LIBRARY OF THE University of California. Mrs. SARAH P. WALSWORTH. Received October, 1894. ^Accessions No . S^/Z/C? . Class No. / Digitized by the Internet Archive in 2008 with funding from Microsoft Corporation http://www.archive.org/details/additionalbourdOOdavirich KEY DAVIES' BOURDON, MANY ADDITIONAL EXAMPLES, ILLUSTRATING THE ALGEBRAIC ANALYSIS: ALSO, A SOLUTION OF ALL THE DIFFICULT EXAMPLES IN DAVIES' LEGENDRE. UWVBRSIT OJf 4km% NEW YORK A. S. BARNES & Co., Ill & 113 WILLIAM STREET, (CORNER OF JOHN STREET.) BOLD BY BOOKSELLERS, GENERALLY, THROUGHOUT THE UNITED STATES. 1866. ID 33 (f^s\V^ Entered according to act of Congress, in the year eighteen hundred and fifty-six^ BY CHARLES DAVIES, In the Clerk's Office of the District Court of the United States, for the Southern District of New York. JONES & DE N YSE , 8TEREOTYPERS AND ELECTROTYPERS, 183 William-Street. 0f THK ViiviasxTY; PREFACE. A wide difference of opinion is known to exist among teachers in regard to the value of a Key to any mathematical work, and it is perhaps yet undecided whether a Key is a help or a hindrance. If a Key is designed to supersede the necessity of investiga- tion and labor on the part of the teacher; to present to his mind every combination of thought which ought to be suggested by a problem, and to permit him to float sluggishly along the current of ideas developed by the author, it would certainly do great harm, and should be excluded from every school. If, on the contrary, a Key is so constructed as to suggest ideas, both in regard to particular questions and general science, which the Text-book might not impart; if it develops methods of solution too particular or too elaborate to find a place in the text ; if it is mainly designed to lessen the mechanical labor of teaching, rather than the labor of study and investigation ; it may, in the hands of a good teacher, prove a valuable auxiliary. The Key to Bourdon is intended to answer, precisely, this IV PREFACE. end. The principles developed in the text are explained and illustrated by means of numerous examples, and these are all wrought in the Key by methods which accord with and make evident the principles themselves. The Key, therefore, not only explains the various questions, but is a commentary on the text itself. Nothing is more gratifying to an ambitious teacher than to push forward the investigations of his pupils beyond the limits of the text book. To aid him in an undertaking so useful to himself and to them, an Appendix has been added, containing a copious collection of Practical Examples. Many of the solutions are quite curious and instructive ; and taken in connection with those embraced in the Text, form a full and complete system of Algebraic Analysis. The Problems comprising the "Application of Algebra to Geometry," at the end of Legendre, are, many of them, quite difficult of solution. The many letters which I have received from Teachers and Pupils, in regard to the best solutions of these questions, have suggested the desirableness of furnishing, in the present work those which have been most approved. They are a collection of problems that have been often solved, and the solutions may be studied with great profit by every one seeking mathematical knowledge. Fishkill Landing, ) July, 1856. J \ V Ol INTRODUCTION ALGEBRA. 1. Otf an analysis of the subject of Algebra, we think it will appear that the subject itself presents no serious difficulties, and that most of the embarrassment which is experienced by the pupil in gaining a knowl- edge of its principles, as well as in their applications, arises from not attending sufficiently to the language or signs of the thoughts which are combined in the reasonings. At the hazard, therefore, of being a little diffuse, I shall begin with the very elements of the algebraic language, and explain, with much minute- ness, the exact signification of the characters that stand for the quantities which are the subjects of the analy- sis ; and also of those signs which indicate the several operations to be performed on the quantities. Algebra. Difficulties How over- come. Language. Characters which repre- sent quantity. Sign«. 2. The quantities which are the subjects of the ^ uanti t>' es - algebraic analysis may be divided into two classes : How divided those which are known or given, and those which are unknown or sought. The known are uniformily repre- sented by the first letters of the alphabet, ez, b, c, d 9 &c. ; and the unknown by the final letters, x, y, z, V, w. &c. How repre- sented. 6 Introduction. May be in- creased or diminished. Five opera- tions. First Second. Third. Fourth. Fifth. Exception. 3. Quantity is susceptible of being increased or diminished ; and there are five operations which ean be performed upon a quantity that will give results differing from the quantity itself, viz. : 1st. To add it to itself or to some other quantity ; 2d. To subtract some other quantity from it ; 3d. To multiply it by a number; 4th. To divide it ; 5th. To extract a root of it. The cases in which the multiplier or divisor is 1, are of course excepted; as also the case where a root is to be extracted of 1. Signs. 4. The five signs which denote these operations Elements are too well known to be repeated here. These, with of the Algebraic the signs of equality and inequality, the letters of the anguage. a ]p} ia k eti anc i tne figures which are employed, make up its words and the elements of the algebraic language. The words p rases ^^ phrases of the algebraic, like those of every How inter- other language, are to be taken in connection with pTete " each other, and are not to be interpreted as separate . and isolated symbols. Bymbols of quantity. 5. The symbols of quantity are designed to repre- sent quantity in general, whether abstract or concrete, whether known or unknown ; and the signs which in- dicate the operates to be performed on the quanti- ties are to be interpreted in a sense equally general. When the sign plus is written, it indicates that the Signs pins an a quantity before which it is placed is to be added to minus. some other quantity : and the sign minus implies the General. Examples. INTRODUCTION. 7 existence of a minuend, from which the subtrahend is to be taken. One thing should be observed in regard Signs have n« effect on the to the signs which indicate the operations that are to nature of be performed on quantities, viz. : they do not at all a quan l r ' affect or change the nature of the quantity before or after which they are written, but merely indicate what is to be done with the quantity. In Algebra, for ex- Examples: In Algebra. ample, the minus sign merely indicates that the quan- tity before which it is written is to be subtracted from some other quantity ; and in Analytical Geometry, that in Analytical Geometry. the line before which it falls is estimated in a contrary direction from that in which it would have been reck- oned, had it had the sign plus ; but in neither case is the nature of the quantity itself different from what it would have been had the sign been plus. The interpretation of the language of Algebra is interpretation of the the first thing to which the attention of a pupil should language : be directed ; and he should be drilled on the meaning and import of the symbols, until their significations and uses are as familiar as the sounds and combina- Its neces81t y« tions of the letters of the alphabet. 6. Beginning with the elements of the language, Element* explained let any number or quantity be designated by the letter a, and let it be required to add this letter to itself and find the result or sum. The addition will be expressed by a + a = the sum. But how is the sum to be expressed? By simply Signification, regarding a as one a, or la, and then observing that one a and one a, make two a's or 2a : hence, O INTRODUCTION. a -\- a = 2a; and thus we place a figure before a letter to indicate how many times it is taken. Such figure is called a Co-efEciont. Co-efficient. Preduct: 7. The product of several numbers is indicated by the sign of multiplication, or by simply writing the letters which represent the numbers by the side of each other. Thus, how indicated. axbxcxdxf, or abcdf, indicates the . continued product of % a, b, c, d, and /, Factor. and each letter is called a factor of the product : hence, a factor of a product is one of the multipliers which produce it. Any figure, as 5, written before a product, as 5abcdf, Co-efficient of is the co-efficient of the product, and shows that the a pro uc . p r0( j uct i s taken 5 times. fiiuai factors : g^ jf t h e num bers represented by a, b, c, d, and what the f 9 were equal to each other, they would each be product becomes. represented by a single letter a, and the product would then become How K expresses aXaXaXdXa = a-; that is, we indicate the product of several equal fac- tors by simply writing the letter once and placing a figure above and a little at the right of it, to indicate INTRODUCTION. 9 how many times it is taken as a factor. The figure Exponent: so written is called an exponent. Hence, an exponent where written. is a simple form of language to point out how many equal factors are employed. 9. The division of one quantity by another is indi- cated by simply writing the divisor below the dividend, after the manner of a fraction ; by placing it on the right of the dividend with a horizontal line and two dots between them ; or by placing it on the right with a vertical line between them : thus either form of expression : -— > b -f- a, or b\a, Division : how expressed Three forms. indicates the division of b by a. 10. The extraction of a root is indicated by the Roots : sign y\ This sign, when used by itself indicates the how indicated lowest root, viz., the square root. If any other root is to be extracted, as the third, fourth, fifth, &c, the index; figure marking the degree of the root is written above where written and at the left of the sign ; as, y cube root, y fourth root, &c. The figure so written, is called the Index of the root. Language far We have thus given the very simple and general operations language by which we indicate each of the five operations that may be performed on an algebraic quantity, and every process in Algebra involves one or other of these operations. OF THE tTHIVBRSITTi 10 INTRODUCTION. MINUS SIGN. languid U- The algebraic symbols are divided into two classes entirely distinct from each other — viz., the ko-w divided, letters that are used to designate the quantities which are the subjects of the science, and the signs which are employed to indicate certain operations to be per- Aigcbraic formed on those quantities. We have seen that all processes : the algebraic processes are comprised under addition, ihei: number, subtraction, multiplication, division, and the extraction Do-not change of roots ; and it is plain, that the nature of a quan- go 6 "antitiL ^J * s not at a ^ cnan g e cl by prefixing to it the sign which indicates either of these operations. The quan- tity denoted by the letter «, for example, is the same, in ever]) respect^ whatever sign may be prefixed to it ; that is, whether it be added to another quantity, sub- tracted from it, whether multiplied or divided by any number, or whether we extract the square or cube or Algebraic an y other root of it. The algebraic signs, therefore, signs: must be regarded merely as indicating operations to how regarded. * ° be performed on quantity, and not as affecting the nature of the quantities to which they may be prefixed. Plus and \\r e say, indeed, that quantities are plus and minus, Minus. but this is an abbreviated language to express that they are to be added or subtracted. Principles of jo In Algebra, as in Arithmetic and Geometry the science- x ' ° ' ' From what all the principles of the science are deduced from tn< definitions and axioms ; and the rules for performing the operations are but directions framed in conformity Example. to such principles. Having, for example, fixed b; definition, the power of the minus sign, viz., that an; INTRODUCTION. 11 quantity before which it is written, shall be regarded as to be subtracted from another quantity,- we wish to What *' e wiah to discover discover the process of performing that subtraction, so as to deduce therefrom a general formula, from which we can frame a rule applicable to all similar cases. SUBTRACTION. 13. Let it be required, for example, to subtract Subtraction. h a — c Process. Difference. from b the difference between a and c. Now, having written the letters, with their proper signs, the language of Al- gebra expresses that it is the difference only between a and c, which is to be taken from b ; and if this dif- ference were known, we could make the subtraction at once. But the nature and generality of the algebraic symbols, enable us to indicute operations, merely, and Operations indicated. we cannot in general make reductions until we come to the final result. In what general way, therefore, can we indicate the true difference ? b -a b — a -f- c Final formula, If we indicate the subtraction of a from b, we have b — a ; but then we have taken away too much from b by the number of units in r;for it was not a, but the dif- ference between a and c that was to be subtracted from b. Having taken ' away too much, the remainder is too small by c : hence, if c be added, the true re- mainder will be expressed by b — a -j- c. Now, by analyzing this result, we see that the sign Analysis of of every term of the subtrahend has been changed ; and what has been shown with respect to these quan- OF THI the result. WVBRsitt; 12 INTRODUCTION, Generaiiza- titles is equally true of all others standing in the same relation : hence, we have the following general rule for the subtraction of algebraic quantities : Change the sign of every term of the subtrahend, or Rule. conceive it to be changed, and then unite the quantities as in addition. Multiplica- tion. Signification of the language. Process. a -b c Its nature. Principle for the signs. ac — be MULTIPLICATION. 14. Let us now consider the case of multiplication, and let it be required to multiply a — b by c. The algebraic language expresses that the difference between a and b is to be taken as many times as there are units in c. If we knew this differ- ence, we could at once perform the multiplication. But by what general process is it to be performed without finding that difference ? If we take a, c times, the product will be ac; but as it was only the differ- ence between a and b> that was to be multiplied by c, this product ac will be too great by b taken c times ; that is, the true product will be expressed by ac — be: hence, we see, that, If a quantity having a plus sign be multiplied by another quantity having also a plus sign, the sign of the product will be plus ; and if a quantity having a minus sign be multiplied by a quantity having a plus sign, the sign of the product will be minus. General case: 15. Let us now take the most general case, viz., that in which it is required to multipy a — b by c — a\ INTRODUCTION. 13 a — b c — d ac — be - ad -f bd ac — be — ad -f bd It* form. First step. Let us again observe that the algebraic language denotes that a — b is to be ta- ken as many times as there are units in c — d ; and if these two differences were known, their product would at once form the product required. First : let us take a — b as many times as there are units in c ; this product, from what has already been shown, is equal to ac — be. But since the multiplier is not c, but c — t/, it follows that this prodtfet is too large by a — b 'taken d times ; that is, by ad — bd : hence, the first product dimin- Second step ished by this last, will give the true product. But, by the rule for subtraction, this difference is found by How token, changing the signs of the subtrahend, and then uniting all the terms as in addition : hence, the true product is expressed by ac. — be — ad -f- bd. By analyzing this result, and employing an abbre- viated language, we have the following general prin- ciple to which the signs conform in multiplication, viz. : Analysis of the result. Plus multiplied by plus gives plus in the product ; plus multiplied by minus gives minus ; minus mul- tiplied by plus gives minus ; and minus multiplied by minus gives plus in the product. General Principle. 16. The remark is often made by pupils that the above reasoning appears very satisfactory so long as the quantities are presented under the above form ; but why will — b multiplied by - Remark. Particular give plus bd? case . 14 INTRODUCTION. How can the product of two negative quantities stand* mg alone be plus ? Minus sign. In the first place, the minus sign being prefixed to t and d, shows that in an algebraic sense they do not t» interpre- otand by themselves, but are connected with other quan- tities ; and if they are not so connected, the minus sign makes no difference ; for, it in no case affects the quantity, but merely points out a connection with other quantities. Besides, the product determined above, being independent of any particular value attributed Fmm «f the to tne letters o, b, c, and d, must be of such a form as product : tQ k e true f or a |j va i ues . anc [ hence for the case in must be true for quantities which a and c are each equal to zero. Making this of any value \ supposition, the product reduces to the form of 4- bd. Signs in The rules for the signs in division are readily deduced from the definition of division, and the principles al- ready laid down. Zero and Infinity. Ideas not abstruse. ZERO AND INFINITY. 17. The terms zero and infinity have given rise to much discussion, and been regarded as presenting diffi- culties not easily removed. It may not be easy to frame a form of language that shall convey to a mind, but little versed in mathematical science, the precise ideas which these terms are designed to express ; but we are unwilling to suppose that the ideas themselves are beyond the grasp of an ordinary intellect. The terms are used to designate the two limits of Space and Number. 18. Assuming any two points in space, and joining INTRODUCTION. 15 Illustration, showing th« meaning of the term Zero. Illustration, showing the meaning of the term Infinity. them by a straight line, the distance between the points will be truly indicated by the length of this line, and this length may be expressed numerically by the num- ber of times which the line contains a known unit. If now, the points are made to approach each other, the length of the line will diminish as the points come nearer and nearer together, until at length, when the two points become one, the length of the line will disappear, having attained its limit, which is called lero. If, on the contrary, the points recede from each other, the length of the line joining them will con- tinually increase ; but so long as the length of the line can be expressed in terms of a known unit of measure, it is not infinite. But, if we suppose the points removed, so that any known unit of measure would occupy no appreciable portion of the line, then the length of the line is said to be Infinite. 19. Assuming one as the unit of number, and ad- mitting the self-evident truth that it may be increased or diminished, we shall have no difficulty in under- standing the import of the terms zero and infinity, as applied to number. For, if we suppose the unit one to be continually diminished, by division or other- wise, the fractional units thus arising will be less and illustration, less, and in proportion as we continue the divisions, they will continue to diminish. Now, the limit or boundary to which these very small fractions approach, is called Zero, or nothing. So long as the fractional number forms an appreciable part of one, it is not zero, but a finite fraction ; and the term zero is only The terms Zero and In- finity applied to numbers. Zero: 16 INTKODUCTION. applicable to that which forms no appreciable part of the standard. illustration. If, on the other hand, we suppose a number to be continually increased, the relation of this number to the unit will be constantly changing. So long as the num- ber can be expressed in terms of the unit one, it is infinity ; finite, and not infinite ; but when the unit one forms no appreciable part of the number, the term infinite is used to express that state of value, or rather, that limit of value. The terms, how employed. Are limits. 20. The terms zero and infinity are therefore em- ployed to designate the limits to which decreasing and increasing quantities may be made to approach nearer than any assignable quantity; but these limits cannot be compared, in respect to magnitude, with any known standard, so as to give a finite ratio. Why limits? 21. It may, perhaps, appear somewhat paradoxical, that zero and infinity should be defined 'as "the limits of number and space" when they are in themselves not measurable. But a limit is that " which sets bounds Definition of to, or circumscribes ;" and as all finite space and finite number (and such only are implied by the terms Space of ^ e pr and and Number), are contained between zero and infinity, we employ these terms to designate the limits of Num- ber and Space. Number OF THE EQUATION. 22. The subject of equations is divided into two parts. The first, consists in finding the equation ; that Subject of equatious : ho\r divided First part: is, in the process of expressing the relations existing INTRODUCTION. IT Solution. Discussion ol an equation Statement : what it is. between the quantities considered, by means of the algebraic symbols and formula. This is called the Statement of the proposition. The second is purely statement. Second part. deductive, and consists, in Algebra, in what is called the solution of the equation, or finding the value of the unknown quantity ; and in the other branches of analysis, it consists in the discussion of the equation ; that is, in the drawing out from the equation every proposition which it is capable of expressing. 23. Making the statement, or finding the equation, ' is merely analyzing the problem, and expressing its elements and their relations in the language of analy- sis. It is, in truth, collating the facts, noting their bearing and connection, and inferring some general law or principle which leads to the formation of an equation. The condition of equality between two quantities is expressed by the sign of equality, which is placed between them. The quantity on the left of the sign of equality is called the first member, and that on the right, the second member of the equation. Hence, an equation is merely a proposition expressed alge- braically, in which equality is predicated of one quan- tity as compared with another. It is the great formula of Algebra. Equality of two quanti- ties : How ex pressed. 1st member. 2d member. Proposition. 24. Every quantity is either abstract or concrete : hence, an equation, which is a general formula for expressing equality, must be, either abstract or con- crete. 2 Abstract*. Concrete^ 18 INTRODUCTION. Abstract equation. Concrete equation. Aii abstract equation expresses merely the relation of equality between two abstract quantities : thus, a -jr b = x, is an abstract equation, if no unit of value be assigned to either member ; for, until that be done the abstract unit one is understood, and the formula merely ex- presses that the sum of a and b is equal to x, and is true, equally, of all quantities. But if we assign a concrete unit of value, that is, say that a and b shall each denote so many pounds, weight, or so many feet or yards of length, x will be of the same denomination, and the equation will be- come concrete or denominate. 'Five opera- tions may be performed. 25. We have seen that there are five operations which may be performed on an algebraic quantity (Art. 3). We assume, as an axiom, that if the same operation, under either of these processes, be performed on both members of an equation, the equality of the members will not be changed. Hence, we have the five following Axioms. First. Second. Third. AXIOMS. 1. If equal quantities be added to both members of an equation, the equality of the members will not be destroyed. 2. If equal quantities be subtracted from both mem- bers of an equation, the equality will not be destroyed. 3. If both members of an equation be multiplied by the same number, the equality will not be destroyed. lNTrlOtrtJCTlOtf. 4. If both members of an equation be divided by the same number, the equality will not be destroyed. 5. If the same root of both members of an equa- tion be extracted, the equality of the members will not be destroyed. Every operation performed on an equation will fall under one or other of these axioms, and they afford the means of solving all equations which admit of solution, l*. Fourth. Fifth. Use of axioms. 26. The term Equality, in Geometry, expresses Equality: Its meaning that relation between two magnitudes which will in Geometry cause them to coincide, throughout their whole ex- tent, when applied to each other. The same term, its meaning in Algebra, merely implies that the quantity, of which equality is predicated, and that to which it is affirmed to be equal, contain the same unit of measure an equal number of times : hence, the algebraic signifi- cation of the term equality corresponds to the signifi- Corresponds w . . equivalency. cation ot the geometrical term equivalency. 27. We have thus pointed out some of the marked characteristics of Algebra. In Algebra, the quantities, classes of about which the science is conversant, are divided, as y — 2, x = 11£. 3. Given 6 - 2x + 10 = 20 - Sx — 2, to find *. Transposing and reducing, x = 2. 4. Given x -f- - x -f - x = 11, to find a:. Multiplying both members by 6. and reducing, 11* = 66; whence, x = 6. 24 KEY TO DAVIES' BOURDON. [79 5. Given 2x — - x + 1 = 5s — 2, to find s. ' ,*> , Multiplying both members by 2, transposing and reducing, -7s = -6; whence, x = - . 6. Given 3as -f - — 3 = bx — a, to find x* m Multiplying by 2, transposing and reducing, 6ax — 2bx = 6 — 3a ; factoring the first member of the equation, we have (6a — 26) x sie'e — 3a ; 6 — 3a whence, x = 6a -26 7 . Given i^i + | = 20 - ?-_!!!. , to find x. Multiplying both members by 6, 3s - 9 + 2s = 120 - 3 «-+ 57 ; transposing and reducing, 8s = 186; .'. x = 23$, ~. s + 3 , s i x — 5 ^ ,, , 8. Given —^— + 3 = 4 J— 1 to find * Multiplying both members by 12, Qx + 18 + 4s = 48 - 3s -f- 15 ; transposing and reducing, 13s =45;' ,'. s = 3^ 79.] EQUATIONS OF THE FIRST DEGREE. 25 A ^ ax — 6 , a 6# bx — a 9. Given — h - = — — , to find x. Multiplying both members by 12, Sax — 36 + 4a = Gbx — \bx + 4a ; transposing, reducing and factoring, (3a - 2b) x = 36, . • . * = 3a — 26 10. Given ■ = 4 =/. to find x. c a Multiplying both members by cd, 3adx — 2bcx — 4cd =fcd ; transposing, reducing and factoring, n n« 8MB — 4 36 — C 11. Given — = 4 — 6, to find x. 7 2 Multiplying both members by 14, IQax - 26 - 216 + 7c = 56 - 146 ; transposing and reducing, i* no i m ~ 56 + 96 -7c 16a* = 56 + 96 — 7c ; . • . * = 16a 12. Given - - _— + - = — , to find z. Multiplying both members by 30, 6x- 10* + 20+ 15 x = 130; iransposing and reducing, liar = 110; .-. x =10. 26 KEY TO DA VIES' BOUEDON. [80 X X X X 13. Given T A ; =/. to find *. abc a Multiplying both members by abed, and factoring, (bed — acd + abd — abc) x =abcdf . * . * — bed — acd -f- abd — abc 14. Given x — 1 — - — = x -f- 1, to "find x. lo 1 1 Multiplying both members by 143^ 143* - 33a; -f 55 + 52a; - 26 — 143* + 143 ; transposing and reducing, 19a? = 114; -. x = 6. 15. Given 5- §S - -^-? = - 12ff , to find x. Multiplying both members by 315, 45* - 280* - 63* 4- 189 = — 3983 ; transposing and reducing, - 298* == - 4172 ; . •. * = 14. 16. Given 2* — = — - — , to find *. D & Multiplying both members by 10, 20* —8* + 4 = 15* — 5; transposing and reducing, — 3* = - 9 ; . • . * = 3. 80^ 86.] EQUATIONS OF THE FIRST DEGREE. 27 bx ~~ d 17. Given 3a; -\ — - = x +- a, to find x. o Multiplying both members by 3, 9x + hx — d ■= 3z 4 3a ; transposing, reducing and factoring, 3a + d (6 + b) x = 3a + d ; 6 + 6 .« ^. (" + A) (« — h) n 4ab — b 2 _ a 2 — bx 18. Given • L - L - ^-t - 3a = — 7 2*4- — t ; a — b a -}• 6 6 to find #. Multiplying both members by a 2 b — b 3 , and performing indicated operations, a 2 bx + 2a6 2 .r -f & 3 .r — a 2 b 2 — 2ab 3 — b* - Sa 3 b + 3a& 3 = 4a 2 b 2 — hab 3 4- b* — 2a 2 bx + 2b 3 x + a 4 — a 2 bx — a 2 6 2 4- b 3 x ; transposing, reducing and factoring, 2b {2a 2 + ab- ft*) a? = a 4 4- 3a 3 6 4- 4a 2 6 2 - Gab 3 + 26* ; dividing by the co-efficient of x, _ a 4 4- 3a 3 6 4- 4a 2 6 2 — Gab 3 4 - 2b* X ~ 2b (2a 2 4- ab - b 2 ) STATEMENT AND SOLUTION OF PROBLEMS. 8. Divide $1000 between A, B, and C, so that A shall have $72 more than B, and C $100 more than A. Let x denote the number of dollars in B's share. Then will x 4- 72 " " " " A's " and x 4- 72 4- 100 " " " " C's " From the conditions of the problem, 28 KEY TO DAVIES' BOURDON. [86. x + tf + 72 4- x+ 172 = 1000; or, 3* = 756, .'. x = 252, or, A's share is 8324, B^fc share $252 and C's share $424. 9. A and B play together at cards. A sits down with $84 and B with $48. Each loses and wins in turn, when it appears that A has five times as much as B. How much did A win? Let x denote the number of dollars that A wins. Then will 84 + x denote what A has at last, and 48 — x what B has at last ; from the conditions of the problem, 84 + x = 5 (48 — x) ; or, 84 + x = 240 — 5x ; whence, # — 26 ; or, A wins 10. A person dying, leaves half of his property to his wife, one sixth to each of two daughters, one twelfth to a servant, and the remaining $600 to the poor : what was the amount of his property 1 Let x denote the whole number of dollars in the property. Then will - " " " " " in the wife's share. 2 x 6 " each daughter's " x ari d Z- " s. « " " " the servant'* " 12 fromHthe conditions of the problem, multiplying both members by 12, transposing and reducing, - * = - 7200 ; or, »sc 7200. 86.] EQUATIONS OF THE FIRST DEGREE. 29 11. A father leaves his property, amounting to $2520, to four sons, A, B, C and D. C is to have $360, B as much as C and D together, and A twice as much as B less $1000: how much do A, B and D receive 1 Let x denote the number of dollars that D receives • Then will x + 360 " " " " B and 2x + 720 - 1000 " " " A " from the conditions of the problem, 360 + x + x+ 360 + 2x + 720 - 1000 = 2520 ; transposing and reducing, Ax an 2080 ; . • . x = 520 or, D's share is $520 ; B's share $880, and A's share $760. 12. An estate of $7500 is to be divided between a widow, two sons, and three daughters, so that each son shall receive twice as much as each daughter, and the widow herself $500 more than all the children : what was her share, and what the share of each child ? Let x denote the number of dollars in each daughter's share ; Then will 2x " * " " son's " and 4x + 3z 4 500 " " " the widow's " from the conditions of the problem, 4* -f 3* 4 4s + $* + 500 = 7500 ; transposing and reducing, 145? = 7000; •. ar = 500. Daughters' share $500; son's share $1000; widow's share $4000. 13. A company of 180 persons consists of men, women ami 80 KEY TO DA VIES' BOURDON. [87. children. The men are 8 more in number than the women, and the children 20 more than the men and women together : how many of each sort in the company ? Let x denote the number of women ; Then will «-f8 " " men ; and- x -f x + 8 -f 20 " children. From the conditions of the problem, #-fa; + 8-f-a;-fa; + 8 + 20 = 180; transposing and reducing, 4.] EQUATIONS OF THE FIBST DEGREE. SIMULTANEOUS EQUATIONS OF THE FIRST DEGREE. ( 2x + 3y = 16 ) 1. Given 1 j-to find x and y. 39 3* — 2# Multiply both members of the first by 2, and of the second by 3 ; 4x 9x whence, by addition, member to member, we have, 13* = 65 ; . * . x = 5, also, y = 2. + 6# = 32 ) -6y = 33) 2. Given 2^ 3y _ 9 T + 1f ~ 120J . to find x and y. Clearing of fractions, and then multiplying both members of the first by 16, and of the second by 5, 128* + 240y = 144) 450* -f- 240y = 305 ) whence, by subtracting, member from member, 322* = 161 ; * = ~, also, by substitution, y = -• Z 3 3. Given + 7y = 99 + 7* = 51 ► to find x and y. Multiplying the first by 343 and the second by 7 ; 49* + 2401*/ = 33957 ) y + 49* = 357 j ^ > 0? TH5 X ■£ ^ *>. 40 KEY TO DAVIES BOURDON. [95. by subtraction, 2400y = 33600; y = 14 ; also, by substitution, x = 7. 4. Given < |- 12 = | + 8 ; to find x and y. Clearing of fractions and transposing, 2x — y = 80 47z -18y = 2100; multiplying both members of the first by 18, and subtracting the result from the second, member from member, 11 x sa 660 ; . • . x = 60 ; by substitution, y = 40. 'x+ y+ 2 = 29 • • • (1)" 5. Given * z + 2y + 3z = 62 • • • (2) 1+ |+ |=10 • • • (3) - to find x, y and *. Combining (1) and (2), y + 2z = 3? • • • • (4) ; combining (1) and (3) 2y + 3* = 54 .... (5); combining (4) and (5) z = l 2 ; by successive substitutions, x = 8, y = 9. {2x + 4y-2z = 22 • - (1)] 6. Given -j Ax — 2y + 5z = 18 • • (2) [\ [6x + 7y — z = 63 Combining (1) and (2), • • (2)[; • • (»)J 95.] EQUATIONS OF THE FIRST DEGREE. 41 lOy- 110 = 26 ... (4); combining (1) and (3), 5y — 8z = 3 . . . (5) ; combining (4) and (5), 5* = 20; .-. « = 4. By successive substitutions, x = 3, y = 7. , + | + 3 = 32 7. Given < .. ; to find a;, y and #. Clearing of fractions, 6a; -f- 3y + 2z = 192 • • : (i); 20a; + 15y + 12* = 900 • • • (2); 15a; + 12y + 10z = 720 . . . (3); combining (1) and (2), 16a? + 3y*= 252 . • • (4); combining (1) and (3), 15a; + % = 240 • • (5); combining (4) and (5), x = 12; by successive substitutions, y = 20, z = 30. 8. Given - 7a;- 4y- 5y — 4y- 2z + 3w = 17 . 2z + *=11 . 3a; — 2t* == 8 . 3?/ + 2* = 9 . 30 + 8w = 33 . • . (1)1 • • (2) • • (3) • • (4) . . (5) ► ; to find x, y t 2, t and u. 42 KEY TO DA vies' bouedon. [98-99. Combining (2) and (4), 4y-4z + 3u= 13 . . . (6); combining (1) and (3), 35y - 6z - hu = 107 . . . (7) ; combining (5) and (6), 12y + 41w = 171 . . . ( 8 ); combining (5) and (7), 35y +11^ = 173 . . . (9); combining (8) and (9), 1303w = 3909; .-. « = 3; by successive substitutions, x = 2, y = 4, 2 = 3, < r= 1. PROBLEMS GIVING RISE TO SIMULTANEOUS EQUATIONS OF THE FIRST DEGREE. 5. What two numbers are they, whose sum is 33 and whose difference is 71 Let x denote the first, and y the second. From the conditions, x -f y = 33 x-y = 7; whence, by combination, x = 20, y — 13. 6. Divide the number 75 into two such parts, that three times f ,n« greater may exceed seven times the less by 15. Let x denote the greater, and y the less From the conditions of the problem, EQUATIONS OF THE FIKST DEGBEE. * 43 X + y = 75 Sx-7y = l5; by combination, lOy = 210 ; .-. y = 21 ; also, a? = 54. 7. In a mixture of wine and cider, % of the whole plus 25 gallons was wine, and -J part minus 5 gallons, was cider : how many gallons were there of each 1 Let x denote the number of gallons of wine ; and y " " cider ; then will x + y " ♦ " mixture. From the conditions, x -p + * = - H^tm clearing of fractions, transposing and reducing, y — x = — 50 — 2y + x =3 15 by combination, y = 35 ; and x = 85 8. A bill of £120 was paid in guineas and moidores, and the number of pieces of both iorts that were used was just 100; if the guineas were estimated at 21s., and the moidores at 27s., how many were there of each 1 Let x denote the number of moidores ; and y " " " guineas ; then, since £120 = 2400s., we have, from the conditions, 44 " KEY TO DAVIES' BOURDON. [99 X + ?/= 100 27* + 21y = 2400 ; by combination, 6y = 300 ; . * . y = 50 ; also, a; = 50. 9. Two travellers set out at the same time from London and York, whose distance apart is 150 mile* ; they travel toward each other ; one of them goes 8 miles a day, and the other 7 ; in what time will they meet] Let x denote the number of miles travelled by the first ; y " " " " " second; then will f " " days " " first ; o and | " " « " " second; From the conditions, whence, by combination, * + y = 150; x y 8 = 7 ; x x = 80 and - as 10, the number of days. 10. At a certain election, 375 persons voted for two candidates; and the candidate chosen had a majority of 91 ; how many voted for each? Let x denote the number of votes received by the first; y " " " " " second; from the conditions of the problem, x 4- y = 375 * = y+ 91; by combination, x = 233, y = 142. 99.] EQUATIONS OF THE FIRST DEGREE. 45 . 11. A's age is double B's, and B's is triple C's, and the sum of all their ages is 140 : what is the age of each % Let x denote the age of A y « " B z " " C from the conditions of the problem) x~ >2y . • • • (1); y= Zz . . . • (2); x + y+z = U0 . • . i (3); from (1) and (2), x = 6z ; substituting, y =: 3*, and a; ss 6z, in (3), and reducing, 10* = 140 ; . • . z =c 14, . .# as 84, y = 42. 12. A person bought a chaise, horse and harness, for £60 ; the Horse came to twice the price of the harness, and the chaise to twice the price of the horse and harness : what did he give for each 1 Let x denote the number of pounds paid for the harness ; y " " " " " horse; z " " " " " chaise; from the conditions of the problem, y = 2x • ♦ ' • (1) z = 2(x + y) • • • • (2) * + #■$- *mM ■• * • • ;(8) from (2) and (1) z = Qx; substituting z sb Gx and y =.2x in (3) 9# ss 60, . * . x B= 6|L also, by substitution, y = 13 J, 2 = 40; hence, the price of the chaise was £40; of the horse £13 6*. Sd. ; and that of the harr .^ss £6 1 3s. 4c?. 46 KEY TO DAVIE8* BOURDON. [99-100. 13. A person has two horses, and a saddle worth £50 ; now, if- the saddle be put on the back of the first horse, it will make his value double that of the second ; but if it be put on the back of the second, it will make his value triple that of the first : what is the value of each horse ? Let x denote the number of pounds the 1st horse is worth; y " " « " 2d " " from the conditions of the problem, x + 50 = 2y y -f 50 = Zx ; whence, by combination, x - 30 y = 40. 14. Two persons, A and B, have each the same income. A saves J of his yearly; but B, by spending £50 per annum more than A, at the end of 4 years finds himself £100 in debt ; what is the income of each? Let x denote the number of pounds in the income of A ; . y u « " " " B; by the conditions of the problem, these are equal ; one only will h* used. Then will 4 -x denote what A spends per year : 5 ic +50" " B " 5 from the conditions of the problem, 4^ + 5o) -.= 4x+ 100; whence, performing indicated operations, transposing and reducing, 4a; = 500 . * . x = 125. 100.] EQUATION8 OF THE FIRST DEGREE. 47 15. To divide the number 36 into three such parts, that J of the first, £ of the second, and J of the third, may be all equal to each other. Let #, y and r, denote the parts. From the conditions of the problem, # + y 4- x = 36 * _ t 2~3 a; _ z 2~4 ; clearing of fractions, and combining, 9x = 72 . * . ar = 8 ; whence, y va 12 and « = 16. 16. A footman agreed to serve his master for £8 a year and livery, but was turned away at the end of 7 months, and received only £2 135. 4d. and his livery : what was its value? Let x denote the value of livery, expressed in shillings : £8 =2 160*., and £2 13*. Ad. = 53£s. ; Then will ( — — — I denote the value of wages 1 month, and 7 (i^ + _*) u . ., 7 by the conditions of the problem, /160 + x\ 7 (-T 2 -) = 53 * + * 1120 + 7x = 640 + 12 x — 5x = — 480 * = 96 .'. value, £4.16*. 48 KEY TO DAVIE8 BOtTBDON. [100- i 17. To divide the number 90 into four such parts, that if the first be increased by 2, the second diminished by 2, the third multiplied by 2, and the fourth divided by 2, the sum, difference, product, and quotient, so obtained, will be all equal to each other. Let #, y, z and «, denote the parts; from the conditions of the problem, x + y + z-\-u = Q0 x + 2z=y —2 x + 2 ss 2z u whence we find from the last three equations, x y = x -f 4, 2^-rl, and u = 2x -f* 4 ; substituting these values in the first equation, flj + « + 4+|+l+2aj + 4 = 90;or44« = 81;.-. x = 18 ; whence, by substitution, y — 22, z — 10, and u = 40. 18. The hour and minute hands of a clock are exactly together at 12 o'clock : when are they next together. 1st Solution. Let x denote the number of minute spaces passed by the hour hand before they come together ; * * and y the number passed by the minute hand ; then, since the latter travels 12 times as fast as the former, and since it has to gain 60 spaces, we have, x — y m 60 * = 12y; 100 J EQUATIONS OF THE FIRST DEGREE. 49 by combination, lly = 60 . • . y = 5 T 5 T ; also, x = 65 T 5 T ; hence, they will be together, 65-^j- minutes after 12 o'clock, or at 1 o'clock, T 5 T minutes, and at the end of every succeeding equal portion of time. 2d Solution. The minute hand will pass the hour hand 11 times before they again come together at 12 o'clock, and the times between any two consecutive coincidences will be equal. Hence each time will be equal to 12 hours divided by 11 = 1-^jhr. = lkr. b^m. 19. 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 be in drinking it alone ? Let x denote the number of days it takes the man to drink it ; y " " " " woman « " then, if the whole quantity of beer be denoted by 1, - will denote the quantity drank by the man in 1 day ; and x _ M li M M y woman j from the conditions of the problem, x^ y 12 1 1 y~30 ; substituting the value of - in the first equation. 1+1-1. x ^ 30 ~ 12 ' 50 KEY TO DAVIES' BOURDON. [100. clearing of fractions, 60 -f 2a? = 5a? ; . • . a? = 20. 20. If A and B together can perform a piece of work in 8 days, A and* C together in 9 days, and B and C in 10 days : how many days would it take each person to perform the same work alone 1 Let the work be denoted by 1 ; Let a? denote the work done by A in one day ; y U « U g « u then will -, - and - respectively denote the number of days that it will take A, B, and C severally to do the work ; from the conditions of the problem, x + y = | . . . . 0) x+z= J . . . . (2) y + z = iV • • • • (3); clearing of fractions, 8a? -f- 8y = 1 • • » • (4) 1 9a? + 9z = 1 • • • • (5) lOy + 102 = 1 • • • • (6); combining (4) and (5), 72y~72z = l • • • • (V, combining (6) and (7), 1440y = 82 y = Tiki substituting in (1) and (3), a? : — T ~~ TJTcT = Tfo \ z == T O ~ 720 = 73 hence, i = 14Ji; -=17Jf5 x 49 y 41 \-*a 100— 101.] EQUATIONS OE THE FIRST DEGREE. 51 21. A laborer can do a certain work expressed by a, in a time expressed by b ; a second laborer, the work c in a time d ; a third, the work e in a time/. Required the time 'it would take the three laborers, working together, to perform the work g ? If a laborer can do a piece of work denoted by a, in a number of days denoted by b, he can do in 1 day so much of the work as is denoted by j ; the second in 1 day can do so much as is denoted by -; and the third so much as is denoted by-; hence, the three u J working together can do a c e _ adf ■+• bcf -f bde 6 + d f = bdf Let x denote the time required to perform the work g ; then, the three can perform the work - in the time 1 : * r x from the conditions of the problem, p __ adf 4- bcf -f bde x~ bdf taking the reciprocals of each member, and then clearing of fractions, we have, _ b dfg X '^ adf±bcj+b * = 224 - This problem is also solved by a single unknown quantity more readily than by two. 23. A number is expressed by three figures j the sum of these figures is 11; the figure in the place of units is double that in the |jlace of hundreds ; and when 297 is added to this number, the sum obtained is expressed by the figures of this number reversed. What is the number ? Let x, y and z denote the digits in their order j then will the number be denoted by 100* -f \§y+z\ from the conditions of the problem, x+ y + z=zll < < - • (1) z = 2* « « < (2) 100* + lOy + z -f 297 ss 100* + lOy -f x - • • (3)} reducing (3), gives 99* ~ 99* = 297 (4); substituting * = 2* in (4), and reducing, 99* = 297; .-. * = 3 ; 101."] EQUATIONS OF THE FIRST DEGREE. 53 whence, by successive substitutions, y = 2, 2 = 6. ^4ns. 326. 24. A person who possessed $100000 dollars, placed the greatei part of it out at 5 per cent, interest, and the other part at 4 pe) cent. The interest which he received for the whole amounted t* 4640 dollars. Required the two parts. Let x denote the greater part ; y " " lesser " From the conditions of the problem, I* — — = interest on x dollars at 5 per cent. ; 100 » i|= " y « 4 « « then, x + y = 100000 .... (1) ^ + iy =4 640 .... (2) 100 ^ 100 ™ clearing (2) of fractions, 5x + 4y = 464000 .... (3) combining (1) and (3), y = 36000, whence x = 64000. 25. A person possessed a certain capital, which he placed out at a certain interest. Another person possessed 10000 dollars more than the first, and putting out his capital 1 per cent, more advan- tageously, had an income greater by 800 dollars. A third, possessed 15000 dollars more than the first, and putting out his capital 2 per cent, more advantageously, had an income greater by 1500 dollars. Required the capitals and the three rates of interest. 54: KEY TO DAVIES* BOURDON. [101. Let x denote the number of dollars in 1st capital; and y the rate per cent. ; then, (x x v^ -i -^ will denote the number of dollars of 1st income ; 2d " 3d " (*+ 10000) (y-H) 100 v (1 u (*+15000)(y+2) 100 ll u from the conditions of the problem, (x + 10000) (y + 1) 100 x x y 100 + 800 (x + 15000) (y + 2) * 100 x x y 100 + 1500; clearing of fractions, performing indicated operations, transposing and reducing, v lOOOOy -f * = 70000 15000y-f 2* = 120000; combining and reducing, 5000y = 20000 . • . y = 4 ; and Oy substitution, a; = 30000 ; 2d. $40000, rate 5 per cent. 3d. $45000, rate 6 " " 26. A cistern may be filled by three pipes, A, B, C. By the two first it can be filled in 70 minutes ; by the first and third it can be filled in 84 minutes ; and by the second and third in 140 minutes. What time will each pipe take to do it in ? What time will be required, if the three pipes run together ? Call the contents of the cistern 1. 101.] EQUATIONS OF THE FIKST DEGREE. 55 Let x denote the quantity discharged in 1 minute by the first ; y " " " " " " second; z " " " " « « third; then will -. - and - denote the number of minutes required x y z for the pipes, separately, to fill the cistern ; and, 1 x 4- y + 2' will denote the number of minutes required for all three tc fill it, running together ; from the conditions of the problem, x+ y=ro • ■ • • (1) *+v=| • • • • (2) y + 2 = no- '•■ ' • (3); clearing of fractions, 70* + 70y = 1 • • • • (4) 84z + 84z = 1 • • • • (5) 140y + 140s = 1 - • • • (6); combining (1) and (2), 840*/ - 840^ r= 2 • . t • (7) ; combining (6) and (7), 1680y = 8, .'. V = tL 210' substituting in (1), and transposing, _L _L JL _L * " 70 210 ~" 210 ~ 105 ; 56 KEY TO DAVIES' BOURDON. [102. substituting in (3), _ J_ 1 J_ X ~ 140 210 ~~ 420' 1 I I , * J T y 210 V 105 T 420 60 hence, - = 105, - = 210 ; - = 420. , * , = 60. x y z ' x -f- y -f z 27. A has 3 purses, each containing a certain sum of money. If $20 be taken out of the first and put into the second, it will contain four times as much as remains in the first. If $60 be taken from the second and put into the third, then this will contain If times as much as there remains in the second. Again, if $40 be taken from the third and put into the first, then the third will contain 2J times as much as the first. What were the contents of each purse ? Let x denote the number of dollars in the first purse. y " " " " second " z " " " " third " then from the conditions of the problem, 4 (x -20) = y + 20, i (y - 60) = z + 60, 2_40 = 23 (* + 40); clearing of fractions, performing operations and transposing, 4x- y= 100 .... (1) 7y - Az = 660 ... (2) 82 -23* = 1240 .... (3); * combining (1) and (2), 28x- 4^ = 1360 .... (4); 102.] EQUATIONS OF THE FIRST DEGREE. 57 combining (3) and (4), 33* = 3960; .\ x = 120 ; by substitution, y = 380 ; z = 500. 28. A banker has two kinds of money ; it takes a pieces of the first to make a crown, and b of the second to make the same sum. Some one offers him a crown for c pieces. How many of each kind must the banker give him ? Since it takes a pieces of the first to make 1 crown, - — the part of a crown in each piece ; and -, the part of a crown in each piece of the second : let x denote the number of pieces taken of the first kind, y " " " " second " from the conditions of the problem, x + y = c x II - + j- = 1, or bx -{- ay z=. ab \ by combination, % by — ay = be — ab; or y (b — a) = b (c — a) ; (a — c) b . a (c — b) .\ y = f— ; whence, x = — -^. r a — b a — b 29. Find what each of three persons, A, B, C, is worth, knowing, 1st, that what A is worth added to / times what B and C are worth, is equal to p ; 2d, that what B is worth added to m times what A and C are worth, is equal to q ; 3d, that what C is worth added to n times what A and B are worth, is equal to r. 58 KEY TO DAVIES* BOURDON. . [102. Let x denote what A is worth, y " " B " z " " C " then, from the expressed conditions, x + l (y + z)=p . . . . (1) y + m(x +z) = q • • . . (2) 9 4. n (x + y ) = r . . . . (3) . wnich, by adding and subtracting Ix, my and nz, may be written under the forms (1_ i) x+ i ( x + 1/ + z ) =p . . . . ( 4 ) (1 — m) y -f m (x -f y + z) = q • • • • (5) (1 — ri) z + n (x -f y -f z) — r • • • • (6) ; dividing both members of each equation by the co-efficient of its first term, •+it1 <«+*,+«)« Hh • ' * ' (T) y + r ^-(*+y + 2) = 1 - ? — .... (8) 1 — m v ' 1 — m x adding these, member to member, and deducing from the resulting equation the value of x -f- y + ^ * + ,-*- + * 1 — / 1 — m 1 — n . • *+y+z= j ' ' (10). 1 + 1 J+i + 1 1 -— / J — m 1 — n Denote the second membei of equation (10) by the single letter s t 1 p I Is 1 — I q ms 1 m 1 — m r ns P- ■Is 1 - ■I 9- ms l - m r — ns 102.] EQUATIONS OF THE FIRST DEGREE. 59 a known quantity. Then by substituting this for the factor * +■ V + z t m eacn °f tne equations (7), (8) and (9), and deduc- >»g the values of ar, y and z, we have, • • • (ii) • • • (12) !,_„ !_„!_„ • * * ( 13 >- Had we represented the polynomial x -j- y 4* i 5>J s> the alge- braic work would be slightly diminished, but the preceding method has been followed in order to show more clearly the process of solu- tion. / 30. Find the values of the estates of six persons, A, B, C, D, E, F, from the following conditions : 1st. The sum of the estates of A and B is equal to a ; that of C and D is equal to b ; and that of E and F is equal to c. 2d. The estate of A is worth m times that of C ; the estate of D is w r orth n times that of E, and the estate of F is worth p times that of B. 1st Solution. Let x denote the value of A's estate ; a — x « it " B's ' y a : ' C's • •'-*•«■ a " D's ' z " a " E's * c — z '• " " F's " from the conditions of the problem, 60 KEY TO DAVIES BOURDON. [102. x = my x — my = • • • • (i) b — y = nz - or, ■ y + nz = J • • • • (2) c — z = p (a — x) j92r — z =z ap — c • • (3) combining (1) and (2), x 4* ffinz = bra • • • • (4) ; combining (3) and (4), and finding the value of z, bmp + c — ap which denote bv P • combining (3) and (4), and finding the value of ar, amnp -j- bm — cmn . , ■ a: = — , which denote by Q\ mnp 4 1 substituting the last value in (1), and finding the value of y, anp -\- b — en mnp -f- 1 which denote by R ; whence, A's estate is equal to ft B's " If « a- ft C's " U ft D's " u " 6-5, E's « a i>, Fs " It " c - P. We might have combined (2) and (3), eliminating 2, and then from this resulting equation, taken with (1), have found the value of ot and y. 2d. Solution By a single unknown quantity, Let x denote the value of C's estate ; then will b - X " " D's « 116.] INEQUALITIES, 61 mx denote the value of A's " a — mx " " B's " p(a — mx) " " Fs " c — ^(a-ma:)" " E's " and from the remaining condition of the problem, b «_ x = n \C 19, to find the smallest limit of x< If we add 6 to both numbers of the inequality, we have 6#>19-f6, or, 6ar>25; dividing both numbers by 5, we have x >5. 14 2. Given, Sx -f- — - x — 30 > 10, to find the least limit of *. Reducing, 8x + ?*— 30 > 10, or, 10;r — 30 > 10 ; adding 30 to both members of the inequality, and dividing by 10, we have, x>4. x x x 13 17 3. Given, ^"~-o+o^"o^ > "o"' t0 ** n< * tne * east ^ m ^ °^ Xi \* *» -» •* 62 KEY TO DAVIES* BOURDON. L 123. Multiplying both members by 6, we have, x — 2x + Zx + 39 > 51; reducing, subtracting 39 from both members, and dividing by 2, we have, #>6. OX CL^ 4. Given, — - -+■ bx — ab > — to find the least limit of x* 5 5 Multiplying both members by 5, we have ax -f- 56rc — 5ab > a 2 ; adding -f *>a6 to both members, and dividing by the co-efficients of x, we have, (a -f 5b) x > a {a -f 56) ; or, x > a. 5. Given, — a# -f- a& < — , to find the largest limit of z> Multiplying both members by 7, adding — 7ab, and dividing by the co-efficients of x, we have, (b — 7a) x < b (b — 7a) or, x < b> SQUARE ROOT OF NUMBERS. ROOT. REMAINDERS. 1. 85. 2. 133. 3. 997. 4. 9256 . »• . . 5437. 5. 8234 . j . . . 13919. 6. 1671 . i , . . 160. 7. 6123 . . . 4913. 132-136.] SIMPLIFICATION OF RADICAL8. 63 60507. 958510 252G04. Note. — "Where the square root consists of six or more places of figures, a very good approximate result may be obtained by finding by the rule half or more than half of the figures, and then having formed the divisor as directed, bring down all the figures and proceed as in simple division till the required number of places of figures is obtained. In the last example, having found the figures 958, we have a remainder 977 ; bringing down the remaining figures and forming the divisor as directed, the process of approximation will be as indicated 1916)977672704(510... . SQUARE ROOT OF POLYNOMIAL ROOT. 2. a 2 + %ax + x 2 . 3. a 2 — ax H- x' 2 . 4. 2z 3 + 3z 2 - x -f- 1. 5. 3a 2 - 2ab + 4b 2 . 6. 5a 2 6 - 4abc -f- Mc 2 — Sate. SIMPLIFICATION OF RADICALS. EXAMPLES. SIMPLIFIED. 1. ynbcMc. 2. y™W. 3. -v/32a 9 6 8 c. 4. -y/256a 2 6 4 c 8 . 5. V / 1024a 9 6 7 c 5 . 6. ^/12SaP b*cH. 5a-\/Sabc. ma?c^/Zb. 4a 4 M- v /2ac^ 16a6V. S2a*b 3 c 2 to find the values of x. So ad Clearing of fractions, transposing and reducing, a 2 - b 2 X 2 - ;nce, by the rule "sr*^ •T 2ab ^V 1 ^ — 2a 2 b 2 -f 6 4 4a 2 6 2 „2_^2 a 8 + 6*. 2a6 * 2a6 ' 2a 2 a taking the upper sign, x = ^ == -, 2b 2 b " lower " x = — 5-j as 2a6 a (fo3^ 1+c x 2 ~ + t + l - — 2. Given h — + * = — ! T + c? to find the values of x. Clearing of fractions, transposing and reducing, 2 1 d2 ~ c l 150.1 EQUATIONS OF THE SECOND DEGEEE. 65 whence, by the rule, d 2 — c x = 2cd ± 7 1 rf 4 -2«P + c 8

90 90 27 _ a . 4. Given — - = — — , to find the values of a?„ x x + 1 a; + 2 Dividing both members by 9, 10 10 3 x x + 1 x + 2* clearing of fractions, transposing and reducing, 2 7 20 X z X = • 3 3' whence, by the rule, 66 KEY TO DAVIES* BOURDON. [150. 7 /20 49 7 17 hence, x as 4, and * a* — —• « — 5 . b o 5. Given — — 2 sb £-, to find the valuei of ar. 8 — a: x — 2 Clearing of fractions, &c, „ 39 28 whence by the rule, 39 _, A~28 io ± V-T 1521 _ 39 31 + 100 ~~ 10 10 hence, x = 7, and rr = — = -. x 2 6 — 1 6 •6. Given or — -+- b ss — = — x 2 -f- -#? to find the values ofx. o o a Clearing of fractions, &c, 2 a2 - b A x 1 x = o ; a whence, by the rule, a* - b L a 4 — 2a 2 6 + & 2 a 2 — 5 a 2 -f& sr *%/* + — s? — ""t^-ir 1 hence, x = a, a; = &— & , 3*3 a 2 6 + « , «" & 2 c to find the values of ar, 7 . Glven ___,+ «,+ 161.] EQUATIONS OE THE SECOND DEGREE* 67 Clearing of fractions) &c., „ 2b a 2 - £ 2 x z x £z whence) by the rule, 6 / c 2 c 2 ~~ c c , 6 -f- « , 6 — a hence) x az —, ana * C ' c 8. Given mx* -f mti s= 2my / n • a; -f *we 2 , to find the values of ts Transposing and reducing, 2m Jn~ mn * 2 y — x s= — ; to — ■ n to — ft whence, by the rule, to x we, — to y^ /_ mn m*?t ^ m\/5t n^/m — /i V to — n (to — lip ~~ to — - n to — n ■ m*/w ■+- n\fm \/mn ( -i/ro~-f- a/w) ^/frm hence £ ssc — — — ~ — — r y *__ j since, to — rt s= (-y/ro + y^n) (^/m — ■ -vAF (Art. 47.) also,, ^_ to yn~- n\/m ^ yrnn (ym— - \/ft) yWi »» — ft (V^+ -\A) (V^— V^) V^+ V* 2c£ Solution. mx 2 -f tow. = 2to y^tT- a; -f wz 2 ; transposing the first term of the second member, we have» 68 KEY TO DAVIES* BOURDON* [15L mx 2 — 2m yn • x + mn ss nx 2 5 observing that the first member is the square of a binomial whose terms are \/m • a: — ^/m -yfn^ we have, •y/m • x — -y/m -y/n a= dfc *y/7i * a:; .t*», yw • a; ^= yV« x = ym yn (^/rrJT^P -y/n) x s= ^/m^/n : hence, s/mn . i« and, x y?/i — -y/rc -y/m 4- -/» , „ 6a 2 , h 2 x ah — 26 2 3a 2 9. Given abx 2 4 = ; x 1 c 2 c c l c to find the values of x. Clearing of fractions, &c, 2 b 2 + 3a 2 _ ah — 2& 2 + 6a 2 * + a*tf P" a** 2 "' whence, by the rule, h \/- 2aJc~ V a6c 2 + 4a% 2 c 2 g -f 3a 2 3a 2 + 4aS - & 2 ■ * 2a6c 2afor ' 4a6 — 2£ 2 2a — 6a 2 -f 4a5 3a + 26 hence, x s — ^— ; = » a; ss ^7-7 sa = • 2abc ac 2abc be ,, ~* 4a: 2 , 2a: , mA mg . 3a: 2 , 58a: 10. Given — -f y + 10 = 19 - y 4- -y* <» find the values of x. 161 J EQUATIONS OF THE SECOND DEGREE. Clearing of fractions, &c., z 2 — Sx = 9 ; whence, by the rule, * = 4rfc v /9-flt> = 4=fc5j hence > a; = 9, x = — 1. 11. Given l±- a __ * = ?_=:.* x — a Clearing of fractions, &c., x , to find the values of x. (M-2)a^ r 6-2 ■ whence, by extracting the square root of both members, -**tflS - — -v/?? 12. Given 2* + 2 = 24 - 5s - 2**, to find the values of * Transposing and reducing, * 2 -f gr = ll whence, by the rule, * ? ±v / 11 + 49 7 15. 4 V T 16 4 4' whence, * = 2, and « = - 11. 2 13. Given *» _ « _ 40 == 170, to find the values of *. Transposing, z 2 -x=z 210 : 70 KEY TO DAVIEs' BOURDON. [151. whence, by the rule, hence, * = 15, and x = —-14. 14. Given 3s 2 -f 2x — 9 = 76, to find the values of x. Transposing and reducing, 2 85 whence, by the rule, * 2 + 3*-3.> 1 /85 , 1 1 16 17 2 hence, x = 5, and x = — — = — 5 -• 3 3 15. Given a 2 -f b 2 — 25a? + x 2 = — — , to find the values of x. n 2 Clearing of fractions, &c, 2i 2bn2 - n2a2 + n2f)2 . w 2 — n 2 " m 2 — n 2 ' whence, by the rule, bn 2 nv r /nW+_nW b 2 n* ■n 2 ~ V m 2 — n 2 m* — 2m 2 w 2 + w* bn 2 n y/ a 2 m 2 + b 2 m 2 — a 2 n 2 n 2 — m 2 ~ n 2 — m 2 whence, x = n2 __ m2 \ bn ± y/a 2 m 2 + b 2 m 2 — a 2 n 2 [ 154.1 EQUATIONS OF THE SECOND DEGREE. 71 PROBLEMS GIVING RISE TO EQUATIONS OF THE SECOND DEGREE. 4. A grazier bought as many sheep as cost him £60, and after reserving 15 out of the number, he sold the remainder for £54, and gained 2s. a head on those he sold : how many did he buy ? Let x denote the number purchased : and x — 15, the number sold ; then will denote the number of shillings paid for 1 sheep, and — the number of shillings received for each. x — — 1«> From the conditions of the problem, 1200 1080 x ~s-15 ' clearing of fractions, &c, x* + 45* = 9000 : whence, by the rule, * = - 4 2 5 ±v / 9 ooo + 2( f =- 45 195 2 2 ; hence. x = 75, and x = — 120, the positive value only, corresponds to the required solution. 5. A merchant bought cloth for which he paid £33 15s., which he sold again at £2 8s. per piece, and gained by the bargain as much as one piece cost him : how many pieces did he buy ? Let x denote the number of pieces purchased : 675 then will, — ■ denote the number of shillings paid fbr each, 72 KEY TO DAVIES* BOURDON. 154. and 48« the number of shillings for which he sold the whole. From the conditions of the problem, 48z-675= — ; then, by clearing of fractions, &c, „ 225 225 x l x — • 16 16 whence, by the rule, 225 / 225 50625 225 255 32 ~ V 16 + 1024 ~ 32 ± 32 ; . , • • 480 ■ using the positive value only, x = — — = 15. 6. What number is that, which being divided by the product of its digits, the quotient will be 3 ; and if 18 be added to it, the order of its digits will be reversed ? Let x denote the first digit, and, y u second " then will 10# -f y denote the number. From the conditions of the problem, 10s + 2/+18 = 10y + *; whence, by reduction, * 10a; -f y = 3xy, y-x = 2-, finding the value of x in terms of y from the second, and sub* stituting in the first, we have, 154.] EQUATIONS OF THE SECOND DEGREE. 73 whence, by transposing, &c., 17 „ 20 by the rule, y 2 ~yy 2 = --3-' and ' 17 / 20 , 289 , 17 7 7* V "T+ir'+T** taking the positive sign, y = 4 ; whence, x 22 2, and the number is 24. 7. Find a number such that if you subtract it from 10, and mul tiply the remainder by the number itself, the product will be 21. Let x denote the number : from the conditions of the problem, (10 - x) x = 21 ; or, x 2 — 10* = - 21 ; by the rule, z = 5±, v /-21+25=:5:fc2; whence, \ x = 7, and x — 3. 8. Two persons, A and B, departed from different places at the same time, and travelled towards each other. On meeting, it ap- peared that A had travelled 18 miles more than B; and that A could have performed B's journey in 15J days, but B would have been 28 days in performing A's journey. How far did each travel? Let x denote the number of miles B travelled ; x + 18 " " " A " l|l " " " A " in one day; t: B 74 KEY TO DAVIES' BOUKDON. [154"55. —X- — denote the number of miles B travelled in one day , * / +18 v " " days A \ 15} / * « from the conditions of the problem, x+ 18 _ x_ x 2 -I- 36a; + 324 _ x 2 # __x__ ~ aT+ .18 ° r ' 28~~ " 15f 5 15f 28 clearing of fractions, and reducing, , 324 2916 X 1 X = • 7 7 By the rule, 162 /2916 *= 7 -V 7 26244 + 49 162 216 7 7 ' hence, using the upper sign, to find the values of x< x \ x 2 b 78 KE? TO DA VIES* BOtJfcDoN. [166. Multiplying botn members by bx, and transposing, b ya 2 — ■ x 1 =z x 2 — ab ; squaring both members, b 2 a 2 — b 2 x 2 -.it x* - » 2afo 2 -f a 2 b 2 ; cancelling 6 2 a 2 , dividing both members by x 2 and transposing, a; V a: multiplying both members by x, and transposing, 2^/ax sb 6 2 x — - re — ■ a s= (b 2 — 1)« — a 5 squaring both members, 4ai 2 + n* l-4n 2 1 - 4n 2 ' whence by the rule, 1 - 3n» / (l-2n 2 + rc*)a 2 (1 - 6tt 2 + 9m*) 1 _ 4^2 a V 1 _ 4w 2 + (1 - 4rc 2 V 2 ~ (1 — 3w 2 ) 2rc 3 a _ (1 - 3/i 2 dfc 2« 3 ) a 1 _ 4 n 2 1 - 4w 2 1 — 4m 2 Taking the upper sign, and dividing both terms of the fraction by 1 +2n, _ (1 — 2n + rv*)a, __ (1 — nfa_ X ** P^ " . 1 - 2n~ ' Taking the lower sign, and dividing both terms by 1 — 2n, — g ( 1 + jfo + n 2 ) ,. (1 + nfa X ~~ 1 + 2n ~ 1 + 2/i ' «0 ± w ) 2 taking the two values together, x as ' • • 7. Given Y — = + — ; =\/ t> to "na *• Jx Jx V 5 Multiplying both members by y^ squaring both members, 2a + 2ya 2 -a: 2 =:y ; r, 2 445 203 X »— • £ — — • 8 8 or, taking the upper sign, x t± — = 81 242 121 " lower sign, x = — - = -j- (X c 5. Given, - — bx 4 -f - x 2 = 0, to find x \ transposing and re'ducing. whence, «» ± Vgj^^p+-|^» reducing, * a =fc ^/ c j V*^" ± EXAMPLES OF REDUCTION OF EXPRESSIONS OF THE FORM OF y« ± Vk Reduce to its simplest form, v/28 + 10 V3. a = 28, 6 * 300, c as 22. 84 KEY TO DAVIES* BOURDON. [172. Applying the formula and considering only the upper sign, y/28 + 10^3 = 5+^ 5. Reduce to its simplest form, wl -f 4-y/— 3. a as 1, 6 as — 48, c = 7 ; applying the formula, &c, */l -f 4V-T =a 2 + ^4 C 2 _ 4s '2 and, s = s — 2 ^ = * — V° 2 — *' 2 i reversing the order of the members of (4), and proceeding as before, we find in like manner. x = s' +yc and, • y — s' — -y/c 2 — s 2 . 4. The sum of the squares of two numbers is expressed by a, and the difference of their squares by b : what are the numbers ? Let x and y denote the numbers. From the conditions of the problem, 88 KEY TO DA VIES' BOUKDON. x 2 -f y 2 = c .• x 2 - y 2 = b . By adding, member to member, • • • (1) • - • (2)-_ 2x 2 =za + b by subtracting, . 4- /° + *. 2y 2 = a — b /a -6 y = ± v-ir- [181. 5. What three numbers are they, which, multiplied two and two, and each product divided by the third number, give the quotients; a, b, c 1 Let x, y and s, denote the numbers From the conditions of the problem, — = a or, xy = az • • • (1) ■^ = b or, yz = bx • • • (2) xz fO\ — = c or, xz =: cy • • • (3). Multiplying (1), (2) and (3) together, member by member, x 2y'z z 2 __ a l C xyz ; dividing both memoers by xyz, xyz = abc • * • (4) ; substituting in (4) the value of xy taken from (1), and dividing both members by a, z 2 = be . • . z = V6c. Substituting the value of yz and dividing by b, 182.] EQUATIONS OF THE SECOND DEGREE. 89 x 2 = ac . * . x = Jac. Substituting the value of xz and dividing both members by c, y 2 = ab .*. y —Jab. 6. The sum of two numbers is 8, and the sum of their cubes is 152 ; what are the numbers % Let x and y denote the numbers. From the conditions, * + y= s (i) *3 + 2,3 = 152 ( 2 ); cubing both members of (1), x* + Sx 2 y + Zxy 2 + y 2 = 512 • • . • (3) ; subtracting (2) from (3), member from member, and dividing both members by 3, x 2 y + xy 2 m 120 ... . (4) ; substituting the value of a; taken from (1), (64 - Uy + y 2 ) y+(8-y)y 2 = 120, or, reducing, y 2 — 8y — — 15 ; whence, y = 4 ± -y/ — 15 -|- 16 = 4 ± 1 . • . y = 5, y = 3 ; whence, from (1) x = 3, a; = 5. 7. Find two numbers, whose difference added to the difference ot their squares is 150, and whose sum added to the sum of their squares, is 330. Let x and y denote the numbers. From the conditions of the problem, 90 KEY TO DAVIES' BOURDON. [182. x z — y 2 + x - y = 150 • • ■ . (1), x 2 + y 2 + x + y = S30 • . • • (2); adding member to member, and reducing, x 2 + x = 240 ; whence, x = — - db y240 -f - — — - ± — - ; or, considering only the positive solution, x = 15; whence, from (1), by substitution, y = 9. 8. There are two numbers whose difference is 15, and half their product is equal to the cube of the lesser number : what are the numbers % Let x and y denote the numbers. From the conditions of the problem, x — y = : 15 • • • (1). xy 2 y\ or x = 2y2 • • • (2); substitul :ing in(l) an( I dividing both members by 2, whence, • y = 1 : 4 2 y V 2~ 15 2 ' 1 n 4 -Jfr 1 16" •• considering only the positive solution, y = 3 ; whence, from (1), x = 18. 182.] EQUATIONS OF THE SECOND DEGREE. 91 9. What two numbers are those whose sum multiplied by the greater, is equal to 77 ; and whose difference, multiplied by the lesser, is equal to 12? Let x and y denote the numbers. From the conditions, (x + y) x z= 77, or *» + zy c= «H • • • (1); (x-y)y=\2, or xy - y* = 12 • • • (2); make x = py; whence, (^+^ = 77, orj y2 = ^_ . . . (3)> 77 (p -1)^ = 12, or, V 2 =y~ ... (4); equating the second members and reducing, f -12^ = -12' whence, jr«^±y^-T*4 77 4225 _ 65 23 12. 576 ~ 24*24' taking the upper sign, _88_ 11^ P ~ 24 ~ 3 ; /Of* O substituting it (4), y = y -g- = g- V^J whence, a; = — -y/2; 42 21 taking the lower sign, p = — = — ; 92 KEY TO DAVIES' BOURDON. [182. substituting in (4), y=z\/-— - = 4; whence, * = 7. 10. Divide 100 into two such parts, that the sum of the!* squar roots may be 14. Let x and y denote the parts. From the conditions, * + y = 100 . • • (1), y^+ -v/y= 14 ... (2): squaring both members of (2) and subtracting (1), member from member. 2 i/xy = 96, or ^/xy == 48, or xy = 2304 : substituting for y its value, 100 — *, 100* - x 2 = 2304, or a; 2 — 100* = - 2304. whence, by the rule, x = 50 ±y 196 = 50 ± 14; hence, a; = 64, x = 36, and y = 36, y = 64. 11. It is required to divide the number 24 into two such parts, that their product may be equal to 35 times their difference. Let x and y denote the parts. From the conditions of the problem, x + y = 24 • • . (1) *y = 35(*-y) . . . (2); substituting in (2) the value, y = 24 — *, 24* - x 2 = 35 (2a: - 24), or, 24* - * 2 = 70* - 840 ; 182.1 EQUATIONS OF THE SECOND DEGREE. 93 whence, v & + 46a; = 840 j by the rule, x sa — 23 ± ^840 + 529 = — 23 =fc 37 ; hence, taking the upper sign, x = 14 ; by substitution, in (1), y s= 10. 12. What two numbers are those, whose product is 255, and the sum of whose square is 514? Let x and y denote the numbers. From the conditions, xy =a 255 • • . (1) x 2 f y* = 514 . . (2), multiplying both members of (1) by 2, adding and subtracting the resulting equation to and from (2), member by member, x* + 2*y 4- y 2 = 1024 . . (3) x*-2xy+i/ = 4: . . ; (4); extracting the square root of both members, x + y = 32, *-y = 2, whence, x = 17 ; y sa 15. « 13. There is a number expressed by two digits, which, when divided by the sum of the digits, gives a quotient greater by 2 than 1 the first digit ; but if the digits be inverted, and the resulting num- ber be divided by a number greater by 1 than the sum of the digits, the quotient will exceed the former quotient by 2 : what is the number 1 94 KEY TO DAVIES' BOURDON. [182. Let x and y denote the digits ; then will 10# + y denote the number. From the conditions. x + y (i) (2); clearing of fractions, and reducing, 8x — y — x 2 — xy = • • • (3), 6y — 4x — x 2 -— xy s= 4 * • • W; by subtraction, 7y — 12z = 4 ; . \ y = - 12x + 4 7 ' substituting in (3), 8.- 18s + 4 *' , ^ + 4a; = ) clearing of fractions and reducing, x 2 — — a? s= — — -, whence, I 20 / 4 , 400 20 18 19 V 19^361 19 19 Taking the upper sign, gives x = 2 ; whence, y — 4. 14. A regiment, in garrison, consisting of a certain number of companies, receives orders to send 216 men on duty, each company to furnish an equal number. Before the order was executed, three of the companies were sent on another service, and it was then found that each company that remained would have to send 12 men additional, in order to make up the complement, 216. How many 183.] EQUATIONS OF THE SECOND DEGREE. 95 companies were in the regiment, and what number of men did each of the remaining companies send % Let x denote the number of companies, and y " " " each should send ; then, y j^. 12 will denote the number sent by each. From the conditions of the problem, sy = 216 • • • (1), (x -3) (y + 12) = 216 . . ." (2) ; Performing operations, subtracting and reducing, 4z-y = 12. .'. y = 4x-12; substituting in (1), 4x 2 — \2x = 216, or x 2 — Sx = 54 ; whence, * = "2 = fc \/ + T == 3'a' 5 taking the upper sign, x = 9 ; hence, y = 24, and y + 12 = 36. 15. Find three numbers such, that their sum shall be 14, th© sum of their squares equal to 84, and the product of the first and third equal to the square of the second. Let rr, y and z denote the numbers. From the conditions of the problem, x +y + * ==14 . . • (1), x 2 -f y 2 + z 2 = 84 • • • (2), xz = y 2 . . . (3). Multiplying both members of (3), by 2, adding to (2) and reducing \ 96 KEl TO DA VIES* BOURDON. [183. z 2 -f- 2zz + z 2 = 84 + y 2 ; .-. x + « = -y/ 84 + y 2 ' ' (4) from (1), ar-f = 14 — y • • • • (5); equating the second members of (4) and (5) and squaring, 84+y 2 = 196-28y + y 2 ; .'• V = 4. Substituting in (1) and (3), x f z = 10 . . . . (6) #z = 16 . *. x ss — . Substituting in (6) and reducing, s 2 - 10« = - 16 z = 5 =fc t^bcs 5 rfc 3, 2 = 8; 8 = 2, and by substitution, ar = 2 ; z = 8. 16. It is required to find a number, expressed by three digits, such, that the sum of the squares of the digits shall be 104 ; the square of the middle digit to exceed twice the product of the other two by 4 ; and if 594 be subtracted from the number, the remainder will be expressed by the same figures, but with the extreme digits reversed. Let x, y and z denote the digits ; then, lOOx -f- lOy ■-+■ x will denote the number. From the conditions of the problem, x* + y 2 + z 2 = 104 . . . (1) y 2 - 2xz = 4 . . . (2) 100* + 10y + z — 594 =100* + 10y 4- x (3) ; 183.] EQUATIONS OF THE SECOND DEGREE. 97 subtracting (2) from (1), member from member, x 2 -f 2z* + z 2 sa 100 . • . x + z = 10; reducing (3) x — z = 6 ; hence, a; = 8, and 2=2. By substitution, y = 6, and the number is 862. 17. A person has three kinds of goods which together cost $230- 5 5 |-. A pound of each article costs as many Jj dollars as there are pounds in that article : he has one-third more of the second than of the first, and 3J times as much of the third as of the second . • How many pounds has he of each article ? Let #, y and z denote the number of pounds of each article. From the conditions of the problem, x 2 y 2 z 2 _5525 5 24 + 24 + 24 ~" 24 ' or, x 2 + y 2 + * 2 = 5525 (1) 4 y 2 =™x 2 . . . (2) 7 14 * = 2 y •*• "*?* and, f-flU • • • (3); substituting in (1) and reducing, x 2 = 225 . • . x = 15, substituting in (2) and (3), y = 20 ; z = 70. 18. Two merchants each sold the same kind of stuff: the second sold 3 yards more of it than the first, and together, they received 35 dollars. The first said to the second, " I would have received 24 dollars for your stuff." The other replied, "And I would have 7 98 KEY TO DAVIES' BOURDON. [183. received 12 J dollars for yours." How many yards did each of them sell ? Let s and y denote the number of yards sold by each. 24 Then will — denote the price the first received per yard, y \ 25 and — will denote the price the second received per yard. From the conditions of the problem, — +^ = 35, or, 48*2 + 25y 2 = 70xy : substituting in the second equation the value of y taken from the first, 48s 2 4- 25 (s 2 + 6x + 9) as 70s 2 + 210s ; reducing, x 2 — 20s = — 75 ; whence, s as 10 =fc <\/25 = 10 ± 5, or, s = 15 ; s as 5 ; substituting, _ y = 18 ; y = 8. J 19. A widow possessed 13000 dollars, which she divided into two parts, and placed them at interest, in such a manner, that the incomes from them were equal. If she had put out the first portion at the same rate as the second, she would have drawn for this part 360 dollars interest ; and if she had ^placed the second out at the same rate as the first, she would have drawn for it 490 dollars interest. What were the two rates of interest % Let s and y denote the rates per cent. -Let z denote the 1st portion ; then will 13000 — z denote the 2d. 183-113-17.] EXTRACTION OP ROOTS. 99 From the conditions of the problem, xz (13000 — z)y ?- /1X m = - — Too - ^ or ' ** = 130<%-^ • • • (1), j^ = 360, or, ^ = 36000 ■ fy. (2), < 13(K ^|-*)« -. 490, or, 13000* - zx = 49000 . . (3). Substituting in (1) the values of zy and zx taken from (2) and (3) and reducing, we find, x =s= y + 1. Substituting this value of # and the value of z taken from (2) in (1), and reducing, we find ' , , 72 36 36 /36 1296 36 ± 42 36 im U Whence, y ^^y^-^ by substitution, 13 ' * * V ~ #k7, and z = 6000, 13000 - z = 7000* EXTRACTION OF ROOTS. 2. Ans. x 2 + 2x — 4. 3. -4n.?> 2z 2 — x -f 2. 5. <4nA 3a^ — 2ta. 6. -4ns» 2^ — 1, 7 TRANSFORMATION OF RADICALS. \ ADDITION. 1. V4SP = 45 v / 3a, and b^Tba^hby/Za - y .% % ilwj. 9^0, 2. 3^4^= 3^/2a, ; 2^2^ .*. Ans. 5^2^ 3. 2V45sn 6^ ; 3V57 .\ ^4rcs. 9^5. OO' Of TH1 100 KEY TO DAVIES* BOTTRfcON, SUBTRACTION. [218-19-20. 1. \fia*b + 16a* =s 2a »/5 -f- 2a, and ^6«-f JW - ^V^ + 2a, .-, .4n*. (2a - 5) %/b+*i> 3. 3^/ia^ = 3^/2^ ; 2 3 y /2a~; ,'. Jnt. ySi, MtfLTHTLlCATlOff, 5. -; § x vr vs* Vsr^ vW : ; :; : f*#? 6. 2-v/l5 -2 6 ^/3375; 3^10^3^/100; .-. Jn*, 6^337500 8. V^V 5 *; 3 y/3^Y^5 V^-V 1 ^ ^w*. j y648000. 9 ' Vr V ?29 ' V2 V 16384' V b - V~ lb > 42 /g . * . Ans. * / — 10. The product by the rule for multiplication is 28 ft „ J% 10 43 . in Ft 43 , 13 ~ DIVISION. 2. 2^3 x ^/i" - 2 ^y^y X ^256 - 2 »/720 x 256, 12 /729 v 256 2i 2 v /7^9x256-f-i 12 / 8x81=£::4 V 8 x 81 ^ l V^ An9 > 220-21.] TRANSFORMATION OF RADICALS. 101 3. y/V*x2>/3 = yV« x2 V 9= \/ 26 V / « = 12 \/^ yfi\/2 X^/3 = 2^/^/4 X V 27 = 2 V 10 ^ / a 1 12 Pv*T 1 12 /2 - - /9 — Ans. Ans. V-5 2 = 10 5. — = =: multiplying both terms by i/a — */6, we have V s + V 6 1 l/a — -y/6 again multiplying both terms by -^/a -f ^/b, we finally obtain V° 3 ~ V aH + V a62 ~~ V 63 Ans. a — b multiplying both terms by */a -f t/^ we have \/a + 2 4 v /o6 + -/^ and i 2-v/2"x (3)3 . , 1. Keduee ■ to its simplest terms. iV* 102 KEY TO DA VIES* BOFEDON. [232-233. Cancelling the ^2, and writing the quotient of 2 -f- J, we havt* { 4(2)^ 3 /3 ) * - ¥—- y to its simplest terms. Raising both terms to the 4th power, we have (1)«3»3» __ ft)* (3) V3_ ys 1 (2)*2(3) a (2) 3 (3)2 • (2) 7 x3 384 /( a Y3 _L . /ai ) "s (JV3 _L -/gj. 3. Reduce / J — ~-~| V to its simplest terms. v ( »%$$>* j Since the square root of the square root is equal to the 4th root, we need only operate on the terms of the fraction : , (F + yi?^ i_+yy' = *+yT 2-/2x(3)* *^*P§ ^6 Multiplying both terms of the fraction by the -\ft$, we have I ih£±^£3$3d+m : hence, 4. Multiply a*+ a 2 4* + aV + aJ + o¥ + 6 , ."2t?- i «i# , Jb , it 1 5 a 3 -f a J & rf + a 2 b* + a z b -f a&* -f a"^ _ A* - a?$ - ah - ab* - gM - jg a 3 — 6 2 . ^4t?s. l?-241-42.] ARITHMETICAL PROGRESSION. $ 5. DLvide a*-a*A"*-A+&* ||a*-6'* a* _ a?b% &-b -ah + b* - ah + 6*. 103 ARITHMETICAL PROGRESSION. 2. a = 2, rf = 7, ?i = 100;" . \ I = a + (n - l)d =695. 3. a = 1, a 7 = 2, n = 100 ; . ' . J = a + (n — 1) a* = 199. Hence, S = %n (a + I) = 10000. 4. Z = 70, c? = 3, n = 21, a = I - (n - 1) d = 10 ; S = }jn [21 -i (n - 1)<] =840'. 5. a =10, d=-$, n = 21 . •. / = a + (w - l)rf = *£■', whence, S = }n [21 - (n - l)rf] = 140. 6. a* = 6, J s 185, S = 2945 _ 2Z + d db ygg + rf) 2 -~8rf£ _ 376 ±4 taking the lower sign, n = 31, a = / — (rc — l)a* = 5. 7. a = 2, Z = 5,- n = llj . • . d = ^* = ~ = 0.3. n— 1 10 8. a = 1, a 7 = 1, n = 7i 7 = n ; ... ^ =i n[2a+0 i -l)d] = (i-±^. 9. a=l, d = 2, »==n; .-. 5 = \n [2a + (» — l)d] = n» 10. a = 4, c? = 4, n = 100 ; . . 8 =s$n [2a + (n - l)d] = 20200 20200yck. = llmt. 840ya 7 s. 104 key to davies' bouedon. [245-46-48-67-68. GEOMETRICAL PROGRESSION. 3. a = 2, r = 3, / = 39366, 8 as ^-^ = 59048. r — 1 4. assl, r = 2, n = 12; (IT* — — ffl . •. £ = = 4095 ; I = ar»-i as 2048. r — 1 5. a=l, r-2, w = 12; . • . S= ar * ~? = 4095 ; 4 195*. = £204 15*. r — 1 6. a = 1, r = 3, n = 10 ; _ ar» — a 59048 .^ ^ .-. S = = —^— 295.24. r — 1 2 / = ar"- 1 = 196.83. INSERTING GEOMETRICAL MEANS. 2. az=2, & = 486, w = 4; .-. r = ^243 = 3: hence, the progression, 2 : 6 : 18 : 54 : 162 : 486. SUMMATION OF SERIES. 4. p = 4, q = 4, n = 1, 5, 9, 13, &c. 4 4 4 4 4 1st auxiliary series, --f^-f---f — + _+... 4 4 4 4 4 5 ' 9 13 ' 17 ' r 4» -3' ^ = i( 4 -i^l)-' if "=»■ *** 274-286.] EXPONENTIAL EQUATIONS. 105 PILING BALLS. l _ 15; ... S = l^+Jl^±3 = 660 - i n = U . ... ig= "<" + | 1 >^ .±l> = 10i6; 1 . Z . o f , »' = 5, S' = 55 ; . • . S — S' = 960. 3. m = 30, - = 30 ; . , 8 = ^^f+H, 23405. 4. m = 26, n = 20 ; . • . S = 8330 ; m = 26, ft' = 8 ; S' = 1140; .-. fif- £'^7190. 5. » = 20; .-.5=1540; ft' = 9 ; .-. S' = 65; ... £-£' = 1475. 6. w = 15; .-.5=1240; »' = 5 ; .-.5' = 55; .-. S- 8' = 1185. 7. w= 52, ft = 40 ; . • . S = G4780 ; m = 52, n' = 18 ; . • . fif = 11001 ; .-.£-£' = 53679. EXPONENTIAL EQUATIONS. These equations may be solved as the preceding ones have been, but it will be better to make use of the table of logarithms (P. 288), together with the principle that the logarithm of a power is equal to the exponent of the power into the logarithm of the root. This principle gives the following solutions : I- 3* = 15; % )(«. fQ taking the logarithms of both members, i o i k lo g * 1.176091 _ Ad slog3 = log 15; .v* = ^ = 5 -^— = 2.46... 106 KEY TO DAVIES' BOUKDON. ' [286-309-310. 2 10* = 3; taking the logarithms of both members, x log 10 = log 3, or x = log 3 = . 477121- 3. 5* = J; taking the logarithms of both members, x log 5 = log f = log 2 — log 3 ; .. log2-log3 _ -.176091 _ log 5 ~ .698970 ~ "~ "^ THEORY OF EQUATIONS. 2. Two roots of the equation, x* — 12a; 3 + 48a: 2 — 68a; + 15 = 0, are 3 and 5 : what does the equation become when freed of them ? ** - 12a; 3 + 48a> 2 - 68a- + 15 Ifj-j X 2 _ [) X 2 4- 21a; — 5 x 2 - 9a; 2 4- 21a; — 5 | x - 5 x 2 — 4a; + 1 =0. Ans. 3. A root of the equation, X 3 _ 6a- 2 -f 11a; -6 = 0, is 1 : what is the reduced equation 1 x s — Qx 2 + 11a; — 6 | x — 1 x 2 — bx 4- 6 =0. Ans. 4. Two roots of the equation, 4a; 4 — 14a; 3 — 5a; 2 4- 31s + 6 = 0, are 2 and 3 : find the reduced equation. 314-325.] GREATEST COMMON DIVISOR. 107 4** - 14* 3 — 5* 2 + 31* + 6 | * - 2 4* 3 — 6* 2 - 17* — 3, 4* 3 — 6* 2 — 17* - 3 j*-3 4* 2 + 6* 4- 1 = 0. -dws FORMATION OF EQUATIONS. 2. What is the equation whose roots are 1, 2, and. — 3 ? (* - 1) (x - 2) (x -f 3) = * 3 - Ix + 6 =t 0. Ans. 3. What is the equation whose roots are 3, — 4, 2 -f- %/**> an( ^ 2- ^fl (* - 3) (x + 4) (* - 2 - V3) (* - 2 + -v/3) ' = ** — 3* 3 — 15* 2 + 49* — 12 = 0. Ans. 4. What is the equation whose roots are 3 -+• \/5^ 3 — ^/b[ and —6? (x-3 - Vo) (x - 3 + a r sb — - ; and applying the above form, we have 3 15 ' 2 A A tt 3 -— = 0. An*> 335-41.] EQUAL ROOTS. Ill 6. Transform the equation 3a; 4 - 13a; 3 + 7x 2 — 8a; - 9 ^ into another, the roots of which shall be less than the roots of the given equation by -• Make, x = u -f- - ; whence, x' = + - ; o o t applying the above formula, we have, \ EQUAL ROOTS. 4. What are the equal factors of the equation x i _ 7*6 + 10 *5 + 22a; 4 - 43a; 3 - 35a; 2 -f- 48a; -f 36 t=z 1 The first derived polynomial is 7a; 6 — 42a; 6 + 50a; 4 -f 88a; 3 - 129a; 2 - 70a; -f- 48, and the common divisor between it and the first member of the given equation, is x* — 3a; 3 — 3a; 2 -f 7a; + 6. The equation a; 4 — 3a; 3 -f 3a; 2 + 7a; -f 6 — 0, cannot be solved directly, but by applying to it the method of equal roots; that is, by seeking for a common divisor between its first member and its derived polynomial, 4a; 3 — 9a; 2 — 6x + 7, we find such divisor to be x -f 1 ; hence, x + 1 is twice a factor of the first derived polynomial, and three times a factor of the first member of the given equation (Art. 268). 112 KEY TO DAVIES' BOURDON. [341. Dividing, x* - 3^3 _ 3^2 _f_ lx + 6 _. o, by (x -f- 1)2 e& x* + 2x + 1, We have, a; 2 — 5a; -p* 6, which being placed equal to 0, gives the two roots xs2 and x x= 3, und the two factors, x — 2 and a: — 3. Therefore, (a; — 2) and (re — 3), each enters twice as a factor of the given equation ; hence, the factors are (x - 2) 2 (x - 3) 2 (x + 1)3. Am. 6. What are the equal factors of the equation x* ~ + 12, + 1, + 2, + 40, + 26, + 22, + 2, + 2, -40, -13, + 8, • « + 8, + 8, -11, -11, -34, + 34, + 15, -15, + 15: - 7, hence, there is but one entire root , which is — 1. 370-378.] NUMBER OF REAL ROOTS. 115 4. What are the entire roots of the equation 9z 6 -f 30a: 5 -f 22tf + 10a; 3 + 17a: 2 — 20a; -f 4 = ! This example is worked like the preceding, and there is but two entire roots. KUMBER OF REAL ROOTS. 3. What is the number of real roots of the equation X 3 _ 5*2 + Sx _ i = o | By finding the expressions which indicate, by their change of sign, the existence of real roots (Art. 302 and Example 1), we have X ss a 3 - 5a: 2 -f 8* - 1, X x =z 3a; 2 — 10a: + 8, X 2 = 2x - 31 X 3 4 - 2295, x 'H — oo gives 1 2 variations, x = + oo gives -f -f- H 1 variation ; hence, there is one real and two imaginary roots (Art. 300). For x as 0, we have f- 2 variations, for x = 1, " + H 1 variation: hence, the real root lies between and -f- 1. 4. What is the number of the real roots of the equation x * — x 3 — 2a: 2 — x — 3 = ? X = a: 4 — z 3 — 2a: 2 — a; — 3, X, = 4a: 3 - 3z 2 - 4x — 1, X 2 = 19a: 2 + 16a: + 49, X 3 b£ 101a; - 174, X< = - 1356277. 116 KEY TO DAVIES* BOURDON. [378-381. For x = — oc , we have -\ \- , 3 variations, for »= oo , " 4* -f + H , 1 variation ; hence, there are two real roots. 5. What is the number of the real roots of the equation x s __ 2z 3 + 1 = ? X = x 5 - 2a; 3 + 1, X x = 5** - 6* 2 , X 2 = 4a: 3 — 5, X 3 ae 24a; 2 - 25a;, X 4 = — 5a; + 6 X> = - 114. For x = — oo , we have - — | f- -f , 4 variations, for «= co , " -f + 4* + > 1 variation % hence, there are three real roots. , CUBIC EQUATIONS. 1. What are the roots of the equation x * _ 6z 2 + 3a; * 18 - • \ (1)? Transforming so as to make the second term disappear, a;3_9. T _28=xO * • • (2) p= — 9 q=z — 28; tubstituting in Cardan's formula, and reducing, x = 4. But the roots of the given equation are greater than those of equa- tion (2) by 2 ; hence x = 6, 381.] CUBIC EQUATIONS. 117 Transposing 18 in equation (1), and dividing both numbers by x — 6, we find x 2 + 3 = .-. x = ± ^^\ hence the three roots are 6, y/ — 3, — V ~ 3. 2. What are the roots of the equation ^3 _ 9;C 2 + 2 8z = 30 .... (1)? Transforming, we find x 3 + s = (2) .-. s = and a; as ± y^H. But the roots of (1) are greater than those of (2) by 3; hence the roots of (1) are, 3, 3 4- y/ — 1 and 3 — */-- 1. 3. What are the roots of the equation a 3 — lx 2 -f 14* = 20 • • (1)? Transforming, we have •, 7 344 , . 7 344 * = -3'* = -2^ substituting in Cardan's formula, we find 8 x = - for the real root. But the roots of (1) are greater than those of (2) by - ; 3 hence, for (1) we have x = 5. Transposing in (1), and dividing both members by x — 5, • x 2 -2x + 4 = 0; 118 KEY TO DAVIEs' BOFKDON. [391. or, x 2 — 2x = — 4 ; whence, x = 1 ± y'— 3; hence the roots required are. 5, 1 + y — 3 and 1 — */ — 3. TRANSFORMATION OF EQUATIONS BY SYNTHETICAL DIVISION. 3. Find the equation whose roots shall be less by 1 than the root&of x 3 - 7* + 7 = 0. 1 + - 7 + 7 || 1_ +1+1-6 + 1 - 6,+ 1 + 1+2 + 2,-4 + 1 l,+ 3 .-. y3_ f .3 3/ 2_ 4y + 1:=0# 4. Find the equation whose roots shall be less by 3 than the roots of the equation a* __ Sx 3 — lbx z + 49a: — 12 = 0. 1 -3 — 15 + 49 - 12 1| 3 + 3 + 0-45+12 + 0-15 + 4,+ + 3+ 9- 18 + 3- 6,- 14 + 3+18 + 6,+ 12 + 3 l,+ 9 .-. y* + 9y 3 + 12y a - 14y = 0. 391-400.] horner's method. 119 5. Find the equation whose roots shall be less by 10 than the roots of the equation ' x* + 2z 3 + Bx 2 + 4x — 12340 = 0. 1+2+ 3+ 4- 12340 1110 + 10 + 120 + 1230 + 12340 + 12 + 123 + 1234,+ + 10 + 220 + 3430 + 22 + # 343,+ 4664 + 10 + 320 + 32,+ 663 + 10 l,+ 42 .-. y* + 42y 3 + 663y 2 + 4664y = 0. 6. Find the equation whose roots shall be less by 2 than the roots of the equation x* + 2x 3 — Qx 2 — lOx = 0. 1+0+ 2- 6 - 10 + || 2 + 2+ 4+12+ 12'+ 4 + 2+ 6+ 6+ 2,+ 4 + 2 + 8 + 28 + 68 + 4 + 14 + 34,+ 70 + 2 + 12 + 52 + 6 + 26,+ 86 + 2 + 16 + 8,+ 42 + 2 1,+ 10; .-. y £ + 10^+422/ 3 +86y2+70y+4=0. HORNER'S METHOD OF SOLVING NUMERICAL EQUATION& 1. x 3 + x 2 + x — 100 = 0. By Sturm's Rule, we find 120 KEY TO DA VIES' BOURDON. [400. variations, variation ; variations, variation ; X = x 3 +z 2 + x - 100, X I = 3a; 2 -f 2x + 1 X 2 = - 4a; + 899 X, = - 2409336. For z = - go , _ + + _ } 2 for x = + ao , -f 4- — — , i hence, there is but one real root. Tor x = 4, - 4- 4 - , 2 for x = 5, 4- 4- H , - 1 hence, the real root lies between 4 and 5. 2. z* — 12z 2 4- 12a; — 3 = 0. By Sturm's Rule, X =x* - 12a; 2 4- 12* - 3, X x = 4a; 3 - 24* -f 12, or z 3 - 6a? 4- 3, X 2 = 2a* — 3a: 4- 1, X 3 = 13a;— 9, • X 4 = 20. a; = — go , H 1 f- , 4 variations, z = 4- co , 4- 4- 4. + 4. j q variation ; For for hence, there are 4 real roots. For for for for for for a;=,-4 xz= -3 x= 3= 4- 1 x= 4-2 a? = 4-3 + - 4- - 4- + - 4- - + + - 4- T + + 4- + 4- + 4- 4- 4- + 4 variations, 3 variations, 3 variations, 1 variation, 1 variation, variation : 400.] horner's method. 121 .hence, one of the roots lies between — 4 and — 3, two between and 1, and the remaining root lies between 2 and 3. 3. s* — 8s 3 + 14s 2 + 4x — 8 = 0. By Sturm's Rule, X = s 4 - 8s 3 + 14a 2 + 4x — 8, n X x = 4x 3 - 24s 2 -f 28s + 4, or s ? - 6s 2 -f 7s + 1, X 2 = 5s 2 - 17s -f- 6, X 3 = 76s - 103, X 4 rr 45475. For a = — co , H 1 f- 4 variations, for z= + ao, + + -f + + variation ; hence, the equation has 4 real roots. For s = -l -r - + - + 4 variations. for s= - + + - + 3 variations, for x= + 1 + + + 2 variations, for x=+2 + - - + + 2 variations, for IB +3 ± + + 1 variation, for x= 5 + 4 + 1 variation, for s= 6 + + + + + variation ; hence, one j root is between — 1 and 0, one between and 1, one between 2 and 3, and one between 5 and 6. 4. s 5 — 10s 3 -f 6s + 1 = 0. By Sturm ,'s Rule, X = s 5 - 10s 3 + Qx + 1, X, = 5s* - 30s 2 + 6, X 2 = 20s 3 - 24s - 5, X 3 = 96s 2 - 5s - 24, X 4 = 43651s + 10920, X 5 = 32335636224. 122 KEY TO DAVIES BOURDON. 1400. For x = — go h h — + 5 variations, for #= + 00 +4- + H — h + variation ; hence, the equation has 5 real roots. h h f- 5 variations, + H h — -f 4 variations, + h — 4- 4 variations, + + h -H 2 variations, 1 — h + 1 variation, h 4- H — h 1 variation; + + + + H — h variation ; hence, one root lies between — 4 and — 3, two roots lie between — 1 and 0, one root lies between and + 1, one ro>t lies between 3 and 4. For x= -4 for x= -3 for x= -1 for **= for * = + l for x= +3 for x— +4 APPENDIX. GENERAL SOLUTU.rf OF TWO SIMULTANEOUS EQUATIONS OF THE FIRST DEGREE. 1. Take the equ&cions, ax + by = c . . . (1), a'x+b'y = c' ... (2); multiply both members of (1) by V and of (2) by 6, then sub- tracting and factoring, we find (ab' — a'b) x = b'c- be' ; b'c — be' ab' - a'b (3). ac — a c In like manner, y = — — • • • (4). By means of formulas (3) and (4) any two simultaneous equations of the forms (1) and (2) may be solved. Thus, 4x + 3y = 31, Sx + 2y = 22 : by comparison with (1) and (2), a = 4, 6 = 3/ c = 31, a' = 3, b' = 2, c' = 22 ; by substitution in (3) and (4). 62 — 60 88 — 93 u 124 APPENDIX. EXAMPLES. ( - + ? =2 ) 1. Given J 3 4 v. to find x and y. ( 3* + 4y = 25 j j By comparison with (1) and (2), « = fc * = *, c = 2, a' = 3, y ek 4, c' s 25 ; by substitution in. (3) and (4), ] 3 — 4 1 3 ~ 4 ( 11a;— 5y == — 1) 2. Given < > to find # and y : (- bx + 16y = 124) J by comparison, a=r 11, b = — 5, c = — 1, a' = - 5, 6' = 16, c' = 124 ; oy substitution, _ 16 +620 1364 — 5 176-25 ? * 176-25 GENERAL SOLUTION OF THREE SIMULTANEOUS EQUATIONS OF THE FIRST DEGREE. 2. Take the equations, ax -f by -f- cz = d • • • (1), a'x -f b'y +c'z =d' • ■ .* (2), a"x 4- b"y 4- c"« = d" • • i (3). From (1) and (2) we obtain, by eliminating z, (c'a - ca')x 4- (e'A - cb') y = c'd - cd' . . . (4). ADDITIONAL EXAMPLES. 125 In like manner, from (1) and (3), (c"a - ca") x 4- (c"b - cb") y = c"d - cd" * • * (5; ; combining (4) and (5) and eliminating y, we find _ (c"b -. cb") (c'd - cd') - (c'b - cb') (c"d - cd") X -{c'a - ca! ) (c"b —cb") - (c"a~ ca") (c'b - c6') ' ' ™ In like manner, _ ( ... (2); multiplying both members of (1) by n, which is entirely arbitrary, we have nax -+- nby ss nc • • • (3) ; adding (2) and (3), member to member, and factoring, (na -f- a f ) x -f (nb -f b') y = nc + c' • • • (4). If it be required to eliminate y, place nb + b' = 0; .-. ft ft* — — j 128 APPENDIX, substituting tnis in (4) and reducing, we find —_c + e ' b + b'c- be' x = ___ _____ ... (5 ). tf it be required to eliminate x, place a' nu + «' == ; .*. n=— — ; a substituting in (4) and reducing, we find t c 4- C a aer — oc y& —-r-— = ab'-a'b ' ' ' «" -f" " a These values of we find y = :2; making n = 12, we find a; = 5. f^ + l = 36] ' 3. Given i >o 1 14 6 , n h to find a; and y, Multiplying by n, adding and factoring, (9n + 14) I + (6ti -n 6) 5 = 36« + 10 ; a; ' y , nmaking n = —-, we have - = S; . • . y = -, 9 y 3 making n = 1, we have 1 # = MISCELLANEOUS GROUPS OF SIMULTANEOUS EQUATIONS OF THE FIRST DEGREE. I. Given j 3z + 5< 2 5y~9 5a 3y 4 J to find a? and y. Combining and eliminating — , x \25 9/ y ""45 12' whence, - = ttt; . • . y = Iff- ADDITIONAL EXAMPLES. Combining and eliminating -> \9 25/ x ~ 27 12 ; 131 whence. 2. Given 1 _ 65 *~~192 . * = 2 $& * 65 + 2y~5 2a;-- 1 -y + 1 £=0 ► to find a? and y ; Clearing of fractions and reducing, * + 6y==: 15 . \ . (1), 2*-5y = -4 . . . (2): combining and eliminating ar, 17y = 34 5 ... yi=2 ; substituting in (I), # — 3 # 3. Given , y-2 X — ~ = B 7 ° ^ * +10 « to find x and y» Clearing of fractions and reducing, 7* - y = 33 . : . (i), 12y — x = 19 . . . (2) 5 combining and eliminating $ 83y = 166; ... y = 2; substituting in (I), a;s:5. j$g APPENDIX. *■ [ 5 \7 4 + 2^24 ] 4. Given 20 - 2y + 5,.40 f to find a; anc Clearing of fractions and reducing, hx + 12y = 148, 25s— 2y = 182; combining and eliminating,^, 155* =1240; ,<. x te 8 ; substituting, we find y = 9. ' 3^3 6 5. Given - ? + f = 8 + 2 ii + 3 = 10 to find ar, y and *. Clearing of fractions and reducing, 3of+-2y+ 2*:72 • • • (1), -ar-f- 3^ + 2*^48 * . - (2), & c + 2* a4 00 1 * • (8) i combining (1) and (2), eliminating y, \\x —Z =: 120 • • • combining (3) and (4), eliminating z r 25j = 300; .•„ #=12; by substitution in (3), ' z = 12, (1), y = 12. (4) J ADDITIONAL EXAMPLES. 133 2 + ! + 7 = 22 • C. Given • . y z 3 + 1 + 2 = 31 to find x, yan» a x , y z Clearing of fractions, • 1 21a: -f 14y -f 6z = 924 . ■ ■ (1), 10* + 6y +15z = 930 . • • (2), *+ 2y + 2 = 114 . combining (1) and (3), eliminating *, • • (3); ■ 15a; -f- 2y = 240 i combining (2) and (3), eliminating *, •• • (4); 5a; -f 24y = 780 . combining (4) and (5), eliminating ar, • • (5); ■ji 70y = 2100; ... $ from (5), a: = 12; from (3), , ' = 30; = 42. - 2ar- 3y -f 2z = 13 . . • (2) 1 t, • (3) | 7. Given - 2v — x = 15 2y+ 2= 7 * 5y + 3t> = 32 . . and v. Combining (2) and (4), eliminating r, 10y + 3ar = 19 . . combining (5) and (1), eliminating ar, • (5); 2% - 62 = _ 1 . . • (6); 134: APPENDIX. combining (6) and (3), eliminating z, 41y = 41; .-. y = li by successive substitutions, 2 = 5, x = 3, v = 9. 8. Given a? y 12 s 3 5 19 4 a; + 2 ' . 24 + y l+^i + 5 v. y z 8 » -> to find ft, y and a. Transposing, reducing, &c, 2i+3l * y 4 A = ± 2 24 x y x z 24 — 5-4-7- + 5-=—. i x y z 24 (2). (3): combining (1) and (3), also (2) and (3), eliminating -, z • ■ (4), _ l0 I + 43l = 4 | x y 24 el 1 " ; * -111™ . . . y '24 («U combining and jeliminating -, oq, 1 468 1 2 234 y= 24' •■• ^ = 2? ° r ^= 12 ' substituting in (5), x = 6 ; whence, 2 = 8. f 10+ 6y~ 4x 4 ] 9. Given < 6ar— 9y+ 3 " 3 126+ 8ar-17y 35 [ 100-12*+ ly ~~ 13 J to find a; and y. ADDITIONAL EXAMPLES. 135 Clearing of fractions and reducing, - 2a: + 3y = - 1 . . . (1), 262*-233y = 931 • • . (2). Combining and eliminating x, 160y = 800; .'. y = 5; by substitution in (1), x = 8. r ax -+- by = c 2 \ 10. Given < a(a -f *)__ - > to find x and y. Clearing of fractions and reducing, ax + by = c 2 • • • (1), ax — by = b 2 — a 2 . (2) ; Sj addition, £2 1 C 2 __ G 2 2as = 6 2 + c 2 -a 2 ; .'. * = — ^— ; by subtraction, a 2 4. C 2 _ ^2 2ty = c 2 + « 2 - &\ .*• y = 26 MISCELLANEOUS EXAMPLES OF EQUATIONS OF THE FIRST, SECOND AND HIGHER DEGREES, CONTAINING BUT ONE UNKNOWN QUANTITY. 1. Given Sx 2 — 4 = 28 -f A to find x : transposing and reducing, x 2 = 16; .-. x= ±4 2 . Given ?£±* - ?1±*? = 1 1? - 5s 2 , to find *; 136 APPENDIX. Clearing of fractions, transposing and reducing, x 2 = 25 ; . • . s = ± 5. 3. Given s 2 -f ab = 5s 2 , to find a; ; transposing and reducing, * 2 = V' '.? s=±-y^. «, a; + 7 a; — 7 7 _ . 4 - Glven irr^-^4:-^ = ^rw wfind *- Clearing of fractions, s* + 14s 3 — 24s 2 - 1022s — 3577 — s 4 + 14s 3 -f 24s 2 — 1022s 4- 3577 == 7s 3 - 343s ; transposing and reducing, 21s 3 = 1701s, or s 2 == 81 ; . •. s = ± 9. /x — 2 /s + 2 5. Given V s~+2 + V s~^2 "= 4 ' to find *> multiplying both members by -y/s -f 2, s 4- 2 / V*T^2 4- -~= = 4ys + 2; ya; — 2 multiplying both members by yx — 2. s — 24-s4-2 = 4 or X 2 = x3; ... ajsdbsV^" 6. Given a; 4- V 5x + 1° = 8 > to find * J ADDITIONAL EXAMPLES. f 137 Transposing and squaring both members, 5a; + 10 = 64- 16* + 0\ whence, x 2 — 21a; = — 54 ; . . , 21 L A " 441 21 , 15 by the rule, x = ~± y> Jb± + — = —±—\ .'. x = 3, x = 18. 7. Given 5 %/& + 7 */x* = 108, to find x : make 3 y /x^ = y ; whence, 3 ^/x i = y 2 ; substituting and reducing, 7 108 y 2 + e y = 5' 5 7 /108 ' 49 7 47 whence, y = _ _ ± ^Z— + _ = - _ ± _; 27 y = 4 and y = — — ; __ — /7 27V from which, a;=±-y/2/ 3 =±8 and x=±yy 3 =zk\/ 1 — ) 8. Given 3s 2 + 10a; = 57, to find x. By division, x 2 -f -— • x = 19 ; whence, o 5 ZtT , 25 5 14 .*. a; = 3 and a;=— 6J« 9. Given (x - 1) (x — 2) = 1, to find x. Performing indicated operations and reducing, x 2 — 3a; = - 1 ; X38 APPENDIX. .S^fHsSfflft**^ 115 10. Given -x 2 — - x = -, to find a:. 2 '3 8 Dividing both members by -, or multiplying by 2, - x = - ; whence, 1 /5 . 1 17 •'• * =1 i, *= — y 11. Given 2* -10 z + 3 Q ~ Z O = 2 ' t0 filld * Clearing of fractions, 2a; 2 - 14* -f 20 - (5s + 24 - a: 2 ) = 20a: - 32 - &r» ; A ' 2 39 28 L reducing, a: 2 — — x = — ; whence, a; 39 "10 v- 28 " 5 + 1521 100 = 39 ±31 10 ' • ' . a; = «* 4 * = 5* 12. Given 1 1 1 o or 5 to find x. Clearing of fractions, 35 (x + 3 - x -f- 1) = x 2 -f 3* — 3 ; reducing, x 2 -{- 2x = 143 ; whence, ar = - 1 ± > /Ii4 = - 1 ± 12; ADDITIONAL EXAMPLES. 139 .'. 3*9*11, X= -13. 24 13. Given x -\ — = 3r — 4, to find x. x-l % Clearing of fractons, x 2 — x + 24 = 3a; 2 — 3a: — 4x + 4 ; reducing, x 2 — 3a; = 10 ; whence, -i*^ . a; = 5, a; = — 2. 14. Given — — - H = — , to find w. x -f 1 # 6 Clearing of fractions, Qx 2 + 6a; 2 + 12a; + 6 = 13a; 2 + 13a? ; reducing, x 2 + x = 6 ; whence, . • . a; = 2, x = — 3. a;— 4 15. Given — = x — 8, to find x 2 + y^ Since * a; — 4 = (yaT— 2) ( (I), 2/+1 = 2v^ squaring both members, y 2 + 2y -f- 1 = 4y ; whence, y 2 - 2y = - 1 5 . • . y = 1 ± -/- 1 + 1, or y = 1 ; substituting in (1), x 2 — x = 1 \ i /rri i ± v^~ 22. Given z 3 — 6z = 6a; + 28, to find £ : transposing, a: 2 — 1 2a; = 28 ; whence, x e= 6 ± -y/28 -f- 36"= 6 =b 8 j .-> » = 14, a;=«~2» ADDITIONAL EXAMPLES. 143 23. Given x* n — 2x 3 « -f x* — 6 = 0, to find v a; = — 1, and the two roots are each equal to — 1 ; hence, the three roots aro -f 2, — > I, and — 1. 26. Given 2a; 2 -f 34 == 20a; 4 2, to find x. Transposing and reducing, a; 2 - 10a; £= — 16; .*•. x m 8, x = 2. 27. Given a;* = 56s 8 + a: 6 , to find x. multiplying both members by x , and reducing, **-** = 56: comparing with trinomial equations (Art. 124), we find 3 , 1 2 n^-, and - =-- - ; hence, by rule, taking the upper sign, x ss 8 =4, " lower " x = (-7) 3 " = ^/49. 28. Given x 3 — 12s 2 -+- 4a; -f 207 = 0, to find & ADDITIONAL EXAMPLES. 145 A superior limit of the positive roots is 13 (Art. 284) ; a supe- rior limit of the negative roots (numerically), is — 7 (Art. 286). By the method of Art. 297, rejecting -f 1 and — 1, which are not roots, we find Divisors, 9, 3, - 3 23, 69, -69 27, 73, -65 3, -9, -1, 0; hence, 9 is a commensurable root. Dividing both members by x — 9, we find X 2 _ % x _ 23 = ; or, x 2 — Sx = 23 ; 3 y/28 + 93 ± yloT 4~ 2 29. Given x 3 + 3x 2 — Qx - -8 = 0, to find x. This is solved in a manner similar to the preceding. A superior limit of positive roots is 4, and of negative roots (numerically), — 7. Divisors 4, 2, h - 1, - 2, -4, -2, - 4, -8, + 8, +4, + 2, -8, -10, -14, + 2, -2, -4, -2, - 5, -14, -2, +1, + 1, + 1, - 2, -11, + 1, +4, + 4, - 1, -11, -1, -2, -1, 0, -10, 0, -1, 0; hence, x = 2, X =z — 1, *=- -4. 10 ! ^> /jp^s^ **/* Of THB r TJ2ri7BE 146 APPENDIX. 30. Given x 3 -f 9x — 1430 = 0, to find x. By the same rule as before, we find 14 for a superior limit of the real positive roots, and from Art. 288 we see that the equation has no real negative roots. By the rule (Art. 297), we have 13, 11, 10, 5, 2, 1, - 143, - 286, - 715, - 1430, - 134, - 277, - 706, - 1421, -110, -130, -101, -121, - 8, - U, ,..., - 1, ..., o, 353, - 1421, ..., - 1421, ..., - 1420, ihence, 1 1 is the only commensurable root. Dividing both members by x — 11, we have x 2 + 11* = - 130; 11 -l. / 10A , 121 - 1 dfc -/- 399 no 31. Given Vx + V x + 7 = > to find x. 'Clearing of fractions and transposing, -vA 2 + 7* = 21 -x) squaring both members and reducing, 49* = 441; .*. x — 9. 32. Given -y/a -f x — y/a —x sa -\/ax, to find x. Squaring both members and reducing, 2a — ax = 2 4/ a 2 — * 2 ; squaring both members, 4a 2 — 4a 2 * + a 2 * 2 = 4a 2 — 4** ; ADDITIONAL EXAMPLES. 147 reducing and dividing both members by a;, *(«2 + 4)=4a»; .'. * = — j 33. Given ' >y/4a f- c ss 2 y^6 -f- a: — -y/^ to find r. Squaring both members and reducing, (a — b) — x = — y^a: -f- a; 2 j squaring both members, hence, (2a — 6) x sa (a — 6) 2 ; . • . x ss -^ £-• 34. Given y"4a -f C -f -/" + * *= 2 V* 2a > to ^ n ^ *• Squaring both members and reducing, 2 -v/4a 2 -f 5oa; -f- a; 2 s2«- 13a ; squaring both members, 16a 2 4- 20a* + 4a; 2 = 4* 2 — 52a* + 169a 2 ; 1 iy reducing, 72ax ss 153a 2 ; »*» « = -— • o 35> Given (TT$ + (T^~$ = "' t0 find * Dividing both terms of the first fraction by (1 4- #), an( * of thd se• to find « and «, (7z 2 -2y 2 = 10 . . (2)) y From (1), we find _ 19 — 3y _ 361 - 114y + 9y 2 X ~ 5~~ ; '• f - ~25~ Substituting in (2), and reducing, 2 _798_ _22T7 # V 13 ~ 13 ; 399^ /— 2277 , 159201 whence, y = _ ±x /_^_ + — _ = 759 hence, y = — , and y = 3 ; , . . 406 by substitution, x = — -, and a: = 2. lo 399 ± 360 13 5 150 APPENDIX. Ix + 4y = 14 . . (1)) 5. Gri*\>*i S > to find a? and y. (y 2 + 4z = 2y+ll (2)) From (1), v = 14 — 4y, or 4a: = 56 — 16y ; subtracting ana' reducing, y 2 _ I8y = _- 45 ; whence, y = db y— 45 + 81 = 9 ± 6 = 15 and 3 ; hence, x = 2 and — 46. ( s 2 + 4y 2 = 256 - 4ary . . (1)) 6. Given \ \ to find x and w. ( 3^ -a: 2 = 39 ... (2)) * Transposing in (1), and extracting the square root of both members* a + 2y=±16; .-. a:=±I6-2y; or, *' = 256=F64y + 4y 2 ; substituting in (2), and reducing, y 2 rp 64y = - 295 ; whence, y = ± 32 ± -y/ - 295 + 1024 = ± 32 ± 27 ; . * . y = ± 59 and ± 5 ; substituting, x = db 102 and ± 6. (a: 2 - y 2 = 24 . . (1)) 7. Given ■{ * )• to find ar and y. (z 2 + a:y = 84 . . (2)) Subtracting (1) from (2), member from member, y 2 -f*y = 60 . . . (3); adding (3) to (2), member to member, x 2 + 2xy+y 2 = U4; . \ x + y = db 12 . . . (4). ADDITIONAL EXAMPLES. 151 Dividing (1) by (4), member by member, x-y = ±2; hence, x = db 7, y = ± 5. -y= 4 . • . (l) *y = 45 . . . (2) ) x = y -f 4, which, in (2), gives 8. Given ' to find** From (1), y 2 + 4 y = 45 ; ... y = -2±v / 45TT= -2 ±7; . • . y = -f 5, and — 9 ; and by substitution, x = + 9, and — 4. W + *y 2 = i2 • • • 0) 9. Given i ( * + ay3=18 . . . (2) Dividing (2) by (1), member by member, and reducing, y to find x and y. 1 +y 3 _ 3 1 - y + V 2 _ 3 y(i+y)~2' or y _ 2 ; y2 -2 y= ~ 1 by reduction, whence, y= 4 ± V — 1 + 16 by substitution, 10. Given . y = 2, and y = - ; x = 2, and a? = 16. 4 4' 1 1 1 — •• — = a . x y L «" jr (2) J )- to find x and y. 152 APPENDIX. From (1), we find, 1 1 1 . 2a 1 - = a , or — = a 2 1- — ; y x y 2 x r x* ' substituting in (2), and reducing, I — a -h i h 1 a2 a2 a ± v^"--^ ■'""s~"2"~V 2 + 4 ~ 2 ; by substitution, - = — • y 2 2 hence, # = , and y = — a ± -v/26 - a 2 a =f V 26 - " 2 11. Given J y 2 y 9 v ' I to find « and y. ( * - y = 2 • . (2) j Clearing (1) of fractions, 9s 2 + S6xy =± 85y 2 . . . (3). From (2), a; = 2 + y ; .'. a 2 = 4 + 4y -f y a ; substituting in (3), and reducing, r 2 — 27 9 y*-TRy = 10 J 10 27 /9 , 729 27 ±33 whence, y =- ± yj- + — = __; 17 by substitution, x = 5 and £ = — • ADDITIONAL EXAMPLES. 153 12. Given J y 2 - x 2 T y T a; 4 w V to find a; and y. * - y = 2 (2) y ' x " x II Make - -\- - = z (3) ; whence, by squaring, &c, y 2 P hence, from (1), by substitution and reduction, z 2 + z _ 35 l /35 1 _ - 1 dbft 5 ,, 7 or, ^ = ^ and --; substituting the positive value of z in (3), and clearing of fractions, 5 x 2 + y 2 = -xy • • • (4), From (2), x = y + 2 ; . • . a 2 = y 2 -f 4y + 4, and #y = y 2 -f 2y ; substituting in (4) and reducing, y 2 + 2y = 8; .'. y = - 1 ± V / 8+T= - 1 ± 3 ; hence, y = 2 and y = — 4 ; by substitution, # = 4 and y = — 2. r * 2 + y 2 -f. z 2 = 84 • • - (1) N 13. Given -j a? + y -f * = 14 • • • (2) I to find ar, y, I xz = y 2 ... (3) J and * Substituting in (1) and (2) the value of y from (3), and reducing x 2 + 2arz -f z 2 = 84 + rz or x + * = -y/84 -f- ars • . . (4). From (2) and (3), x + z _ 14 _ y^" . . (5). 154 APPENDIX. Equating the second members, 14 - i/xz— -/84 -f xz ; squaring both membeis, 196 - 28 i/x~z -{- xz = 84 + xz ; •educing, -yfxz _- 4, or xz = 16 • • • (6) ; hence, from (5), x + z = 10 • • • (7) ; 1 (\ aobstituting in (7) for z its value — > and reducing, x* - 10* = - 16 ; ^.httfee, x — 5 ± -/— 16 + 25 = 5 ± 3 ; or, a; = 8, a: a= 2 : by substitution, z = 2, 2 = 8, and y == ± 4. 14. Given ^ • , ' f. to find x and y. ( x 2 .+ y 2 = 41 • • (2) ) From (1), by transposition, V* + y = 12 - (* + y) ; squaring both members, x + y = 144 - 24 (a? + y) + (* + r )»j reducing, (a; + y) 2 — 25 (x -f- y) = — 144 ; , 25 / 7777*^5 , 25 ±7 ... iC 4. y = + _ ±v /_i44 + — =+-^— ; whence, a: -f y = 16, or a- -f y =a 9. Tlie first value does not satisfy (1), unless the radical have the negative sign ; adopting, therefore, the second value, from which x = 9 — y, or x 2 = 81 — 18y + y 2 , • ADDITIONAL EXAMPLES. which in (2) gives, after reduction, w- y 2 - 9y = - 20 ; whence, 9 y = 2 -20 + 81 9±1 4 - 2 ' .*. y = by substitution, x = 4 and a; = 5. 15. Given ir -.Jf m 3 • • • 0) . (2) 155 to find x and y. Cubing both members of (2), subtracting from (1), member from member, and dividing both members by 3, we have x 2 y — xy 2 = 30, or (.* — y) xy = 30 • « • (3) ; dividing (3) by (2), member by member, -" 1A . 10 *y = 10; .'. y = — ; substituting in (2), and reducing, x 2 -Sx = 10: 3 / 9 3 7 whence, x = - ± yj 10 + - = - ± -; . • . a; = 5, x = — 2 ; by substitution, y = 2, and 3/ = — 5. f*»+'f*9/iyr= 208 • • (1) ) 16. Given I [ to find x and y, (y* + y^/*V = i053 • • (2) ) * These equations may be written, **+*V= 208, or **(**+y*)= 208 • . . (3), y* + y*** = 1053, y*(y* +**) = 1053 - - (4> 156 APPENDIX. Dividing (3) by (4), member by member, ** 208 16 i- 1053 81 y extracting the 4th root of both members, y (5); substituting in (3), A/1 1 A y 2 4- 77T V 2 — %®8 ; whence, 20S — y 2 = 208, or j/ = 729; ... ^^±27, and by substitution, x = ± 8. MISCELLANEOUS PROBLEMS. 1. A courier starts from a place and travels at the rate of 4 miles per hour ; a second courier starts after him, an hour and a half later, and travels at the rate of 5 miles per hour : in how long a time will the second overtake the first, and how far will he travel ? Let x denote the number of hours travelled by 2d courier : then will x + 1 \ " 5x and 4(x + 1^) " From the conditions of the problem, 5x -— 4 (x + 11) ; . • . x = 6 and hx = 30. 2. A person buys 4 houses for $8000 ; for the second he gave half as much again as for the first ; for the third, half as much again as for the second ; and for the fourth, as much as for the first and third together : what does he give for each ? (( a 1st " miles a 2d u a u 1st ADDITIONAL EXAMPLES. 157 Let x denote the amount paid for 1st house : then will x + % " " " " « 2d " 2x + ~ " " « " " 3d * 4 Sx -f % " " " " " 4th " 4 From the conditions of the problem, 8* -=8000; .\ * = 1000 x + * = 1500 2x -f S .-.= 2250 4 3* -f ? = 3250 4 3. A and B engaged in play : after A had lost $20, he had one third as much as B ; but continuing to play, he won back his $20, together with $50 more, and he then found that he had half as much again as B : with what sums did they begin ? Let x and y denote the sums with which A and B began. Then from the conditions, ™ V + 20 x + 50 = (y - 50) x 1 J Whence, Zx -~ 60 as y + 20 ; . '. a: = »70 2a; rf 100 = Zy - 150 ; y = 130. 4. A can do a piece of worl in 10 days, which A and B together can do in 7 days ; in how many days can B do it alone ? Let x denote the number of days. 158 APPENDIX. Since A and B together can do -J- of the w t>rk in 1 day, A can do y\ of it, and B, - of it in 1 day ; hence, from the relations existing, x 10 x l\ a; 70' 5 5. A person has $650 invested in two parts t the first part draws interest at 3 per cent, and the second at 3. J- per cent, and his total income is $20 per annum : how much has he invested at each rate ? Let x denote the number of dollars at 3 per cent : then will 650 — x denote the number at 3£» From the conditions, 3x 050 -* „, _ 20 . whence, Sx + 2275 — 3£ c sb 2000 ; . •. 5 s 275, or * = 550 ; . • . 650 - x = 100. 2 6. A boatman rows with the tide, in the channel, 18 miles in 1-J hours ; he rows near the shore against the tide, which is then only three-fifths as strong as in the channel, 18 miles in 2J hours : what is the velocity of the tide per hour in the channel t Let x denote the velocity of the tide in the channel : Sx then, -r- " " " " " near shore ; 5 and 1 18 — —J -r 1? will denote the rate of rowing, neglecting tide also, (lS + ^-f-2* " « « « f / 10 Sx\ 2 /,. , 27*\ 4. henee, (l8 - -) X - - (l8 + w ) X gi or, 12 -x = 8 + ^5 ;*,'_'•**•§. ADDITIONAL EXAMPLES. 159 7. A garrison had provisions for 30 months, but at the end of 4 months the number of troops was doubled, and 3 months afterwards it was reinforced by 400 troops more, and the provisions were ex- hausted in 15 months : how many troops were there in the garrison at first 1 Let x denote the number of men at first ; then will 30# denote the number of months that one person could subsist on the provi* sions, or the number of month", y rations in the garrison. 4x denotes the number of monthly rations used in 4 months, Gx " " " " " the next 3 " (2*+400)8 " " " " " " 8 " hence, 2Gx + 3200 = 30#, or 4x m 3200 , . i . x tc 800. 8. What is the number whose square exceeds the number itself by 6? Let x denote the number. From the conditions, 1 ±5 . • . x 2= 3 and — 2. 9. Find two numbers such that their sum shall be 15, and the sum of their squares 117. Let x and y denote the numbers. From the conditions of the problem, x +y = 15 • • • (1), x 2 + y 2 = 117 • i . (2). From (1) x sbe 15 — y, or x 2 z- 225 — 30y + y 2 } 160 APPENDIX. substituting in (2) and reducing, y* - 15y - — 54, 15 ^ / 7 A , 225 15 3 „ y=-2-±y-54 + - r = y±-, or y = 9 and * *x 6, or a; ss 9 and y = 6. 10. A cask whose contents is 20 gallons, is filled with brandy ; a certain quantity is drawn off into another cask of the same size, after which the latter is filled with water : the first cask is then, filled with this mixture ; it then appears that if 6J gallons of this mixture be drawn from the first into the second cask, there will be equal quantities of brandy in each. How much brandy was first drawn off? Let x denote the number of gallons first drawn off. Then will 20 — x denote the quantity remaining as well as the quantity of x water added to the second cask ; — j will denote the quantity of brandy in each gallon of the mixture, and x x 2 * X 20' ° r 20 will denote the quantity of brandy returned to the first cask, which will, therefore, contain gallons of brandy. Each gallon of this new mixture will contain jfc of the brandy in the cask, or 400 — 20a; + * 2 . 400 ' hence, 6f gallons will contain 400 — 20* -f a? 8 60 ADDITIONAL EXAMPLES. 161 gallons ; and after this is drawn off, 10 gallons must remain ; hence, 400 — 20a? -f x 2 400 — 20* -f- x 2 20 60 whence, 800 — 40* -f- 2s 2 = 600, or, z 2 — 20* = 100; = 10; * = 10 ± v-HW+Too, or * = 10. 11. What number added to its square will produce 42? Let x denote the number. From the conditions of the problem, x 2 + * = 42 ; 1 ± 13 •. af = — -J+ V42 + } = » ; .*. * = 6 and * = ~7. lit 12. The difference of two numbers is 9, and their sum multiplied by the greater gives 266 : what are the two numbers ? Let * and y denote the numbers. From the conditions of the problem, ****** • • • (1), * (* + y) = 266 • (2). From (1), y = * — 9 ; substituting in (2), * (2* — 9) = 266, or x 2 — - x = 133 ; , 9 ' / " 81 9 ± 47 whence, £=-±^133 + - = —— ; .*. * = 14, * = — 9£; whence, y = 5, y = — 18^. 13. A person travelled 105 miles : if he had travelled 2 miles 11 162 APPENDIX. per hour slower, he would have been 6 hours longer in completing the journey : how many miles did he travel per hour 1 Let * denote the number of miles travelled per hour. Then will denote the number of hours. * From the conditions, 105 105 + 6, or 105* = 105* - 210 + 6* 2 - 12* x —2 x reducing, * 2 — • 2x = 35 ; .-. * = 1 ± V35 + 1 = 1 ± 6 ; .-. x = 7. 14. The continued product of four consecutive numbers is 3024 : what are the numbers % Let * denote the least number. From the conditions of the problem. *(* +1) (* -f- 2) (x + 3) = 3024, or * 4 -f 6* 3 + ll* 2 -f- 6* — 3024 ±s 0. A superior limit of the real positive roots is 9 (Art. 284). Ne- glecting the divisor 1, and all negative divisors, we may proceed by the rule (Art. 297), as follows : 9, 8, % 6, 4, 3, 2, -336, -378, -432, -504, -756, - 1008, - 1512, -330, -372, - 426, -498, -750, - 1002, - 1506, "j **> "? - 83, **? - 334, - 753, ~» **» "i - 72, i - 323, - 742, ••» '*» *■? - 12, '*> "i - 371, "» "i *'j - 6, "j **> - 365, "» "i "*> - h **» "•. **» "» ", ••> — 0, .-, .-, Hence, 6 is the required value of *, and the numbers a?e 6, 7, 8 and 9. ADDITIONAL EXAMPLES. 163 15. Two couriers start at the same instant for a point 39 miles distant ; the second travels a quarter of a mile per hour faster than the first, and reaches the point one hour ahead of him : at what rates do they travel ? Let x denote the number of miles per hour of first courier. 39 Then will — denote the number of hours he travels. x From the conditions, x x -+- -J- 4 4 reducing, x 7, -f - x = -— - ; 1 J. /39 . 1 - 1 ± 25 16. The fore-wheels of a wagon are 5 J- feet, and the hind-wbeek 7£ feet in circumference ; after a certain journey, it is found that the fore- wheels have made 2000 revolutions more than the hind-wheels : how far did the wagon travel ? Let x denote the number of feet. From the conditions of the problem, . ' . 4-^ = 2000; 1197 multiplying both members by -55-, Tfir- -5i* 2394000 32 598500 8 ' 57 * -42 a; = 598500, 15x = 598500, X = 39900. 164 APPENDIX. 17. A wine merchant has 2 kinds of wine ; the one costs 9 shil- lings per gallon, and the other 5. He wishes to mix them together in such quantities that he may have 50 gallons of the mixture, and so that each gallon of the mixture shall cost 8 shillings. Let x and y denote the number of gallons of each, respectively. From the conditions, ^+ y = 50 . . . . (1), 9* + by = $(x + y) . . (2); substituting for x -f y its value in (2), 9z + 5y = 400 . . . . (3); combining (1) and (3), Ay = 50 ; . •. y= 12-J, and x = 37£. *18. A owes $1200 and B, $2500, but neither has enough to pay his debts. Says A to B, "Lend me the eighth part of your fortune, and I can pay my debts." Says B to A, " Lend me the ninth part of your fortune, and I can pay mine :" what fortune had each % Let x and y denote the number of dollars in the fortunes of A and B. From the conditions of the problem, x + 1 = 1200, x or &r + f =2 0600, 8 y + ^ ~ 2500, or x -fify = 22500 j 9 combining and eliminating ar, 71y = 170400 \ .'. y^ 2400, x = 900. 19. A person has two kinds of goods r 8 pounds of the first, and 9 of the second, cost together $18,46; 20 pounds of the first, and 16 of the second, cost together $36,40 : how much does each cost per pound % ADDITIONAL EXAMPLES. 165 Let x and y denote the cost of a pound of each in cents. From the conditions of the problem, 8* + 9y = 1846, 20* + 16y = 3640 ; combining and eliminating x, ISy = 1950 : . •. y = 150, and x = 62. 20. What fraction is that to the numerator of which if 1 be added the result will be J, but if 1 be added to the denominator the result will be J- 1 Let x denote the numerator, and y the denominator. From the conditions of the problem, x+ 1 1 x 1 or Sx -f- 3 = y, or 4x = 1 + y ; l +y 4 hence, by combination, x =. 4 and y = 15. Ans. ^-. 21. A shepherd was plundered by three parties of soldiers. The first party took -J- of his flock and i of a sheep ; the second took £ of what remained and J of a sheep ; the third took £ of what then remained and J of a sheep, which left him but 25 sheep : how many had he at first 1 Let x denote the number of sheep. Then, after being plundered by the 1st party, he would have -(l + i) = — 4 — s^Pi after being plundered by the 2d party, he would have Sx — 1 Ar — 1 . 1\ x — 1 4 V 12 r 51 166 APPENDIX. after being plundered by the 3d party, he would have x — I (x — 1 1\ ar— 3 -p?+i> 2 from the conditions of the problem, x — 3 ■ = 25, or x — 3 = 100 ; :\ * = 103. 22. What two numbers are those whose product is 63, and the square of whose sum is equal to 64 times the square of their dif ference ? Let x and y denote the two numbers. From the conditions of the problem, *y = 63 (i), (* + y)» = 64 (« - y)» • , (2); extracting the square root of both members of (2), x + y = 8(x — y), or 7* = 9y; .-. x = $ y ; substituting in (1), %y 2 = 63 ; . • . y 2 = 49 and y = 7, also x = 9. 23. The sum of two numbers multiplied by the greater gives 209 ; their sum multiplied by their difference gives 57 : what are the two numbers 1 Let x and y denote the numbers. From the conditions of the problem, {x + y) x = 209, or x 2 + xy= 209 . . (1), (x + y )(x-y)^ 57, or x 2 - y 2 = 57 . . (2); subtracting (2) from (1), member from member, xy + y 2 = 152 . . (3) ; adding (3) and (1), member to member, x 2 + 2*y + y 2 = 361 ; .-. x + y = 19-, ADDITIONAL EXAMPLES. 167 209 hence, from (1), x = — - = 11 ; also, y = 8. iy 24. Three numbers are in arithmetical progression ; their sum is 15, and the sum of their cubes is 495 : what are the numbers % Let ar, y and z denote the numbers, From the conditions of the problem, y — x = z — y . . (1) * +y +z = 15 . . (2) x 3 + y 3 + z 3 = 495 . . (3) ; from (1), 2y = z -f x, which in (2), gives y = 5 ; substituting in (2) and (3), z + x = 10 . . (4) z 3 + x 3 = 370 . . (5) ; dividing (5) by (4), member by member, Z 2 __ zx + X 2 £J 37 , c ^ . squaring both members of (4), Z 2 + 2zx + x* = 100 . . (7) ; combining (6) and (7), 3^ = 63, or zx = 21: .'. z = — ; substituting in (4), re + — = 10, or z 2 - 10a; = - 21 : .-. x == 5 db or ff + JfsU . . (6); oombining (3) and (6), * = 9. y = 4. ADDITIONAL EXAMPLES. 169 27. What two numbers are those the square of the greater of which being multiplied by the lesser gives 147, and the square of the lesser being multiplied by the greater gives 63 1 Let x and y denote the numbers. From the conditions of the problem, x 2 y = 147 . . (1) xy* = 63 . . (2); multiplying (1) and (2), member by member, x 3 y* = 9261 . . (3), or xy = 21 . . (4) ; dividing (2) by (4), member by member, y = 3 ; in like manner, x = 7. This method of solution might be applied to the equations of the preceding example. 28. There are two numbers whose difference is 2, and the product of their cubes is 42875 : what are the numbers ] Let x and y denote the numbers. From the conditions of the problem, x-y = 2 . . (1) *V = 42875 . . (2); extracting the cube root of both members of (2), 35 xy = So; .'. y = —' y x substituting and reducing, x* — 2x = 35, x = 1 ± -/35 + 1 = 1 ± 6 ; . • . x = 7, and — 5, y = 5, and — 7. 170 APPENDIX. 29.. A sets out from C towards D, and travels 8 miles each day ; after he had gone 27 miles, B sets out from D towards C, and goes each day ^ of the whole distance from D to C; after he had travelled as many days as he goes miles in each day, he met A : what is the distance from D to C? Let x denote the number of miles from D to C. Then, — will denote the number of miles B travels per day, also the number of days that he travels ; x 2 hence, — — denotes the number of miles travelled by B, 27 + Sx " " " " " A. From the conditions of the problem, x 2 Sx clearing of fractions and reducing, x 2 — 240z ta — 10800 ; .-. x - 120 ± -/— 10800 + 14400 = 120 ± 60 ; whence, x = 60, x = 180. 30. There are three numbers ; the difference of the differences of the 1st and 2d, and 2d and 3d, is 4 ; their sum is 40, and their con tinned product is 1764 : what are the numbers 1 Let a?, y and z denote the numbers. From the conditions of the problem, (#-y)-(y-*)= 4 . . (l) x + y + z= 40 . . (2) xyz = l7U . . (3); ADDITIONAL EXAMPLES. 171 combining (1) and (2), eliminating x and s, 3y = 36; .-. y = 12 ; substituting in (2) and (3), x + z = 28 . . (4) xz = 147 . . (5) ; combining (4) and (5), * ss 7, or 21 ; y = 21, or 7. 31. There are three numbers in arithmetical progression : the sum of their squares is 93, and if the first be multiplied by 3, the second by 4, and the third by 5, the sum of the products will be 66 : what are the numbers ? Let x denote the first number, and y their common difference. From the conditions of the problem, x* + (x + yY + (x + 2yy = 9Z . . (1) 3a + 4(s + y) + 5(* + 2y) = 66 . . (2); performing indicated operations and reducing, 3* 2 4- 5y 2 + 6*y = 93 . . (3) 12* + 14y = 66, or 6* -f- 7y = 33 . . (4). ■^ 33 — 6a? From (4), y = = ; 1089 — 396* + 36z 2 a 33* — 6* 2 .-. y 2 = _ , and xy = _ . substituting in (3) and reducing, 2 198 296 * -^5*-~~ 25 ; 172 APPENDIX. 99 / 296~ 9801 99 ± 49 whence, * = - ±W - — + — = 25 ' 148 .*. x = ——-, x = 2. 25 Taking the second value of x, we find y = 3, and the numbers are 2, 5 and 8. The problem supposes the numbers entire, therefore the 1st value of x is not used. 32. There are three numbers in arithmetical progression whose sum is 9, and the sum of their fourth powers is 353 ; what are the numbers ] Let x, y and z denote the numbers. From the conditions of the problem, 2y = x + z . . (1) x + y + z = 9 . . (2) a;Hy 4 + 2 4 = 353 . . (3). From (1) and (2) we find y = 3 ; substituting in (2) and (3), x + z = 6 . . (4) a* + z 4 = 272 . . (5) ; raising both members of (4) to the 4th power, X 4 _|_ 4x 3 z _j_ q x 2 z 2 + 4 XZ 3 + 2 * _ 1296 . . (6) ; adding equations (5) and (6), member to member, and dividing by 2 x* 4 2x 3 z 4 3z 2 2 2 + 2z2 3 4 - 4 = 784 . . (7); extracting the square root of both members, x* 4 xz 4- s 2 = 28 . . (8) ; ADDITIONAL EXAMPLES. 173 squaring both members of (4), x 2 + 2xz -f z 2 =a 36 . . (9); from (8) and (9) we find M&8 . » (10); from (4) and (10) we get # =s= 2, or 4 ; « = 4, or 2: hence, the numbers are 2, 3 and 4. 33. How many terms of the arithmetical progression 1, 3, 5, 7, &c, must be added together to produce the 6th power of 12? The Gth power of 12 is 2985984. From Art. 176 we have the formula, d~2a dfc y/(d-Za) 2 + §dS n m g Ir. the present case, o=l, d — 2, and 8 aa 2985984 ; i • • -v/ 1 * 5 X 2985984 IMi substituting, n ss — — 2* 1728. 34. The sum of 6 numbers in arithmetical progression is 48 ; the product of the common difference by the least term is equal to the number of terms : what are the terms of the progression 1 Let x denote the 1st term, and y the common difference. "From the conditions of the problem, c Gx -f 15y ss 48, xy zz 6 ; . . y -= - ; x substituting and reducing, x 2 — 8x ss -» 15; .-. X ae 4 dfc V- 15 -f- 16 = 4 dfc 1, or a; = 5, a; = 3 } whence, y = £ , y = 2 ; 174 APPENDIX. hence, the series is 3.5.7.9.11.13, or 5.6±.7£ . 8f . 9f .11. 35. What is the sum of 10 square numbers whose square roots are in arithmetical progression the least term of which is 3, and the common difference 2 ? Let x denote the sum. The progression of roots is 3.5.7.9.11. 13. 15.17.19.21, and the series of squares, 9 . 25 . 49 . 81 . 121 . 169 . 225 . 289 . 361 . 441. 1st order of diffs, 16, 24, 32, 40, &c, 2d order of diffs, 8, 8, 8, &c, 3d order of diffs, 0, 0, &c. Frgm Art. 210, making ' S' = x, a = 9, 7i = 10, c?! = 16, d 2 — S, d 3 = 0, <&c. x == 90 -f 45 x 16 + 120 x 8 --= 1770. 36. Three numbers are in geometrical progression whose sum is 95, and the sum of their squares is 3325 : what are the numbers? Let #, y and z denote the numbers. From the conditions of the problem, y 2 = xz . . (1) x 2 + y 2 + * 2 = 3325 . . (2) x +y +z = 95 . . (3); combining (1) and (2), x* + 2xz + z 2 = 3325 + m . . (4) ; ADDITIONAL EXAMPLES. 175 combining (1) and (3), x + y'xz + z = 95 . . (5) from (4) and (5), x 4- t = -y/3325 + a* a; 4- z = 95 — ^$bb ; hence, -y/3325 -f #2 = 95 — y»T; squaring both members, 3325 + xz — 9025 — 190 ^/xz + *z ; . • . y/x~z = 30, or xz -. 900 . . (6) ; substituting in (5), a; + z as 65 . . (7) • from (6) and (7), * = 20 and 45, y = 45 and 20. 37. Three numbers are in geometrical progression : the difference of the first and second is 6 ; that of the second and third is 15 : what are the numbers 1 'Let x. y and z denote the numbers. From the conditions of the problem, y 2 = xz . . (1) x — y = — 6 ; .". x = y — 6 y — z=i — 15; .'. 2=^4-15, and xz = y 2 -f- 9y — 90; substituting in (1) we find y = 10 ; . • . x =z 4, and 2 = 25. 38. There are three numbers in geometrical progression ; the sum 176 APPENDIX. of the first and second is 14, and the difference of the second and third is 15 : what are the numbers? Let #, y and z denote the numbers. From the conditions of the problem, y 2 as xz . . (1) # + y = 14; .*. x =e 14 — y z —y = 15; .-. z = 15 + y, and substituting in (1), g2 t= 210 — y — y 2 ; ^ + | = 105; and '.» 5+ ,'„..-':«=.., 21 ~ 2" } taking the 1st value of y, we find x 3s 4i * = 25. 39. A, B and C purchase coffee, sugar and tea at the same prices; A pays $11,621 for 71 pounds of coffee, 3 pounds of sugar, and 2} pounds of tea; B pays $16,25 for 9 pounds of coffee, 7 pounds of sugar, and 3 pounds of tea; C pays $12,25 for 2 pounds of coffee, 5J pounds of sugar, and 4 pounds of tea : what is the price of a pound of each 1 Let x, y and z denote the number of cents that the coffee, sugar and tea cost, respectively. ^rom the conditions of the problem, 7J&+ 3y + 2l2 = 1162^ * • 0) 9*+ 7y + 3z= 1625 • • (2) 2 z + 5Jy + 4 2 = 1225 • . (3) ; ADDITIONAL EXAMPLES. 177 clearing (1) and (3) of fractions, SOx f 12y + 9* =4650 • • (4) 4 x + lly + 8z =2450 • • (5). From (2) and (4), Zx — 9y = — 225, or x — Sy = — 75 • • (6) ; from (2) and (5), 60s + 23y = 5650 . - (7); from (6) and (7), y = 50; by substitution, s = 75, z = 200. 40. Divide 100 into 2 such parts that the sum of their square roots shall be 14. Let x denote the first part. From the conditions of the problem, * = 12 °- 47. Two travellers set out together and travel in the same direc- tion ; the first goes 28 miles the first day, 26 the second day, 24 the third day, and so on, travelling 2 miles less each day ; the second travels uniformly at the rate of 20 miles a day : in how many days will they be together again ] Let x denote the required number of days. The distance travelled by the first in x days is [(Art. 176), since a = 28, d = — 2, and n = #], denoted by £s[56 — (s - I) 2 ], or 29* -z 2 ; and the distance travelled by the second is denoted by 20a?: hence, we have 29x — x' 1 =z 20*, or x = 9. 48. A farmer sold to one man 30 bushels of wheat and 40 of barley, for which he received 270 shillings. To a second man he sold 50 bushels of wheat and 30 of barley, at the same prices, and received for them 340 shillings : what was the price of each 1 Let x denote the number of shillings for 1 bushel of wheat, and y " " " " " " barley. 182 APPENDIX. Prom the conditions of the problem, 30* + 40y = 270 • . (1) 50x + 30y = 340 • • (2) ; whence, HOy = 330, or y = S; hence, x = 5. 49. There are two numbers whose difference is 15, and half their product is equal to the cube of the lesser number : what are the numbers 1 Let x and y denote the numbers ; from the conditions of the problem, x — y = 15 xy = 2y 3 ; or, substituting and reducing, v 2 1 15 2 ' * = 2y 2 ; 1 /15 1 1± 11 •'• y = 4 ± V^ + l6 = ""4— ; 5 hence, y = 3, and — - ; also, x = 18, and 25 50. A merchant has two barrels and a certain number of gallons of wine in each. In order to have an equal quantity in each, he drew as much out of the first cask into the second as it already- contained ; then again he drew as much out of the second into the first as it then contained : and lastly, he drew again as much from the first into the second as it then contained, when he found that there was 16 gallons in each cask: how many gallons did each originally contain 1 ADDITIONAL EXAMPLES. 183 Let x denote the lumber of gallons in the first cask, and y the number in the second ; x — y will denote the quantity in the first cask after the first drawing, and 2y the quantity in the second cask ; after the second drawing, 2y — (x — y) or Sy — x will denote the quantity in the second, and 2x — 2y the quantity in the first cask ; after the third drawing, 2x — 2y — (Sy — x) or Sx — 5y will denote the quantity in the first cask, and 6y — 2x the quantity in the second. From the conditions of the problem, Sx - 6y = 16 6y - 2x = 16. By combination, x = 22; y = 10. 184 APPENDIX. PROBLEMS FROM LEGENDRE. Problem VI. — In a right angled triangle, having given the base and difference between the hypothenuse and perpendicular, to find the sides. Let a = the base AB, b = the difference between AC and BC, • x = the hypothenuse AC, and y = the perpendicular CB. Then, x — y = b, by the conditions, .... z 2 = a 2 + y 2 (Bk. IV. Prop. 11), . . . From equation (1) we have, x = y4-5; by squaring, x 2 = y 2 -f- 2by -f b 2 . Substituting this value of x 2 in equation (2), y 2 + 2by + b 2 = a 2 + y 2 ; a 2 -b 2 hence, 2by = a 54 — b 2 , or, 2/ 26 ; substituting this value in equation (1), we readily find a 2 + b 2 x = — — 26 Problem VII. — In a right angled triangle, having given the hypothenuse and the difference between the base and perpendicular, to determine the triangle. Let a = the hypothenuse AC, b = difference between AB and BC, PROBLEMS FROM LEGENDRE. 1* 35 x = the base AB, and C y = the perpendicular BC. / Then, x — y = ft, by the conditions, • • (1) / a 2 L x 2 + y\ (Bk. IV, Prop. 11) . (2) A JB From equation (1), we have, x — y + ft; and by squaring, x 2 = y 2 + 26y + 62, Substituting this value of x 2 in equation (2), we have, a 2 = y 2 + 2by + b 2 + y 2 -, and by transposing and reducing, .... « 2 - b 2 V 2 + by = — - — ; - ft + \/2a 2 - ft 2 /ft 4- t/2a 2 - ft 2 \ hence, y = =g ; y = - ^ v -- J ; Substituting either of these values in equation (1), we readily find the corresponding value of x. Note. — The positive value of the unknown quantity generally fulfils the con- ditions of the problem, understood in its arithmetical sense. The negative value will always satisfy the conditions of the equation: with its sign changed, it may be regarded as the answer to a problem which differs from the one proposed only in this : that certain quantities which were additive have become subtractive, and the reverse. Problem VIII. — Having given the area of a rectangle inscribed in a given triangle, to determine the sides of the rectangle. By a given triangle, we mean one whose sides are all known, or given. 1st. To find the perpendicular and segments of the base: 186 APPENDIX. Let ABC be a triangle, in which the three sides are given : viz., b = AB, a = AC, and c = BC. Let CD be drawn perpendicular to AB, and £ let y = CD, and x z= AD ; then will b — x = DB. r Then, a 2 == x 2 -f- y 2 (1) c 2 =y 2 + x 2 — 2bx + b 2 . . • (2) ; substituting in equation (2) the value of x 2 + y 2 = a 2 , from equation (1), and transposing, c 2 = a 2 — 2bx -f 5 2 ; hence, a 2 + 5> _ c 2 a: = 26 and substituting this value in equation (1), we find the altitude of the triangle. 2d. To find the sides of the rectangle. Suppose the rectangle to be inscribed in the c triangle ACB. Let d = the area of the rectangle, x = its base, * y = its altitude, and A h = CD, the altitude of the triangle. Then, by similar tr AB b and angles, CD : : GF : CJ; that is, h : : x : h — y\ hence, hx = bh-by • • • (1), xy = c?, by the conditions, (2). d From equation (2), we have x = - ; PROBLEMS FROM LEGENDRE. 187 substituting this value in equation (1), we have, h- = bh — by; clearing of fractions, h d ~ bhy — by 2 ; or, -hd hy »=+ 2 + w- b h 2 - 4M whence, J 1 b h 2 - 4hd b ' 2 2V b Substituting these values of y in equation (1), we find the corres- ponding values x. Problem IX. — In a triangle, having given the ratio of the two sides, together with both the segments of the base made by a perpendicular from the vertical angle, to determine the triangle. Let ACB be a triangle, and CD a line drawn perpendicular to the base AB. Let Then, AC = x CB= y CD= i AD= a DB= b c— ratio. - = c, by the conditions y x 2 — a? = z 2 (Bk. IV., Prop. 11, Cor. 1) . . y 2 — b 2 = z 2 (Bk. IV., Prop. 11, Cor. 1) . . Subtracting (3) from (2), member from member, (i) (2) (3). y 2 + J 2 — a 2 = ; cr, x 2 = y 2 + a 2 — i 2 (4). 188 APPENDIX. From equation (1) we have x = cy ; hence, x 2 = c 2 y 2 • • • (5) j combining (4) and (5), we have C 2 y 2 _ yl _j_ a 2 _ ^ and ( C 2 _ 1) y2 _ a 2 _ J2 . '» a 2 -^ 2 , /a 2 - 6 2 hence, y 2 = — — p ; and y = ±\J ^—-y ; substituting these values in equation (1), we find the corresponding values of x. Problem X. — In a triangle, having given the base, the sum of the other two sides and the length of a line drawn from the vertical angle to the middle of the base, to find the sides of the triangle. Let ABC be a triangle, and CD the line drawn from the vertex C to D, the middle point of the base. Let AC = x \ BC =y AD = b CD -a AC -f BC =V Then, x -f y = s, by the conditions • • (1) and x 2 -f y 2 = 26 2 + 2a 2 , (Bk. IV., Prop. 14) • • (2) ; equation (1), by transposing and squaring, gives y 2 = x 2 — 2sx + s 2 ; substituting this value in equation (2), we have 2x 2 — 2sx + s 2 = 2b 2 + 2a 2 ; PROBLEMS FROM LEGENDRE, 189 transposing and reducing, -we have 26 2 -f- 2a 3 - s 2 s? — sx zz ; whence, s + V*t> 2 + 4a 2 — * 2 % ^— -v/46 2 -f4a 2 — « 2 x = - , and x as Problem XI.— In a triangle, having given the two sides about the vertical angle, together with a line bisecting that angle and terminating in the base, to find the base. Let ACB be a triangle, and CD a line bisecting the angle ACB, and terminating in the base at D. Let AC =za C BC = 6 A\ CD = c /W AT> = x DB = y. I \ \ A i) 1 Then, x ! y : : a : b (Bk. IV., Prop. 17) ; whence, bx 23 ay ' • • • (i); also, a x b - c 2 + *y (Bk. IV., Prop. 31) • • (2). Multiplying the first equation by x, and the second by a, and sub- tracting, member from member, we have bx 2 — a 2 b =2 — ac 2 j whence f a ( a h _ C 2) A (ab - c 2 Problem XII. — To determine a right angled triangle, having given the lengths of two lines drawn from the acute angles to the middle points of the opposite sides. 190 APPENDIX. Let ABC be a right angled triangle, and AD, CE, two lines drawn to the middle points D and E of the opposite sides. Let AD = a CE "= b AB = 2x BC = 2y Then, 4x 2 + y 2 = a 2 (Bk. IV. Prop. 11) . (1) and, 4y 2 -f x 2 — b 2 " " " " . . (2). Multiplying equation (1) by 4, and subtracting (2) from it, we have, /AaT 15z 2 = 4a 2 — b 2 ; when, x =.- dt \I A and substituting in equation (2) y ts sfc */ - -& 2 4 46 2 - rf 15 Problem XIII. — To determine a right angled triangle, having given the perimeter and the radius of the inscribed circle. 1st. Let ABC be a right angled triangle, O the centre, and r the radius of the inscribed circle. Let p = the perimeter ; let x denote the length of the equal tangents drawn from A (Bk. III. Prop. 14, sch.) ; v the length of the equal tangents drawn from B ; and y the length of the equal tangents drawn from C. Then, AC = x -f- y> AB sc # -|- v 9 and BC = y + v ; then, AC + AB -f BC = 2x + 2y + 2v =p ; transposing and reducing, x + y = ^—7; — = a > a known quantity • • • (1). PR0BLEM8 FROM LEGENDRE. 191 Then, AB 2 + to = AC 2 : that is, x 2 + %vx + v 2 -f y 2 + 2vy +v 2 = a 2 • • • • (2). 2nd. Observe that the double area of each of the triangles AOB, BOC, and AOC, is equal to its base multiplied by the radius of the inscribed circle ; and hence, the sum of these products is equal to the sum of the bases multiplied by r ; that is, = r x p, a known quantity. But the base AB x BC is also equal to double the area of the triangle ABC ; hence, (x + v) x (y -f v) = r x p ; that is, xy + vx + vy + v 2 — r X p . . . . (3) ; Multiplying both numbers of equation (3) by 2, we have, 2xy -f- 2w -h 2vy -\- 2v 2 = 2rp . . . (4) ; subtracting equation (4) from (2), we have, x 2 — 2xy + y 2 = a 2 — 2rp ; whence, by extracting square root, x — y = dfc -y/a 2 — 2rp = 5 (5) ; combining (1 ) and (5) we readily find the values of x and y , a + b , a — b * = - — , and y = -2— Problem XIV. — To determine a triangle, having given the base } the perpendicular, and the ratio of the two sides. Let ACB be a triangle, and CD perpendicular to the base AB. Let AC = y, then, r == ratio ; ry as CB AB — b DB — b -+X AD — X CD 2S h. 192 APPENDIX. Then, y 2 = x 2 + A 2 ; and r 2 y 2 a b 2 - 26* -f x 2 + A*. Multiplying the first equation by r 2 , and subtracting, = 6 2 — 2bx + a; 2 + h 2 — r 2 * 2 — r 2 A 2 ; whence, (1 _ r 2 ) x 2 - 2bx a (r 2 - 1) A 2 - 6 2 ; or, 2b (r 2 - 1) A 2 — 6 2 1 -r* whence, x = b a y[(r 2 - 1) h 2 - b 2 ] (1 - r 2 ) + 6 2 1 -r 2 Problem XV. — To determine a right-angled triangle, having given the hypothenuse, and the side of the inscribed square. Let ACB be a right-angled triangle, and FDEB an inscribed square. Let AC a A, and DF a s : also, denote AB by x, and BC by y; then CE=y — s. Then, AB : BC : : DE : EC ; that is, a; : y : : 5 : y — 5 ; whence, xy ~ sx = sy; or, ary = sy + *r a .9(2- + y) • • (1); also, x 2 + y 2 = A 2 (Bk. IV., Prop. 11) If to equation 2, we add twice equation (1), we have x 2 + 2xy + y 2 = 2s (x -f- y) + A 2 ; or, (* + y)2-2s(z-by)=A 2 . . (3), which is an equation of the second degree, in which the unknown quantity is x + y ; hence, x + y = s + A 2 - 2s 2 - 2s -y/AM^P (8), (9). Problem XVI. — To determine the radii of three equal circles, de scribed within and tangent to, a given circle, and also tangent to each other. Let O be the centre of the given circle, and A, B and C, the centres of the equal in- scribed circles. Denote the radius of the given circle by R, and the equal radii of the inscribed circles by r. Joining the centres A and C, C and B, B and A, by Straight lines, we have the equilateral triangle ABC, each of whose sides is 2r. Draw COD and prolong it to E, and it will be perpendicular to AB. 13 19± APPENDIX. Then, in the right-angled triangle ACD, we have CD 2 = AC 2 -AD 2 , or CD 2 = 4r 2 - r 2 = 3r 2 , or CD = r y /S. But since is the point at which lines drawn from the vertices of the angles to the middle points of the opposite sides, in both tri- angles, intersect each other, it follows, from Cor. of Prob. 21, that CO = f CD = \r V§"; hence, OE = R = r + Jr y^ ; hence, finally, 3R R ~3 + 2y^~ 1 +2v4 Problem XVII. — In a right-angle triangle, having given the peru 'meter and the perpendicular let fall from the right angle on the hy- pothenuse, to determine the triangle. Let ACB be a right-angled triangle, right angled at C ; and let CD be drawn perpendicular to the hypothenuse AB. Let p = the perimeter, and h = the perpendicular CD. Denote A!C by ar, and CB by y ; tken, AB = -y/z 2 -f y 2 , and A AC-f-CB-f ABrrrjt?; that is, z + y + tJx 2 -f y 2 =p • xy = double the area of ACB • h -y/z 2 + y*~= double area • Again, and hence, xy-h V* 2 + y 2 (1). (2), (3); (4) PROBLEMS FROM LEGENDRE. 195 transposing in equation (1) and squaring, x 2 -f 2xy + y 2 =p* — 2p -/a 2 -f y 2 + x 2 + y2j or 2zy=^-%?V^ r T7 r . . (5) . multiplying both members of equation (4) by 2, 2zy^2Ay*2 + y2 . . . (6) ; combining equations (5) and (6), 2hy/x 2 + y« = ^2 _ 2 p^/ X 2 + y 2. fo^ - V^T7 = 2 -^ • . . (7); substituting this value of ^/x 2 -f y\ in equation (4), h P 2 squaring both members of equation (7), we have x 2 4- v 2 — P ■ - adding and subtracting 2 times each member of (8), y y 4(h+ P y TO Extracting the square roots of equations (9) and (10), r | V - P(P + M) . ,,„ y 8(A + j>) I"'' 196 APPENDIX. hence, P (P + 2A) -+• jo y^>2 __ 4^^ _ 4^2 (13), (14). - 4^/, 4^2 4(A+^) Problem XVIII. — To determine a right-angled triangle, having given the hypothenuse and the difference of two lines drawn from the two acute angles to the centre of the inscribed circle. \ Let ABC be a right-angled triangle, right-angled at 13, and AO, OC, two lines drawn from the acute angles to the centre O of the in- scribed circle. Let AC = £; AO - OC = d, OC~*; then, AO =s x -f d Produce AO, and from C draw CD per- pendicular to the prolongation, meeting it at D. Then, since the sum of the angles BAC and ACB is equal to a right-angle (Bk. I, Prop. 25, Cor. 4), and since the lines AO and CO bisect these angles (Bk. Ill, Prob. 14), OAC -f ACO is equal to a haif a right-angle. Since the outward angle COD is equal to the sum of the inward angles (Bk. I, Prop. 25, Cor. 6), it is equal to half a right-angle; and hence, OCD is equal to half a right-angle, and hence OD and CD are equal. Denote either by z ; then, Then, x 2 ±= z 1 -f- z 2 ; and z =s x \/- -(2-V^)^ 2 Let OC = s = m, AO = m + d, CD = z = m-y/j, AB = y, and CB = «. Then, since the triangles ACD and AOG are similar, h : my/% : : m + a : OG = r = — h ~> and h : w 4- d + *»y J : : m + d : AG — AB = y = AG + GB h (ra+eZ) (m + d+m-T/2) = 6 ; and u = -y/h 2 - b'K Problem XIX. — To determine a triangle, having given the base,, the perpendicular, and the difference of the two other sides. Let ACB be a triangle, and CD perpendicular to the base AB. Let AB = 6, CD = «, AC - BC = d, BC = x, AC = x+d, and DB = z, and AD = b—z. Then, and AC= V a2 + ( b - z f ; CB r= -y/a 2 4- z l ; hence, ^/ a 2 + (6 _ 2)2 _ -/a 2 4- z< Transposing and squaring, we have a*+(b- z) 2 = d* 4- 2<2 ^/a 2 4- * 2 4- a 2 + * 2 J 198 APPENDIX, squaring b — z } in the first member, we have a 2 + b* — 2bz + z 2 = d 2 + 2d v^ 2 + * 2 + a 2 + s 2 ; whence, by reducing, (4» - V __ ^2 + c 2 — a 2 ) 2 D B rt*f6-J-c 2(&-f-6 + c) pROBf^.M .VXTEL — To de!enn:i>t .1 tight-angled triangle, having given the sid*, of the inscribed tqw-.t yid the radius of the inscribed circle. Let ABC be a right-angled tiiangV, y*i*j» «\ square and circle both inscribed. Let AB = x, BC = y, and AC • z: denote the side of the square by s, and tht avl ; i * of the circle by r. Then, x + y - z = 2r (Prob. XIII) i (1) x 2 + y 2 = z 2 (2) t V and x + y = * + -v/2 2 + s 2 (Prob. XY) • • <2> Combining equations (1) and (2), and transpose g, 2 -f 2r — 5 = -yjz 2 -f * 2 ; squaring. z 2 + 4r 2 -f- s 2 + 4rz — 2$z — 4rs = z 2 + s 2 t 202 APPENDIX. . 2rs — 2r 2 whence, z = — t . . . (4) . combining equations (4) and (1), x -f y = o ; = m i and equation (2) gives 2 , (2rs-2r 2 ) 2 • + y (2r - s) 2 = ' whence > ar — \ (m + ^/2n 2 — m 2 ), y = 1 (m — y^i 2 — m 2 ). Problem XXIV. — To determine a right-angled triangle, having given the hypothenuse and radius of the inscribed circle. Let ABC be a right-angled triangle, and O the centre of the inscribed circle. Let AC = h, AB = x, BC = y, and r = the radius of the circle. x + y = A+2r(Prob.XIII) . (1) j£-^ts£ (2). j£ Since the perimeter is equal to 2A -f- 2r, and since four times the area is equal to the perimeter into 2r (Prob. XIII), 2xy — 4r 2 -f Mr • . . (3) ; subtracting equation (3) from (2), and extracting the square root, x — y = -y/A 2 — 4r 2 — 4rh • • • (4) ; combining (4) and (1), h + 2r + -yA 2 — 4r 2 -4rh 2 ' A + 2r — -vA 2 — 4r 2 — 4rA PROBLEMS FROM LEGENDRE. 203 Problem XXV. — To determine a triangle, having given the base, the line bisecting the vertical angle, and the diameter of the circum- scribing circle. Let ABC he a triangle, EF the diameter of the circumscribing circle perpendicular to AB, and CD the line bisecting the vertical angle : this line will pass through E, the middle point of the arc AEB. AB = b CD = a EF s d KO = x BC = y I DE = v Auxiliaries. EO = z; OY=d-z DO = « z{d-z) = ^- (Bk. IV. Prop. 28) ; wnence, d ± Jd 2 - b 2 Z = T-7Z = l> By drawing CF, we have two similar right-angled triangles, EOD and EFC ; hence, d : a -f- v : : v : Z; or, v 2 + av = dl ; whence, — a ± J a 2 + 4dl v = £■ = m also, u — ^/v 2 — z 2 = -\/m 2 — I 2 = n. Then, *y* S a* f fj + ») (|- - ») (Bk. IV. Prop. 31), and, 5 , 6 204: APPENDIX. + n whence, x=yx and y = x x b t substituting these values in the preceding equation, and reducing, 1 / (4a 2 + b 2 - 4n 2 ) (6 H- 2») 6 — 2w y= 4\/ 1 /(4a 2 4- A 2 - 4 m 2 ) (6 - 5>n) (b + 2») PROBLEM OF THE THREE POINTS. CONSTRUCTION. From a station P there can be seen three objects, A, B and C, whose distances from each other are known ; viz., AB = 800 yards, AC be 600 yards, and BC = 400 yards : there are also measured the horizontal angles, APC = 33° 45' and BPC = 22° 30'; it is required to find the three distances, PA, PC and PB. Lay off from a scale of equal parts AB = 800 yards ; then, with the known sides AC and CD construct the triangle ACB. At A lay off an angle equal to BPC = 22° 30', and at B lay off an angle equal to APC — 33° 45', and through the point D, where these lines intersect, and the two points A and B, describe the cir- cumference of a circle. Join C and D, and produce it till it meets the circumference at P. Then, since the angle DAB = CPB, standing on the same arc DB, and the angle ABD = APC, standing on the same arc AD, it follows that P will be the position of the third point. PROBLEMS FROM LEGENDRE. 205 CALCULATION. In the triangle ABC, we have given the three sides ; hence, we can find the angles CAB and CBA (Case IV. Trigonometry). In the triangle ADB we know the base AB and the angles at the base ; hence, by Case I., Trigonometry, we can find the sides AD and BD. From the found angle CAB, subtract the known angle DAB = 22° 30', and we have the angle CAD. Then, in the triangle CAD, we know two sides and the included angle, and hence, can find the angle ACP by Case III. Then, in the triangle ACP, we know the side AC ss 600 yards, and the two angles ACP and APC ; hence, we can find PA and PC, after which we can easily find PP. PA=71(U93 yards; PC= 1042.522 yards ; PP= 934.291 yards. f»S^ THE [TJjriVBESITY; THZ END. UNIVERSITY OF CALIFORNIA LIBRARY BERKELEY THIS BOOK IS DUE ON THE LAST DATE STAMPED BELOW Books not returned on time are subject to a fine nf Ssssl-s t„~ d : if •«"«««- ™% &s iUL 31 jy 24 15m-4,'24