- tion are necessarily unequal, (255). Rule XXII. For the Solution of a Pure Equation , (256). How a Surd Quantity in an Equation maybe rationalized, (257). — How two Surds in an Equation may be rationalized, (258). — Of an Equation con- taining a Fraction whose terms are both irrational , (259-). Rule XXIII. For the Solution of a Complete Equation of the Second Degree , (260). How any Binomial of the form ax 2 -\-bx maybe made a Perfect Square (261). Rule XXIV. To Reduce an Equation of the form ax 2 -\-bx to a Simple Equa~ tion, (262). Of Equations having a Quadratic Form with reference to a Power or Root of the Uuknown Quantity, (263). — Of the Solution of two Equations, one or both of the Second or a Higher Degree, Containing two Unknown Quantities, (264).— Solutions by means of an Auxiliary Unknown Quantity, (265). — Solutions by means of two Auxiliary Unknown Quantities, (266). — General Law of the Coefficients of Equations, (267). — Determination of the Integral Roots of Equations, (268). — Solution of Equations by Ap proximation , (269). — General Method of Elimination, (270). CHAPTER XI. General Description of Problems. — Miscellaneous Problems. — 205 . 217. A Determinate Problem — How a Determinate Problem is represented, (270). — An Indeterminate Problem— How represented, (271). — An Impossible Problem — How represented, (272). — Greatest Product of any two Parts into which any Quantity may be divided, (273). — Signification of the Different Forms under which the value of the Unknown Quantity may be found in an Equation, (274). — Inequa- tions,^ 275). — For what purpose Inequations are employed, (276). — Inequalities iu the same, and in a contrary sense, (277). — Inequalities between negative quantities, (278). — Transformation of Inequations, (279). ' ALGEBRA 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 numbers two, three, Ac., are repetitions of the unit one. (3.) Quantity 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 of days, hours, Ac 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 length breadth , and height, depth, or thickness. (5.) Algebra is 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 unknown 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 he understood as known in value ; while the quan- tity x or y will be unknown or required. Quantities represented hv 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 he added t< the quantity a. (3.) The sign — minus prefixed to a quantity, denotes that the quantity is to he subtracted, or taken subtract ively. Thus a — b, a minus b, denotes that the quantity b is to he sub- tracted from the quantity a. (9.) The sign X into between two quantities, denotes that the two quantities are to he multiplied together. Tims ax 5, a into b, denotes that the two quantities a and b are to be multiplied together. A point (.) between two literal quantities, or a numeri al and a 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 of several numbers maybe denoted b - points or wven 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 he multiplied together. Thus ab denotes a and b multiplied toge- ther ; and axy denotes a, x, and y multiplied together. (10.) The sign - 4 - by between two quantities, denotes that the quantity before the sign is to he divided by the one after it. Thus a -4-5, a by b, 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- -b. 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 (<2 + 5).c, or (<2 + 5)c, a plus 5 in a parenthesis into c, denotes that the sum of a and 5 is to be multiplied into c. The same thing would be denoted by a+bxc, a plus 5 under a vinculum into c. — In af-bc, only 5 would he multiplied into c. The vinculum, and the expression affected by it, are sometimes set vertically. Thus a x is equivalent to (<2 + 5 — c) x + £* — c or af-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. Suppose the letters a, b, c to represent the numbers 3. 4 5, respectively ; then what is the numerical value of <2 + 5 — c ? What is the value of and a negative quan tit.y by the sign — , prefixed to it. In the polynomial a-\-b — c, the quantity b is positive, while c ia negative. A quantity with neither + nor — prefixed to it, is understood tc be positive. Thus in the preceding polynomial, the first term a is understood t$ be +« ; for it may be regarded as 0 + a. (25.) A negative quantity has an effect contrary to that of ar equal positive quantity, in the expression, or calculation, into which it enters. This is evident with regard to the addition and subtraction of the same quantity ; the effect of subtracting it is contrary to that of add- ing it. The effect of multiplying by a negative quantity is contrary to that of multiplying by an equal qjositive quantity. For example; 3a X 2, 3a multiplied by positive 2, denotes that 3a is to be repeated additively, and is equivalent to 3a + 3a. But 3a X — 2, 3a multiplied by negative 2, denotes that 3a is to be repeated subtractively, and is equivalent to — 3a — 3a. The same thing is true with respect to dividing by a negative, and an equal positive quantity. The positive and negative signs are sometimes used to distinguish quantities as estimated in contrary directions from a given point or line. Thus if Motion in one direction be denoted by the sign +. then motion in the contrary direction will be denoted by the sign — . Ii North Latitude be considered as positive, South Latitude will be negative. 9 CHAPTER II. ADDITION— SUBTRACTION.— MULTIPLICATION.— DIVISION. ADDITION. (20 1 Algebraic 'Edition consists m finding the simplest expres sion foi the value ot two or more quantities connected together by th« sign 4- or — , and this equivalent expression is called the sum of tht quantities. The simple t expression for the value of 5a-f 3a, 5 times a plus 3 times a, is 8a ; then 8a is the sum of 5a and 3a. The simplest expression for the value of 5a — 3a, 5 times a minus 3 times a, is 2 a ; then 2 a is the sum of 5a and — 3a. In the second example, observe that adding — 3a is equivalent to subtracting 3 a ; the Adding of a negative quantity being the same as the Subtracting of an equal positive quantity. Addition of Monomials. (27.) Similar terms with like signs, are added together by taking the sum of their coefficients , annexing the common literal factor , and prefixing the common sign. Thus 3 ax-\-2ax is 5ax ; just as 3 cents + 2 cents is 5 cents. And — 3a — 2a is — 5a ; for 3a to be subtracted and 2a to be sub- tracted make 5a to be subtracted. HdP* What is the Sum of 4a and 3a ? Of ax ar 1 5ax ? Of — ° and —7xl Of 5a 2 , 2a 2 , and 4a 2 ? Of — 2b, —3 b, and — 5b r> (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 0 ; that is, 0 is the simplest express : -ev for the value of 3a — 3a, c*-- 3 is therefore the sum of the two terrr . So 5a-\-x — 5a .s equal to x; for 5a and - -5a cancel each other, a r x is therefore the sum of the three given terms 10 ADDITION. What is the Sam of 2 ax and — 2 ax, that is, the simplest ex- pression for the value of 2ax— 2ax1 What is the Sum of 76 and \ — 76? Of a, 3x, and — a? Of 3 y 2 , — 3 y 2 , and 5b 2 ? Of ax 2 , 5, and — 5 ? Of —abc, -\-a 2 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— 3a is 4 a; just as 7 vents — 3 cents is 4 cents; that is, the sum of la and — 3 a is 4 a. And 3a— la is —4 a. For —la is equal to —3a — 4a (27); then 3a — la is equal to 3a — 3a — 4a; 3a and — 3a cancel each other (2S), and leave —4a. KF 5 " What is the Sum of 5ab and — 3a6 ? Of 3a 2 c and — a 2 c? Of 9 ax 2 and — 4aa; 2 ? Of — a 2 x and 3a 2 z? Of 5ac and — 5 oc ? What is the simplest expression for the value of 2b — 5b, that is, the Sum of 2b and — 5b ? How do you reason in finding that sum ? What is the Sum of 3a 2 and — 8a 2 ? and how do you reason in finding it? What is the Sum of ax 2 and — 2 ax 2 ? and how do you reason in finding it? What is the Sum of — 11a 2 and 5a 2 ? and how do you reason in finding it ? (30.) The Sum of two or more dissimilar terms can only he indi- cated by arranging them in a Polynomial, so that each term shall have its given sign prefixed to it. Thus the Sum of ax and be 2 can only be indicated as axfibc 2 ; and the Sum of ax and — be 2 as ax — be 2 . What is the Sum of 3a and 5b ? Of 5a and — b ? Of 3a 2 and 2a: 2 ? Of ax and by ? Of —36 and — 4 y ? What is the Sum of 36, 56, and 6c? Of a 2 , bx, and 3a 2 ? Of 4c and —3i / ? Of —ab, 3 xy, and —2c ? Of 3a 2 , 46, and 2c ? Of 5a, — 5, and 56 ? The preceding principles, and the following Rule, provide for all the cases in Algebraic Addition. ADDITION. 11 RULE I. (31 ) 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 4a 2 +2bc— xy, 2a 2 —3bc+5y, bc—a 2 + 3xy, and cy—2a 2 —5bc. 4a 2 + 2bc — xy 2 a 2 — 3 bc-\-5y — a 2 + bcA^xy — 2 a 2 — 55c+q/ 3a 2 —5bc-\-2xy+cy-\-5y The sum of the positive terms 2a 2 and 4a 2 is 6a 2 , and the sum of the similar negative terms —2a 2 and —a 2 is — 3a 2 , .(27). Adding together these two similar sums, 6a 2 and —3a 2 , the residt is 3a 2 , (29). The sum of the positive terms be and 2 be is 35c, and the sum of the similar negative terms — 55c and — 35c is —85c. Adding together these two similar sums, 35c and — 85c, the result is — 55c. The sum of the similar terms 3 xy and —xy is 2 xy. The results thus found, namely, 3a 2 , — 55c, and 2 xy, and the dissimilar terms cy and 5y, are connected in a Polynomial. EXERCISES. 1. Add together a5 + 3c 2 — 2x, oab — c 2 + 5.r, 5c 2 -2ab-{-y, and 4a5— c 2 +x. Ans. 6a5 + 6e 2 + 4a;+?/. J 2. Add together 4a 2 — 65 + 3, 3a 2 + 65 — 7, 55 + a 2 — 5c, and a 2 — 5c+10. Ans. 9a 2 + 55 + 6— 25c. 3. Add together 25 2 +5c+a;, 35 2 — 25c + ?/, 5c — 35 2 + 5.r, and 5 2 + 5c+3y. Ans. 35 2 +5c+6x+4?/. 4 Add together 5a — 65 3 + 3, 25 3 — 4a — 7, 7a — 5 3 +c, and a+ 35 2 + 2c+4. Ans. 9a — 55 3 + 3c+35 2 . 5. Add together a 2 5 — 5c+zi/, 4a 2 5 + 8c — xy, a 2 5 + c+3:c?/, and b 2 +3c— 13. Ans. 6a 2 5+7c+322/+a 2 — 13. B 12 ADDITION. 6. Add together a 3 + 2a 2 —35c, 2a 3 — a 2 Abe, — a 3 — a 2 +£ 2 , and 3a 3 + 3a 2 + 36 2 . Am. 5a 3 + 3a 2 -26c+46 2 . 7. Add together 2x+ab 2 —3y 2 , 2ab 2 — 3x-\-iy 2 , — ab 2 + 5x—y 2 , and 5.c+a5 2 — y. Aws. 6x-\-3ab 2 — 3?/ 2 — ?/. 8. Add together 4a — 35 2 +ca; 2 , b 2 — 3a+3, 2a+36 2 — cx 2 , and 2 b 2 — 2a — 9. A?is. a + 36 2 — 6. 9. Add together lb — 2 a 2 —xy,5a 2 — 6b-\-3xy, b— 3a 2 +4, and a? — b — 3 xy. Ans. b-\-a 2 — zy+4. 10. Add together 3 b 2 — 2a s + l3, 3a 3 — 2 b 2 —5, Aab-\-7a 3 — 3, and 2 b 2 — a 3 -\-ab. Am. 3d 2 + 7a 3 + 5 + 5aA 11. Add together ax 2 — 2 yfb, 2y-\-2ax 2 — 3 b, Aax 2 —y—b, and 2b—3ax 2 Ab. A?is. Aax 2 —y. 12. Add together 2c 2 +a 2 +35c, 5c 2 — 3a 2 — 2 be, c 2 + 2a 2 — be, and bAbc— 3. Am. 8 c 2 -\-bc-\rb — 3. 13. Add together abfa 2 c — 5, 3 ab — 3a 2 c+7, 2a 2 c—2ab — 3, and a5 + a 2 c+5. Ans. 3a6+a 2 c+4. 14. Add together 3 b* — 2 a 2 x-\-b, — b 2 -\-3a 2 x — 3 b, b 2 — a 2 x\-c, and 3 b 2 -\-b — 3t - . Ans. 6b 2 — b — 2c. 15. Add together 2a 2 + 3 — ac, 3 a 2 — 7+ac, 3ac— 5a 2 + 9, and 3ac+4 — a 2 . Ans — a 2 + 9 + 6ac. 16. Add together £ 2 c+ 2 — y 2 , y 2 — 3b 2 c— 10, 2 b 2 c — 3 + 2 y 2 , and b 2 — ?/ 2 + 5. Ans. — 6+y 2 + 6 2 . 17. Add together a 3 +Sc — £c, 2 a 3 — £c+fc, 3a 3 + 3ic — \c, and a 3 + 5c+c. Ans. 7a 3 + 46c+c. 18. Add together b 3 — 3a 2 +fn 2 , 26 2 + 5a 2 +d-c 2 , b 2 — -^c+2, and 5b 2 + 3a 2 -2c-\-3y. Ans. 6 3 + 5a 2 + l-|c 2 + S6 2 — 2^c+2+3y. (32.) When the same Polynomial is enclosed in two or more paren- theses , (11), this polynomial enters into a calculation in the same man- ner as the common factor of similar monomials. Thus the Sum of 3(a + b) and 5(a + b), is evidently 8(a+S), 19. Add together 2(a — 3Z>) + c, 3(a— 3Z>)— 3c, — 4(a— 35) + c, and 5(a— 3&) + 3c. A?is. 6(a— 36) + 2c. 20. Add together 3a 2 + 2(a— c 2 ), a 2 — 3(a— c 2 ), 4a 2 +5(a— c 2 ) and — 7a 2 + (a— c 2 ). A?is. a 2 + 5(a— c 2 ). 21. Add together 5(a + 5— c) + 3& 2 , 3(a-\-b—c) — 2b 2 , 2(a-\-b—c) and 3b 2 + 2?/, -\-b 2 . Ans. 10 (a + i — c)-\-5b 2 -\-2y. 22. Add together 2aZ> 2 + 3a(£ + c 2 ), ab 2 — 2a(&+c 2 ), a(b+c 2 ) and — ab 2 — 3a(5 + c 2 ). Ans. 2ab 2 — a(b+c 2 ). 23. Add together 5 + -|{c— d+m)-\-2b, 1 +f(c— d+m)— b + 2, and 2 {c—d+m)+%b. Ans. 8 + 3i(c— cf+w) + 1^-6. SUBTRACTION 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 3 a from 5a is 2a, because 2 a added to 3a produces 5a, (27). But the difference found by subtracting — 3 a from 5a is 8a ; be- cause 8 a must be added to — 3a to produce 5a, (29). In the second example, observe that subtracting — 3a has an effect contrary to that of subtracting 3a in the first example, (25) ; the Sub- tracting of a negative quantity being the same as the Adding of an equal positive quantity. Subtraction of Monomials. (34.) A monomial is subtracted from another quantity, by changing the sign of the monomial, and then adding it, algebraically, to the other quantity. Thus 5a subtracted from 2a gives — 3a, because — 3a must be added to 5a, to produce 2a, (29) ; and this difference, — 3a, will be found by changing 5a to — oa, and adding the — 5a to the 2. a [TP* What difference will be found by subtracting 5a from 9a ? and why is that difference true? By subtracting 9a from 5a? What difference will be found by subtracting 3 ab from 4 ab 1 and why is that difference true? By subtracting — 5x from — 10a;? and why is that difference true ? By subtracting lax from — 3 ax. (35.) When the two quantities are dissimilar, a monomial is sub- tracted by changing its sign, and then connecting it with the other quantity, in a Polynomial. Thus 2x subtracted from 3a gives 3a — 2x ; and — 2x subtracted from 3a gives 3a + 2cc, (34), (30). [TP What difference will be found by subtracting 2a from 3a; ? By subtracting —35 from 5axl By subtracting ax from 105? By subtracting 4 y from —9a: 2 ? 14 SUBTRACTION. From the preceding we derive RULE II. ( 36 .) For the Subtraction of Algebraic Quantities. Change the sign of each term to be subtracted, + to — and — to + , or conceive these signs to be changed, and then proceed as in Alge- braic Addition. EXAMPLE. To subtract a 2 + 2a£ — 3 je — z from 3a 2 — 5ab-\-5x — y. 3 a 2 — 5 ab-\-5x — y a 2 -\-2ab — 3x — z 2 a 2 — 7ab-\-8x—y-\-z Having set the polynomial which is to be subtracted under the one from which the subtraction is to be made, we conceive each term in the second line to have its sign changed; then — a 2 added to 3a 2 makes 2a 2 ; — 2 ab added to — 5 ab makes — lab ; and ox added to 5x makes 83. These results, together with the — y and -\-z, are con- nected in a polynomial. EXERCISES. 1 From 4a5 + 3c 2 — xy, subtract 2 ab — c 2 -\-3xy — 2 z. Ans. 2ab-\-4c 2 — 4:ry-|-2z. 2 From 5a 2 -|-3Z>c — cd-\-ox, subtract 2a 2 — bc-\-3cd. Ans. 3a 2 +4ic — 4cd-\-3x. 3 From la 2 b — abc-\- Axy, subtract a 2 b-\-5abc—xy-{-niz. Ans. 6a 2 b — Gabc-{-5xy — mz 4 From 8a 3 — 4 a 2 b — 2Z>c+10, subtract 3a 3 +a 2 & — o. Ans 5a 3 — 5a 2 b—2bcf 15 . 8. From 3b 2 c-\-abx-\-2xy 2 , subtract b 2 c— 3abx+2xy 2 — 3. .A?zs. 2b~ c 4ubx-\- o . 6. From 4a 2 c — 3c 2 6 + 67/ + 20, subtract 3c- 2 6+6?/+3iy. Ans. 4a 2 c — 6c 2 6 + 20 — 3 by. 7 . From — 3a-\-5b 2 —1 xy, subtract 3b 2 -\-2a — xy+bc. Ans. — 5a+2b 2 —Gxy—bc. 8. From 2ab-{-3bc — 2az 2 + 15, subtract lab-\-3bc — 10. Ans. — Sab — 2ax 2 + 25. 9. From Sax — 3Z>c+2& 2 , subtract 6 ax — be — 3 b 2 -\-y 2 . Ans. — ax — 2 bc-\-5b 2 —y 2 . 10 . From a 2 b-\-3abc-\- b 2 c— 7 , subtract ha 2 b — abc-\-\b 2 c Ans. |a 2 6+4«6c+*4b 2 c— 7 ADDITION AND SUBTRACTION. 15 When one quantity is to be subtracted from the Sum of two or more quantities, the operation will be expedited by changing the signs of the subtractive quantity, and then adding all the quantities toge- ther. 11. From the Sum of 2a+35c — 1, and 3 a — 4&C+3, subtract 4a — be— 2+i/. Ans. a + 4 — y. 12. From the Sum of az 2 — &b 2 A3 ay 2 , and 2aa: 2 + 66 2 — a,y 2 , subtract — 2>ax 2 -\-b 2 + 2a?/ 2 — 5. Ans. 6 ax 2 — 6 2 + 5. 13. From the Sum of 2>ab-\-bc — 5x 2 , and ab — 3 bc-\-% 2 — 3, sub- tract 5bc — 3.r 2 — 4 Ay 2 - Ans. 4 ab — Ibc — £2 + 1 — y 2 . 14. From the Sum of 5x — 3zy-\-7 y 2 , and 5xy — 3 x — 4 y 2 — 9, sub- tract —xAxy — 3y 2 + 7 . Ans. 3z+£?/+6?/ 2 — 16. 15. From the Sum of 2 a — 5 + 3c + 3ax + 3 — 2y. 18.. From the Sum of 3a 2 Axy 2 — 2by, 5a 2 -\-3xy 2 — 3 by, and 3a;?/ 2 +4a 2 + 5y, subtract 2a 2 — xy 2 — 56y+5- Ans. 10a 2 + 8xy 2 Aby — 5. 19. From the Sum of a 2 b 2 — 3y 2 -\-5xy, 3a 2 b 2 A3y 2 — 2 xy, and 5?/ 2 +3a 2 6 2 — xyA 6, subtract a 2 b 2 -\-xy — y 2 — 3. Ans. 6a 2 £ 2 + 6?/ 2 +au/+9. 20. From the Sum of 5y 2 — 3a* — 2 be, 4 ax — 2y 2 -\-5bc, and 3 ax-\-y 2 — 5c+5, subtract ax— bc-\-3y 2 — 1. Ans. i/ 2 + 3aa; + 3ic + 6. 21. From the Sum of a 2 y — x 2 -\-bc — d, ox 2 — 3bc-\-5d, and a 2 y-\-3x 2 — 4 be — d, subtract ox 2 — 2bcA1d. Ans. 2a 2 y — 4 be — 4 d. 22. From the Sum of 5i 2 + 3a*+2y, 3b 2 —ax — y-\- 1, and 4a* — b 2 —5yA 2, subtract 36 2 +2a* — y+10. Ans. Ab 2 +4a*— 3y — 7. 23. From the Sum of 7a+6* 2 — y 2 A 1, * 2 +3 y 2 — c — 3, and lx 2 Ay 2 — 3c + 5, subtract 5a + 2x 2 — 2_y 2 + 5c+2. Ans. 2a+12* 2 + 5y 2 — 9c+l. 24. From the Sum of 2 (a — x)Ay, 3 (a — x) + 2y, and 5(a — x ) — y. subtract 4(a— x)Ay— 2, (32). Ans. 6(a— x)AyA2- 25. From the Sum of 26(a + c) 2 +3, Z>(a+c) 2 — 1, and 3 6(a + c) 2 — 2, subtract the sum of b(a-{-c) 2 Ab, and 5b[aAc) 2 — b. Ans. 0. 16 ADDITION AND SUBTRACTION. Indicated 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 he 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 becomes a-\-b. The subtraction of a polynomial 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 is to be subtracted. When the subtraction is performed, the expression becomes a — b — c — d. Change of Signs in a Polynomial. (38.) The signs of the value of a Polynomial, will evidently be changed, by changing the signs of all the terms of the Polynomial. The value of a Polynomial is therefore not affected 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-\-b — c is equivalent to a — ( — 5 + c), or a — (c — b)\ for if (c — b) be subtracted, as is required by the sign — prefixed to it, the result will be a-\-b — c. 26. Under what different forms may the value of the polynomial ab-\-2cd — 3a; + 5 be expressed? 27. Under what different forms may the value of the polynomia 1 ax — 3y-f-46 2 — 5c — 7 be expressed? MULTIPLICATION. 17 MULTIPLICATION. (39.) Algebraic multiplication consists in finding the Product of one quantity taken as many times, additively , or sub tr actively, as there are units in another quantity. The quantity to he 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 by positive 3, is 5a-\-5a-\-5a, equal to \5a ; the multiplicand 5a being repeatedly added. When the Multiplier 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 sub- 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, 3acx2xis equal to 6 acx\ for the factors may be taken in the order 3 X 2 acx, which becomes 6 acx by substituting G for 3x2. 03^ What is the product of 3a X 45 ? and how do you reason in finding that product? What is the Product of 5a 2 b x 2a; ? and how do you reason in finding it? Of7ac 2 Xxy ? Oixy 2 x5l Of 3 X la 2 b 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 3a 2 xX2ax, or 3a 2 x 1 x2 a 1 x 1 , is equal to 6a 2 axx, (40)., and this product becomes G a 3 x 2 by substituting a 3 and x 2 for their equiva- lents a 2 a and xx, (13) and (16). 2 * 18 MULTIPLICATION. Observe that the exponents of a and x in the product 6 a 3 x 2 , art the sums of the exponents of the same letters, respectively, in the two monomials which are multiplied together. DP 1 * What is the Product of 4a 3 X 2a 2 ? Of 3 ax x 5ax 2 ? Of 7ax2a 3 Z>? Of 4 acxacb'l Of a 2 b 3 x5bl Of 8 a 2 cbxc 1 . Of a 2 bxa 3 b 2 ? 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 signs produce + , and unlike 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 — 3a X — 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, their product is negative because it results from repeatedly subtracting a positive., or adding a negative quantity. Thus in 3a X — 2, the 3a is to be taken twice subtractively, and the product is therefore — 3a — 3a, equal to — 6a. [CP 1 * What is the Product of — 2x X — 3 ? and how do you reason on the sign of the product? Of — 3 abx — 2 ? Of — 5a 2 X — 3? Of — x X — 5 ? What is the Product, of 5 axX — 4 ? and how do you reason on the sign of the product ? Of 4a 2 cX — 3? Of 7a: 2 X — 5 ? OfZ» 2 x— 9. What is the Product of — 2 ah x3 ? and how do you reason on the sign of the product ? Of — 3a 2 at X 4 ? Of —3a 2 X 2 ? Of — c 3 X 5-1 ? What is the Product of 4x 2 X — 7 ? and how do vou reason on the sign of the nroduct ? Of 5ayx —3? Of —3a X — \o ? Of 6b 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 0 ; for it is evident that 0 repeated any number of tunes, produces only 0 ; and the product is the same when the multi- plicand and the multiplier are taken the one for the other . Thus 5x0 is 0; axOisO; («+5 — c) X 0 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 sign prefixed to each, (40), (41), (42). EXAMPLE. To multiply 3a5 + 5c 2 — 2 xy — 5 by 2 a 2 x 3 ab-\-bc 2 — 2 xy—p 2 a 2 x 6a 3 bx-\-2a 2 bc 2 x — 4 a 2 x 2 y — 10a 2 a; It will be most convenient to multiply from left to right. The first term of the multiplicand is positive, and the multiplier being also positive, the first term, 6 a 3 bx, of the product is positive. In like man- ner the second term, 2 a 2 bc 2 , of the product is positive, &c. EXERCISES. 1. Multiply 2a 2 — 35 + e — 2z/ by 3. Ans. 6a 2 — 95 + 3c— 6?/. 2. Multiply 3 x 2 —xy — y 2 4 by 2. Ans. 6a; 2 — 2 xy — 2y 2 fQ. 3. Multiply 4 5 + c 2 — 3?/+l by — y. Ans. — 4 by — c 2 y-\-3y 2 — y 4 Multiply —« + 35 — 2a; 2 —3 by —2a;. Ans. 2 ax — 65a; + 4a; 3 + 6a". 5. Multiply ab 2 — 5 3 + 2c — y by — 5. Ans. — 5ab 2 -\-5b 3 — 10c+5y, 6. Multiply — 3+aa: — 1+5;/ by— a. Ans. 3a—a 2 x-\-a—aby. 7. Multiply lx-\-by 2 — 4c+d 2 by 4. Ans. 28a;+452/ 2 — 16c+4i5 2 . 8 Multiply a 2 b—c 2 fx— 2 by 3a. Ans. 3a 3 b— 3ac 2 -\-3ax— 6a. B* 20 MULTIPLICATION. RULE IY. (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 2 + 4ac— c 2 by 3 a — 5c 2a 2 -f4ac — c 2 3a —5c 6a 3 -f- 12a 2 c — 3ac 2 — I0a 2 c-20ac 2 + 5c 3 6a 3 -f- 2a 2 c — 23ac 2 + 5c 3 . 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, first, 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 a 2 — 2a + 5 by a-f-3. 10. Multiply 2x 2 -|-3x — 1 by x — 5. 11. Multiply a 2 — 2 az-j-.i 2 by a + x. 12. Multiply x 2 -\-3xy-\-y 2 by x — y. 13. Multiply h 2 — 3 be — c 2 by 2 b-\-c. 14. Multiply 2a 2 + 3ac — c 2 by a — 2c. 15 Multiply 2 b 2 — 2 bx-\-x 2 by 2b — x. 16 Multiply b 3 -\-b 2 -\-b by b 3 -\-b 2 . 17 Multiply x 2 — 2c 2 -\-2y by c 2 -\-y. 15. Multiply a 2 -f bxpy by a 2 — bx. Am. a 3 -fa 2 — a+15 Ans. 2z 3 — lx 2 — 16x-(-5. Ans. a 3 — a 2 x — ax 2 -\-x 3 Ans. x 3 + 2 x 2 y — 2xy 2 —y 3 Ans. 2 b 3 — 5b 2 c — 5bc 2 —c 3 Ans. 2a 3 — a 2 c— lac 2 + 2c 3 A?is. 4 b 3 — &b 2 xd-4bx 2 — x : ' Ans. b 6 -\-2b 5 -j-2b* Ab 3 . M/iS. c 2 a' 2 — 2c 4 -\- x 2 y-\-2y 2 Ans. a*-\-a 2 y — b 2 ,i 2 — bi y. DIVISION. 21 I 3 Find the Product of (2a 2 — 4a5-|-26 2 )(3a — 2b). Ans. 6a 3 — I6a 2 6+14a6 2 — 46 3 . 20 Find the Product of (3a; 2 — 2au/+2z/ 2 )(2a; 2 -\-3,xy). Ans. 6a : i -\-5x 3 y — 2x 2 y 2 -\-3xy 3 . 21 Find the Product of (2a 2 +3a5 — b 2 ){a 2 — ad + 6 2 ). Ans. 2a i -\-a 3 b — 2a 2 6 2 +4a6 3 — 6*. 22 Find the Product of (3a; 2 — 2xy+5)(x 2 + 2xy— 3) A7is. 3a; 4 +4 x 3 y — 4a; 2 — 4a; 2 y 2 + 1 Gxy — 15. 23. Find the Product of the three polynomials (a+6), (a— b), and (a 1 \-ab-\-b 2 ) Ans. a 4 +a 3 6 — ab 3 —b^. DIVISION- (46.) Algebraic division consists in finding a factor or Quotient, which, multiplied into a given divisor, will produce a given dividend. Thus 6a 2 z — 2a, 6a 2 x divided by 2a, gives the quotient 3 ax, be- cause this factor, multiplied into the divisor 2a, produces the dividend 6 a 2 x. Division of Monomials. (47. ) The Quotient 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 \3a 2 x 2 y-^5o J 2 x gives the quotient 2axy, because this quo- tient, multiplied into the divisor 5a 2 x, produces the dividend 1 0a 3 z 2 y ; and 2 axy 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. (UP 3 What is the Quotient of 6a 3 x 2 y— 3ax 1 and how do you prove that quotient true ? What is the Quotient of 4 ab 2 c-t-bt and how do you prove that quotient true ? What is the Quotient of ia 2 hz — 2a '? and how do you prove that quotient true ? 22 DIVISION. When any Exponent is reduced to 0. (48.) Any quantity whatever with exponent 0 is equivalent to unity ; and a factor with this exponent may therefore he canceled. For example, a 2 —a 2 gives the quotient a 0 , by subtracting the ex- ponent of the divisor from that of the dividend. But any quantity contains itself once, and therefore a 2 —a 2 also gives the quotient 1 ; and these two quotients, a 0 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 10 a 3 b by 5a 3 . Subtracting the exponent in the divisor from the exponent of a in the dividend, we find the quotient 2 a°b The factor a 0 is equivalent to 1, and may therefore be canceled without affecting the value of the quotient. The quotient of 10 a 3 b — 5a 3 may therefore be given under the two different forms 2 a°b and 2b. [FP* Under what two different forms may the Quotient of 6a 2 z 2a 2 be represented ? Of 8«6 2 — 2b 2 1 Of 10 abc — 5al Of 9 ax 2 — x 2 ? Of 7 abc 2 -h7 c 2 ? Of 5ax-^axl Of 3 aby 2 —ay 2 l Of 10a: 2 - x 2 ? Sign of the Quotient. (49.) The Quotient 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 +:r, because +aX + x is -j- ax; — ax-. a gives +x, because — a X is — ax ; -\-ax-. a gives — x, because — aX — x is +ax; and — ax~-\-a gives — x, because +«x — # 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 rule 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 03=* What is the Quotient of 4ax be divided by 3a;, the quotient will he l + 2a ; the binopiial may therefore be resolved into the factors 3a;(l + 2«), 3a; into the binomial l + 2a. Resolve 2« + 4aa; — 6a 2 a; 2 into component factors. Resolve a 2 x — 3ax 2 -\-8ay 2 into component factors. Resolve 4 a 2 -\-a 2 b — 5a 2 y into component factors. Resolve 2a 3 — 3ux-\-l a*y into component factors. Resolve 5ax-\-5a 2 x — I0a 3 a 3 into component factors. Resolve lab — I4ab 2 — liabx 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 polynomi d 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, (55). The following divisions will be found, on trial, to terminate with- out remainders. 28 COMPOSITE QUANTITIES. (56.) The Difference of two quantities will divide the difference of any powers of the same degree of those quantities. Thus a — b will divide a 2 — b 2 , or a 3 — b 3 , or a 4 — b 4 , &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, oi those quantities. Thus a-\-b will divide o 3 +d 3 , or a 5 +5 5 , or a 7 -\-b 7 , &c. , also a-\-b will divide a 2 — b 2 , or a 4 — d 4 , or a 6 — b 6 , See.; The factors of certain Binomials and Trinomials may also be known from the manner in which the products and squares of binomials are formed. This will be seen in the following propositions. (58.) The Product of the sum and difference of two quantities is equal to the difference of the squares of those quantities. Thus (a+i) ( a — b) is equal to a 2 — b 2 ; and this last binomial may therefore be divided by either a-\-b or a — b. This proposition, it will be observed, is included in the two pre- ceding ones, (56.) (57.) ttsP* What is the Product of (a+x)(a — x) ? What is the Pro- duct of ( sum of the squares minus twice the product of the two quantities. Thus (a — b) ( a — d), that is, the square of a — b , is equal to a 2 -\-b 2 — 2 ab or a 2 — 2ab-\-b 2 . This trinomial may therefore be divided by a — h. [CP* What is the Product of ( a — x)(a — x). or the square of a — xl What is the Square of a — yl Ofy — 2? Ofd— x? Of 1 — y! COMPOSITE QUANTITIES. 29 Tlie preceding principles will be found applicable to the following EXERCISES. 1. Resolve a 2 — x 2 into component factors. Ans. ( a — x ) (a + a:) 2. Resolve a 3 +?/ 3 into component factors. Ans. (a-{-y)(a 2 — ay-\-y 2 ) 3. Resolve a 3 — a; 3 into component factors. Ans. ( a — x)(a 2 -\-ax-\-x 2 ) 4. Resolve a 4 — y i into component factors. A?is. (a 2 -\-y 2 ) (a-\-y) (a — y). 5. Resolve a 3 — 8a: 3 into component factors. ai«s. (a — 2x)(a 2 -\-2ax-\-4x 2 ). 6. Resolve i —&y 3 into component factors. Ans. (1 — 2y) (l + 2y+4?/ 2 ) 7. Resolve l+27a: 3 into component factors. Ans. (l+3a:)(l — 3a:+9a: 2 ). 8 Resolve 8a 3 — 27 y 3 into component factors. Ans. (2a — 3y) (4a 2 -\-6ay+9y 2 ). 9. Resolve a 3 x 3 ~\-c 3 y 3 into component factors. A?is. (ax-\-cy)(a 2 x 2 -axcy-\-c 2 y 2 ). 10. Resolve a 4 — 16a: 4 into component factors. Ans. (a 2 -f-4a: 2 ) (a-\-2x) (a — 2a:). 11. Resolve a 2 -|-2aa:-f-a: 2 into component factors. Ans. (a + a:) (a + a:). 12. Resolve a 2 — 2ay-\-y 2 into component factors. Ans. (a—y)(a—y). 13. Resolve a 2 -\-4ax-i-4x 2 into component factors. Ans. ( a-\-2x ) (a+ 2a:). 14. Resolve 9a 2 — 6 ay-\-y 2 into component factors. A?is. (3a— y) (3a — y). 15. Resolve 4a 2 + 12ax-\-9x 2 into component factors. Ans. (2a + 3a;) (2a+3x) 16. Resolve a 2 a; 2 — 2ax-\- \ into component factors. Ans. (ax — 1) (aa:— 1) 17. Resolve \-\-4xy-\-4x 2 y 2 into component factors. Ans. ( 1 + 2a:y) ( 1 + 2xy) 18. Resolve 4a 2 x 2 — \2abxy-\-9b 2 y 2 into component factors. Ans. (2 ax — 3by) (2ax — 3 by) 19. Resolve 9a i x i -\-24a 2 cx 2 y-{- 16c 2 ?/ 2 into component factors. Ans. (3a 2 x 2 + 4q/) (3a 2 x 2 +4 cy). 20. Resolve 16a 4 3 2 — 32a 2 czy 2 -f- 1 6c 2 ?/ 4 into component factors. Ans. (4 a 2 x — 4cy 2 )(4a 2 x—4cy 2 ) 30 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 two quantities For example, let it be required to decompose the trinomial a 2 — a — 12. The first term is the square of a ; the second term is the sum oj 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 a 2 — a— 12 is equal to (a 2 x-{-ia' 3 x and d 3 y — 5a 2 y 3 1 Of 5a 3 c+ 10a 3 c 2 and 10a 2 b — 5a 3 bxl Of 2 ax 2 — 6 a 2 y and 4a 2 a:+8a?/ 2 — 2a l 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 ?>axy 2 and 6a 2 y 2 z are 3, a, and y 2 ; and 3 ay 2 is the greatest common measure of the two quan tities. COMMON MEASURE. R 2 But as no general Rule can be given for resolving a Polynomial into the 'polynomial factors of which it may be composed, we make the Rule for finding the greatest common measure depeud 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-\-b ; n being some integral number, and b less than a. It is piain that any measure of a will also measure na , n times a ; ) and, measuring na, if it measure ?ia-\-b, it must also measure b. Hence there can be no common measure of a and na-\-b which is not < a common measure of a and b ; the greatest common measure, there- | fore, of a and b, will be the greatest common measure of a and { na-\-b. But b is the remainder after dividing a into na+b ; hence the proposition is true for two quantities ; and in like manner it may tx» ! proved for three or more quantities. RULE VII. (GG.) To find the Greatest Commo?i Measure of two Quantities. ] 1. Divide one of the quantities into the other, and the remainder i into the divisor, and so on, until there is no remainder : the last _ risor i will be the greatest common measure required. — But, 2. When any factor is contained in each term of the divisor, with- i 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 I term of the dividend — multiply the several terms of the latter by some ij quantity which will thus render its first term divisible, without a remainder. Note — T li;s Rule might be readily extended to three or more quantities; I 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 4a 2 x — 5ax 2 -\- x 3 and 3a 3 — 3 a 2 x-\-ax 2 — x 3 . 4 a 2 x — 5ax 2 -\- x 3 3a 3 — Sa 2 x-\- ax 2 — x 3 4a 2 — 5ax -\-x 2 4 12a 3 — 12a 2 a:+ 4 ax 2 — 4a; 3 ( 3a 12a 3 — 15a 2 a:+ 3 ax 2 19aas 2 — 19x 3 a — x 3a 2 x+ ax 2 — 4a: 3 4 12a 2 a:+ 4aa: 2 — 16a: 3 ( 3a; 12a 2 a: — 15aa: 2 -f- 3a: 3 19aa: 2 — 19a: 3 4a 2 — 5ax-\-x 2 ( 4a — x 4 a 2 — 4aa: — ax-\-x 2 — ax-\-x 2 In this example the factor x is contained in each term of the divisor 4 a 2 x — 5ax 2 -\-x 3 , without being contained in each term of the dividend ; we therefore cancel this factor, and take 4a 2 — 5ax-\-x 2 for a divisor. Then, the first term 4a 2 of this divisor not being a measure of the first term 3a 3 of the dividend, we multiply the dividend by 4, which renders the first term 12a 3 divisible, without a remainder. We also multiply the remainder 3 a 2 x-\-ax 2 — 4a: 3 , still regarded as a dividend, by 4, to make the first term divisible, without a remain- der. In the next remainder, 19aa: 2 — 19a: 3 , 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 19a: 2 , into the formei divisor, which now becomes the dividend. The divisor a — x completes the operation, and is the greatest com- mon measure required. 3 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 4 a 2 x — 5ax 2 -\-x 3 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 affected 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). In the preceding Example, we might have taken the first remain der, 3 a 2 x-\~ax 2 — 4a: 3 , for a divisor, and the first divisor for the (livi dend. The operation from that point would be as follows ; — oa 2 x-\-ax? — 4.r 3 3 a 2 -\-ax — 4a: 2 4a 2 — 5ax-\- x 2 3 12a 2 — 15«a:-i- 3.r 2 ( 4 12a 2 + 4aa" — 16a: 2 — 19aa:-(- 1 9a- 2 — 19aa:+ 19a: 2 — • a + x 3a 2 + ax — 4a: 2 ( — 3a — 4ar 3a 2 — 3 ax 4 ax — 4a: 2 4aa: — 4a: 2 COMMON MEASURE. 3 The greatest common measure is here found to be — a-\-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 3 — 3a 2 *-|-a* 2 — a: 3 4a 2 * — 5a* 2 -f- x 3 3 a 12a 3 * — 15a 2 * 2 + 3a* 3 (4* 12a 3 * — 12a 2 * 2 +4a* 3 — 4* 4 — 3a 2 * 2 — a* 3 -|-4* 4 — 3a 2 — a* +4* 2 3a 2 * 2 — a* 3 -)-4* 4 1 3a 3 — 3 a 2 *+ a* 2 — * 3 ( — a 3a 3 + a 2 * — 4a* 2 — 4a 2 *+ 5a* 2 — * 3 3 — 12a 2 *+15a* 2 — 3* 3 ( 4a: — 12a 2 * — 4a* 2 + 16* 3 '9a* 2 — 19* 3 a — * 19a* 2 — 19* 3 — 3a 2 — a*+4* 2 ( — 3a + 4* — 3a 2 + 3a* — 4a*+4* 2 — 4a*+4* 2 C 36 COMMON MEASURE. In this case the factor x is contained in each term of the dividend, without being contained in each term of the divisor. This factor, therefore, does not enter into the composition of the greatest common measure, (64), and might he 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 2 x — x 3 and a 2 x-\-ax 2 — 2x 3 . The factor x is contained in all the terms, and the two polynomials without this factor are a 2 — x 2 and a 2 -\-ax — 2x 2 . The greatest common measure of these will be found to be a-i ; then, mu 1 '. plying x into a — x, we find ax — x 2 for the greatest common measure of the given polynomials. EXERCISES. 1 . Find the greatest common measure of a 3 —x 3 and a i —x i . 2. Find the greatest common measure of a 2 — ax — 2x 2 and a 2 — 3ax-f-2x 2 . 3. Find the greatest common measure of 3a 2 — 2a — 1 and 4a 3 — 2a 2 — 3a+l. 4. Find the greatest common measure of x 2 + 2 ax-j-a 2 and x 3 — a 2 x. 5. Find the greatest common measure of 2x 3 — 16s — 6 and 3s 3 — 24x— 9. Ans. a— x Ans. a—2x* Ans. a — 1 i Ans. x+a Hns. x 3 — Sx— 2 COMMON MULTIPLE. 37 6. Find the greatest common measure of a 2 — 5ax-\-4x 2 and a 3 — a 2 x-\-2>ax 2 — 3a: 3 . Ans. a—x. 7. Find the greatest common measure of a 2 — 2 ax-\-x 2 and a 3 — a 2 x — ax 2 -\-x 3 . Ans. a 2 -2ax- s rx 2 . 8. Find the greatest common measure of 6a 2 + 7aa; — 3a; 2 and 6a 2 + llax-j-3x 2 . Ans. 2a+3ar. 9. Find the greatest common measure of lx 2 — 23xy-\-Gy 2 and 5x 3 — \8x 2 y-\- Wxy 2 — Gy 3 . Ans. x — 3 y. 10. Find the greatest common measure of a 4 -\-a 2 y 2 -\-y 4 and a 3 — 2a 2 y-\-2ay 2 — y s . Ans. a 2 — ay-\-y 2 . 11. Find the greatest common measure of 3a 5 + 6a 4 a;+3a 3 a: 2 and a 3 z~l-3a 2 x 2 ~{-3ax 3 -\~x 4 . Ans. a 2 + 2ax-{-x 2 . 12. Find the greatest common measure of a: 3 — ax 2 — 8a 2 x-\-Ga 3 and x 4 — 3ax 3 — 8a 2 x 2 -\-18a 3 x — 8a 4 . Ans. x 2 + 2ax — 2a 3 . COMMON MULTIPLE. (70.) One quantity is called a multiple of another, if it can be neasured by the other, that is, divided by it, without a remainder ; fmd A common multiple of two or more quantities is any quantity that ■an be divided by each of them, without a remainder. Thus 8a 3 a: 2 is a common multiple of 2a 2 a: and 4a; 2 . CCP* Name a Common Multiple of 3 ab and 5c 2 . Of ax 2 and 2 ay. )f 3a 2 and 4a; 2 . Of 5bc and a; 2 . Of y 3 and 5a: 3 . Least Common Multiple. (71.) The least common multiple of two or more quantities, is th< mailest quantity that can be divided by each of them, without a re rainder. Thus 6aa: 2 is the least common multiple of 6a and 6a; 2 . 38 COMMON MULTIPLE. tO 55 ’ What is the Least Common Multiple of 2 ab and 3b 2 ? Of 4x and 4 y 2 1 Of 5a and 3 a 2 b1 Of ab and 2 be 2 1 Of 7 ax and ax 2 ] Of 3a 2 and 6 ay 2 1 Of axy and 5x 2 y 2 ? (72.) The least common multiple of two or more quantities is com- posed of the smallest selection of factors that includes the factors of each given quantity. For example, take the quantities 3 ab 2 c and 6a 2 bxy, Resolving these two quantities into their prime factors , we have 3 abbe 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 2 b 2 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. If, therefore, the product be divided by the greatest common mea- sure, the quotient will contain only those factors which are conunon to the two quantities, and those which dice peculiar to each of them ; and these are the factors of the least common multiple, (72). RULE VIII. (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 hvo or more of them by any prime quantity, greater than unity, thal will divide them, without a remainder, — placing the quotients and the 1 undivided quantities in a line below. 2. Divide any two or more of the quantities in the lower line, a- . before ; and so on, until no two quantities in the lowest line can be ; divided. The product of the divisors and quantities in the line, will he the least common multiple required. 3. If no two of the given quantities can be divided as above, th- product, of all the quantities will be their least common multiple. COMMON MULTIPLE. 39 EXAMPLE. R To find the least common multiple of 2ax 2 , 5a 2 y, and 3 y — 4 y 3 3 J 3ax 2 6 a 2 y 3y — 4 y 3 aj ax 2 2 a 2 y 3y—iy 3 y) 2 ay 3 y—fy 3 X 2 2 a 3-4 y 2 In the first operation we divide 3 ax 2 and 6 a 2 y hy 3, and set down 3 y — 4 y 3 without dividing it ; and in like manner in the second opera- tion. In the third operation we set down x 2 without di\ r iding it. Then 3 ayx 2 x2ax(3 — 4 y 2 ), equal to 18 a 2 x 2 y — 24 a 2 x 2 y 3 , is the .east common multiple of the three given quantities. 1 J This Rule depends on proposition (72) : the divisors and quantities n the lowest line, are the smallest selection of factors that includes the factors of each given quantity. EXERCISES. 1. Find the least common multiple of ax 2 , 2a 2 y, 4 y-{-y 2 , and ax 2 4a 2 , Ans. 8 a 3 x 2 y + 2a 3 x 2 y 2 -\-3>2a 2 x 2 y-\-8a 2 x 2 y 2 . 2. Find the least common multiple of 2 ay 2 , Aay 2 , 2x — 4a: 2 , and a: 3 + . Reduce — to its lowest terms. Ans. a — b a 2 — ab-\-b 2 a 2 + a: 2 Ans x 2 +2 ax+a 2 Ans. 6. Reduce 7. Reduce a 3 — ay 2 a 2 -\-2ay-\-y 2 x—a x+a ' to its lowest terms. Ans. a 2 —ay a+y a * —x“ a 3 — 2>a 2 x-\-2>az 2 — x 3 to its lowest terms. Ans. a-\-x 2 ax 2 —a 2 x — a 3 8. Reduce — to its lowest terms. 2x 2 + 3ax+a 2 a 3 + 2a 2 x-j-3a 2 x 2 y. Reduce — — — to its lowest terms. 2a 4 —3a 3 x — 5a 2 x 2 Ans. a 2 — 2ax-\-x 2 ’ . ax — a 2 Ans. x-\-a a,-\-2x-\-3x 2 2 a 2 —oax — 5x 2 ' , _ ,, . 6a 2 +7aa? — 3x 2 . 10. Reduce — — - to its lowest terms. 6a 2 + llaz+3z 2 Ans. 6a— z a ^ — |- a ^ x ^ 'X/ ^ 11. Reduce — — — — 7 to its lowest terms. a 4 -{-a 3 x— ax 3 — x 4 3a + ce a 2 —ax+x 2 a“ — x“ t* i 5a 5 + 10a 4 w+5« 3 w 2 . 1 2. Reduce — to its lowest terms. a 3 y-\- 2 a 2 y 2 + 2 ay 3 -\-y 4 5a 4 -\-5a 3 y A?is. a 2 y+ay 2 +y 3 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-\-b 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 same for each. For example, to reduce to a common denominator the Fractions a , be and , 2 a — 2x '3a — 3x 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 3 a 6 a — 6x and 2 be 6 a — 6x (81). 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 tivo or more Fractions to a Common Denomi- nator. 1. Multiply each numerator by all the denominators except its oicn, for new numerators ; and multiply all the denominators together, for a common denominator . 2. If the Least Common Denominator he required, — Find the least common multiple 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. EXAMPLES. 1. To reduce — , — , and - — - to a common denominator. 3x G y-\- 2 For the new numerators, we have a . 6 . (y-\- 2), equal to &ay -\- 1 2 a ; b.3x.(y+2), “ 3bxy-\-6bx; and c . 3x . 6 “ 18ca:. A.nd the common denominator is 3x . 6 . (y-\- 2), equal to 18a;?/+36;2. 48 FRACTIONS. The given Fractions are thus reduced to 6aw+12a 3bxy-\-§bx 18 cx I8xy+3§x ’ lQxy+3&x ’ \Qxy+3bx ’ res P ectlve 2. To reduce the same Fractions to the least common denominator. The least common multiple of the denominators 3x, 6, and y-\- 2 will he found to be bxy-\~\2x , (74), which is the required denominator. Dividing this Multiple by each given denominator, and multiplying the quotients by the given numerators, respectively, we find the new numerators, 2ay-\-4a, bxy-\-2bx, and 6 cx. The Fractions reduced to their least common denominator, are then 2ay-\-4a bxy-\-2bx 6 cx d>xy+ I2x ’ Qxy+\2x ’ 6xy+l2x ’ res P ectlvel y- 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. Thus in the first example, — has both its terms multiplied by 6 oCC and y-\- 2 , — producing the new terms bay-\-\2a and 18xy+36x. Hence the values of the given fractions are not altered in reducing them to a common denominator, (81). EXERCISES. 1. Reduce — , O b 2x and to a common denominator. x — 4 2. Reduce — a i n 2 ax 2 — 8 ax 3bx— 125 6cz ,lS ' 6x 2 — 24x ' 6x 2 — 24x ’ 6x 2 —24x a a-\-c , and to a common denominator. y 2 i —y 2y 2 — 2y 2 a 2 —a 3 y a z y 2 -\-a 2 cy 2 a 2 y 2 —a 2 y % a 2 y 2 —a 2 y % ’ a 2 y 2 —a 2 i/ a FRACTIONS. 49 3. Reduce * ’ C , (2+5 , and — — to a common denommator. Ans. ax a b — 1 i. Reduce — , — , and to a common denommator. 2 3a: l + » _ a 3 b a-\-b 5. Reduce — , — - , and y 2 x 2 xy Ans. „ _ -i a b ab 6. Reduce — - , - , and -5— — - 2 x 2 y x 2 -{-x 3 Ans. 3a 2 y 3 cxy 2 3axy-\-\5xy 3xy 3 ' 3 xy 3 ’ 3 xy 3 mon denommator. 3ax 2 A^abx 2 2a-\-2ab 6 bx — 6 x 6 x-j- 6 bx ’ 6z+6 bx ’ 6a:+6 bx to a common denominator. 2 a 3 x 3 y bxy 2 2 ax 2 y J r 2 bx 2 y 2 x 3 y 2 ’ 2 x 3 y 2 ' 2 x 3 y 2 to the least common denominator. ay-\-axy 2 bx 2 -\- 2 bx 3 2 aby 2 x 2 y-\- 2 x 3 y ’ 2 x 2 y-\- 2 x 3 y ’ 2 x 2 y-\- 2 x 3 y 7 Reduce — , — , and — to the least common denominator. 4 3 y 6 y 2 3 x 2 y 2 Aay 2 a — 2x Am ‘ Ity 2 ’ l2p ’ 0 _ , a b c 8. Reduce — , - , and — y i 5 yh—y 2 Ans. to the least common denominator. 5ay — 5a by 3 — by 2 5c 5y 3 — 5y 2 ’ 5 y 3 — 5y 2 ’ 5y 3 — 5y 2 'll CL (l A 9. Reduce - . — ^ , and to the least common denominator. 4 2x 2 2 + x Ans. 2 x 2 y-\-x 3 y 4a + 2aa; 4 a 2 x 2 rjQ 10. Reduce — , — , and 3 ’ 3 y Ans. 8a: 2 + 4z 3 ’ 82 2 +4£ 3 ’ 8x 2 +4x 3 ' to the least common denomi- 6y—3y 2 2 xy — xy 2 — 2 y-\-y 2 2 a 2 — a 2 y &y—3y 2 5y — 3y 2 ’ 6y-3y 2 nator. 50 FRACTIONS. Integral and Mixed Quantities Reduced to Improper Fractions. (89.) An integral quantity is one which does not contain any fractional expression; as 3 ax 2 , or 2ab—5xy. (90.) A mixed quantity is partly integral and partly fractional ; CL (C as Sax 2 -\ , or 2 ab . 0 c 1+2/ (91.) An improper Fraction is a fraction whose value may he ex- pressed by an integral or a mixed quantity. 5 Thus — is an improper fraction whose value is 2 a-\ . RULE XI. (92.) To Reduce an Integral or a Mixed Quantity to an Impro- per Fraction. 1. Under an integral quantity 7 , regarded as a numerator, set 1 for a denominator. Or multiply the integral quantity 7 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 numeratoi- 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 2 to a Fraction whose denominator shall be a — 2x. 3 a 2 3 a 2 is the same as — — ; and by multiplying both terms of this fraction by the proposed denominator, we have 3a 2 equal to 3a 3 — 6a 2 x a. — 2x , (81). 2 . To reduce 3a 2 to an improper Fraction Multiplying 3a 2 by the denominator 2, and subtracting the nume- rator a — x 2 from the product, — observing to change the sigfis of the 'uimerator, (36), — we have 3a 2 - a — x‘ equal to 6a 2 — a-\-x 2 2 2 FRACTIONS. 51 The reason of this operation will be evident, if we consider that, by multiplying the oa 2 by 2, we reduce the integral part to a common denominator with the fraction annexed to it, according to the first part of the Rule. 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 2 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 4 ax 2 to a Fraction whose denominator shall be y-{- 2. Aax 2 y-\-8ax 2 Ans. y + 2 2 Reduce a 2 -\- 3x to a Fraction whose denominator shall be 2 y 2 . 2a 2 y 2 -{-6xy 2 Am. 2y 2 3. Reduce 5xy to a Fraction whose denominator shall be 3 + x 2 . \Sxy-\- 5x*y Ans. 3 + z 2 4. Reduce y 2 —5 to a Fraction whose denominator shall be 3ax 2 . . 3 ax 2 y 2 — \5ax 2 Am - — m — 5. Reduce a+5 to a Fraction whose denominator shall be a — b. a 2 -b 1 Ans. a — b 6. Reduce Say 2 to a Fraction whose denominator shall be a 2 — b 2 ^ 5a 3 y 2 —5ab 2 y 2 a 2 —b 2 7 Reduce * 2 -J-l to a Fraction whose denominator shall be 3 a 2 y. Ans. 3a2 * 2 ?/+ 3 * 2 ? 3 a 2 y 8 Reduce ab 2 — x to a Fraction whose denominator shall be 1 — y . ab 2 — x—ab 2 y-\-xi Ans. — — - 52 FRACTIONS. 9. Reduce a 2 + #+- to an improper Fraction. Am. a 2l±^±l . y 10. Reduce ax 2 to an improper Fraction. y + 1 a+x Am. y + 1 11. Reduce 5a +35 — to an improper Fraction. o Ans. 1 4 a -f~ 9 b — x 12. Reduce 2a— 4c-| to an improper Fraction. Ans. 1 y2 13. Reduce 2-\-y 2 — — — — to an improper Fraction. 11a — 20c — y 2 Atis. 7+5y 2 /£ 3 14. Reduce a 2 -)-# 2 — — to an improper Fraction. a—x ' Ans. ax 2 —a 2 x a — x q2 CLCC 15. Reduce a — x-\ to an improper Fraction. Ans. a 2 —x 2 4iT — 4 16. Reduce 1 + 2z — — to an improper Fraction. 5a; 5y 2 oi«S. z+lOx 2 - 1 4 5x 17. Reduce 2x-f 2 y -\-~ — to an improper Fraction. /wX — oy Ans. 4x 2 — 2 xy—y 2 18. Reduce a-\-x — — — - — ; — , to an improper Fraction. 2x — oy a 2 -ax-\-x 2 Arts. 2a- 3 a 2 — ax- fa; 2 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 the 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 9a 2 c + 3ab — 2b + y 3 a Dividing the numerator by the denominator, we find the integral quo- tient to be 3 ac-\-b, and the remainder — 2 b-\-y. Setting the denominater under this remainder, and connecting the fraction so formed to the integral quotient, by the sign +, the result is 3 ac+b+-^±Z . Or, changing the signs in the numerator — 2 b-\-y, and connecting the fraction by the sign — , the result, under a somewhat simpler form, • o 7 2b— y is 3 ac-\-b — — . 3 a 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 3 ac-\-b, — the divisor 3a not being contained in the remainder — 2 bfy, the division of these terms is indicated by setting the divisor under them. The fractional part of the result must evidently he 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 -f- 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, •wall he in its lowest 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. a—x Ans. a 2 +ax-\-x 2 . IQx 2 5x-\-3 2. Reduce — to an integral or a mixed quantity. OX Ans. 2x— 1+— . ox 4 3. Reduce to an integral or a mixed quantity. a—y A?is. a 3 -\-a 2 y-\-ay 2 +t/ 3 £p4 — 3x^7/^ | " CLX 4. Reduce ^ = — — to an integral or a mixed quantity. x* — Siy 1 _ ax Ans. z 2 + -v — — , • ^ 4a 2 x 2 — 3ay— 2b . . 5. Reduce ' to an integral or a mixed quantity. 2a 2 Ans. 2x 2 - x 2 — 3 y 2 tntity. 3(2?/+ 2b 9ax 2 — 2x+3 . . b. Reduce to an integral or a mixed quantity. 2a 2 6a Ans. 3x 2 - 2x—3 I yi 3 7. Reduce — - — to an integral or a mixed quantity. a+y Ans. a 2 —ay Ay" 1 ■ $2 y2 — 4 : 8. Reduce to an integral or a mixed quantity. x+y Ans. x — y- Q-d'ij — - j- 9. Reduce — — - — to an integral or a mixed quantity’. 2 a-y B H „ Ans. 2 y+- x+y :a —y FRACTIONS. 55 ADDITION OF FRACTIONS- (94.) The Sum of two or more Fractions is found by means of a common denominator. mi i o r a t b . a-\-b 1 hus 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 b he divided by x, the whole 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. EXAMPLE. b c To add together 2 a -\ — and 3 a . x y Reducing these mixed quantities to improper fractions, they be come 2 ax-\-b , oay — c — and — - — - . x y Reducing these fractions to a common denominator, we have 2axy-\-bij 3 axy — cx xy xy Placing the sum of these numerators over the common denomma tor, the result is, 5axyfby—cx by — cx , equal to 5a-\ . xy xy 56 FRACTIONS. Otherwise, by adding the integral and th e 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 + — , (84). y i —c Then, reducing — and — to a common denominator, they become x y by , —cx — and . xy xy Adding these fractions together, and adding together the integral parts 2 a and 3a, we obtain ly—cx o a-j , as beiore. 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 XERCJSES. . . ,, . 3a 2a . 3a+2 1. A.dd together — , — , and — - — T U O d 2 2 (jP* 2. Add together 2x-\ , and ix -\ — — 4 o 3. Add together ?/ 2 + ^ , and 3 y 2 — . O ut 26-3 4. Add together — , 2a 2 , and 2x 5 Ans. 2a + 6a+8 12 Ans. 6x4- 13a 2 20 ' Ans. 4 y 2 -\- 5x 12 Ans. 2a 2 + 5a 2 -\-Abx — 6x lQx ' 5. Add together 3x 2 -j- and —j— Ans. 3 x 2 +a+ — . 4 FRACTIONS. 57 6. Add together x 2 y, — , and x 2 y-\- — . o « x a 2 — 3a: 7 Add together 2a 2 — — and — - — _ . ,, a 2 2>a a 2 — 4 8. Add together — , — , and - — - — . y y y 2 „ 2a 2 + 3 Ans. 2 x 2 y-\ — Ans. 2« 2 + 2 — in ip 4-fj* 14. Given ■ — = — |-£ to find the value of x J./«S. £=12 A)is. £=5 .r = 4. |i 15. Given - X - ^ -- = - — — to find the value of r. A. cS. J. ~ -i SIMPLE EQUATIONS. 7 ? 16. Given — — 16= — i — - to find the value of x. 3 5 4 L7 Given — — — = x — 3 — to find the value of x. 7 3 18. Given — =1+ - to find the value of x. 3 2 19. Given x — — =3* — 1 ^ to find the value of x. Ans. £=41 Ans. x—8. Ans. x=7. 20 . Given — — +5= to find the value of x. id O O Ans. x= Ans. x—6. op 2 21 . Given — + 6 £ = — — - to find the value of x. 3 5 Ans. x =—£- 3 __ „ . 3x — 5 „ 2x — 4 „ , . . „ 22. Given x — =12 — to find the value oi x. 2 3 Ans. £=65 „„ r,- 21— 3£ 2(2£+3) „ 5.-C+1 „ . . 23. Given ' =6 to find the value of x. 3 9 4 Ans. £=3. An Equation in which the unknown quantity is found in every term, with different exponents in different terms, may often he reduced to a simple Equation by dividing it by some power of the unknown quantity. (113...5). Thus if 2a 3 = 10a; 2 , — by dividing by a ; 2 we have 2a;=10 ; hence £ = 5. Ans. x—5. x 2 3a ; 2 24. Given -f 3a; = 7£ — _ 1 — to find the value of x. 5 5 •m . 25 . Given ^ | 1 „ fi „d the value of 2 . Ans. £= 10 -^,. 26. Given — -f = — 5 — — — to find the value of z. 3x 2x x 2 — 1 U£ 74 SIMPLE EQUATIONS. Literal Equations. 27. Given ax — c— x—b a-\-c to find the value of x. Clearing the equation of fractions, by multiplying it by the deno- minator a-\-c, we have a 2 x — ac+acx—c 1 =x — b. By transposition a 2 x-\-acx — x=ac-\-c 2 — b. Dividing by the coefficient of x, ac+c 2 — b x — ~ 2 ~, 7 * a z +ac— 1 dx cix 28. Given x-\ —b to find the value of x. c c Ans. x— be cib 1 29. Given be ——d to find the value of x. x x Ans. x~ a-\-c~\~d ab— 1 30. Given 3a: — a=x — — — to find the value of x. Ans. X— „ 2>ax 7 — 2bx-\-ax 31. Given 4 abx 2 — to find the value of x. be Ad ' 3 a-\-d ~e+y 32. Given .4/is. x— x(a—b) ab x _ . , , _ — — =a - 1 — to find the value of x. Ans. X— a — 2 u 12ab-Za' 3o(5 + 4) 6(a— 6)-|-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 — ox — 5, the two members will be equal whatever be the value of x. (113. .3) SIMPLE EQUATIONS. 75 PROBLEMS In Simple Equations of one unknown Quantity. (119.) A Problem is a question proposed for solution ; and the solu- tion of a problem by Algebra consists va. forming an Equation which shall express the conditions of the problem, and then solving the equation. The general method of 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 x or y , supposing that value to have been 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 2z+13 = 75. The value of x in this equation is the number required. Ans. 31. 2. Find a number such that if it be multiplied by 5, and 24 be subtracted from tbe product, the remainder will be 36. Ans. 12. 3. What number is that to of which if 25 be added, the sum ob- tained will be equal to the number itself minus 39 l Ans. 96. 4. Find a number such that if \ 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 halftone 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 part ; then 165 — x will represent the greater ; and the equation will be 165 — x 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 54. 9. What sum of money is that to which if $100 he added, -I of the amount will he $400 ? Ans. $500. 10. A prize of $100 is to he divided between two persons. — the share of the first being £ of that of the other. What are the shares ? Ans. $43f ; $561. 11. A post is J of its length in the mud, of it in the water, and 15 feet above the water. What is the length of the post ? Ans. 36 feet. 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 1 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 -g- of his money, \ he had $15 more 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 ? Ans. $95. 17. After paying away of my money, and then -3 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 1 than the third ; what are the parts? A?is. 12, 15, and 10 feet. 19. A can perform a piece of work in 12 days, and B can perform 1 the same in 15 days. In what time could both together do the work ? Let x represent the number of days. Then since A could do ^ of the work, and B ^3 of it, in 1 day, cc — will represent the part of the work A could do in x days, 15 will represent the part of the work B could do in x days The equation is ^ = 1 , the entire work. Ans. 6 | days. SIMPLE EQUATIONS. 77 20. If A could mow a certain meadow in 6 days, B in 8 days, and L *4 5 days, in what time could the three together do it 1 Ans. 2 Ay days. 21. Out of a cask of wine, which had leaked away a third part, 20 gallons were afterwards drawn, 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 ? Ans. $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 15 miles a day; and B 20 ; in what time would the two come together again 1 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 woman 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 1 = — , the part both would drink in 1 day. i 30 12 f J Ans. 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 1 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 1 Ans. minutes past one 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. $222| ; $444f ; S333f. 32. A person being asked the hour, answered that the time past noon was § 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 1 A?is. 8f gal. ; Ilf 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. Iff hours. 36 If $1000 he 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. $210ff ; $315ff; $4731f. 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. Ans. 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. 583f 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 ? Am. $4S00 ; $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, i A’s claim is \ of B’s, and C's is f of B s ; what sum must each of the i creditors receive ? Ans. $1292^; $2584^-; $1/^3^. 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 he 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 I 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 I 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 sum , or tlse in the difference, of the new equations. .EXAMPLE. Given the equations 2x + 3y=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; + 6y=46, and 15a; — 6^/=30. (113. ...4). Adding together the corresponding sides of these equations, we find 193=76. (113. ...2). which gives X—4. Putting 4 for x in the first of the two given equations, we obtain 8 + 3y= 23, which gives iy = 5. 80 SIMPLE EQUATIONS. Observe that if 6y had the same sign in the two equations which were added together , this term 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. EXAMPLE. Given, as before, 2x+3y=23, and 5x — 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, 2x—2o — 3 y ; dividing both members by the coefficient of x, 23 — 3 y 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. 115 — \5y ; —2y—10. Then find Clearing this equation of fractions, transposing, See., we shall y=5. Putting 5 for y in the first equation, 2z+15=23; which gives x=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. 81 Elimination by Comparison. (124.) If the value of the same letter he 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, 2a:+3?/=:23, and 5x — 2?/=l0, 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, _ 23—3 y x- — . Transposing and dividing in the second equation, 10 + 2 y X ~ 5 Putting the first of these values of x equal to the second, 23—3?/ 10 + 2?/ , „ 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 of fractions, if they contain any ; the necessary 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 fro m 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) 2x+?/ = 10, would be called equation (A). 5 82 SIMPLE EQUATIONS. EXERCISES. 1. Given 22 + 3 ?/ = 29, and 2>x — 2y=\\, to find the values of x and y. Ans. 2 = 7 , and y— 5. 2. Given ox — 3?/=9, and 22 + 5 ?/= 1 C, to find the values of i and y. Ans. 2 = 3, and y—2. 3. Given x+2?/=l7, and 32+?/= 16, to find the values of x and y. Ans. 2 = 3, and y = 7. 4. Given 42+?/ = 34, and 10^ — 2 = 12 , to find the values of x and y. Ans. x = 8, and // = 2. 5. Given 32+4?/=88, and 62 + 5 ?/= 128, to find the values of x and?/. Ans. 2 = 8 , and 2 /= 16. 6 . Given 72 + 3?/=42, and 8 y — 22=50, to find the values of x and y. Ans. 2 = 3, and y—1. 7. Given 8 y — 32=29, and 6 y — 42=20, to find the values of x and y. Ans. x=l, and y= 4. 8 . Given 62 — 5y = 39, and lx — 3?/=54, to find the values of x and y. Ans. 2 = 9, and y— 3. 9. Given 12x — 9z/ = 3, and 122+ 16?/=228, to find the values of x and y. Ans. 2 = 7 , and ?/= 9. 10. Given 52+7?/=201, and 82 — 3y = 137, to find the values of 2 and y. Ans. 2 = 22, and ?/=13. 11 . Given ly-\- y =99, and 72+ y =51, to find the values of 2 and y. A7is. 2 = 7, and?/=14. 12. Given =7, and — - + ^ = 8 , to find the values of 2 Ans. 2 = 6 , and ?/=] 2. 4 ?/ 9 y 13. Given 64 — = — , and 77 — = — to find the values of do 6 10 and y. 13. 2 and y. 14. G of 2 and y. Ans. 2 = 60, and y=30. 14. Given 21 — 6 ?/= , and 23—52= ■ , to find the values ,4 3 Ans. 2 = 4, and y=3. _. 32+4y . 2 , 62 — 2 ?/ y 15. Given — - =10 , and — - =14 — + , to find the 5 4 3 6 values of 2 and y. 16. Giv< of 2 and y. A)is. 2 = 8 , and y=i. _ . 22 3?/ 22 3x 2y 7 , . , 16. Given 1 — - = — , and 1 — + =4 — . to find the values 3 5 5 5 3 10 +;?s. 2 = 3 , and y= 4. 10—2 y— 10 . 2?/+4 22 +?/ 2+13 ' 1,,d = s^ = — 17. Given d — 2 SIMPLE EQUATIONS. olivll LiJii U A 1 IOjNS. g<^ to find the values of x and y. Ans. x—1, and y=lO 18. Find the values of x and y in the equations -m , c c in , 6 a-— 35 80 + 3 * 4x+3y-8 5 lo 6 7 Ans. *=10, and y—15. 19. Find the values of * and y in the equations y — a a — * x ; — —c, and?/ — — — = d . b J b Clearing the equations of fractions, we have (A) bx — y-\-a = bc : and (B) by — a-\-^~bd. By transposition in these two equations, we find (C) bx—y=bc — a, and (D) bij-\-x=bd-\-a. Multiplying equation C by b, in order to eliminate y, (E) b 2 x — by—b 2 c — ah. Adding together equations D and E, (122.) b 2 x-\-x=b 2 c-\-bd — ai + — + — = — y + z 10 ' By subtracting the second equation from the first, we shall elirm- • nate x, and then by adding the third equation we shall eliminate z Ans. A 14f-| days, B 17^, C 23^- x . 27. From two places, which are 154 miles apart, two persons set out at the same time to meet each other, one traveling at the rate of , 3 miles in 2 hours, and the other at the rate of 5 miles in 4 hours ; in how many hours will they meet ? Ails. 56 hours. 28. In a naval engagement, the number of ships captured was 7 more, and the number burned was 2 less, than the number sunk. Fif- a teen escaped, and the fleet consisted of 8 times the number sunk ; of how many ships did the fleet consist ? Ans. 32. 29. A and B together could have completed a piece of work in 15 days, but after laboring together 6 days, A was left to finish it alone, i which he did in 30 days. In how many days could each have per- formed the work alone? Ans. 50, and 21y days. 30. On comparing two sums of money it is found, that ? of the first is $96 less than of the second, and that ■§■ of the secoud is as much as of the first. What are the sums ? $720, and $512. 31. A privateer, running at the rate of 10 miles an hour, discovers a ship 18 miles off, making way at the rate of 8 miles an hour. In how many hours will the ship be overtaken ? Ans. 9 hours. SIMPLE EQUATIONS. 89 32. In a composition of copper, tin, and lead, 1 of the whole minus 16 pounds was copper, ^ of the whole minus 12 pounds was tin, and ~ of the whole plus 4 pounds was lead ; what quantity of each was there in the composition? Ans. Copper 128, tin 84, lead 76 pounds. 33. The sum of $660 was raised for a certain purpose by four per- sons, the first giving ^ as much as the second, the third as much as the first and second, and the fourth as much as the second and third. What were the several sums contributed ? Ans. $60, $120, $180, $300. 34. Two pedestrians start from the same point, and go in the same direction ; the first steps twice as far as the second, but the second makes 3 steps while the first is making 2. How far has each one gone when the first is 300 feet in advance of the second ? Ans. 1200, and 900 feet. 35. A merchant has cloth at $3 a yard, and another kind at $5 a yard. How many yards of each kind must he sell, to make 100 yards which shall bring him $450 ? Ans. 25, and 75 yards. 36. In the composition of a quantity of gunpowder, the nitre was 10 pounds more .than ■§■ of the whole, the sulphur 4-J pounds less than •g- of the whole, and the charcoal 2 pounds less than J- of the nitre. What was the amount of gunpowder? Ans. 69 pounds. 37. Four places are situated in the order of the letters A, B, C, D. The distance from A to D is 34 miles ; the distance from A to B is § of the distance from C to D ; and ^ of the distance from A to B, plus of the distance from C to D, is 3 times the distance from B to C. What are the distances between A and B, B and C, C and D ? Ans. 12, 4, and 18 miles. 38. A vintner sold at one time 20 dozen of port wine, and 30 of sherry, for $120 ; and at another time 30 dozen of port, and 25 of sherry’ at the same prices as before, for $140. What was the price of a dozen of each sort of wine ? Ans. $3, and $2. 39. A person pays, at one time, to two creditors, $53, giving to one of them jj of the sum due to him, and to the other $3 more than | of his debt to him. At another time he pays them $42, giving to the first of what remains due to him, and to the other ^ of what remains due to him. What were the debts ? Ans. $121, and $36. 40. A farmer has 86 bushels' of wheat at 4s. 6 d. per bushel, with which he wishes to mix rye at 3s. 6 d. per bushel, and barley at 3s. per bushel, so as to make 136 bushels, that shall be worth 4s. a bushel. What quantity of rye and of barley must he take ? Ans. 14, and 36 bushels. 5 * 90 SIMPLE EQUATIONS 41. A composition of copper and tin, containing 100 cubic inches, weighs 505 ounces. How many ounces of each metal does it contain, supposing the weight of a cubic inch of copper to be 5j- ounces, and of a cubic inch of tin 4j ounces ? Ans. 420, and 85 ounces. 42. A General having lost a battle, found that he had only one- half of his army plus 3600 men left, fit for action; of his' men plus 600 being wounded, and the rest, who were J of the whole army, either slain, taken prisoners, or missing. Of how many men did his army consist? Ans. 24000. 43. Two pipes, one of them running 5 hours, and the other 4, filled a cistern containing 330 gallons ; and the same two pipes, the first running 2 hours, and the second 3, filled another cistern contain- I ing 195 gallons. How many gallons did each pipe discharge pel hour ? Ans. 30 and 45 gallons. 44. After A and B had been employed on a piece of work for 14 days, they called in C, by whose aid it was completed in 28 days. Had C worked with them from the beginning, the work would have | been accomplished in 21 days. In how many days would C alone have accomplished the work ? Ans. 42 days. 45. Some smugglers discovered a cave which would exactly hold J their cargo, viz., 13 bales of cotton and 33 casks of wine. A revenue I cutter coming in sight while they were unloading, they sailed away i with 9 casks and 5 bales, leaving the cave two-thirds full. Howmanv bales or casks would it contain ? Ans. 24 bales or 72 casks. 46. A gentleman left a sum of money to be divided among four servants, so that the share of the first was 1, the sum of the shares of the other three ; the share of the second -j of the sum of the other three ; and the share of the third ^ of the sum of the other three ; and it was also found that the share of the first exceeded that of the iast by $14. What was the whole sum ? and the share of each ? Ans. Whole sum $120 ; shares $40 ; $30; $24, $26. 91 CHAPTER VI. RATI 0— P R 0 P 0 R T I 0 N— Y A R I A T I 0 N . RATIO. (127.) The ratio of one quantity called the antecedent to another of the same kind called the consequent , is the quotient of the former divided by the latter. Thus the ratio of 12 to 4 is 3, since 12 is 3 times 4 ; and the ratio of 5 to 13 is ^g-, since 5 is five thirteenths of 13. The antecedent and consequent together are called the terms of the ratio. Sign of Ratio. (128.) A colon ( : ) between two quantities denotes that the two quantities are taken as the antecedent and consequent of a ratio. Thus 3 : 5, the ratio of 3 to 5 ; a : b, the ratio of a to b. (129.) The value of a ratio may always be represented by making the antecedent the numerator , and the consequent the denomirvztor ol a Fraction. Thus 3 : 5 is equal to -- ; and a : b is 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 y. The inverse ratio of the first quantity to the second, is the direct ratio of the seco?id 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 |-, (127). Hence inverse is often called reciprocal ratio. The term ratio used alone always means direct ratio. E* 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 y is — ; xy = the product of the simple ratios — and — , or — and — . x y y x (132.) The ratio of the first to the last of any number 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 interve?iing 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 a b x . . abx a . . , . — , — , — ; and their product is — , = — , which is a : y. o x y bxy 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 b is the ratio of a 2 to b 2 ; and the triplicate ratio of a to b is the ratio of a 3 to b 3 . The subduplicatc ratio of quantities is the ratio of their square roots, and the subtriplicate ratio is the ratio of their cube roots Equimultiples and 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 3 a and 3 b are equimultiples of a and b, while, conversely, a and b are equisubmultiples of 3a and 36. (135.) Equimultiples, or equisubmultiples , of two quantities have the same ratio as the quantities themselves, (SI). PROPORTION. 93 I PROPORTION. (136.) Proportion consists in an equality of the ratios of two or more antecedents to their respective consequents — hut 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 by 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 a, b, x, y are in proportion, CL X 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 fourth '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 mean proportional between 8 and 2. Direct and Inverse Proportion. (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 -§-, (130). The term proportion used alone always means direct proportion. 94 PROPORTION— 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 G : 3 = 8 : 4, or =| 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 : 3yt4 : 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 1 : ^-=4 : 8, by substituting A and The reason of this is evident from the nature of inverse 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 dir eel ly as the Principal, since the Interest will be dou- j bled, or tripled, Ac., if the Principal be doubled, or tripled, Ac. 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 pet cent., since the Time will be doubled, Ac. if the Rate be halved, Ac. 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 the same number ; hence its value will remain unchanged. 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 say, 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. a-~ — denotes that x varies as the reciprocal of y \ or x varies re- s' ei [.locally or inversely as y. V — denotes that x varies directly as y, and inversely as z \ z 1 that is, x varies as the quotient of y divided by z. 96 THEOREMS IN PROPORTION. Theorems in Proportion. (145.) A Theorem is a proposition to be demonstrated or proved. — A Corollary is an inference drawn from a preceding proposition 01 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 he the common denominator ; a c . then the ratio of — to — is d d a c ad d ' d cd a c > (127), and — - is the ratio of the numerator a to the numerator c. c Secondly. Let n be the common numerator ; , , . . n n . then the ratio oi — to — is a c n n cn c a c an a ’ Q and — is the inverse ratio of the denominator a to the denomirtatoi a c, (130). Therefore, two fractions having a common denominator, Ac. (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 he greater than its con- sequent, the other antecedent will be greater than its consequent ; il equal, equal ; and if less, less. Let a : b : : x : y ; then — = — , (lo/). by -Now if a be greater than b the first ratio will be greater than a unit, and consequently the second ratio will he greater than a unit, and therefore x will be greater than y. In like manner if a he equal to b, x will be equal to y , Ac. Hence, in any Proportion, if one antecedent, Ac. THEOREMS IN PROPORTION. 97 Theorem III. (149.) When four quantities are in Proportion, the product of the two extremes is equal to the 'product of the two means. Let a : b : : x : y ; then is ay=bx. For since the quantities are in proportion, = (137). by Clearing the Equation of fractions. ay — bx . (115). Therefore, when four quantities are in Proportion, &c. (150.) Cor. 1. A fourth proportional to three given quantities, is f uid by dividing the product of the second and third by the first. bx Thus from the equation ay — bx , 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 2 . (152.) Cor. 3. A mean proportional between two given quantities, is tqual to the square root of their product. Thus from the equation ax=b 2 , we find b — [ax) 2 . Theorem IY. (153.) When the product of two quantities is equal to the product of two other quantities, either pair of factors ?nay be made the extremes , and the other the means , of a Proportion. Let ab=xy ; then will a : x : : y : b. Dividing both sides of the given equation by b, xy a= T‘ Dividing both sides of this last equation by x, _ y_ x b Hence a : x : : y : b. (137). In like manner a and b may be taken for the means , and x and y for the extremes. Therefore, when the product of two quantities, &e 98 THEOREMS IN PROPORTION. Theorem V. (154.) If three quantities are in Proportion, the first will he to the third 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 2 : b 2 ; or a : x : : b 2 : x 2 . From the given proportion, we find ax—b 2 , ( 151 ). Multiplying both sides of this equation by a, a 2 x—ab 2 . Converting this equation into a Proportion, a : x : : a 2 : b 2 , ( 153 ). And by multiplying both sides of the first equation by x , we shall, in like manner, find a : x : : b 2 : x 2 . Therefore, if three quantities are in proportion, &c. Theorem YI. (155.) Four quantities in Proportion are also in proportion by in- version , — that is, when each antecedent and its consequent are inter- changed with each other. Let a : b : : x : y ; then is b : a : : y : x. From the given proportion we find ay=bx, ( 149 ). Making h and x the extremes, and a and y the means, b : a :: y : x, ( 153 ). Hence, four quantities in proportion are also in proportion, See. Theorem VII. (15G.) 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 : b :: 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, Sec. THEOREMS ET PROPORTION. 99 Theorem YIIT. (] 57.) In any Proportion, if the tivo 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 he in pro- portion. • Let a \ b : : x : y ; and let n he any numerical quantity ; then will an : b : : nx : y, &c. From the given proportion we have ay—bx. Multiplying hoth sides of this equation hy n, any—bnx. Converting this equation into a Proportion, an : b : : 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 the 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-fiy : 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+y)=.x{a+b). Converting this last equation into a Proportion, a-\-b : a : : x-\-y : x, (153). By adding both sides of the first equation to by, it may be proved, n like manner, that a-\-b : b : : x-\-y : y. Therefore, four quantities in proportion are also in proportion, &c 100 THEOREMS IN PROPORTION. Theorem X. (160.) Four quantities in Proportion are also in proportion by dim sion, — that is, the difference of the first and second terms is to the first or second, as the difference 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 find ay=hx. Subtracting both sides of this equation from ax, ax — ay=ax — bx. Resolving each member of this equation into its factors, a(x—y)=x(a—b). Converting this last equation into a Proportion, a— b : a : : x—y : x. By subtracting both sides of the first equation from by, it may b* proved, in like manner, that a — b : b : : x — y : y. Hence, four quantities in proportion are also in proportion, &c. Theorem XI. (161.) 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, &cc. ; then is a-\-c : b-\-d : : x : y. From the given proportion we shall find ay—bx, and cy—dx, (149). Adding together the corresponding members of these two equations, aiy-\-cy=.bx-\-dx. Resolving each member of this equation into its factors, ( a + c)y=(b+d)x , Converting this equation into a Proportion, a-\-c : bfd : : x : y. By adding xy to both sides of the third equation, we shall, in like manner, find that a + c+a? : b-\-d+y : : x : y. Therefore, when any number of quantities are in proportion, &c. THEOREMS IN PROPORTION. 10 ] Theorem XII. (162. ) If two Proportions have an antecedent and its consequent , or the two antecedents, or the two consequents, the same in both , the re- maining terms will be in proportion. Let a : b : : x : y, and a : b : :w: z ; then will x : y : : w : z. For the ratio of x to y is equal to the ratio of w to z, since each of these ratios is equal to the ratio of a to b ; hence x \y : : w: z. The two given proportions have an antecedent and its consequent the same in both. If the two antecedents were the same in both, the demonstration would be the same, after interchanging 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 : v, a \b :: iv : x, and a : b :: y : z, form the continued proportion, a : b :: u : v :: w : 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 — b : a : : x—y : x. (160). These two proportions have the consequents a and x the same ir, 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 XIY. (164.) The •products of the corresponding terms of two or more Proportions, are in proportion. Let a : b :: x : y, and c : d : : w : z ; then is ac : bd : : xw : yz. From the two given proportions, we have ay — bx, and cz=dw. Multiplying together the corresponding members of these equations. ac .yz-Sbd . xw. Converting this equation into a Proportion, ac : bd : : xw : yz, (153). In like manner the demonstration may be extended to three or more proportions. Hence, the products of the corresponding terms, &c. (165.) Cor. Like poivers or roots of proportional quantities, are in proportion. For if a : b : : x : y, by multiplying each term by itself, we shall have, according to the Theorem, a 2 : b 2 : : x 2 : y 2 ; and multiplying these by the given terms, we shall have a 3 : b 3 : : x 3 : y 3 , k c. Theorem XY. (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 : : rx : sy. From the two given proportions, we find an \ bp w nrx : psy, (164). Dividing the antecedents in this proportion by n, and the consequents by P< a : b : : rx : sy, (1 58). This proves the first affirmation in the Theorem. By inter- changing the extremes in the two given proportions, and afterwards the means, (156), the other two affirmations in the theorem may be demonstrated. Therefore, for any factors in an antecedent and its consequent, kc ! 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 unknown 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. Py adding the second equation to the first ; and also subtracting the second from the first, we have 2 x= s+d, and 2y= s — d, , . , . s+d s—d which give X — — and y= — ^ — . From these general values of x and y, we learn that the greater of two numbers is equal to of (the sum + the difference ), a,nd that the less is equal to of (the sum — the difference), of the two numbers ; hence the following rule : B (1G8.) To find two numbers from their sum and difference , — Add the difference to the sum, and divide by 2, for the greater of the two numbers ; subtract the difference from the sum, and divide by 2, for the less number. For example, if the sum of two numbers be 500, and their differ- snce 146, m , . 500 + 146 The greater number is — — =323 ; 2 , . , , . 500-146 and the less number is — ■ — =177. 2 1 (169.) An algebraic Formula is an equation between the symbols 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 formulas for finding two numbers from the sum and difference of the numbers. 104 PROBLEMS. PROBLEMS In Proportion , Percentage , Interest, Spc (170.) A Proportion occurring in the solution of a Problcrs , may bt converted into an Equation by putting the product of the two extrema 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 2, C’s share. Then by the conditions of the problem, we have a:+y + 2=1000 ; x : y : : 2:3; and y : z : : 3 : 5. By converting the two Proportions into Equations, we find ox — 2y, and 5y=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 : : x : ^ # (150). Hence B’s share is represented by — , and C’s by ~ ; and the equation of the problem is 3a: 5a: _ x-\ — - — | — — =$1000. 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 bs cs a-\-b-\-c’ a-f-5+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. 105 tions of ¥ It may bo here remarked that algebraic Formulas express general 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. EXERCISES. 1. Divide $950 between two persons so that their shares shall bo to each other as 3 to 5. This problem might be solved without employing proportion, by ob- serving that the first share will be § of the second. Ans. $3561; $593|. 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+b 3. Divide the sum of $3000 between A, B, and C, in the propoi- tions of 1, 2, and 3. Ans. $500 ; $1000 ; $1500. 4. Divide the sum of $7600 between three persons, in the propor- i and l . 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. $327 T 3 T ; $572 T 8 T . 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 1 Ans. $137£; $1831; $2291. 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 1 Ans. $8|; $6f ; fill; $13$. 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\ to 2, and the share of the second to that- of the What were the shares 1 Ans. $3150; $2520; $2160. 10. A, B, C, and D together have $3000 ; A’s part is to B’s as 9 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 third as 3^ to 3. 106 PROBLEMS. 11. Three persons contributed funds in a joint speculation as foi 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 2 x 5 to 4 x3 and 2 x 5 to 5 x4, (158). By still further reductions, on the same principle, the ratios become 5 to 6 , and 1 to 2. Ans. $142$; $171$; $285$. 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 ? Ans. $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; S720 ; $840. 14. An estate consisting of 1000 acres of land is to be divided be- tween three persons, so that the first share shall be to the second as 2 to 3, and the first to the third as 1 to 2 . What are the shares ? Ans. 222-| ; 333$ ; 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 ? Ans. $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 sens. What are their several shares ? Ans. $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 b, and the second the Capital c for the Time d. . abs , cds Ans. — and — 3 . ab-\-cd ab-\-ca PROBLEMS. 107 Problems in Percentage. (171.) Percentage is an allowance at a certain rate per hundred ; and this rate is called the rate per cent., from the Latin centum , which means a hundred. The ratio of percentage is the rate per cent, h- 100, and is there- fore equal to the rate per unit. The basis of percentage is the sum or number on which an amount of percentage is computed. From these definitions it follows, that (172.) The basis of percentage X the ratio of percentage produces the amount, of percentage ; and, conversely, that the amount the ratio produces the basis of percentage. HHP" 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. Ans. 100a I - '"' 100 -|-r 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- jlbhase. 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 pods valued at $7500, which includes this amount and the premium or the insurance. What is the sum insured ? Ans. $7653.06’. 22. W r hat is the Formula for finding a sum of money which, di- ninished at the rate of r per cent., shall leave the sum a? 100 a Ans. — — . 1 00 — r 23. The profits of a manufacturing company this year amount to 13096, which is 31 per cent, less than their profits last year. What jvas the amount of profits last year ? Ans. $3200. 24. What must be the percentum of profit at which a quantity of aerchandise, bought for $3750, must be f profit shall be $1500 ? ?old, that the whole amount 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 1 100(a — s) Ans. — — . 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 ? Am. s 28. W r hat 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 I capital at the beginning of the year 1 A)is. $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 w r as sold for $250. — which was at a loss of 16|- per cent. For what sum was the cloth purchased 1 Ans. $300. I 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 ? A?is. 50 per cent. 33. An agent receives $2030 to invest in merchandise — himself to I retain a commission of 1^- per cent, on the amount of the purchase. What is the sum to be invested 1 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 ■§• per cent ? -d.«s. $3600. I 35. What amount of stock in a Savings Bank, at an advance of c I per cent., could be purchased for $4200 ? and what amount in another at a discount of 5 per cent., could be purchased for $1995 ? $4000, and $2100. j PROBLEMS. 109 Problems in Interest, Spc. (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 which 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. [CP“ The Student may be required to write the Rides 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 6x then xX -ttv or is the Interest for one year ; 100 100 and by adding 5 years’ interest to the Principal, Ox we have x-\- 100 — $ 1000 . Ans. $769^. 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? . 100a Ans - TMTS- 38. What Principal would amount to $2500, in 10 years, allow- ing the rate of interest to be 7 per cent. ? Ans. $1470.588’. 39. At what Fate 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 Fate per cent, of interest at which the sum s would amount to the sum a in t years ? 100(a— s) Ans. st 41. In how many years would $6000 amount to $7470, allowing the rate of interest to be 7 per cent? Ans. 3-| 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 be at r per cent ? 100 <- s > . sr 43. What Principal would produce as much interest in 3^ years, as $500 would in 4 years, the rate of interest in both cases being 6 per cent 1 Ans. $57 ll . 44. At what Hate per cent, of interest would $525 pi'oduce the same amount of interest in 5 years, that $700 would produce at 5 per cent, in 3 years 1 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. The 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 — *) X j-g-Q is the Interest on the remainder of the debt ; and we have therefore the Equation, a: + (300 — x) x T ^ o =200. Ans. $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 3 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. Ans. 100 s—nrt rt(n-s) and — ™ 100 — rt ’ 100 — rt What would be the Rule for finding the amount of payment that could be made on the debt ? Ill CHAPTER YII. ARITHMETICAL, 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 difference 2. And 15, 12, 9, 6, 3, is a progression in which the quantities de- crease by the continual suht-raction 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 he the first term, and d the common difference ; then in an increasing progression the series will be, a , a-\-d, a-\-2d, a-\-c>d , a + 4c?, &c. ; and in a decreasing progression the series will be, a, a—d , a — 2d, a — 3 d, a — 4 d, &c. I In these series th e fifth or last term a ±^d, a plus or minus 4 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 difference between the two extremes the number of terms less one. 112 PROGKESSION. The Sum of the Two 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 , a-\-d, a-\-2d, «+3cZ, a + 4r7, &c. Of these five terms the sum of the first and the last, is a -}“ ( a -f 4 d') = 2a 4 d. The sum of the second and the fourth, which are equidistant from the extremes, is ( a-\-d)-\-(a-\-3d ), also =2a-\-\d. We see moreover, that the sum of the two extremes is equal to twice the middle term, a + 2 d. In like manner the proposition will he 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 sum 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-\-t$d, a-\-3d a~\-2d a-{~ d a 2«+3c7’ 2a-f-3c7’ 2a-\-od’ 2a-\-‘3d' The sum ( 2a.-\-od)-\-(2a-\-od ) &c. of the tico 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. 113 Formulas in Arithmetical Progression. (181.) In an Arithmetical Progression, let a be the first term, d the common difference, n the number of terms, l the last term, and S the sum of all the terms. Then A. ... . I =a±d(n— 1) (176); B . . . . S=±n(a+l) (180). The sign + is to be prefixed to d(n— 1) when the progression is an increasing one, and — when decreasing. In these two F ormulas 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). HARMONIC AL PROGRESSION. (182.) An harmonic al progression is a series 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—b : c — d ; or a \ d :: b — a : d — c. The numbers 16, 8, 3, 2, are in Harmonical Proportion, since 16 : 2 :: 16 — 8 : 3— 2 . I - * The first and last terms of the Progression or proportion are called the two extremes, and all the intermediate terms the means. a* 114 PROGRESSION. An Harmonical converted into an Arithmetical Progression. (183.) The reciprocals of the terms of an Harmonical progression, are in Arithmetical progression. Let a, b, c, be three consecutive terms of a decreasing harmonica', progression. Then a : c :: a — b : b—c, (182); Converting this proportion into an equation wc have ab — ac—ac—bc. Dividing each term in this equation by abc, and reducing + h a eer eral quotients to their lowest terms, we find 1 _ _ _ 1 _ _ 1 c b b a Transposing the first and the last term of this equation We thus find that the difference between the reciprocals of a and b, 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 b b b iV; in which the common difference of the terms is jj. (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 — be, we shall find b = — — ; a+c and b is the harmonical mean between a and c. The harmonical mean between 3 and 6. is 3x6x2 36 3 + 6 - T ~ ' PROGRESSION. 115 GEOMETRICAL PROGRESSION. (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. ■ I Tims 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 Hence the successive terms of a Geometrical Progression consist of the first term multip>lied continually into the ratio, that is, multiplied into the successive powers 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 power of the ratio which is expressed by the number of terms less one. Let a he the first term, and r the ratio of the progression ; then, multiplying a continually into r, the series will be a, ar, ar 2 , ar 3 , ar 4 , &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 * which is expressed by the number of terms minus 1. 1 (187 ) Cor. The last term oi a Geometrical Progression -f- the first term, gives that power of the ratio which is expressed by the number of terms less one. Thus nr 4 H-a=r 4 ; the number of terms being yiTe. F* 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 square of the middle term when the number of terms is odd. Let a be the first term, and r the ratio of the progression; then, multiplying a into the successive poivers of r, the series is a, ar , ar 2 , ar 3 , ar 4 , &c. Of these five terms the product of th e first and the last, is axar 4 =a 2 r 4 . The product of the second and the fourth, which are equidistant from the extremes, is arxa? 3 , also =a 2 r 4 . We perceive moreover that the product of the two extremes is equal to the square of the middle term or 2 . In like manner the proposition will he found true for any number of terms. (189.) Cor. A geometrical mean, or a mean proportional, between j two given terms, is equal to the square root of the product of those 1 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 first term and the product of the j last term X the ratio, — the difference between the ratio and a unit. Let S represent the sum of the terms, and we shall have S=a-\-ar-\-ar 2 -f-ar 3 -\-ar 4 , See. Multiplying both sides of this equation by the ratio r, Sr=zar-\-ar 2 -f ar 3 -\-ar 4 + ar 5 . Subtracting the first equation from the second, we find Sr — S= ar 5 —a ; which gives — — . ° r— 1 In the numerator of this value of S, observe that ar 5 is the last term ar 4 of the progression X the ratio r. The demonstration will evidently apply io any number of terms. PROBLEMS. 117 (191.) The sura of an infinite number of terms in a decreasing Geometrical Progression, is equal to the first term divided hy 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 he 0, and the expres- sion for the sum of the terms, found above, will become The divisor in this case is 1 — r, because the ratio of the progression being a proper f raction, is less than a unit. Formulas in Geometrical Progression. (192.) In a Geometrical Progression, let a he the first term, r the ratio, n the number of terms, l the last term, and s the sum of all the terms. Then 0 . . . . l=ar n ~', (186), D . . . .S= 1 ^, (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, l, s ; hence if any t/iree of these quantities be given, the values of the other two 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 terms ? .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 fifth term 70, what will the fourth term be ? Ans. 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. 11^; 17-^-; 23f. 7. If the fourth term of an Arithmetical progression be 37, and the eighth term 60, what are the intermediate terms Ans. 42f; 48J, 54£. 8. What is the sum of 25 terms of an increasing Arithmetical pro- gression in which the first term is •£, and the common difference of the terms also £ ? (176). Ans. 162£. 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 + 22* is the last term, (176) ; 2 + 22 * and - — X 23 is the sum of the terms, (ISO). 2 Hence an Equation may be formed from which the value of * will I be found. — Or we might substitute the numbers 1, 23, and 100, for a, n, and s in Formulas A and B, ( 1 S 1 ), and find the value of d, as one of the two unknown quantities. Ans. fj. 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 ? 4. 11. 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 Iris rate of travel J } of a mile per day. How far will they be apart ai llie | end of 40 days 1 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. 11 miles 840 yards. PROBLEMS. 119 13. Find the third term of an Harmonical progression, whose first and second terms are 12 and 15 respectively. If x represent the third term ; we shall have 12 : x :: 15 — 12 : a;— 15. Ans. 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 proportion whose first, second, and third terms are 2, 3, and 8 respectively ? Ans. 16. 16. If the first and third terms of an Harmonical progression be 25 and 40 respectively, what will the second term be ? Ans. 30y|. 17. The first and fourth terms of an Harmonical progression, are 10 i and 20 respectively. What are the two intermediate terms ? This problem may be solved by finding two arithmetical means between yg and yy, 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 l and the sum of all the terms? Ans. 54, and 80. 20. The first term of a Geometrical progression is f the ratio of the progression is and the number of terms 4. What is the last i term ? and the sum of all the terms ? Ans. gy, and |y. 21. What is the sum of an infinite number of terms in the Geome- ’ tncal progression whose first term is 100, and ratio \ ? Ans. 133-J-. 22. What is the sum of an infinite number of terms in the Geome- trical progression 'whose first term is 300, and ratio y? Ans. 450. 23. If the first and third terms of a Geometrical progression are 8 and 72 respectively, what is the second term ? - The second term is equal to the square root of 8 X72, (189). 1 • Or, considering the third as the last term of the progression, 72-f-8 = 9 is the square of 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. 150 120 PROBLEMS. 25. If the first and fourth terms of a Geometrical progression are 3 and 24 respectively, what are the two intermediate terms ? Am. 6 and 12. 26. If the seventh and tenth terms of a Geometrical progression are 6 and 750 respectively, what are the intermediate terms? Am. 30 and 150. 27. What is the sum of an infinite number of terms in the series h i, ■$, &c., in which the ratio of the progression is evidently -f ? 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 distance it can reach? Am. 4000. 29. If 10 yards of cloth he sold at the rate of $1 for the first yard $2 for the second. $4 for the third, and so on, what would he the price of the last yard ? and what would the whole amount to ? . Am. $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 would the last acre amount to? Am. $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? Am. $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, Ac. ; 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. 414|^ miles. 121 CHAPTER YIII. PERMUTATIONS AND COMBINATIONS.— INVOLUTION.— BINOMIAL THEOREM.— EVOLUTION. PERMUTATIONS. (193.) Permutations are the different orders of succession in which a given number of things may he taken — either the whole number to- gether or the whole number taken two and two, or three 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 tivo, are ab, ba, ac, ca, be, cb. Number of Permutations. (194.) If n represent a given number of things, the number of 'per- mutations that can be formed of them, will be equal to n(n — 1 ){n — 2 )(n — 3 )(n — 4), and so on, until the number of 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 tico and two, in which a stands first. In like manner we should find n— 1 permutations of n letters, taken two and two, in which b stands first ; and so for each of the n letters. Hence we shall have n[n — 1) permutations of n letters taken two and two. Suppose now that the n letters are to be taken three and three. By reserving a, and proceeding with n — 1 letters as before, we should find ( n — 1 )(n — 2) permutations of n — 1 letters taken two and two ; and by writing a before each of these permutations, we have 122 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 a letters. Hence we shall have n[n — 1 )(n — 2) permutations of n letters taken three and three. Suppose now that the n letters are to be taken four antifour. By reserving a, and pursuing the operation in the same manner as before, we should find n[?i — 1 )(n — 2 )(n — 3) permutations of n letters taken four and four. Thus the demonstration proceeds ; the number of 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 atove demon strated, — suppose it were required to determine the number of Permu- tations, or di fferent orders of successio?i, 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. 03?=* 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 which may be formed out of a given number of things, by taking the same number in ea_h collection — without regard to the order of succession. Thus the different Combinations which may be formed out of the three letters a , b, and c, by taking two at a time, are ab, be, ac. Observe that ab and ba are not different combinations, hut different ■permutations, of the letters a and b. 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 n represent a given number of things, the number of com- binations that can be formed out of them, will be equal to n{n — 1 )(n — 2 )(n- 1 . 2 . 3. 4 and so on, 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 rmmerator in the preceding expression, is the number of permutations of n things taken four and four, (194). On the same principle, the denominator 4x3x2xl is the num- ber of permutations of four things taken all together. Now since there can be hut one combination of 4 things taken all together, the number of permutations of n things taken four 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 four and four, is 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. 324 PROBLEMS. PROBLEMS In Permutations and Combinations. 1. In how many different ways might a company of 10 person! seat themselves around a table ! (194). Ans. 3628800. 2. How many different numbers might be expressed by the 10 nu- meral figures, if 5 figures he used in each number! Ans. 30240. 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 1 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 1 Ans. 5040. 6. How many different combinations of two colors could be formed out of the 7 prismatic colors ! (195). Ans. 21. 7. How many different combinations of 5 letters may he formed out of the 26 letters of the Alphabet! Ans. 65780. 8. How many different combinations of 2 elements might be formed I 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 1 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 1 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 tlfe solar system ? Ans. 40320. 12. A company of 20 persons engaged to remain together so long j as they might be able to combine in different couples in their eve- ning walks. What time will be required to fulfil the engagement ! M;iS. 1 90 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 he taken from a class of 15 students. How many different selections of the five might be made 1 and in how many different ways might the 5 sue- 1 ceed one another in the delivery of their speeches 1 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 I might be made 1 and in how many different ways might the 6 chosen be disposed with regard to the order of succession ? Ans. 1102052400; and 720 INVOLUTION. 12D INVOLUTION. (197 ) Involution consists in raising a given quantity to any re- quired power. 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 2 , the second power or square of a ; aaa is a 3 , the third power or cube of a ; and so on. Observe that one multiplication of a into itself produces the second power of a ; two multiplications produce the third power, and so on ; also that the number of times the quantity becomes a factor in raising a Power, is equal to the exponent of the Power. (198.) A higher poioer 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 rpquired power. Thus a 2 Xfl 2 produces a 4 ; a 2 x a 3 produces a 5 , &c. Powers of Unity , Monomials, Fractions, Spc. (199.) Every power of unity is unity , since any number of Is mul- tiplied together produce only 1 ; thus lxlXl &c. =1. (200.) A monomial is raised to any required Power by raising its numerical coeffcie?it to that power, and multiplying the exponents of its other factors by the exponent of the power. Thus to find the third power of 4ax 2 , we raise 4 to its third power, which is 4 X 4 X 4 = 64, and multiply the exponents of a and x by 3 : we thus obtain 64a 3 a: 6 . Observe that multiplying the given exponents hy the exponent of the required Power, is only a brief method of performing the requisite multiplications in the Involution of Molromials. (201.) A fraction is raised to any required Power, by raising its numerator and denominator, separately, to that power. d 2 d 6 Thus the third power of — is - 3 - . (98). | A mixed quantity may be raised to any required Power by involv- ing the equivalent improper fraction. 126 INVOLUTION. The Sign to be Prefixed to a Power. i (202.) Every even 'power of a quantity is positive ; while every odd power has the same sign as the quantity f rom which it is derived. Thus if a he positive , all its powers, as aa, aaa, and so on, will evidently he positive ; hut if a be negative we shall have — a. — a=a 2 ; a 2 . — a— — a 3 ; — a 3 . — a = a 4 ; a 4 . — a— — a 5 , &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 3 ax 2 . Ans. 9a 2 * 4 2. Find the cube of — 2 a 2 x. Ans. — 8a 6 * 3 . 3. Find the square of — 4a* 2 . Ans. 16a 2 * 4 . 4. Find the cube of 3a 3 *. Ans. 27a 9 x 3 . 5. Find the square of ab 2 c 3 . Ans. a 2 b*c 6 . 6. Find the cube of —a 2 x 2 y. Ans. — a 6 x 6 y 3 . 7. Find the square of \ax 2 . Ans. ^a 2 * 4 . 8. Find the cube of \ay 3 . Ans. £ a 3 y 9 . 9. Find the square of — | -ab n . Ans. %a 2 b 2n . 10. Find the cube of — |ri* 2 . Ans. — fpz 3 x 6 . 11. Find the 4th power of 2a". . Ans. 16a 4n . 12. Find the 4th power of — \a 2 . Ans. jL a s . 13. Find the 4th power of — 3* 2 . Ans. 81* 8 . 14. Find the 5th power of \x 2 . .Ans. ^2* 10 - 15. Find the 6th power of — ax 2 . Ans. a 6 * 12 16. Find the 6th power of 2 y 2 . Ans. 64»/ 12 17. Find the 7th power of — a 2 y n . Ans. — a 14 i y"'*. 18. Find the 7th power of \y. Ans. 19. Find the 8th power of a 2 b n . Ans. a 16 i s ". 20. Find the 8th power of — %x 2 . jins, ale* 16 BINOMIAL THEOREM. 127 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 of a-\-b, is (a-\-b){a-\-b)=za 2 -\-2ab-\-b 2 . And the 3d power or cube of a-\-h, is (a+6)(a-f Z>)(a + d) = a 3 + 3a 2 5 + 3a6 2 +6 3 . 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 affected. In developing ( a-\-b ) 2 by the Binomial Theorem we should obtain the 2d. power or square of ( a-\-b ). JL In developing (a-^b)' 1 we should obtain the square root of (a-\-b ) ; and so for other exponents. Development of {a±b) n ; n representing any exponent. 1. The Exponent of a in the first term of the development, is the same as the exponent of the Binomial, and decreases by the subtrac tion of 1, continually, in the succeeding terms. 2. The Exponent of b entering as & factor, commences with 1 in the second term of the development, and increases by the addition of 1, continually, in the succeeding terms. 3. The Coefficient of the first term is 1 ; that of the second term is the same as the exponent of the Binomial ■ — and the coefficient of the second, or of any term, X the exponent of a in that term, and -h by the exponent of b increased by 1, gives the coefficient of the succeeding term. 4. The Signs of the terms in the development of (affb) will all be + ; while for ( a — b) they will be alternately + and — . From these principles we have the following 128 BINOMIAL THEOREM. (205.) Formula for the Development of the Binomial (a + 5)". (a4 -b) n =a n -\-na n 1 6 + *(*-!) _—**» , n(n-l)(n-2) n 2 2><3 “ ' 3 5 3 , icc When the exponent n is a positive integer , as 2, 3, or 4, &c., the development will terminate at the term in which the exponent of b becomes equal to the exponent of the Binomial. For the exponent of a in that term will be 0, and 0 will thus be- come a factor in finding the coefficient of the next term ; hence the uext term will be 0, (43). example I. To find the 4th power of (« + 5), by the Binomial Theorem ; that is, to develop (a + 6) 4 . The literal factors without the coefficients, will be a 4 a 3 b a 2 b 2 ab 3 b 4 By computing and introducing the coefficients, with the sig>is, we have a 4 + 4a 3 §+6a 2 6 2 +4a6 3 + 6 4 . The student will readily perceive the application of the preceding principles to the several terms of this Power. Demonstration of the Binomial Theorem. The principles which have been given for the development of (aFb) n , will be demonstrated under the supposition that the exponent n is a positive integer. — A general demonstration would be equally applicable to negative or fractional exponents ; such demonstration is unnecessary here, and is too abstruse for the present stage of our sub- ject. If we multiply together the binomial factors a-\-b, a+c, a+d, a-\-e, and decompose the product terms which contain the lower powers of a, the final Product may be represented as follows ; +6 a 3 -\-bc a 2 -\-bcd + C -\-bd -\-hce -\-d -\-be -\-bde -\-cd -\-cde -j-ce -\-de -\-bcde BINOMIAL THEOREM. 129 In this Product observe that the coefficients of the powers of a in the successive terms, are as follows ; The Coefficient of a 4 is 1 ; the coefficient of a 3 is the sum of the second terms, b, c, See., of the binomial factors composing the Product ; the coefficient of a 2 is the sum of the products be, bd, See., of the second terms of the binomial factors combined two and two, (196) ; and the coefficient of a is the sum of the products bed, bee, See., of the second terms of the binomial factors combined three and three. Suppose now that c, d, e, & c. are each equal to b, and the number of binomials equal to n. The Product of these factors will be the wth power; of (a -\-b ) ; that is, it will be the development of («+£)”. The Exponent of a in the first term of the Product will evidently be n ; and, as exemplified in the preceding multiplication, this expo- nent will decrease by 1, continually, in the succeeding terms. Hence we shall have a n , a ”~ 1 , a n 2 , See. in the consecutive terms. The Coefficient of a n will evidently be 1. The coefficient of a" -1 i in the second term, will be n times b. Hence the second term will be nba n ~' i or na n ~ x b. The Coefficient of a™" 2 in the third term, will be b 2 taken as many times as there are combinations of two letters in n letters, (196). Hence the third term will be n{n — 1) 1X2 b 2 a n 2 n{n— 1) or — - — — ^ a n 1x2 - 2 b 2 . The Coefficient of a” -3 in the fourth term, will be b 3 taken as many times as there are combinations of three letters in n letters. Hence the fourth term will be n(n-l)(n-2) 1x2x3 or n(n — l)(n— 2) 1x2x3 3 b 3 . The several terms thus obtained agree with Formula (205) ; I and the law of development thus indicated, will, in like manner, be found applicable to any number of terms. With regard to the Signs , — if is evident that when a and b are positive, all the terms in any power of (a + 5) will be positive, (42). When b is negative, all the odd powers of b will be negative, (202) ; and these negative powers multiplied by the positive powers of a, will cause the 2d, 4th, 6th, Sec. terms to become negative. 7 BINOMIAL THEOREM. rjo EXAMPLE II. To find the 5th power of the binomial a — x. The literal' factors without the coefficients, are a s , a i x, a 3 % 2 , a 2 z 3 , ax*, x 3 . By computing and inserting the coefficients, with the signs, we have a 5 — 5a 4 x+ 10a 3 x 2 — 10a 2 x 3 -f-5a:z 4 — x 5 . EXERCISES On the Powers of Polynomials. 1. Find the square of a — x. Ans. a 2 — 2 aP-\-x 2 . 2. Find the cube of a-\-y. Ans. a 3 p3a 2 yp3ay 2 -\-y 3 . 3. Find the cube of ap2b. By applying the principles of the Binomial Theorem, we obtain a 3 P3a 2 .2bp3a{2b) 2 P(2b) 3 ; which may be developed into Ans. a 3 p8a 2 bp\2ab 2 p8b 3 . 4. Find the square of 3« + y. Ans. 9a 2 p6aypy 2 . 5. Find the cube of 2a — 3x. Ans. 8 a 3 — 36ffl 2 x-f54 Ans. 2 a 3 y 3 . 11 Find the square root of -\a 4 y 2 . Ans. ±ha 2 y. 12 Find the cube root of — 8 a 3 x 3n . Ans. — 2 ax*. 13. Find the square root of $jcy 4 . A7is. d=f.r T y 2 . 14. Find the cube root of — -£aa: 3 . j. Ans. —\a 3 x. 15. Find the fourth root of 16a 4 y 2 . al;zs. ±2 ay 2 . 16. Find the cube root of — fjx 2 . •1 Ans. — 17. Find the fourth root of pt-ay 2 . 1 1 Ans. d= 4a 4 yh 18. Find the fifth root of — a 10 a: 2 y. 2 1 Ans. — a 2 x s y 5 . 19. Find the sixth root of a 3 x 2 y. 1 4 I ±a'-x 3 y 6 . 20 * Find the sixth root of ha 2 y 3n . 1 1 > Ans. z tb 6 a''y i ' EVOLUTION. 135 Roots of 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 2 J r 2ab-\-b 2 . The required Root, we already know, is a-\-b, since the square of this binomial is the given trinomial : — our object is to show that the root a + 5 would be found by the foregoing Rule. a 2 + 2ab + b 2 (a + b a 2 2 a-\-b) 2ab + b 2 2 ab + b 2 ■' The first term a of the root, is the square root of a 2 , the left hand term of the given polynomial. Subtracting the. square of a , we have the remainder 2 ab + b 2 . We have now to discover a divisor of this remainder, which will give, for a quotient, the next term b of the root. 2a, that is, twice the root already found, divided into 2 ab, gives b ; and b annexed to 2 a makes the divisor 2 a-{-b. This divisor multiplied by b, produces 2ab-\-b 2 , — 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 4 — 4a 3 +6a 2 — 4a+l- Ans. a 2 —2a+ 1. 2. Find the square root of the polynomial a 4 -j-4a 3 z-h6a 2 x 2 -\-4ax z -{-x* . Ans. a 2 + 2ax+z 2 . 3. Find the square root of the polynomial 1 — §y-\-\2>y 2 — 12y 3j r4y*. Ans. l—3y-i-2y 2 4. Find the square root of the polynomial 4x 6 — 4a: 4 + 12a; 3 + a: 2 — 6x+9. Ans. 2x 3 — ar+3. 5. Find the square root of the polynomial 4a 4 + 12a 3 x+ 13a 2 x 2 ~i~6ax 3 +x 4 . A?is. 2a 2 + 3aa:+a; 2 . 6. Find the square root of the polynomial 1 — 6a+13a 2 — 12a 3 -f-4a 4 . Ans. 1— 3 2 + Z> 3 ( a+5 a 3 3a 2 + 3a6+5 2 ) oa 2 b-\-'3ab 2 -\-b 3 3a 2 6+3a6 2 +d 3 The first term a of the root, is the cube root of a 3 , the left hand term of the given polynomial. Subtracting a 3 , we have the remainder 3a 2 6 + 3ad 2 +6 3 . We have now to find a divisor of this remainder, which will give, for a quotient, the next term b of the root. 3a 2 , that is, 3 times the square of the root already found, divided into 3 a 2 b, gives b. Adding 3 ab, and also b 2 , to 3a 2 , the completed divisor is 3a 2 +3a6 + b 2 . 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 a 6 — 6a 5 + 15a 4 — 20a 3 + 15a 2 — 6a+l. Ans. a 2 — 2a+l. 2. Find the cube root of the polynomial 1 — 6x+21x 2 — 44x 3 + 6ox 4 — 54x s + 27x 6 . Ans. 1 — 2x+3x 2 . 3. Find the cube root of the polynomial 8 — 12x+30x 2 — 25x 3 + 30x 4 — 12x 5 + 8x 6 . Ans. 2— x+2x 2 . 4. Find the cube root of the polynomial 8 + 36?/ 2 + 24 i y+32?/ 3 + 6y s +?/ 6 + 18?/ 4 . Ans. 2+2 ypy 2 5. Find the cube root of the polynomial 9a 4 — 3a 5 — 13a 3 +a 6 — 12a + 8+ 18a 2 . Ans. a 2 — a + 2. 6. Find the cube root of the polynomial 21x 4 — 44x 3 — 54x + 63x 2 — 6x 5 +x 6 + 27. Ans. x 2 — 2x+3. 7. Find the cube root of the polynomial y 6 — 9«/ 5 +39i/ 4 — 99?/ 3 + 156^ 2 — 144i/+64. A/iS. y 2 —3y-rl 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 left, — these periods will correspond, respectively, to the units, tens, hundreds , &c., 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 ; apd 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 the first part + 3 times the square of the first X the second + 3 times the first X the square of the second + 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 + 6) 3 =a 3 + 3t 2 u-{-otu 2 +w 3 , (224). Hence the cube 64, — which is 64 thousand , — of the 4 tens, sub- tracted, leaves 27125 = 3£ 2 w+3£w 2 +2< 3 . By the Rule 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 times (-4 tens) 2 =4800. 144 EVOLUTION. In dividing we omit the 00, and at the same time exclude the 25 in the corresponding places ef the dividend 27125. By the Buie just referred to, the quantity to he added to complete the divisor 3 t 2 , is (3t+w)w=(3x4 tens +5)x5 = 625 ; that is, the product which arises from annexing the last figure in the root to 3 times the other part of the root, and multiplying the result hy the last figure. Suppose now that the given number contains three periods , and its root, consequently, three figures. Begarding the two figures already found as constituting the first part of the root, the next incomplete divisor, hy Buie XIX, would be I 'St 2 +{3t-\-u)u-\-(2>t-\-u)u-\-u 2 ; which consists, first, of the last complete divisor 3Z 2 + (37 +m)m; secondly, of the product (3 t-\-u)u which was added to 3 1 2 to complete that divi- sor ; thirdly, of the square of the last figure u in the root. In like manner the Buie may he shown to be applicable, whatever be the number of figures in the required root. Cube Root of Decimals. (226.) In extracting the Cube Boot of a Decimal Fraction, the pe riods must be taken from the decimal point towards the right, and 0 or 00 must be annexed, if necessary, to complete the last period. The last period must be complete, 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. EVOLUTION. 145 EXERCISES On the Cube Root of Numbers. 1. Find the cube root of 262144, and of . Ans. 64; and 2. Find the cube root of 2406104, and of Ans. 134; and 3. Find the cube root of 22906304, and of -££\XT 2 - Ans. 284 ; and 4 . Find the cube root of 479.2735, and of Ans. 7.825’; and 20.309’. 5. Find the cube root of 5371.3745, and of 3059^g-. Ans. 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 Ans. 16.34’ ; and 8. Find the cube root of 20796875, and of 3511^4j£. Ans. 275 ; and 15.20. 9. Find the cube root of .202262003, and of f-g-f-f. Ans. .587 ; and A-|-. 10. Find the cube root of 103823, and of 2460-|. Ans. 47 ; and 13.5. Extraction of the nth. Root. (227.) Any Root whatever of a Polynomial might he 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 square 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. [ 146 EVOLUTION. RULE XXL ( 228 .) To Extract any Root of a Polynomial. 1 Uunderstanding the order of the root to he denoted by n, — ar- range the Polynomial according to the powers of one of its letters, and take the nth root of its left hand term, tor the first term of the root. 2. Subtract the nth power of the root found from the given Poly- nomial ; and divide the remainder by n times the (n — 1) power of this root, for the second term of the root. 3. Subtract the nth 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 nth power of the root becomes equal to the given Poly- nomial. This Rule may also he 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 4 = 16 4x 2 3 =32 ) 144980 .... Exclude 980 in dividing. 23 4 = 279841 4x23 3 =48668 ) 251390625 . . Exclude 625 in dividing. 235 4 = 304980 0 6 25 •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 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 2 are perfect squares, having the exact square roots 2 and 3 a ; while 2 a and 8a; 3 are imperfect squares, since they have no exact square roots. In like manner 8 and 27a 3 are perfect cubes; while 9 and 25a: 2 are imperfect cubes, since they have no exact cube roots. The Polynomial a 2 -\-2ab-\-b 2 is a perfect square, whose root is a-\-b ; (218); and a 3 — oa 2 xp3ax 2 — a: 3 is a perfect cube, whose root is a — x. • Irrational or Surd Quantities. (230.) A Rational quantity is one which can be accurately express- ed without any indicated root ; as 2, 3a, or -fa;. An Irrational or Surd quantity is one which can be accurately expressed only under the form of a root, — being the indicated root ol an imperfect power. i Thus 2 2 is an irrational quantity, since the exact square root of 2 cannot he determined. Also a * and a;^ are irrational quantities. By the common Rule for the square root of numbers, we should find the square root of 2=1.414213’ ; hut 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 mots — are also called Radical quantities, from the Latin Radix, a root 148 IRRATIONAL OR SURD QUANTITIES. Radical Sign. (231.) The ladical sign y/ prefixed to a quantity, denotes the square root of the quantity ; and is therefore equivalent to the expo- nent -J. Thus -J a is the square root of a ; equivalent to a 2 . With the index 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 t/ a to a 4 . The radical sign may always he thus superseded by a fractional exponent; and this should be done whenever any obscurity wouli arise in calculation from using this sign. O C ^ For example, in extracting the square root of y a, we substitute the exponent and find the required root to be a&-~ 2 = a ^; ( 208 ). A quantity preceding«the y/, without + or — interposed, is a co- efficient or multiplier of the Surd ; and when no coefficient is express- ed, a unit is understood. . ' Thus 5 V a is 5 times the square root of a ; and V ax is the same as 1 V ax. Similar and Dissimilar Surds. (232.) Similar Surds are such as express the same root of the same quantity ; otherwise, the Surds are dissimilar. Thus f 3 and 5y/3 are similar ; while y/ 3 and 5 "\/3 are dissimi lar. So (a + 6)“ and oV a -{-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 ro.pt, &c., by placing the corresponding power of the quantity under the exponent or sign of the root. To express 3 a under the form of the square root , we pla ‘.e the square of 3 a, which is 9a 2 , under the exponent or sign of the square root, thus 3a=(9a 2 )^ or V$a 2 . IRRATIONAL OR SURD QUANTITIES. 149 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 1 2 * x 2 is equal to (ax) 2 ; the product of the squares of a and x, is equal to the square of the product of a and x. 1 i i And a 2 x^ is equal to (ax) 2 ; the product of the square roots of a 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 hy squaring ax, — and extracting the square root of ax , — we have (az) 2 =a 2 x 2 , (200) ; and (ax)' 2 —afx 2 , (208). These methods of illustration may he applied to any other power or root as well as to the square and square root. Thus the product of the cube 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 ; — VlX V25=Vl00 ; or2x5 = 10. y/8x 1/27^-^216; or 2x3=6. (235.) The Exponent of a quantity may he 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- strated, (213). As examples, 1 2 . 3 . 4 . a 2 = cj 4 —a 8 —a 8 ; 1 2 . 1 4^=4 4 = 16 4 =2. The following are the principal Transformations required in the calculation of Irrational or Surd quantities. L£)0 IRRATIONAL OR SURD QUANTITIES. Product of a Rational and an Irrational Factor. (236.) A rational multiplier or coefficient 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. Thus aV‘i=V a 2 oa 2 , by putting the coefficient a un- der the form of the square root , (233), and transferring the sign -f, which is equivalent to the exponent \ , from the two factors to the pro- duct 3a 2 , (234). In like manner 2 y/5 = y/4 X = V^IO. As an example of the utility of this Transformation, — suppose it were required to find an approximate value of 2y/o. By the common method of extracting the square root, we might find an approximate value of V 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 y/ 5 ; and the error thus arising will be greater as the coefficient of the Surd is greater. By taking V2Q, instead of 2\/5, the approximate value will be found independently of the preceding source of error. Surds Reduced to Simpler Forms. (237.) A Surd is simplified 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 V / 50 = 'v/25x -v/2 = 5y/2; or, 50^= 25^ X 2^=5 X 2^, (234). In dike manner, V 4a 2 +8« 3 — V 4 a 2 V \-\-2a = 2a\ l-\-2a. In these examples the given Surds are square roots, and the factors 25 and 4a 2 are accordingly perfect squares. As an example of the cube root, we have y 54= y/ 27 X y/ 2 = 3\/ 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 3Y^^~^VTGadV2a—oXiaV2a = l2aV2a. IRRATIONAL OR SURD QUANTITIES. 151 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 ff , the Surd will be reduced to its simplest form. Thus in each of the examples above given, the Surd is reduced to its simplest form. The second example would also give V 4a 2 + 8 a 3 = V 4 V a 2 -f- 2a 3 =2V a 2 + 2a 3 ; which is not the simplest form, since a 2 is still 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 power , and then resolving into factors, and proceeding as be- fore, (237). For example, = Vie = VIS ^ 6 = I ^ : 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 fraotional 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/d to the same root. Reducing the exponents ^ and | to a common denominator, we find y/5 = 5%, and ^4 = 4^, (235). Cubing the 5, and squaring the 4, according to the numerators of their new exponents, (211), we find i ,, i / 5 = 125 6 , and V4 = 16 6 . The square root and the cube 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 -f A.a. (236). 2. Find the Product of 2y[3x. 3 Find the Product of 4 -fab. 4. Find the Product of a\f^x: 5. Find the Product of 3x^7 . 6. Find the Product of ax^f 10. 7. Find the Product of 7 a -\- 1. 8 Find the Product of 2%/ 1 — x. 9 Find the Product of ( a-\-b)f2 . 10 Find the Product of (a — x)\/3. 11 Find the Product of (a+ 1)%/ x. 12. Find the Product of ( a — 1 )\/y. Atis. y/36a Ans. \f~2Ax. Ans. -\/l6 ab. Ans. oa? x. Ans. V 63z 2 Ans. V lOa 3 ^ 3 - Ans. -\/49a+49. Ans. ^/S — tx. Ans. -y/2 (a + 6) 2 . ■ Ans. \ / 3(a—x) 3 ’ Ans. Ans. \fy[a— l) 3 13. Reduce -f 24 5ay & to its simplest form. (237). To discover the greatest square factor of 245, we will divide this number successively by the square numbers, 4, 9, 16, 25, 36, 49, 64, &c., and take the largest divisor that leaves no remainder. Such divisor will be found to be 49 ; — 19)245(5. Then -f 24 5ay s = -\/49 y* f 5ay~7y 2 \/ day. 14. Reduce y/4 a 2 b to its simplest form. 15. Reduce 3'fa 3 x to its simplest form. 16. Reduce \/8y to its simplest form. 17. Reduce 2\/8 ax 2 to its* simplest form. 18. Reduce \/27a 4 to its simplest form. . 19 Reduce 5y/48 a 3 y to -its simplest form. 20. Reduce 2y/64a 4 to its simplest form. Ans. 2 a-fb. Ans. 3a 'fax A>is. 2 \/y Ans \xf2a Ans. 3 a\/ a. Ans. 20a-y/3ay. .Aws. 8a\/a. ' EXERCISES. 21. Reduce 3-\/l28x 3 to its simplest form. 22. Reduce ey / 250 y to its simplest form. 23. Reduce 2x-f 432 to its simplest form. 24. Reduce 3y\/ 13b to its simplest form. 25. Reduce \/8-{-l2a 3 to its simplest form. 26. Reduce [a-\-b)\/8 lx 3 to its simplest form. t 27. Reduce 3 a 2 x — 2a 3 to its simplest form. 28. Reduce (a — x)\/l92 a 3 x to its simplest form. 29. Reduce 4-f 4x 2 + 8x 3 to its simplest form- 153 Ans. 24x\/2x Ans. ba\/2y Ans. 24x \/ 3 . Ans. 9? f\/5. Ans. 2^2+3 a 3 . Ans. 3 x(a-\-b) Ans. 3a-\f x— 2a. Ans. 4a{a — x) \/3x. Ans. 8x-fl-{-2x. 30 /27 Reduce 3 W ^ to an i?tfegral Surd in its simplest form. (238). V- 31 Reduce 2 V 25 to an integral Surd in its simplest form. 32 Reduce 4 4ax 3 ir Ans. — -/H Q to an integral Surd in its simplest form. 33. Reduce a 2 V 3 y 3 34. Reduce 5x \J ^ v 36 Qx . Ans. — -yfWax. 11 v to an integral Surd in its simplest form. . . 2a 2 3 , - — Ans. — — -v/ 18a. % to an integral Surd in its simplest form. bx Ans. — -f 1 -\-y 2 . 2 f „ 35 Reduce — v to an integral Surd in its simplest form. 5 v a+x 18 Ans - 5(5+^ -‘ /2 (»+4 8 104 EXERCISES 36. Reduce 5^/2 and 3\/4 to Surds of the same root. (239). By reducing the exponents of the Surd factor to the common deno- minator, we find 13 * 1 . 1 2 . 1 _ 2^=2 6 =8 e ; and 4*=4 6 = 16 6 . Ans. 5^/8, and 3 y/ 16. Prefixing the rational co-efficients, we have 5 ^ 2 = 5 %/$, and 3V 4 = 3 Vl6. 37. Reduce 2y/ 5 and 5y/3 to Surds of the same root. Ans. 2^/25, and5y27. 38. Reduce a\/ 5 and x\/2 to Surds of the same root. Ans. a\/ 125, and x^/4. 39 Reduce 1 0 y/ 1 0 and 2^/'3x to Surds of the same root. Ans. lOy/lOO, and 2^32 40. Reduce 7 ^/3 y and 2 y\/ xy 2 to Surds of the same root- Ans. 7 1 'v/81 i y 4 , and 2y l ^/x 3 y 6 41. Reduce a 2 x \/\ and 3 y/ a 2 x to Surds of the same root. Ans. a 2 x\/\, and J \/a i x 2 . 42. Reduce — •(/ 2 and 5y\J — to Surds of the same root. %x £ oc Ans ■ Yx ^ 4 ’ and 5y * 43. Reduce 2 xy/ a and-^yv^ to Surds of the same root. Ans. 2x\/ a~ , and 4 ; yb y 44. Reduce k y/lO and 31 v 7- to Surds of the same root. Ans. |\/ 100, and 31^/- 3 ' 45. Reduce 77-^/J-andv-f/ — to Surds of the same root. 2 yv a 2x* y 2 • A*m. ~ 1 2/It, and - 1 - l2 /±! 2y V (J 4 2* vys ADDITION AND SUBTRACTION OF SURDS. 155 ADDITION AND SUBTRACTION OF SURDS. (240.) 1. Tlie 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). EXAMPLE. To Add together Sy/SOa and of 125 a. Reducing the surds to their simplest forms, we find 5f80a=5f 16\/Sa=20-\/5a ; and of \25a=.3f25f 5a=\5f 5a. The two given Surds have thus become similar , (232.) Adding together the coefficients 20 and 15, we find the Sum o5f 5a. _ I The Difference of the two given Surds, is (20 — 15) f 5a — 5 f 5a. The Addition or Subtraction of similar Surds is evidently nothing more than the addition or subtraction of similar monomials ; thus 20 times f 5a+ 15 times f 5a is 35 times f5a ; just as 20a+15a is 35a. EXERCISES On the Addition and Subtraction of Surds. 1. Find the Sum of 3f27 and 2f 48, 2. Find the Difference between f 50 and fl2. 3. Find the Sum of if 28 and 6-^63. 4. Find the Difference between 2\f2 and 5f 18. 5. Find the Sum of -y/ 180 and ^405. H Am. 17 f3. Arts, f 2 . Ans. 32 f 7. Ans. 6f2. Ans. \5f5. 156 ADDITION AND SUBTRACTION OF SURDS. 6. Find the Difference between -\/l 8 and 2 -\/50. Ans. 7^2. 7. Find the Sum of -\/ 12 a 2 and -\/2ria 2 . Ans. 5aV 3. 8. Find the Difference between 3-\/24x 2 and \/54 x 2 . Ans. 3z\/6 9. Find the Sum of 4^/3 a and -\/48 a. Ans. 8\^3 a 10 Find the Difference between V" 4 a 3 and -\/ 9 a 3 . Ans. ay* a. 11. Find the Sum of 3 t/ 4 and 7^/4. Ans. 10-^4 12 Find the Difference between 9-\/200 and -\/288. Ans. 18^2. 13 Find the Sum of 4-^54 and 2 ^250. Ans. 22^/2. ! 14. Find the Difference between 5\/9x 3 and 3-v/x 2 . Ans. 12x-[/x. 15. Find the Sum of 2^/161 2 and 54 a. A?is. l\/2a. 16. Find the Difference between 3 -\/ 10 and 5 \/l9. Ans. 2\/l0- 17. Find the Sum of 5-\/98x and 10-\/2x. Ans. 45-\/2x j 18. Find the Difference between 3a\/ 5 and a^/5. Ans. 2a\/5. 1 19. Find the Sum of a \/y 2 and 35 \/y 2 . Ans. [a-\-3b)y/y 2 . 20. Find the Difference between 5\/5 and 2 ay/5. Ans. (5 — 2a)\/5 21. Find the Sum of (a+ 1) 2 and -y/4rt+4. Ans. 3 (a-f-1)* . \ 22. Find the Difference between y 14-a: and 3 (1 + ^) 2 - Ans. 2 (1+ x) 1 f 23. Find the Sum of 2 (a— a;) 2 and V9a — 9x). 1 Ans. 5 (a—x)1 ; , , i 24. Find the Difference between y 2+?/ and 4 (2/+ 2) 3 . Ans. 3 (24-3/)* 25. Find the Difference between 4 (l + a: 2 ) 2 and 4 v^x 2 + l. 4 1 Ans. 4 (l-)-x 2 — a; 2 + y). MULTIPLICATION ANl> 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 same fractional exponent or radical sign. 2. Surds of different roots may he reduced to equivalent ones of the same root, (239), and then multiplied, or divided, as above. But 3. Any two roots of the same quantity 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 from that of the dividend. example. • To find the Product of2'y/l0x3'v/2. Since it is immaterial in what order the four factors are taken, we may take them in the order, 2x3Vl0x a/2; which gives the Product 6 a/20, (234), =12 a/5, (23?^ The Quotient of 2 a/10-^-3a/ 2 is -|a/ 5, 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 a/ 8 X 3 y/5. Ans. 30a/10. 2. Find the Quotient of 6 a/54-^3a/2. Ans. 6 a/3. 3. Find the Product of a/108 X 2 a/6. Ans. 36 a/2. 4. Find the Quotient of 2a/96-^- a/54. Ans. 2§. 5. Find the Product of 3 ffoax X 4:\/20a. Ans. 120a a/x. 6. Find the Quotient of4A/l2an-2A/6. Ans. 2a/2 a. 7. Find the Product of a/3 ax X 3 a /ax. Ans. 3axff3 8. Find the Quotient of 12x 2 =3a/ 4. Ans. 2a: a/3. 9. Find the Product of %/\8 X 5a/4. Ans. lO-yAh 10. Find the Quotient of 4 3 v /72^2yi8. Ans. 2 z ^j 4. 158 MULTIPLICATION AND DIVISION OF SURDS. 11. Find the Product of 5 a X 3 \/ a. By reducing the surds to the same root , we obtain 5yV and 3%/ a?, (239). These are to be multiplied together as before. Ans. 1 5\/a s . But the given roots of the same quantity a. may also be multiplied into each other, by adding together their fractional exponents ^ and -J. Thus, 5a'- t x3a :i = l5a 6 — l5^/a 5 , 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 X 2 x. 13. Find the Quotient of 2q/34-5^/3. 14. Find the Product of 4q/3 X 2^/2. 15. Find the Quotient of 8-y/a — 4 V a. 16. Find the Product of ffaxX3y / 'ax. 17. Find the Quotient of 4 x 3 -^2,ff x. 18. Find the Product of 5 X^V 10. 19. Find the Quotient of a 3 \. Ans. 6^^- Ans.*^ \/3. Ans. 8-^108 Ans. 2 1 \/a. Ans. 3Va 3 a' 3 . Ans. 2x\J%. Ans. i-\/250. Ans. a 2 \/^. 20. Find the Product of (3 + 2-y^) X (2 — 5). 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'v/o 2— 5 G + 4-/5 — 3-y/5 — 10 6+ 5 — l0 = -v/5 — 4. E ach term of the multiplicand is multiplied by each term of the multiplier, and the partial products are then added together. Observe that 2 q/5 X — 5— — 2 \/25 = — 10. 21. Find the Product of (2 + 3^2) x(l+5-\/2). M«s. 13-/2 + 32. RATIONALIZATION OF SURD DIVISORS. 159 22. Find the Product of (4- ff?>) X (2 + 3^3). Ans. 1 0 -\/ 3 — 1. 23. Find the Product of (5 + 2-y/6) X (1 + 2 -y/ 6) . Ans. 12-\/6 + 29. 24. Find the Product of (1 — 4-^/7) X (3 — 3 -\/l). Ans. 87-1 5 -v/7. 25. Find the Quotient of ( -\/ 20 + 1 2 ( -\/ o + -y/ 3) . Ans. 2. Rationalization of Surd Divisors. (242 ) In computing 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 -7- = — — — , in which the divisor is rational. V3 -t/9 3 The value will therefore be found by taking 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 iind 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. A , X-J- i 2. Thus a 3 multiplied by a 3 , or a 3 Xa 3 =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 sisn changed. Thus (-/ 3+-/ 2)x(/3— / 2) = 3 — 2=1. 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. 160 RATIONALIZATION OF SURD DIVISORS. 3. A trinomial containing irrational square roots will produce a binomial Surd by being multiplied into itself with a sign changed ; 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 2+-v/3 1. Reduce ~ — to a Fraction having a rational denominator. . 2^9+3 5 _ 3 2. Reduce t0 a Fraction having a rational denominator. 15 — 5 \/ 5 3. Reduce 4. Reduce 5. Reduce 2 a 2-fl Ans. to a Fraction having a rational denominator. Ans. \a-\-2afl -3 3x ■fa — V x to a Fraction having a rational denominator. Ans. 3xf a-\-3xf x a—x 10 t+ Vio to. Reduce a-V 2 4/2 to a Fraction having a rational denominator. Ans. iq^-io-y/io a 2 — 10 ‘ to a Fraction having a rational denominator Ans. VS -2 7 . Reduc-e- f 3 + f 2 .-\- 1 to a Fraction having a rational denominator Arts. 4+2^2— 2-^/6. INVOLUTION AND EVOLUTION OF SURDS. 161 INVOLUTION AND EVOLUTION OF SURDS. (244- ) The Powers and Roots of irrational quantities are obtained, or indicated , according to the general principles of Involution and Evolution which have been established in the preceding Chapter. We present here however 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 2 — b is a perfect square. To determine the method to he pursued in this case of evolution, we must find Formulas for the Square Root of and ^/24. 5. Find the Cube, and also the cube root, of 3a q/3. Ans. 81a 3 q/3, and y / 27a 3 . 6. Find the Square, and also the square root, of x 2 q/ 5. Ans. a; 4 q/25, and x \/lT. 7. Find the Square, and also the square root, of 3 q/n+T. Ans. 9 (a + 1), and^/9 (a+l). 8. Find the Square, and also the square root, of 4 q/l — x. Ans. 16 ^/(l — x) 2 , and 2^1 — x 9. Find the Cube, and also the cube root, of 2 q/a 2 — x. Ans. 8 q/(a 2 — x) 3 , andj/4 (a 2 — x). 10. Find the Square, and also the square root, of 3 y /2 — a. Ans. 9 12 -a). and ft 9' (2 -a). 11 * 164 EXERCISES. 11. Find the Square of 2+/3, and of 2-/3. These squares may he found by multiplying each binomial into itself, or, more readily, by applying the propositions relating to the squares of the sum and the difference of two quantities. Ans. 7 + 4/3, and 7-4/3. 12. Find the square of 3 + 2/5, and of 5 — 3ff~- A)is. 12/5 + 29, and 43 — 30/2^ 13. Find the Square of V" 5 + 2/ a, and of V 3 — 5/ x. Ans. 5 + 2/a, and 3— 5/z. 14. Find the Square of / 2+3/ 5, and of / a — 2x x. Ans. 6/10 + 47, and a — 4^/ax+4z 3 . 15. Find the Cube of ^1+5/4, and of Va — 3/a. Ans. 1 + 5/4, and a — 3/a. 16. Find the Square of /3 + a/3, and of /3 — 3/a- Ans. 3 + 6a + 3a 2 , and 3 — 6/3^_|_g a> 17. Find the Cube of 2 + 2/a, and of 2 — a/2. Ans. 8 + 24a + (24 + 8a)/a, and 8+ 12a — (12a + 2a 3 )/2. 18. Find the Cube of (/ a _ and of (/l — Ans. /a+/ 2, and v 1 — /yT 19. Find the Square roo£ of a + 2 /ax +x, (218). Ans. /a+ / x. 20. Fijrd the Square root of a + 2/a+l, and of 3 + 2a/3 + a 2 . Am. /a + 1, and /3 +a. 21. Find the Square root of x — 2/x+l, and of 5 — 2x/ 5 + .r 2 . Ans. -/x — 1, and /5— x. 22. Find the Square root of 23 + 8/7, ( 245 ). Formula A. Ans. 4+/7. 23. Find the Squaie root of 194 8/3, and of 7 — 2/l0- +as. 4+ /3, and /o— v 2- IMAGINARY QUANTITIES. 165 IMAGINARY QUANTITIES. (246.) An even root of a negative quantity is impossible, and the symbol of such a root is therefore called an imaginary or impossible quantity, in contradistinction to real quantities. Thus yj — 4, the square root of — 4, is imaginary, since there is no quantity whose square is — 4 ; and for a like reason -y/^16 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 fir om 5 shall leave 9. If x represent the required number, the Equation of the problem will be 5 — a; 2 =9. From this equation, x 2 = — 4 ; and hence x = yj — 4 The value of x being imaginary or impossible, shows that the problem is impossible ; 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. Ifa6 IMAGINARY QUANTITIES. (243.) An Imaginary Quantity may always be resolved into the like root of ( — 1) multiplied into the like root of a positive quantity which is equal to the negative quantity in the given imaginary. Thus —4 being equal to— 1x4, we have y/ — 4 = y/ — 1 Xy/4 1 and — a being equal to — 1 X a, we have y / — a = y/ — 1 x y/a.(234-) By means of this transformation it may be shown, that (249 ) The Product, of two imaginary squire roots is the negative square root of the product of the two quantities under the -y/, if the given roots are preceded by like signs, -f- or — ; otherwise, it is the vositive square root of that product. For example, -y/ —a . y / —b= — yfab; and, -y / —a . — -y/ — b— + i/ab. On general principles of calculation, the Product of the square roots of —a and —b, is equal to the square root of the product ab . which would be -\-yf ab, or else — -y f 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 composition of the quantity whose root is considered. When a 2 , for example, is known to have been derived from ax a, the square root of a 2 is a, and not — a. V To determine the sign of -y fab in the first example, we have \/ — a . y/ — b — yj — 1 . y /a . y/ — 1 . y / b, (24S) = y/ — 1 ■ y / — 1 . y/ ay/ b — {V — l) 2 y^ab= — ly/~ab, or — y fab. In the second example, we have yf—a . — y/^Q = y/^l . y/a . — y/^1 . y /b, (248), = y/ — "l . — y/ — 1 . y/ ay/ b — — (y/ — l) 2 y fab— — ( — l)y rabzii-fyfab. In this example the square (y/ — l) 2 has the sigu — before it, be cause it results from multiplying y ' — 1 by — \ — 1, (42); and — ( — 1) becomes +1 by changing the sign in subtracting. 167 CHAPTER X. QUADRATIC AND OTHER EQUATIONS. (250.) A Quadratic Equation, or an equation of the second de- cree, is one in which the highest power of the unknown quantity is its square, or second power, as 3z 2 = 12; or 3x 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 oxy-\-y=a is an Equation of the second degree, its first term containing the unknown factors xy. And x 2 y — y=b 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 hut 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: 3 = 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 3x = 10 is an o.ffected, and also a complete quadratic , x 3 — 2.r 2 =7-5, or x 3 — 3.r=110, is an affected cubic; and x 3 -\-oX 2 | 4.t; = 28 is a complete cubic. 168 GENERAL PROPERTIES OF EQUATIONS. Roots of Equations. (252.) A Root of an Equation is a value of the unknown quantity 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 3*= 15, the value of the unknown quantity x is 5 ; then 5 is the root of the equation. It is evident that in a Simple Equation there can be but one valu« of the unknown quantity that will satisfy the equation. An equation of the 1st degree has therefore hut one root. General Properties of Equations. 1. Divisors of an Equation. (253.) If a he a root of an Equation of any degree, containing hut one unknown quantity, x ; the equation — with all its terms transposed to one side — will be divisible by a: — a. Let a be a root of the Equation x 2 -\-mx=s ] then will the equation x 2 +mz — s = 0 be divisible by x -a For let r be the remainder , if any, after the quotient q has bee* obtained ; then will x 2 -\-mx — s = (x — a)q-j-r — 0 ; the dividend being equal to the remainder added to the product of the divisor and quotient. But a being a value of x, (252), we have x— a = 0 ; then the {x—a) X q is 0, (43) ; and consequently r = 0 ; that is, the division will leave no 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 x—a — 0, or that a is a value of r. GENERAL PROPERTIES OF EQUATIONS. 169 2. Number of Roots of an Equation. (255.) Every Equation containing but one unknown quantity, has just as many roots 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 3 -\-mx 2 -\-nx-s. Transposing s to the first side of the equation, we have x 3 +mx 2 +nx — s=0. Dividing this equation hy x—a, (253), we shall obtain an equation of the second degree, which may be represented by x 2 -\-px — q= 0. Let b 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 — 0. 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 3 +ffli 2 -\-nx — s. The original equation is thus resolved, representatively, into (x — a)(x — b){x — u) =r0. Since this equation is divisible by each of these three binomial fac tors, it follows that a, b, and u are three roots of the equation, (254). And since the given equation — being of the 3d degree — cannot he 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 fac.toxs, each containing a root of the equation. Two or more of these roots may be equal to each other. In the Equation x 2 — 5x + 6 = (x — 2)(x — 3) = 0, the two roots are 2 and 3. In the Equation x 3 — 7a; 2 + l6a: — I2 = (a: — 2)(x — 2)(x — 3) = 0 the three roots are 2, 2, and 3. 170 . 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.) Ear the Solution of a Pure Equation. 1. Reduce the Equation to the form x n =s ; in which x n must he positive. 2. Extract that root of both sides of the equation which corresponds to the power of the unknown quantity e x A M p l e . To find the value of x in the Equation x 2 t-5 = 2x 2 — 23. 4 Clearing the equation of its Fraction, we have z 2 + 20 = 8a: 2 —92. Transposing, and adding similar terms, we find —7x 2 = — \12. Dividing both sides of this equation by the coefficient — 7 , x 2 = 16. 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 Equation of the second degree, the two values of the unknown quantity will always be equal , with con tranj 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 -fx-j- y/ 1 -\-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. J 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 root. To rationalize x in the Equation Vx+1 —a—b. By transposition, V ze+ 1 =2 — ^ x—\y/2>2- Ans. z=50 6. Find the value of x in the equation 3 + l/^+lX V x — 4 = 10. Ans. x=±\/65. 7. Find the value of x in the equation a-{--\/x — 3 X V a;+3=4«. 10 8. Find the value of x in the equation V^+v^+5‘=pj=- 9. Find the value of x in the equation 4-v/6.r — 9 \/6x— 2 4-\/6.r+6 -v/6x+2 *10. Find the value of x in the equation Vx+c b V Ans. x=±2>-fa 2 + 1. \Zx+c—V x — c = 3 . Ans. x —\ | Ans. cb 2 +c Ans. x— 2b ' SOLUTION OF COMPLETE EQUATIONS. 173 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 he 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 2 -\-bx=s ; in which x 2 must he positive. 2. Awd the square of half the coefficient of x, in the second term, to each Mde of the equation : — the first side will then be a perfect square. 3. Ei wact the square root of each side, and the result will he a Simple Equation, — from which the value of x may readily he found. example. To find the value of x in the equation 2>x — 5 5x- 2x-\ — =5x- 2 i x ~ 3 Clearing the equation of its fractions, transposing, and adding sim- ilar terms, we shall find — 3x 2 f-14tX= — 5. Dividing both sides of this equation by — 3, „ 14* 5 X ~ ~ = 3' Adding the square of half the coefficient of x, in the second term, to each side, — which is called completing the square , 2 14a; 49 _ 5 49 _ 64 X 3~ + '9' - 3 + T — ~9' Extracting the square root of each side, 7 8 x — — = ± - . 3 3 m 7 8 7 8 1 Whence x= — — =5 ; or *= = . 3 3 ’ 3 3 3 ( 255 ). 174 SOLUTION OF COMPLETE EQUATIONS. Either of these two values of x, 5 or — will satisfy the given equation ; and each of the binomials x — 5 and 2 +-^ will divide the equation * 2 “^r-§=°- (253) - ! By performing the division it will he found that the left hand side of this equation, is the product of the two binomials ; that is, (x 5)(x-\-jsr)=x 2 ~ ~ =0. On the preceding Rule we remark, 1. The method of completing the square in the first member, re- ji suits from the composition of the square of a Binomial. Thus the square of the binomial a-\-b is a 2 -\-2ba-\-b 2 , in which b t is half the coefficient of a in the second term. 2. If in reducing the equation to the required form, the first term 1 1 should become — x 2 , the signs of all the terms in the equation must be: I changed (117), before completing the square, — otherwise the root of 1 the first side would be imaginary , (246). 3. The square root of the first side of the equation — after the eom-1 pletion of the square — will always be the square root of the first termfi I f- or — half the coefficient of x in the second term, according as the I second term is + or — Hence, without completing the square on the first side, we may 1 shorten the solution, by observing that 4. In an equation of the form x 2 -f-6.r=s, the value of x is half the co-efficient of 2 in the second term, taken with a contrary sign, + the square root of (the second member of the equation + the square of said > half co-efficient). 14# 5 Thus from the equation x 2 T — — — , we immediately find *=Z+ x /E^_l+ x / 6 i = Z+? 3 V 3 ' 9 3 — -V 9 3-3 EXERCISES. 175 EXERCISES On Quadratic Equations with One Unknown Quantity. 1 . 2. 3 ' 4 . 5 - 6 , 9 . 10 . 11 . 12 . 13 . 14 . 15 Find the value of x in the equation x 2 — 15 = 45 — 4a;. Find the value of x in the equation 2 2 + 10 = 65 + 62. Find the value of x in the equation 22 2 + 82— 30 = 60. Find the value of x in the equation 3a: 2 — 32+9=8^. Find the value of in the equation 5a: 2 + 4x — 90 = 114. Find the value of x in the equation +c 2 — ^2+2 = 9. Find the value of x in the equation 22 2 + 122+36 = 356. Ans. 2=10 or — 16 F ind the value of x in the equation . .„ 36— x 4a: — 46= . Ans. 2=6 or — 10. Ans. 2 = 11 or — 5. Ans. 2=5 or —9. Ans. X—\ or I-. Ans. 2 = 6 or — 6|. Ans. a: = 4 or — 3+ Find the value of x in the equation 8tc 2 + 6 = 72+ 171. Find the value of x in the equation 5 ,_ 23 = 2 -^. x Find the value of x in the equation 32 2 + 6 = 32 + 5 ^-. Find the value of x in the equation 2 2 x T “ 3 + 2 °*= 42 * Find the value of x in the equation 8 — x x — 2 ( 2^—11 ~2~ = ~6 + 2 — 3 Find the value of x in the equation lx — 8 2+4 = 13 . x Find the value of x in the equation 2 — 3 a 9 (b — a) Ans. 2=12 or — f-. Ans. 2=5 or — 4^. Ans. 2=5 or — 1. Ans. 2=-^* or J. Ans. 2=7 or — 6^. Ans. 2=6 or Ans. 2=4 or — 2 2 Ans. 2=3 b, or 3 (a—b) 176 EXERCISES. Another Method of Solving Quadratic Equations. (261.) A Binomial of the form ax 2 -\-bx will he made a perfect square by multiplying it by 4 times the coefficient of x 2 , and adding the square of the given coefficient of x. Thus {ax 2 -\-bx)\a-\- b 2 =4a 2 x 2 -\-\abx-\-b 2 — {2ax-\- b) 2 , (59). To apply this principle to solution of the Equation 3 a: 2 — 2x=G5. Multiplying both sides of the equation by 4 times the coefficient 3, a nd adding the square of the coefficient 2 to both sides, we have 36a: 2 — 24* + 4 = 780 + 4 = 784. Extracting the square root of both sides, we find i 03 | 2 6a: — 2= ±28; which gives x=z~ — — — 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: 2 , + or — the given coefficient of x in the second term, according as this term is -f- 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 2 +bx=s to a Simple Equation. 1. Double the coefficient of x 2 , and divide the first member by x. 2. Multiply the second member by 4 times the given coefficient oi x 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 3a: 2 — ox-50, we immediately obtain 6a; — 5 = ± -v/50 X 12 + 2o= ± -y/625. : When the coefficient of x 2 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 Qmadratic 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 2 , or in completing the square, according to that method. EXERCISES. 16. Find the value of x in the equation 2>x 2 -\-2x — 9 = 76. 17. Fin'd the value of x in the equation 2a: 2 — 14a:+2 = l8. 18. Find the value of x in the equation x 2 — 122+50 = 0. 19. Find the value of x in the equation \x 2 — 12^-=4-a;+ 15^-. 20. Find the value of x in the equation 3a: 2 — f=5i+2a;. 21 Find the value of x in the equation 5x 6 22. Find the value of x in the equation . t+1 =13-— . x 23. Find the value of x in the equation * 2 - — +1=0. 177 Ans. 2 = 5 , or — 5|-. Ans. x=8, or — 1 . Ans. x=6dz\/ — 14, Ans. 2 = 8 , or —7. Ans. a;=|±^yi9. Ans. x=%, or + Ans. 2 = 4 , or —2. ju y + — — 34|=0. Ans. 2 = 9 , or —Ilf. 24. Find the value of x in the equation n X 2 X , , cc 2 Ans. a^-jLijL/97. Ans. 2=3, or 1 { 2 . Ans. x—6±^ — 4. 2 3 3 25. Find the value of x in the equation 10 14 — 2a; — o± x x 2 9 ‘ 26. Find the value of x in the equation x 2 — — + 32+3 = 13. 27. Find the value of x in the equation „ 17.2 \lx x 2 -j = 4. Ans. X— — 4, or — 8 4 4 2 28. Find the value of x in the equation, 3a: — 3 3a: — 6 ox — =2*-) — . Ans. x =4, or -1. a: — 3 2 29. Find the value of 2 in the equation 2+4 7 — x 42+7 ~ 3 ^c~3 = ~ 1 ^ — 9 — ‘ ^ nS ‘ x ~ 21 ’ or 5 30. Find the value of a: in the equation x+a 32 x — a = 0 . Ans. x=d=ay ' — + 178 EXERCISES. Equations in which the Unknown Quautity is contained in a Surd Expression. 31. Find the value of x in the equation 5 + V' a;3 + 36 = 15 - By transposition, y'a 3 +36 = 15 — 5 = 10. Squaring both sides, a; 3 + 36 = 100; (257). from which, a: 3 =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 ;r = 4, which is one value of x. 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 x 3 — 64 = 0 by a: — 4, we find x 2 +4a:+ 16 = 0, or x 2 -j-4x= — 16 ; hence x— — 2±y/ — 12. We have thus found 3=4, or — 2-b-y/ — f2, or — 2 — \/^12 ; the first value being real, the other two imaginary . Each of these imaginary values of x, as well as the real value 4, will satisfy the equation a; 3 =64. Thus x being equal to —2+ V — 12, we have, (247), (249), x - 2 = ( - 2 + \2 ) 2 = 4 - 4 /Hl2 - 1 2 = - S - 4 /^12 . and a: 3 =( — 8 — 4-y/ — 12)( — 2-fi -y/ — 12) = 16 + 48 = 64, In like manner the other imaginary value of x may he verified. 32. Find the value of x in the equation 6 + V3*+4 = ll. Ans. x=7. 33. Find the value of x in the equation 24— V2x 2 + 9 = \5. Ans. x= ±6. 34. Find the value of x in the equation 20 — ■ V /a; 3 + 40=4 Ans. .r=6 or— 3+V —27, or —3 — 7-27. EXERCISES. 179 35. Find the value of x in the equation x+ y / X = 20. From this equation Ave shall find a; = 25 or 16. The value 25 will not satisfy the original equation if the square root of x be restricted to its positive value ; but this root is ± y'.c, and 25 satisfies the equation for the negative root. Two values of a: may be found in each of the next three Exercises ; hut only that value is given in the Ans. Avhich corresponds to positive roots in the given Equation. 36. Find the value of x in the equation 2Vx+ VxAA = l3. 37. Find the value of x in the equation 2Vx-{-V2x+l = ■\Z{2x-\- 1) 38. Find the value of x in the equation 4 V 2 + 16=7 V^ar+lfi — x — 6. 39. Find the value of x in the equation Ans. 2=16. Am. 2=4. Am. a=9 =V4 q- V 2x 3 -f- 2 2 . Ans. 2=12, or 4. 40 . Find the value of x in the equation 2+y'(2 2 — 9) ( 2 - 2)2 = By rationalizing the denominator, (243... 2), we shall find ( 2 - 2)2 = ( - x ~ V ( x2 ~ 9 ) . 2 _(^+ v (* 2 - 9)) 2 Extracting the square root of each side of this equation, we have 2-2 = x + qA* 2 — 9) Ans. 2=5, or 3 41 . Find the value of x in the equation 2 2 — 6z+9= Ans. 2 = 5. or 4 x — -\/{x 2 — 16) 42 . Find the value of x in the equation ^/2 3 + 37x(2 3 + 37p = 64. Ans. 2 = 3, or— f±^A/ — 3. 180 EXERCISES. Equations of a Quadratic Form with reference to a Poioer or Root of the Unknown Quantity. (263.) Any Equation containing the unknown quantity * in but two terms — with its exponent in one double its exponent in the other — is a Quadratic with reference to the lower power of x ; and the value of such power may be found accordingly. To find the value of * in the equation * 72 +4*^=21. 1 . i The higher fractional power x' 1 is the square of the lower a: 4 ; and the equation is therefore quadratic with reference to a: 4 . Completing the square, we have *^+4* 4 +4=21 +4 = 25. Extracting the square root of each side, *^+2= ±5 ; o. from which'* 4 = 3, or — 7. By raising each of these values to the 4th power, we find *=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 4* 4 is therefore — 28. When * is in a fractional power, in the following Exercises, only f that value will be given in the An's, which satisfies the given form of the Equation. — Imaginary values of x are also omitted. Ans. *=±4. Ans. *=^/3. j Ans. *=4^/6. Ans. *=3-l-j-. Ans *=S 43. Find the value of * in the equation x 4 — 2* 2 + 6 = 230. 44. Find the value of x in the equation z 6 + 20* 3 — 10 = 59. 45. Find the value of * in the equation 2* 4 -* 2 + 20=23. 46. Find the value of * in the equation 3aQ— 5* 4 = — 1+ 47. Find the value of x in the equation 6x3— 5*3+1184 = 0. EXERCISES 181 48. Find the value of x in the equation (a;-j-12)^+(a:+12) 4 = 6. In this equation we must regard the binomial (£+12) as the un- known quantity ; and to simplify the operation we may represent this binomial by y. Then ?/' 2 +?/ 4 = 6 which gives?/ 4 = 2 or — 3. By restoring the binomial value of y, we have (a:+12) 4 = 2, or —3. Ans. x=4. 49. Find the value of x in the equation (2a;+6)^— 6 = — (2*+6)^. 50. Find the value of x in the equation x 2 + 11 + • v /a: 2 + ll=42. 51. Find the value of x in the equation (2x — 4) : =1 + (2x — 4) 4 ' Ans. x—5. Ans: x= ±5 Ans. x=3, or 1. 52. Find the value of x in the equation 4a: 4 - -4a: 3 + - =33. This equation may be reduced to the form of a quadratic , thus. — The first two terms of the square root of the first side will be found to be 2a; 2 — x ; and the remainder will be — x 2 ^ , which may be put under the form — i-(2x 2 — x). Now the square of the root found, and the remainder , are together equivalent to the first side of the equation ; hence we have (2x 2 —x) 2 —\(2x 2 —x)=33. Ans. x=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 the part of the root thus found 182 EXERCISES. 53. Find the value of z in the equation x 3 — 8a 2 + 19a; = 12. Transposing the 12, and multiplying by x, we have x 4 — 8a 3 + 19a; 2 — 12a = 0. We have now a Biquadratic equation which may be reduced to the form of a Quadratic by the method just exemplified. We fi.ad the first two terms of the square root of the first side of the Equation, by the common Rule. x 4 — 8a; 3 + 19a 2 — 12a: (a 2 — 4a; /y.4 2a; 2 — 4a; /-8a 3 + 19a 2 / — 8a; 3 + 16x 2 3a; 2 — 12a:. The remainder, 3a: 2 — 12a:, may be resolved into 3 (a: 2 — Ax), 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. a=4, 3, or 1. 54. Find the value of x in the equation x 4 — 2a 3 +a = 51l2. Ans. a = 9 55. Find the value of x in the equation a 4 + 2a 3 — 7x 2 — 8a:= — 12. Ans. a=2, — 3,1, or — 2 56. Find the value of a; in the equation x 4 — 10a 3 + 35a 2 — 50a: + 24 = 0. Ans. a=l, 2, 3, or 4. 57. Find the value of a in the equation x 4 — 12a 3 + 44a 2 — 48a = 9009. A?is. a=13 60. Find the value of x in the equation a 4 — 8aa 3 + 8a 2 a; 2 + 32a 3 a=s. Atis. x—2a±^J 8a 2 ± v/s+ I6n+ PROBLEMS. 183 PROBLEMS In Pure Equations and Affected Quadratics containing but One Unknown Quantity. 1. Find two numbers such that their product shall be 750, and the quotient of the greater divided by the less, 3^. Let x represent the greater of the two numbers ; 750 then will represent the less ; and the Equation will be 750 x 2 x-. or = 34. x /o0 a From this equation we shall find 2 = 50, or — 50. Each of these values will 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. A?is. 50, and 15. 2. Find a number such that if \ and of it be multiplied together, and the product divided by 3, the quotient will be 298|-. 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 ; 9 # then 4 : 9 :: x : — , the number of yards. But without forming a Proportion, the number of yards is readily known to be of the price per yard. Ans. 27 yards, at 12s. 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 he 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 $5^ 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 x 800 — x — • : :: 9 : 25: 800 — x 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? Ans. 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? Ans. 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 linen ? Ans. $30. 14. A grazier bought as many sheep as cost him $100. After re- serving 5 of the number, he sold the remainder for $135, and gained $1 a head on them : how many sheep did he buy ? 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 ? 75, and 32. PROBLEMS. 185 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 ? Let x represent A’s capital; then 26 — x will be his gam ; and since the gain is in the compound ratio of the capital and the time it was employed, we have 12^+16x30 : 12a; :: 18 : 26— £r. 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. 4650. 19. A company at a tavern had £8 15s. 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 1 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 y' 5 — 10, and 30 — 10^/5. 22. A laborer dug two trenches, one of which was 6 yards longer than the other, for £l7 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 subtracted from the greater, the product of the ' sum and the remainder will 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 £ll, 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 eent. 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 ^9 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. 9 * 186 SOLUTION OF TWO EQUATIONS. Solution of Two Equations — One ok Both of the Second ok 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 neio 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 the 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 2x+y=10, and 2a; 2 — a?^+3z/ 2 =:54. From the first equation, we have 10 —?/ nr „ , 2 100— 20y+y 2 ), , 10?/— ?/ 2 Then 2* 2 = -1 J J ' and xy= J J . 4 2 By substituting these values in the second equation, we find 2( 100—20 yfy 2 ), 10 y-y 2 J — — + 3y 2 = 54. 4 o J 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 loiver terms ; and the solution of the equation be thus some- what simplified. Ans. x=3. or y= 4, or — J. EXAMPLES AND EXERCISES. 187 2. Find the values of x and y in the equations %-\-y — 9, and x 2 -\-y 2 =45. Ans. x—3, oi 6 ; y—6, or 3 [fF == ' Whenever x and y may he interchanged with each othei', without changing the form of the given equations — as in the preceding example — the two values of one of these letters may be taken, in re- verse order, for the- two values of the other. 3. Find the values of x and y in the equations zy —28, and x 2 -\-y 2 — 65. Ans. x=±l, or ±4; y=-± 4, or ±7- 4. Find the values of x and y in the equations 5. Find the values of x and y in the equations x-\-2y=l, and x 2 -\-3xy — y 2 = 23. Ans. x—3, or 15§; y= 2, or — 4 S 6 . Find the values of x and y in the equations x-j-4y=14, and 4x — 2y-\-y 2 — \\. Ans. x—2, or — 46 ; y—3, or 15. 2 1 Ans, x=8, or 17g • y= 6, or — 133 . 7. Find the values of x and y in the equations 8. Find the values of x and y in the equations 2 J x x+2 Ans. x—2, or 5 ; y= 6 , or 3. 9. Find the values of x and y in the equations 5 (x-\-y)=13 (x — y), and x-\-y 2 = 25. 188 EXAMPLES AND EXERCISES Solutions by Means of an Auxiliary Unknown Quantity. (265-) When the number of unknoivn factors is the same in every unknown term of the two Equations, the solution will often be facili- tated by substituting for one of the unknown quantities the product of the other into a third unk?iown quantity. 10. Find the values of x and y in the equations x 2 -\-xy=5 4, and 2xy-\-y 2 =45. The number of unknown factors in each of the unknown terms in these equations, is two. If we assume x to be equal to vy, and substitute this product for 2 , the given equations will become v 2 y 2 -\-vy 2 =54, and 2 vy 2 -\-y 2 =45, From the first of these equations we have ?/ 2 = . — — _ — and from the second y 2 — v 2 +v, 2v+l Putting these two values of y 2 equal to each other, 54 _ 45 v 2 -\-v 2v-\-Y By solving this equation in the usual manner we shall find v—2 , or - — | ; then by substituting these values, successively, in either of the expressions for y 2 , we shall find the values of y ; and since X—Vy, we may also readily obtain the value of x. Ans. a?=d=6, or q=9-\/ — 1 ; i/=± 3, or ±15-\/^l. If in the preceding example an expression for the value of x or y were obtained from either equation, and substituted in the other, the resulting equation, when cleared of radical signs, would be of the fourth degree; and we have accordingly found four values for each of the unknown quantities, (255 ) 11. Find the values of x and y in the equations xy= 28, and x 2 -\-y 2 = 65. Ans. a:=±7,or±4; y=±4, or±7. 12. Find the values of x and y in the equations x 2 -\-xy = 12, and xy — 2y 2 — l. 8 1 Ans. x— ±3, or ± — — ■> y=±l, or ± — - \/ 6 y \/ 6 SOLUTION OF EQUATIONS. 189 13. Find the values of a and y in the equations 4a; 2 — 2xy=\2, and 2y 2 -{-3xy—8. Ans. *,= ±2, or • y=dt 1, or±8-/i 14. Find the values of x and y in the equations 3 y 2 — x 2 =39, and a 2 +432/ = 25 6 — 4 y 2 . Ans. x= ±6, or ± 102 ; y — ±5, or =p59. Solutions by Means of Two Auxiliary Unknown Quantities. (266.) When the unknown quantities are similarly involved in 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 unknown quantities. 15. Find the values of x and y in the equations x-\-y = 12. and ^L+— =18. y x If we assume x equal to v-\-z, and y equal to v — z, we shall have x-\-y—2v = 12 ; and hence v=6. Then 3 = 6 + 2, and y= 6 — z. Substituting these values of x and y in the second equation, we have (6+ z )2 + (6- Z ) 2 _ 1R 6 — z 6 -} -z Clearing this equation of its fractions, (6 + s) 3 + (6— 2) 3 = 18 (36— z 2 ). By developiug both sides of this last equation, and proceeding with the solution in the usual manner, we shall find z=±2. Having now found the values of both v and z, the values of x and y are easily obtained. Ans. a = 8or4; y ~ 4 or 8. 16. Find the values of x and y in the equations a+?/= 10, and a; 3 + ?/ 3 =280. Ans. 3=4 or 6 ; y= 6 or 4. 17. Find the values of x and y in the equations x-\-y = 11, and a: 4 +y 4 =2657. Ans. a-=4 or 7, y — 1 or 4 18. Find the values of x and y in the equations x-\-y= 10, and a 5 -\-y s — 17,050. A';s. a = 3 or 7, y — 1 or 3 190 MISCELLANEOUS SOLUTIONS AND EXERCISES. Miscellaneous Solutions and Exercises. ] 9. Find the values of a and y in the equations 6, and x 2 -\-x=\8 — y 2 — y. From the second equation, x 2 -\-y 2 -\-x-{-y=\8. Adding twice the first, x 2 -\-2xy-\-y 2 -\-x-\-y = ?>t). This last equation may he put under the form (o5+7/) 2 +(a:+7/) = 30 ; which is quadratic with reference to x-\-y, and from which we may therefore find the value of x-\-y. Ans. 35 = 2, or 3 ; or — 3 =f q/3 ; y — 3, or 2 ; or — 3± y/3. 20. Find the values of x and y in the equations 3"-) ?/ = 6, and x 2 y 2 +435^ = 96. Ans. 35 = 2, or 4- ; or 3 -\/ 21 ; y=4, or2 ; or3±y/21 21. Find the values of 35 and y in the equations x 3 y 2 =2y 2 , and 835 s — t / 2 = 14. 3 2 1 L 2 Dividing the first equation by y 2 , we find x 3 =2y 2 , or y 2 =^x 3 , 1 J and by substituting this value of?/ 2 iu the second equation, that equa- tion will become quadratic with reference to a: 3 . Ans. a; = 2744, or 8 ; y = 9604, or 4. 22. Find the values of x and y in the equations x 2 y — 25?/= 6, and x 3 y—y= 21. Dividing the first equation by the second, we have 35 2 y xy 6 x 3 y y — 21 This equation will be simplified by reducing each of its two frac- tional members to its lowest terms. 1 Ans. x=2, or 2 ! 2/ = 3, or — 24. 23. Find the values of x and y in the equations % 3j rzy 2 =39, and x 2 y-{-y 3 = 26. Ans. 35=3 ; y= 2. MISCELLANEOUS SOLUTIONS AND EXERCISES. 191 24. Find the values of x and y in the equations x — y— 4, and a: 3 — t/ 3 =316. Dividing each side of the second equation by the corresponding side of the first, we shall find, x 2 -\-xy-\-y 2 =79. Squaring the first equation, and subtracting, we have 3xy=63 Ans x=7, or — 3; y~ 3, or — 7 25. Find the values of x and y in the equations x 2 — y 2 —5, and a: 4 — r/ 4 =65. Ans. x— ±3, y~± 2 26. Find the values of x and y in the equations x-\-y— 60, and 2 (a 2 -\-y 2 ) — 5xy. The second equation may be put under the form x 2 -\-y 2 — 2^xy — (S , and the solution will be facilitated by subtracting this from the square of the first equation. Ans. x=40, or 20 ; y^= 20, or 40. 27. Find the values of x and y in the equations x 2 y 2 xy = 8, and \- — = 9. y x Multiplying the two equations together, we find x 3 - \-y 3 =72 ; and, multiplying this equation by x 3 , we have x 6 -{-x 3 y 3 = 7 2x 3 . From the first equation, x 3 y 3 =8 3 =512 ; and if this number be substituted in the preceding equation we shall have a quadratic with reference to x 3 . Ans. x=4, or 2; y=. 2, or 4. 28. Find the values of x and y in the equations xij— 25, and x 3 +y 3 = lOxy. 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. 29. Find the values of x and y in the equations x 2 — y 2 — (_c+r/) = 8, and ( x — r/) 2 (x+?/) = 32. Dividing each equation by x-\ -y, we have 8 NO 32 x — v — 1= , and (x — 1/) 2 = • x+y x+y Transposing — 1, and squaring, we obtain , / 8 \ 2 32 {X ~ y) -C+y) +1 ) =x+y ] from which will result a quadratic with reference to x-\~y. Ans. x—5 ; y= 3. 192 MISCELLANEOUS SOLUTIONS AND EXERCISES. 30. Find the values of x and y in the equations x-\--\/xy-{-y—l , and x 2 +xy+y 2 =21. Dividing the second equation by the first, we have x — -\/xy+y~3. W e shall now obtain two equations of simpler forms by adding the third equation to the first, and subtracting it from the first. , Ans. x=\, or 4 ; y— 4, or 1 31. Find the values of x and y in the equations x-\-y , — z — 2a cy 12 -g-=V*y+4, and Vxy=^^+-j- From the first equation, by transposition, — x4 -y * xy m2 — 4 ‘ Clearing this equation of its fraction, and squaring, . 4xy=(x+y ) 2 — 1 6 (z+?/) + 64. By equating the two values of sfxy, from the second and third equations, 2xy 12 _x+y x+y + 5 ~ 2 or 2 xy x + y x+y~ 2 32 " T" Clearing this equation of its fractions, 64 4,xy={x+y) 2 — ~(x+y). We shall now obtain a simple equation by subtracting this last equation from the fourth equation. Ans. x = 2, or 18; ^ = 1 8, or 2. 32. Find the values of x and y from the equation and proportion xy 2 — x—3 ; x 2 ?/ 4 — x 2 : x 2 +x 2 y 2 +x 2 y i : : 5:7. In any proportion the difference of the first and second terms is to the first, as the difference of the third and fourth is to the third, (160). Hence from the given proportion, we shall find 2 x 2 +x 2 y 2 : x 2 y i — x 2 . : 2:5. Dividing the first antecedent and consequent by x 2 , (15S) 2 +y 2 : y* — 1 : : 2:5. MISCELLANEOUS SOLUTIONS AN! EXERCISES. 193 33. Find the values of x and y in the equations x 2 -\-xy-\-y 2 = 13, and x* -\-x 2 y 2 -\-y* =91. Dividing the second equation by the'first, x 2 — xy-\-y 2 — 7. Adding the third equation to the first, and dividing the result by i! : and also subtracting the third from the first, x 2 - \-y 2 = 10 ; and 2xy=&. 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. a;=:±3; y— ± 1. 34. Find the values of x and y in the equations x 2 y—x 2 y 2 — x 2 , and x 2 y 2 -\-x 2 =x 3 y 3 — x 3 Dividing each equation by x 2 , we have y—y 2 — 1, and y 2 -\-\—xy 3 — x. The value of y is to be found from the first of these equations, and substituted in the second. Ans. x=ly'o ; y=\ ^/5. 35. Find the values of x and y in the equations xy—X 2 — y 2 , and x 2 -j-y 2 =z 3 — y 3 . If we assume y to be equal to xv, and substitute this product for y in the two equations, (264,) we shall have x 2 v—x 2 — x 2 v 2 , and x 2 -\rx 2 v 2 =a? 3 — x 3 v 3 . These equations may be solved in the same manner as those in the preceding example. Or the value of i may be found, in terms of y, from the first of the given equations, and substituted in the second. Thu» 3: =I±V / ¥=I±^- 2 v 4 2 2 V Then for the second equation, Ans. x—\{5± y—±\^5. MISCELLANEOUS SOLUTIONS AND EXERCISES. i 4 36. Find the values of x and y in the equations x 3 — x 2 y — xy 2 -\-y 3 =576, and a; 3 -\-x 2 y-\-xy 2 -\-y 3 =2336 Subtracting the first equation from the second, 2x 2 y+2xy 2 — \l&Q. Adding this to the second equation, we obtain 2 3 -f-3x 2 t/-|-3a:?/ 2 -|-t/ 3 =4096. By extracting the cube root of this equation, we shall fun the value of x-\-y, which may be substituted for x-\-y in the third equ. tion. -Arcs. 2=11; y~5 37. Find the values of x and y in the equations au/ = 320, and x 3 —y 3 =61 (2 — y) 3 . Dividing the second equation by x — y, we have x 2 +xy-\-y 2 =61 ( 2 — y ) 2 . By converting this equation into a proportion, (153,) x 2 -{- xy-\-y 2 : ( x — y) 2 : : 61 : 1. The solution now proceeds by developing the term (x — y) 2 , — sub- tracting each consequent from its antecedent, and forming a proportion of the consequents and remainders, Ac, (160). Ans. 2=20 ; y= 16. 38. Find the values of x, y, and z, in the equations, x 2 -\-y 2 -\-xy=.Zl ; x 2 -\-z 2 -\-xz— 49; y 2 +z 2 + 2/2 = 61. Subtracting the first equation from the second, and decomposing, (z—y)(z+y) + {z—?/)x, or (z+y+x)(z-y) = 12 ; 12 from which z-\-y-\-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 2 y=x-\-z. By substituting 2 y for x-\-z in the sixth equation, we shall fiud y 2 — yx—\. The value of 2 from this equation, is to be substituted in the first equation. Ans. 2 = 3; y~ 4; c = <3. PROBLEMS. 195 PROBLEMS In Quadratic Equations of One or More TJnknowr. Quantities. 1. The area of a rectangular lot of ground is 384 square rods, and its length is to its breadth as 3 is to 2. Required the length and breadth of the lot. The area , 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 be taken in the same denomination in multiplying : the area will be found in the corresponding denomina- tion of square measure. Let x represent the length, and y the breadth of the lot ; then by the conditions of the problem, xy — 384, and x : y \ . 3:2. 2x Or, if x represent the length, - will represent the breadth, and o we shall then have 2x 2 = 384. 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. 10y/5— 10, and 30— lOV^. 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 il their product be increased by 40, the digits will be reversed. What is the number ? Ans. 46. 196 PROBLEMS. 7. The sum of two fractions is li, and the sum of their recip 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 2x v ( 1 84 ). 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 i of the whole, being in meadow, there remain for ploughing 1296 square rods. W T hat 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 x and y represent the two extremes, the mean term will be '\fxy, (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- fi metrical mean by 13, and the geometrical mean exceeds the harmoni- | cal mean by 12. What are the numbers? Ans. 234 and 104. |j 31. Three merchants made a joint stock, by which they gained a sum less than that stock by $80. A’s share of the gain was $60, and i his contribution to the stock was $17 more than B's ; also B and 0 : 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. W ? hat are the numbers ? Assume x to represent the first term, ond y the ratio of the pro- gression. Ans. 5, 10, and 20, 33. Two persons set out from different places, and traveled towards i each other. On meeting, it appeared that A had traveled 24 miles I more than B, and that A could have gone B’s journey in 8 days, while |i B would have been 18 days in performing A’s journey. 1/V hat distance was traveled by each ? Ans. 72, and 48 miles. Jj 34. The joint stock of two partners was $416. A’s money was in j the business 9 months, and B’s 6 months. When they shared stock II and gain, the first received $22S, and the second $252 ; what was each I j man’s amount of stock? Ans. A s $192, B s $224. ‘J 35. The sum of $700 was divided among four persons, A, B. C, and j I D, whose shares were in geometrical progression ; and the difference It between the greatest and the least was to the difference between the .1 two means as 37 to 12. What were the several shares ? Ans. $108, $144, $192, and $256. ■ } CUBIC AND HIGHER EQUATIONS. 199 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, — while 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 any complete Equation containing but one unknown quantity x, are all arranged, according to the descending I powers of x, hi the first 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 pro- ducts of all the roots combined two and two, 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 a, b, and — c ; then x — a' x — b, and x 4- care th e divisors of the equation, and the equation may be ac- cordingly resolved into (x — a) ( x — b) (.r+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 3 + (c — a — b) x 2 -\-(ab — ac — be) x-fabcz^Q. In this equation the coefficient of x 2 is the sum of the roots a, b, and — c, with their signs changed ; the coefficient of a; is the sum ol the products of the roots combined two pnd two, with their sign? 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 z 2 3x — 10 = (a; — 2) (z-j-5) = 0, whose roots are 2, and — 5 ; or to c 3 — 19a:-)- 30 ~(x — 2) (a-(-5) ( x — 3) = 0, whose roots are 2, — 5, & 3 In the second equation it will be observed that the second term \ containing x 2 , is wanting, since its coefficient, that is, the sum of th< i roots 2, —5, and 3, is 0 ; and that 19a: therefore corresponds to the third term, (267. ..2). ' I Determination of the Integral Roots of Equations. (268 ) If an equation containing hut one unknown quantity a:, wit! all its terms transposed to one side, be divisible by a:-)- 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 ch 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 + 3x’ 2 — 4.r=12. The divisors of the known term 12 are 1, 2, 3, 4, 6, and 12; and it will he found, on trial, that the equation x 3 -)-3x 2 — 4a: — 12 = 0, is divisible by a: — 2, x-\-2, and cc -4- 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. x — 2 ) z 3 -\-3x 2 —4a: — 12 ( x 2 +5a:-)-6. By dividing the given equation by x — 2, we thus find r- 2 -|-5x-(-6 = 0, or x 2 +5a:=: — 6, which gives x= — 2 or —3 By this method the last two routs are found the same as before. CUBIC AND HIGHER EQUATIONS. 201 Solution of Equations by Approximation. (2G9.) The following method of solution may he 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 knoivn term of the equa- tion ; then, The difference between the tivo results, Is to the difference between the two assumed numbers, A.s the difference between either result and the knoivn term, Is 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.... ....125 16... .... 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... . . . a; 3 . . . ...79.507 17.64 ... ...18,49 4.2 ... . . . x . . . ... 4.3 95.928 10 102.297 202 CUBIC AND HIGHER EQUATIONS. Forming a proportion between the difference of these two results, and the difference between greater result and the known term 100, 6.369 : ,1 :: 2.297 : the correction .036. By subtracting this correction from the greater assumed number 4 3, we have 4.264 for a nearer value of x. 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, the error in the less result 84 is (100 — 84)=: 16, and this being less than the error in 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 i 4.2 for the first two correct figures ; the next solution gives 4.264 ; j 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 Equation, we would divide a ,3 + a 2 +a; — 100 = 0 by Z— 4,2644, &c. = 0, (253). "We should thus obtain a quadratic equation, from which the other two values of x might be determined, according to the usual method. "When all the Boots 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, j (267. ..1). Thus the sum of the three roots of the equation in the preceding Example, with their signs changed, would be unity, which is the co- efficient of the second term a 2 . CUBIC AND HIGHER EQUATIONS. 203 EXERCISES On Affected Cubic and Biquadratic Equations. I . Find the values of x in the equation x 3 — 6a; 2 + 11* = 6, (268). 2. Find the values of x in the equation x 3 — 9x 2 + 26x = 24. 3. Find the values of x in the equation x 3 — 3x 2 — 6x= — 8. Ans. x=l, 2, or 3. Ans. x=2, 3, or 4. Ans. x=l, 4, or — 2. 4 Find the values of x in the equation 2x 3 — e^ 2 — Sx=— 24. Ans. x=2, 3, or — 2. 5. Find the values of x in the equation x 4 +2x 3 — 13x 2 — t4x+24 = 0. Ans. 1, —2, 3, or —4 6. Find an approximate value of x in thP t_ -pation x 3 + 10x 2 + 5x=260, (26^). 7. Find an approximate value of x in the equ Mon x 3 — 15x 2 + 63x=50. 8. Find an approximate value of x in the eqv tion x 3 — I7x 2 -t-54x = 350. 9. Find an approximate value of x in the eqi?’ ton x 4 — 3x 2 — 75x= 10000. 0. Find an approximate value of x in the equation 2x 4 — 16x 3 + 40x 2 — 30x+ 1 = P. 1. Find an approximate value of x in the equation Qx 2 — 15) 2 +x^/x = 90. Ans. £=4.117’. Ans. x= 1.028.’ aL/is. x= 14.95’. Ans. £=10.23’. An* an-' 1.284’ Anz. x=l a J'’ K 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 remainder into the divisor , and so on, as in finding the (freatest 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 x 2 -{-xy—\0, and xy + 2y 2 — 24. xy+2y 2 — 24 x x 2 -\-xy — 10 j x 2 y-\-2xy 2 2\ — xly f x 2 y-\-xy 2 — lOy xy 2 — 24x+10 y, or x{y 2 — 24) + 10_y. x 2 +xy — 10 y 2 — 24 x iy 2 — 24)+ \0y 1 a ,2 (y 2 — 24) + a;(y 3 — 24_y) — 10y 2 +240(.r+y ) x 2 [y 2 — 24)+10 xy x(y 3 — 24w) — 1 Oxy— 1 0y 2 + 240 x{y 3 — 24 y) + 10 y 2 — 10 xy — 20y 2 +240 x{y 2 -24) + 107/ y a?/+2y 2 — 24 ) x[y 3 — 24 y) + 10 w 2 {y 2 — 24 / x(y 3 — 24?/) + 2y 4 — 48y 2 — 24y 2 + 576 — 2y 4 + S2y 2 — 576~0. In the remainder — lOxy— 20?/ 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 2 — 24 for the quotient, and multiply the divisor by this binomial. The first remainder is equal to 0, because the divisor and dividenc are each equal to 0 ; and it follows hence that each subsequent re \ mainder is equal to 0. The operation will be much more simple if we divide the firs equation by the second : the result will be the same. 205 CHAPTER XI. «w GENERAL DESCRIPTION OF PROBLEMS. INEQUATIONS. 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 different conditions to be expressed, or unknown quantities to he determined, (120.) All the Problems which have hitherto been proposed in this work, are determinate j 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 Indeterminate 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. Example 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 z, we shall have only the two Equation^ y—x—5; y+z= 12. By subtracting the first equation from the second, we have 206 6ENERAL DESCRIPTION OF PROBLEMS. X + Z = 7 . 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 x, and determine thence the corresponding value of z. Thus if x—\, Z— 6| : if x=\, z— 6J; if *=|, z= 6\, &c.; and ifom 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— 6, 7, 8, 9, 10, or 11. \ Example II. To find a number such, that ^ of it, diminished by y of it, and by 5. shall be equal to j ^ °f excess of 5 times the number above 60 The equation of this problem will be 3a; x 5x— 60 T~3~ 5 ^ 12 . Clearing the equation of its fractions, 9a; — 4x — 60=5.r — 60 ; or 5x — 60=5x — 60. This last is an identical Equation, which 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. lEus, by transposition, 5x — 5x— 60 — 60. By adding similar terms, and retaining x as a symbol in the farst member, we liave 0.r=0 ; which gives a f =§. Hence § is a symbol of an indeterminate quantity. The same thing will appear from considering that the quotient cf 0 -r~ 0 is any quantity ichatever ; inasmuch as the divisor Ox any quantity will produce the dividend 0, (43). GENERAL DESCRIPTION OF PROBLEMS. 207 3? Impossible Problems. (272.) An Impossible Problem is one in which there is some condi- tion, expressed or implied, which cannot 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 im, aginary . 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 x-\-y= 10; x — y — 2; xy= 20. From the first and second equations the values of x and y will be found to be x = 6, and y — 4. The third equation cannot, therefore, be fulfilled ; that is, the problem is impossible. If the third equation were xy= 24, 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 4au/=96, or xy= 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_L 53+x t7 + a;= ___. From this equation we shall find x= — 5. This number, added to 17 and 53, gives 12 and 48, and 12=i of 48. The problem is impossible in an arithmetical sense, according to which 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. 208 GENERAL DESCRIPTION OF PROBLEMS. Example III. • To find a number such, that if 83 he increased by 3 times that number, £ of the sum will be equal to 13f . If x represent the number, the equation will be 82+3* 6 ; : 1 3j - Clearing the equation of its fractions, we find 82 + 3a;=82; which gives 3a?=82 — 82 = 0 ; and a?=^=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 of 82 itself is equal to 13|-. Example IV. To find a number such, that the sum of \ of it and §- 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 „ 11a; „ 4+T~ 2= 72 +3 - From this equation we shall find 3x-\-8x — 24 = llx-\-3&-, or 11a; — lla;=0a;=60 ; and a?= 6 0 ° = ao , infinity ; (50). The result shows that it would require a number infinitely great, to fulfil the conditions of the problem. The problem is therefore possible. Example V. To divide the number 24 into two such parts, that their product shall be 150. If x represent one of the two parts, 24 — x will represent the other and the equation will be 24x— x 2 = 150 or x 2 — 24z= — 1 50, (1 17); which gives ;r= 12± -\/l44 — 150, = 12±i/=6. In this value of x, the part V — 6 is imaginary, that is, it is an impossible quantity, (246) ; hence the problem is impossible. GENERAL DESCRIPTION OF PROBLEMS. 209 That the preceding problem is impossible, will also appear from the following general proposition ; viz : (273-) The square of one half of any quantity is greater than the product of any tivo unequal parts of the quantity. Let s represent any number, and d the difference between any two parts of the number ; then the respective parts are o I . fl e rl p2 /72 ■ - - and ^ (168), and their product is — - — This product will vary directly as its numerator s 2 — cl 2 (147); and will therefore have its greatest value when <7=0. In that case the pro- duct becomes j.s 2 , which is the square of f s , half the given number. It may also be remarked here, that the sum of the two equal factors of any quantity, is less than the sum of any two unequal factors into which the quantity can be resolved. For, s representing the sum, and d the difference, of the two factors of a quantity, those factors are S —rp- and 4 S ^ - (168), and their product is s 2 — d 2 4 This product will retain a constant value , if s 2 and d 2 be equally di- minished, i for then the value of s 2 — d 2 will remain constant. In this diminution, s 2 will have its least value when d 2 is made 0 ; so that s, the sum of the two factors, will have its least value when a — 0, and the two factors will then be equal to each other. Signification of the Different Forms under which the Value of the Unknown Quantity may he 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 -j), 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 adclitively, instead of subtractively, 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- prived is impossible. 10 * 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 meu 4. A’s money was equal to f of B’s. A paid away $50 less than | of his, and B $50 more than of his, when it was found that the latter had remaining only A 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 inauv by 70. What was the number of his palisades ? Ans. 180 palisades. 6. From two tracts of land of equal size, were sold quantities in the proportion 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 | of an hour; but it takes him ll 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§ 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? H/zs 48, 72, and 120 sheep MISCELLANEOUS PROBLEMS. 10. The sum of $1170 is to be divided between three persons, A, B, and 0, in proportion to their ages. Now A’s age is to B’s as 1 to lg-, and to C’s as 1 to 2 ; what are the respective shares ? Ans. $270 ; $360 ; $540. 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 -g- of the quantity in the second were to be drawn off, the contents of the two casks would be equal. How many 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. 43 m. 38^-s. past eight. 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 S 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 a mass 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 finish it, which he does in 36 days more. In what time could each do it separately ? Ans. 24, and 48 days. K’ 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 cauuht. 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+<1 P+ut demonstration. It contains the elements of the science in their proper integrity tnd proportions. Its author is a learned man and a practical instructor, as the itithor of every school-book should be. The style is a model for a text-book, com lining in a high degree perspicuity, precision, and vivacity. In a word, it is the very test elementary work on Astronomy with which we are acquainted. This notice is echoed by a large number of academies, who are promptly intro lucing the book. Elements of Meteorology; designed for Schools and Ac ademies. By John Brocklesby, A. M., Professor of Mathematics and Natural Philosophy in Trinity College. Hartford 84 cents. The subject of Meteorology is of the deepest interest to all. Its phenomena every •vnere surround us, and ought to be as familiarly known to the scholar as his arith- i die or philosophy. 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