Nae wee ext toe Se ees ve aes Does 7a MATHEMATICS L'BEARY Return this book on or before the _ Latest Date stamped below. University of Illinois Library | way 17 4972 MAY 4 oe Reco OCT 4 2 ; 19 f §. * ti ‘% Ro etit. a} vy TVeLis ed Py SAN etn BEY Pech L161—H41 UNIVERSITY OF ILLINOIS LIBRARY NOV 22 ALGEBRA FOR BEGINNERS. ALGEBRA “OR BEGINNERS BY ae | eke ay vv at ‘abe 's! “HALL anv st Rs KNIGHT, ’ Autuors or “ELEMENTARY ALGEBRA FOR Scnootns,”’ ‘* Hiener ALGEBRA,”’ ‘‘ ELEMENTARY TRIGONOMETRY,’ Ero. Ero. REVISED AND ADAPTED TO AMERICAN SCHOOLS BY FRANK L. SEVENOAK, A.M., M.D.,. PROFESSOR OF MATHEMATICS AND ASSISTANT PRINCIPAL IN THE STEVENS SCHOOL, ACADEMIC DEPARTMENT OF THE STEVENS INSTITUTE OF TECILNOLOGY New Work MACMILLAN & CO. AND LONDON 1895 All rights reserved CopyricuT, 1895, By MACMILLAN & CO. Norwood Wpress : J. S. Cushing & Co. — Berwick & Smith. Norwood, Mass., U.S.A. _ f") = fae fl ated Ba ad Bs SS H\4a JATNEMATICS LIBRARY PREFACE. Tur rearrangement of the Elementary Algebra of Messrs. Hall and Knight was undertaken in the hope of being able to give to our advanced secondary schools a work that would fully meet their requirements in this important study. Many changes were made and ad- ditional subject-matter introduced. The Algebra, for . Beginners, however, so fully meets the needs of the tselass of students for which it was written, that we have made only such changes as seemed to bring out more clearly important points, and better adapt it to American schools. With reference to the arrangement of topics, we quote from Messrs. Hall and Knight’s preface to a former edition : Y “Our order has been determined mainly by two con- *’ siderations: first, a desire to introduce as early as pos- ‘sible the practical side of the subject, and some of its most interesting applications, such as easy equations and ~“eproblems; and secondly, the strong opinion that all %sreference to compound expressions and their resolution nto factors should be postponed until the usual opera- tions of Algebra have been exemplified in the case of simple expressions. By this course the beginner soon va. 351058 thang & Mid = ‘ oy & - v1 PREFACE. becomes acquainted with the ordinary algebraical proc- esses without encountering too many of their difficulties ; and he is learning at the same time something of the more attractive parts of the subject. “ As regards the early introduction of simple equations and problems, the experience of teachers favors the opinion that it is not wise to take a young learner through all the somewhat mechanical rules of Factors, Highest Common Factor, Lowest Common Multiple, Involution, Evolution, and the various types of Frac- tions, before making some effort to arouse his interest and intelligence through the medium of easy equations and problems. Moreover, this view has been amply sup- ported by all the best text-books on Elementary Algebra which have been recently published.” The work will be found to meet the wants of all who do not require a knowledge of Algebra beyond Quadratic Equations —that portion of the subject usually covered in the examination for admission to the classical course of American Colleges. FRANK L. SEVENOAK. JUNE, 1895. = : arn ' OHAP. PAGE I. DEFINITIONS. SUBSTITUTIONS . : : : : ; o pis NEGATIVE QUANTITIES. ADDITION OF Likk TERMS : i yi II. Simphe Brackets.» ADDITION. .. + + DS ee ce LL " IV. SUBTRACTION . ; “. . ‘ . " ° 2 16 MISCELLANEOUS EXAMPLES I. . . . . ° ; 19 V. MULTIPLICATION ; ; ; - P c : : 21 wae, Vi, DIVISION, . : - ‘ a : : ‘ ° ‘ 31 « ¥ fp __NI1...REMOVAL AND INSERTION OF BRACKETS . F $ A 38 Seo MISCELLANEOUS EXAMPLES II, . : : : r 42 VIIl. BReEvIsION oF ELEMENTARY RULES . : : ‘ 3 44 IX. SiMPLE EQUATIONS . ; Se ‘ : : - 52 X. SyYMBOLICAL EXPRESSION . ‘ . : : : ‘ 59 XI. Propiems LEADING TO SIMPLE EQUATIONS . F A 64 >. XII. Hicuest Common Factor, Lowest Common MULTIPLE or Simple EXPRESSIONS. FRACTIONS INVOLVING SIMPLE EXPRESSIONS ONLY. : : ; . : (fs XIII. SimuULTANEOUS EQUATIONS : ‘ : P H ; 17 XIV. PROBLEMS LEADING TO SIMULTANEOUS EQuaTIons : 85 XV. INVOLUTION ; : ‘ ; : , ‘ ; 89 XVI. EvoOLUTION ; ; : : 5 : * : : 93 ry XVII. RESOLUTION INTO FACTORS. CONVERSE USE OF FACTORS 101 MISCELLANEOUS EXAMPLeEs III. ‘ ; ; . Peg BE XXV, XXXVI. CONTENTS. PAGE HIGHEST COMMON FACTOR OF COMPOUND EXPRESSIONS 116 MULTIPLICATION AND DIVISION OF FRACTIONS ; ; 122 LOWEST COMMON MULTIPLE OF CoMPOUND EXPRESSIONS 126 . ADDITION AND SUBTRACTION OF FRACTIONS . : fF Wea) MISCELLANEOUS FRACTIONS . .. : ; : . 139 HARDER EQUATIONS . : 5 ; ; : ; . 146 HARDER PROBLEMS . ; ; ° . ° : . 153 Misce.taNrous Exampims IV. ee oe QUADRATIC EQUATIONS . : ‘ ‘ ° ne LG2 PROBLEMS LEADING TO QUADRATIC EQUATIONS. Lis MISCELLANEOUS EXAMPLES V, : ‘ 4 : alee ALGEBRA. CHAPTER L DEFINITIONS. SUBSTITUTIONS. 1, ALGEBRA treats of quantities as in Arithmetic, but with greater generality ; for while the quantities used in arithmetical processes are denoted by figures which have one single definite value, aleebraical quantities are denoted by symbols which may have any value we choose to assign to them. The symbols employed are letters, usually those of our own. alphabet ; and, though there is no restriction as to the numerical values a symbol may represent, it is understood that in the same piece of work it keeps the same value throughout. Thus, when we say “let a=1,” we do not mean that a must have the value 1 always, but only in the particular example we are considering. Moreover, we may operate with symbols without assigning to them any particular numerical value at all; indeed it is with such operations that Algebra is chiefly concerned. We begin with the definitions of Algebra, premising that the symbols +, —, x, +, will have the same meanings as in Arithmetic. 2. An algebraical expression is a collection of symbols; it may consist of one or more terms, which are separated from each other by the signs + and —. Thus 7a+5b—38c-—4#+2y is an expression consisting of five terms. Note. When no sign precedes a term the sign + is understood. 3, Expressions are either simple or compound. A. simple expression consists of one term, as 5a. A compound expression consists of two or more terms. Compound expressions may be € H.A. A 2 ALGEBRA. [CHAP. further distinguished. Thus an expression of two terms, as 3a — 26, is called a binomial expression ; one of three terms, as 2a—3b+c¢, a trinomial; one of more than three terms a multi- nomial. 4, When two or more quantities are multiplied together the result is called the product. One important difference between the notation of Arithmetic and Algebra should be here remarked. #3;In Arithmetic the product of 2 and 3 is written 2 x 3, whereas in Algebra the product of a and 6 may be written in any of the forms axb, a.6, or ab. The form ab is the most usual. Thus, if a=2, b=3, the product ab=axb=2x3=6; but in Arithmetic 23 means “twenty-three,” or 2x 10+3. 5, Each of the quantities multiplied together to form a pro- duct is called a factor of the product. Thus 5, a, b are the factors of the product 5ab. 6, When one of the factors of an expression is a numerical quantity, it is called the coefficient of the remaining factors. Thus in the expression 5a), 5 is the coefficient. But the word coefficient is also used in a wider sense, and it is sometimes convenient to consider any factor, or factors, of a product as the coefficient of the remaining factors. Thus in the product Gubc, 6a may be appropriately called the coefficient of be. A coefficient which is not merely numerical is sometimes called a literal coefficient. Note. When the coefficient is unity it is usually omitted. Thus we do not write la, but simply a. 7, If a quantity be multipled by itself any number of times, the product is called a power of that quantity, and is expressed by writing the number of factors to the right of the quantity and above it. Thus axa is called the second power of a, and is written a? ; LS URIS ere seas setter, POWE? OL Cau. wherne metic: fe a’ ; and so on. The number which expresses the power of any quantity is called its index or exponent, Thus 2, 5, 7 are respectively the indices of a”, a®, a’. Note. a? is usually read ‘“‘a squared”; a® is read ‘fa cubed”; a* is read ‘‘a@ to the fourth” ; and so on. When the index is unity it is omitted, and we do not write a', but simply a. Thus a, la, a’, 1a4 all have the same meaning. I] DEFINITIONS. SUBSTITUTIONS. 2 8. The beginner must be careful to distinguish between coefficient and index. Lzxample 1, Whatis the difference in meaning between 3a and a?? By 3a we mean the product of the quantities 3 and a, By a? we mean the third power of a; that is, the product of the quantities a, a, a. Thus, if a = 4, 34:2 3X GS KAS a=axaxa=4x4x4= G4. Example 2. If b =5, distinguish between 4b? and 2b4+, Here 4b =A b= 4x xX B= 1003 whereas OU Ue D0 = DOR Oo MOK D = 1250. Hxample 3. If «x =1, find the value of 52%. Here De = Mo woe x om Le heel =D; Note. The beginner should observe that every power cf 1 is 1. 9, In arithmetical multiplication the order in which the factors of a product are written is immaterial. For instance 3x4 means 4 sets of 3 units, and 4x3 means 3 sets of 4 units; in each case we have 12 units inall. Thus 3x4=4x3, In a similar way, Ox4 xX Daat x ox Dat SD Mo and it is easy to see that the same principle holds for the product of any number of arithmetical quantities. In like manner in Algebra ab and ba each denote the product of the two quantities represented by the letters a and b, and have therefore the same value, Again, the expressions abc, ach, bac, bca, cab, cba have the same value, each denoting the product of the three quantities a, 6, c. It is immaterial in what order the factors of a product are written; it is usual, however, to arrange them in alphabetical order. Fractional coefficients which are greater than unity are usually kept in the form of improper fractions. Example 4. If a= 6, # = 7, z= 5, find the value of ‘as Here Paws = 3x 6x7 x d= 273, 4 ALGEBRA. [crrap. EXAMPLES I. a. Ifa=5, (bes oe lee 3, y= 12"2-2 and thevalue of Le ed. Ne ibe Oreos Ate. Ny eae Geese Tye ab Se ce Ne a ARGs Oe Lee Lp, Melons 18, 62°. TAS be. 15, Fe: 1b re 6, p =4,q=7, r=5, 2 =1, find the value of - 16 Sap. Lye opG: IG SCE, 1D eOTe: 90, S8aqa. D1. por. OD SA0T eS DG 1 OTe: Pe le TANS | bas DOs op. WA Hei 08> Dapon 29 556xro ith, een lfih=5, k=3, 2 =4, y =], find the value of 1 1 The 1 pclos'f \ps. Ve, La, 1 ike. 1 ys, Bl. G6, 82, abe 88, fy 8A, hk, 85, Sy Chee co yeeoe op ke yee og tis «Ag Shee * 8 * OF “7125 sa ys 10, When several different quantities are multiplied together ~ a notation similar to that of Art. 7 is adopted. Thus aabbbbeddd is written a?b’cd>. And conversely 7a*cd? has the same meaning as 7xaxaxaxcxdxd. Example 1. If c=3, d=, find the value of 16c*a?. Here. 16ctP S16 x 34x? = (16 x5") 3 | 2000 81 62000: Note. The beginner should observe that by a suitable combination of the factors some labour has been avoided. Example2, If p=4, q=9, r=6, s=5, find the value of oa Slps* ager 32q7r3_382x9x6?_ 32x 9x6x6x6 _3 Slp? = 81x4® S8lx4x4x4x4x4 4 11, If one factor of a product is equal to 0, the product must be equal to 0, whatever values the other factors may have. A factor O is usually called a zero factor. For instance, if v=0 then ab’ry? contains a zero factor. Therefore ab?xy? =0 when «=0, whatever be the values of a, b, 7. Again, if c=0, then c?=0; therefore ab’c}=0, whatever values aand b may have. Note. Every power of 0 is 0. ot Tet DEFINITIONS. SUBSTITUTIONS. EXAMPLES I. b. feos pec =) 1 0, 2=7, find the value of 1, 3bp. O- Bax. 3, op. 4, Gaqz. 5, bpz. 6, 3079. pease Segre. 9, gz’. 10, 50? pa. Vive tee oeop 7. - lomeeas 1S ares 15, Saiq’. ieee = 2.90 = 0,-p = 3,'¢ = 45.97=.5, find the value of 3k? 513 m 3m? 16p? 16, p? he qr’ 18, Bye 19, 4/ $ 20. Iq" . 5m? 60 8r3 Imq 8lqtr? als es a OO: 3q? 23. 25lq? 94, 4p? 25, 400p> ma JN q® kr* 5m” ee 26, aT. G08, a9, S80, 12, We now proceed to find the numerical value of expres- sions which contain more than one term. In these each term can be dealt with singly by the rules already given, and by combining the terms the numerical value of the whole expres- sion is obtained. 18. We have already, in Art. 8, drawn attention to the importance of carefully distinguishing between coefficient and index ; confusion between these is such a fruitful source of error with beginners that it may not be unnecessary once more to dwell on the distinction. Hxample. When c=5, find the value of ct ~ 4c + 2c? — 8c?. Here (oe OGD KO on dee 4 = 2) « Qe? 2 5) 2 6 DX = 250 + BC = 8X07 KD Hence the value of the expression = 625 — 20 + 250-75 = 780. 14, The beginner must also note the distinction in meaning between the swm and the product of two or more algebraical quantities. For instance, ab is the product of the two quan- tities a and 0b, and its value is obtained by multiplying them together. But a+b is the sum of the two quantities a and 8, and its value is obtained by adding them together. 6 ALGEBRA. [CHAP. 1. Thus if a=11, 6=12, the sum of a and b is 11412, that is, 23 ; the product of a and b is 11x12, that is, 132. 15, By Art. 11 any term which contains a zero factor is itself zero, and may be called a zero term. Example. Ifia=2,b=0, x=5, y =3, find the value of 5a’ — ab? + 2au°y + 8bay. The expression = (5 x 2?) -0+ (2x 5*x 3)+0 = 40+ 150 = 190. Note. The two zero terms do not affect the result. 16, In working examples the student should pay attention to the following hints. 1. Too much importance cannot be attached to neatness of style and arrangement. The beginner should remember that neatness is in itself conducive to accuracy. 2. The sign = should never be used except to connect -,e . . . quantities which are equal. Beginners should be particularly careful not to employ the sign of equality in any vague and mexact sense. 3. Unless the expressions are very short the signs of equality in the steps of the work should be placed one under the other. 4. It should be clearly brought out how each step follows from the one before it; for this purpose it will sometimes be advisable to add short verbal explanations ; the importance of this will be seen later. EXAMPLES I. c. lig=4,b=21,0=3;f—o0.9 = 1, d= 0, find themalue of 1, 3/+5h- 7b. 2, Tco-9h+2a. 38, 49-5c-9b. 4, 3g-4h+7c. 5, 3f-29-6. 6, 9b-3c+4h. 7, 3a-9b+e. 8, 2f-39+5a. Q, 3c-—4a+7b. 10, 3/+5h-2c-4b+a. ll, 6h-7b-5a-7f+99. 12, 7c+5b —4a+Sh+3g. 13, 9b+a-394+4f+7h. 14, fgtgh-ab. 15, gb-3he+fb. 16, fht+hb-3he. Ll]. f7*= 3a" 4.2¢7, 18, 0?-2h?+3a?, 19, 3b?-2b?+ 4h? - 2h. CHAPTER II. NEGATIVE QUANTITIES. ADDITION OF LIKE TERMS. 17, Iv his arithmetical work the student has been accus- tomed to deal with numerical quantities connected by the signs + and — ; and in finding the value of an expression such as 17+72-—344+6-41 he understands that the quantities to which the sign + is prefixed are additive, and those to which the sign — is prefixed are subtractive, while the first quantity, 13, to which no sign is prefixed, is counted among the additive terms. The same notions prevail in Algebra ; thus in using the expres- sion 74+3b—4c—2d we understand the symbols 7a and 36 to represent additive quantities, while 4c and 2d are subtractive. 18, In Arithmetic the sum of the additive terms is always greater than the sum of the subtractive terms; if the reverse were the case the result would have no arithmetical meaning. In Algebra, however, not only may the sum of the subtractive terms exceed that of the additive, but a subtractive term may stand alone, and yet have a meaning quite intelligible. Hence all algebraical quantities may be divided into positive quantities and negative quantities, according as they are expressed with the sign + or the sign — ; and this is quite irrespective of any actual process of addition and subtraction. This idea may be made clearer by one or two simple illus- trations. G) Suppose a man were to gain $100 and then lose $70, his total gain would be $30. But if he first gains $70 and then loses $100 the result of his trading is a loss of $30. The corresponding algebraical statements would be $100—$70 = +30, $70 —$100 = — $30, 8 ALGEBRA. [CHAP. and the negative quantity in the second case is interpreted as a debt, that is, a sum of money opposite in character to the positive quantity, or gain, in the first case; in fact it may be said to possess a subtractive quality which would produce its effect on other transactions, or perhaps wholly counterbalance a sum gained, (1) Suppose a man starting from a given point were to walk along a straight road 100 yards forwards and then 70 yards backwards, his distance from the starting-point would be 30 yards. But if he first walks 70 yards forwards and then 100 yards backwards his distance from the starting-point would be 30 yards, but on the opposite side of it. As before we have 100 yards— 70 yards= +30 yards, 70 yards —100 yards= — 30 yards. In each of these cases the man’s absolute distance from the starting point is the same; but by taking the positive and negative e signs into account, we see that —30 is a distance from the starting point equal in magnitude but opposite in direction to the distance represented by +30. Thus the negative sign nay here be taken as indicatir : l of direction. may here be taken as indicating a reversal of direct (iii) The freezing point of the Centigrade De ae is marked zero, and a temperature of 15° C. means 15° above the freezing point, while a temperature 15° below the freezing point is indicated by — 15° C. 19. Many other illustrations might be chosen; but it will! be sufficient here to remind the student that a subtractive quantity is always opposite in character to an additive quantity of equal absolute value. In other words subtraction ts the reverse of addition. 20, Derinition. When terms do not differ, or when they differ only in their numerical coefficients, they are called like, otherwise they are called unlike. Thus 3a, 7a; 5a°b, 2u*b ; 3a°b?, —4a°b? are pairs of like terms; and 4a, 3b ; “Te? 9a°b are pairs of unlike terms. Addition of Like Terms. RuleI. The sum of a number of like terms is a like term. Rule II, Jf all the terms are positive, add the coefficients. u.] NEGATIVE QUANTITIES. ADDITION OF LIKE TERMS. 9 Example. Find the value of 8a + 5a. Here we have to increase § like things by 5 like things of the same kind, and the aggregate is 13 of such things ; for instance, 8 lbs. + 5 lbs. = 13 lbs. Hence also, 8a+5a = 13a. Similarly, 8a+5a+a+2a+6a = 22a, Rule III. Jf all the terms are negative, add the coefficients numerically and prefix the minus sign to the sum. Example. To find the sum of -3a, -—5a, —7x, —2. Here the word sum indicates the aggregate of 4 subtractive quantities of like character. In other words, we have to take away successively 3, 5, 7, 1 like things, and the result is the same as taking away 3+5+7+1 such things in the aggregate. Thus the sum of —3x, —5x, -—7x, -—x is — 162. Rule IV. Jf the terms are not all of the same sign, add to- gether separately the coefficients of all the positive terms and the coefficients of all the negative terms; the difference of these two results, preceded by the sign of the greater, will give the coefficient of the sum required. Example 1. The sum of 17% and —8z is 9x, for the difference of 17 and 8 is 9, and the greater is positive. Example 2. To find the sum of 8a, -9a, —a, 3a, 4a, —lla, a. The sum of the coefficients of the positive terms is 16. PS ee rege cts ef cis yer es Puce vckies negative.......... 21. The difference of these is 5, and the sign of the greater is nega- tive; hence the required sum is — 5a. We need not however adhere strictly to this rule, for the terms may be added or subtracted in the order we find most convenient. This process is called collecting terms. ‘21, When quantities are connected by the signs + and -, the resulting expression is called their algebraical sum. Thus lla—27a+13a=-— 3a states that the algebraical sum of lla, —27a, 13a is equal te — 8a, 22. The sum of two quantities numerically equal but with opposite signs is zero. ‘Thus the sum of 5a and — du is 0. 10 ALGEBRA. (CHAP. II. EXAMPLES II. Find the sum of COTS Si Con 11, 13, 15, A: 19, 2a, 3a, 6a, a, 4a. Gb, 11b, 8b, 9b, 5b. 2p, p, 4p, Tp, 6p, 12p. —2x, —62, —10a, - S82. Uy 49) ou, OY ae —2ly, —5y, -8y, - 18y. -—4s, 3s, s, 2s, -2s, —-s. 32, 102, — 7x, 12x, Oa. 2ay, —4xy, — 32y, xy, Txy. abe, - 8abe, 2abe, —5abce. Find the value of 21, 23, 25. 27. 29. — 9a? + lla? +3a2-—4a%. ¢ ~ Lia? + 3a*>— 8a? — Fa? + 2a? a?b? — 7a7b? + Sa2b? + 9a2b?. 2p°q? - 81 p3q? + 17 pq’. 9Jabed — llabcd — 4labed. AY 2, oe, OL, O2. 6c, 7c, 3c, 16c, 18c, 101c. d, 9d, 3d, 7d, 4d, 6d, 10d. — 3b, -—13b, —19b, — 5b. —l%c, —34¢, —9c, — 6c. —4m, —138m, -17m, — 59m. LLY Oy yoy, TY: Sab, —6ab, 5ab, —4ab. Spq, —8pq, 8pq, —4p¢. — xyz, -Qxyz, Txy2z, — xyz. 3b? — 2b? +76? — 96°. 92° — 32° — Oa? — 923. ae — lla®x + 3a2x — Qu?a. Tmin —15m4n + 3m4n. l3pqu — d5xpq —19qpz. CHAPTER IM: SIMPLE BRACKETS. ADDITION. 93, WHEN a number of arithmetical quantities are connected together by the signs + and —, the value of the result is the same in whatever order the terms are taken. This also holds in the case of algebraical quantities. Thus a—b+c is equivalent to a+c-—), for in the first of the two expressions 6 is taken from a, and ¢ added to the result ; in _ the second ¢ is added to a, and 6 taken from the result. Similar reasoning applies to all algebraical expressions. Hence we may write the terms of an expression in any order we please. Thus it appears that the expression a—6 may be written in the equivalent form —b+a. To illustrate this we may suppose, as in Art. 18, that a rep- resents a gain of a dollars, and —b a loss of 0 dollars: it is clearly immaterial whether the gain precedes the loss, or the loss precedes the gain. _ 24, Brackets ( ) are used to indicate that the terms enclosed within them are to be considered as one quantity. The full use _ of brackets will be considered in Chap. vir.; here we shall deal only with the simpler cases. 8+(13+5) means that 13 and 5 are to be added and their sum added to 8. It is clear that 13 and 5 may be added separately or together without altering the result. Thus 8+(18+5)=84+ 13+5=26. Similarly a+(b+c) means that the sum of b and ¢ is to be added to a. | Thus a+(b+c)=a+b+e. 8+(13—5) means that to 8 we are to add the excess of 13 over 5; now if we add 13 to 8 we have added 5 too much, and must therefore take 5 from the result. Thus 8+(13—5)=8+4+13 -5=16. . Similarly a+(b-—c) means that to a we are to add 6, diminished vy c. : Thus at+(b-—c)=at+b-e., 12 ALGEBRA. [oHap. In like manner, a+b—c+(d—e-f)=a+b—-c+d-—e-f. By considering these results we are led to the following rule : Rule, When an expression within brackets is preceded by the sign +, the brackets can be removed without making any change in the expression. 25, The expression a-(b+c) means that from a we are to take the sum of band c. The result will be the same whether b and ¢ are subtracted separately or in one sum. Thus a—(b+c)=a-—b-c. Again, a—(b—c) means that from a we are to subtract the excess of b over c. If from a we take b we get a—b; but by so doing we shall have taken away ¢ too much, and must therefore add etoa—b. Thus a—(b-—c)=a—b+e. In like manner, a—b—(c—d-—e)=a—b—c+d+te. Accordingly the following rule may be enunciated : Rule. When an expression within brackets, is preceded by the sign —, the brackets may be removed if the sign of every term within the brackets be changed. Addition of Unlike Terms. 26, When two or more like terms are to be added together we have seen that they may be collected and the result expressed as a single like term. If, however, the terms are unlike they cannot be collected; thus in finding the sum of two unlike auantities a and 6, all that can be done is to connect them by the sign of addition and leave the result in the form a+b. 27, We have now to consider the meaning of an expression like a+(—6). Here we have to find the result of taking a negative quantity —b together with a positive quantity a. Now —b implies a decrease, and to add it to @ is the ¢ /.1e in effect as to subtract 6; thus a+(-—b)=a—6b; that is, the algebraical sum of a and —b is expressed by a—b. 98, It will be observed that in Algebra the word swm is used in a wider sense than in Arithmetic. Thus, in the language of Arithmetic, a—6 signifies that 6b is to be subtracted from a, III. } ADDITION. 13 and bears that meaning only ; but in Algebra it is also taken to mean the sum of the two quantities a and —b without any regard to the relative magnitudes of a and 0. Example 1. Find the sum of 38a —-5b+2c, 2a+3b-d, -—4a+ 20. The sum = (3a — 5b + 2c) + (2a +386 - d) +(—4a+ 2b) = 38a -5b4+2c+2a4+3b-—d-4a+2b = 8a+ 2a -4a—-5b4+3b+2b+2c-d =a+2c-d, by collecting like terms. The addition is however more conveniently effected by the following rule : Rule. Arrange the expressions in lines so that the like terms may be in the same vertical columns: then add each column beginning with that on the left. 3a — 5b + 2c The algebraical sum of the terms in the 2a +3b ayy first column is a, that of the terms in the ep ee second column is Zero, The single terms in the third and fourth columns are a +2c-d brought down without change. Hxample 2. Add together -5ab+6bc-TJac; 8ab+3ac-2ad; ~2ab+4act+5ad; be-3ub+4ad. — 5ab + 6bce — Tac Sab + 3ac —-2ad Here we first rearrange the ex- “Ge LIES pressions so that like terms are in r the same vertical columns, and then ~ 8ab+ be +4ad add up each column separately. — 2ab + Tbe +7ad EXAMPLES III. a. Find the sum of 3a+2b-5c¢; —4a+b—Te; 4a-3b+6c. 32+2y+625; w2-3y-32; Qa+y—3z. 4p+3q+dr; -2p+8q—-8r; p-qt+r. 7a—5b+8c3; lla+2b-c; 16a+5b-2c. 8l-2m+5n; -614+7m+4n; -1—4m-8n. 5a-—7b+3c—4d; 6b-5c+3d; b4+2c-d. 2a+4b—5x; 2b-5a; -38a+2y; -G6b+Sa+y. Wa-Sy-Tz; 4e+y; 62; b5a-38y+2z, °. e « ° ONaoorRrwhr . 14 ALGEBRA. [CHAP, 9, a—-2b+7c¢+3; 2b-8c4+5; 38c+2a; a-8-Te. 10, 5-x-y; 7+2xe3; 3y-22; -4+a”-2y. 1], 25a-15b+c; 4c-10b+18a; a-—c+20b. 12, 2a-3b-2c+2x; 5x24+3b-7c; 9c-6x—2a. 18, 3a-—5c+2b-2d; b+2d-a; 5c+38f+3e —2a - 3d. 16a Pe ls POU EK Ps BOO ie eon 15, l7ab-13kl-Say; Tay; 12kl—5ab; 3xy—4kl-ab. 16, 2ax-38by-—2cz; 2y-ax+7cz3; axw-—4cz+7Tby; cz—Gby. 17, Saxv+cz-4by; Thy-S8ax-cz3; -3hby+9ax. 18, 3+5cd; 2fg-3st; 1-5ced3; -442st—fy. 19, 5cea+38fy-2+2s; -2fy+6-9s; -3s-4+4+2cx—-fy. 90, -—38ab+7cd-—5qr; 2ry+8qr—cd; 2cd-3qr+ab—2ry. 29, Different powers of the same letter are unlike terms ; thus the result of adding together 2x”? and 3x7 cannot be ex- pressed by a single term, but must be left in the form 2x? + 327. Similarly the algebraical sum of 5a7b?, —3ab°, and —O* is 5a°b? — 3ab—b*, This expression is in its simplest form and cannot be abridged. Example. Find the sum of 6x? — 52, 2x7, 5a, —2a°, - 32, 2. The sum = 62° — 54 +22? + 5a — 223 — 3x2+2 = 623 — 2Qu3 + Qu? — 8a? -5a+5x+2 = 47? — 7? +9, This result is in descending powers of 2. 30, In adding together several algebraical expressions con- taining terms with different powers of the same letter, it will be found convenient to arrange all expressions in descending or ascending powers of that letter. This will be made clear by the following example. Example 1. Add together 32° +7 + 6” - 52°; 27-8 - 9a; Ada —2a° 4327; 32a°-9e—2?; w-a?-774+4. 303 — 522+ 6a+7 | In writing down the first expression Tr ee we put in the first term the highest ss power of xz, in the second term the ~ 20° + 3a? + daz next highest power, and so on till the 30° — x2? -—9x last term in which x does not appear. ia dg ites peed The other expressions are arranged in the same way, so that in each column 32° — 20° -Ta+3 we have like powers of the same letter. III. | ADDITION, 15 Example 2. Add together 8ab? - 2b? +a; 5a*b - ab? - 3a? — 2b? + 3ab? ea? 5b? 3b? — ab?+ 5a*b -3a3 + 8a? ab?+ 9a?b —-2a3 + 3ab? + 14a7b + 4a? ; 8a2+5b?; 9a2b - 2a? + ab? Here each expression contains powers of two letters, and is arranged according to descend- ing powers of b, and ascending powers of a. EXAMPLES III. b. Find the sum of the following expressions : SON OORONH v7 + 3xy-38y°?; -—3a?+ay+2y?; Q2x?-3xyt+y". 2427-2443; —247+5x4+4; 5a —a2+a-—-1; 2x? -Qr+5; xz” — Qa -—6. ~— §2° + 5a — 4. a®—a*b+5ab?+b?; -—a®—10ab?+6°; 2a°b+5ab?- b. 3a? —~9x?—-lle+7; 2u?-5x?4+2; 5a2+152°-TFas 8-9. x°—52°+8a;3; Ta°+4a°+5e; 822-92; 2a°- Tx? — 4a. 4m? +2m2-5m+7; 3m?+6m?-2; -—5m?+3m; 2m-6. ax®—4be?+cx; 8ba?-2cxr-d; bx*+2d; Qax?+d. py — 99. It; = 2py" + 3qy — 5y® + 20y2+8y-1; -2y+5- Gr; Tay-4r; 3py*. 7Ty?; —8y?-4+2y8-y. 2-—a+S8a?-a®; 2Qa?-3a2+2a-2; -—38a+4+7a3—5a? 1+2y - 3y?-5y?; | -1+4+2y?- a7a3 —3a®x2-++-0; 52+ 7a®x? ; xv — 4darty — Bary? 3 Baty + 2a3y3 y; 5y>+3y?+4. 4aPu? — a?x? — 5x. -6xy?; 3°42 + 6ayt - y?. a®—4a*b+6abe; a?b-l0abe+c?; 63+3a7b+abce. ap’ -6bp?+Tep; 5-6cep+5bp?; 3-2ap®; 2p--7. e7 —20° +11c8s; —2c7 —8c8+5c5; 4c8-10c>; 4c? -c®. 4h3—-7+3h4-2Qh; Th-3h?+2-ht; Qh4+2Qh?-5. 30° +4+2y?-5et+2; Tai-5y?+7x-5; 9a3+11-84%+4y?; 6x — y? - 182° — 7. 7+ Qry + 3y"; 327+ 2yz+y*; ey + 22 + yx — 62? - 4y? — 22, e274 32242923 22-3xy - 38yz; CHAPTER IV. SUBTRACTION. 31, THE simplest cases of Subtraction have already come under the head of addition of dike terms, of which some are negative. [Art. 20.] Thus 5a-38a= 2a, 3a —Ta= —4a, —da46a= — Qa. Since subtraction is the reverse of addition, +b—b=0; . a=at+b—b. Now subtract —b from the left-hand side and erase —b on the right ; we thus get a—(-—b)=a+b. This also follows directly from the rule for removing brackets. [Art. 25.] Thus 3a —( --5a)=38a+5a = 8a, and —3a—(-—5a)= —3a+5a = 2a. Subtraction of Unlike Terms. 32. We may proceed as in the following example. Kxample. Subtract 8a—2b—c from 4a—3b+5e. The difference The expression to be subtracted is =4a—3b+5c—(8a—2b—c) | first enclosed in brackets with a minus =4a—38b)+5c—8a+2b+¢ sign prefixed, then on removal of the =4a—8a—3b42b+5c+¢ brackets the like terms are combined =a—b+6e. by the rules already exclaimed in Art, 20, CHAP. Iv.] ~ SUBTRACTION. 17 It is, however, more convenient to arrange the work as follows, the signs of all the terms in the lower line being changed. 4a, — 3b + 5e The like terms are written in —3a+2b+e the same vertical column, and each by addition, a— b+6c column is treated separately. Rule. Change the sign of every term in the expression to be subtracted, and add to the other expression. Note. It is not necessary that in the expression to be subtracted the signs should be actually changed; the operation of changing signs ought to be performed mentally. Example 1. From 5a?+ay take 2a? + 8axy - Ty. 5at+ xy In the first column we combine mentally 52? and — 22°, the algebraic sum of which is 3x7, In = the last column the sign of the term — 77 has to 3a" — Tay +7y? be changed before it is put down in the result. 2a? + Say — Ty? Hxample 2. Subtract 3x?-2x from 1-2. Terms containing different powers of the same letter being wnlike must stand in different columns. _ 73 co) In the first and last columns, as there is nothing to be subtracted, the terms are put — down without change of sign. In the second ~ 23-327 +2e+1 and third columns each sign has to be changed. 3x? — 2x The re-arrangement of terms in the first line is not necessary, but it is convenient, because it gives the result of subtraction in descend- ing powers of x. EXAMPLES IV. Subtract a+2b—c from 2a+3b+c. 2a—b+c from 3a—5b--c. 3a--y-—% from 2 —4y+4 32. x+8y+8z from 10x —7Ty - 6z. —m—3n+p from -—2m-+n- 3p. 3p -2q+r from 4p—7q+3r. a—7b-3c from -—4a4+35)+4+8c. -a-b-9c from -a+b-9e. H.A. B CONDO FPwONH 18 ALGEBRA. [CHAP. Subtract 9, 38x-5y-7z from 2x+3y —4z. 10, -—4¢-2y+1lz from —~a#2+2y- 132. ll, -2x-5y from 2+ 3y- 2z. 12, 3x-y-8z from 2+2y. 138, m-2n-p from m+2n. 14, 2p-3q-r from 2q-4r. 15, ab-2cd-ac from —ab—3cd+2ac. 16, 3ab+6cd -3ac—5ld from 3ab+5cd —4ac -- 6bd. 17. -xyt+yz-—2x from 2xy+2. 18, -2pq-3qr+4rs from qr -4rs. 19, -mn+llnp-—8pm from — 1lnp. 20, x°y -2Qxy?+ 3xyz from 2x?y + 38xy? - xyz. From 91, x?-38x?+2 take —2°+322-—-2. 99, -2x?—x? take 2° -2?-x. 93, a?+l—-B8abc take b?—2abe. 94, -8+4+6bc+b°c? take 4 —- 3be — 5b7c*. 25, 3p*—2p°q + Tpq? take p*gq — 8pq?+q’. 96. 7+x2-2 take 5-x+27+2°%. QT, —44+2°y-ayz take —3-2x°y+llaryz. 98, —Sa’x?+5x?+15 take 9a?x? - 8x? -5. 99, p®+r-38pqr take 1°+¢9°+ 3pqr. 80, 1-32? take 2° -32?+1. “3, 2430-72? take 322- 32-2. 82, x? +1la?+4 take 8a?- 52-3. 83, a®+5-2a? take 8a*+3a? -7. 04, 24+323-2?-8 take 244+ 32?-2#+2, v85, 1-2x4+32? take 7x? —42?4-32+1. 36. xtyztyzx take — 3y?2u — Qary2? — xyz. 37, 4a%a?-38aat+a® take 3a°x?+7a72x3 — a. 88, l-w2+a°-at-az take zt-l+x-2". 89, —-Smn?+15m2n+n3 take m3 —n?+8mn? —7m?n. 40, 1-p? take 2p? - 3pq?- 2q'. 33, The following exercise contains miscellaneous examples on the foregoing rules, IV. | SUBTRACTION. 19 MISCELLANEOUS EXAMPLES I. 1, When x=2, y=3, 2=4, find the value of the sum of 52’, -3xy, 2’. Also find the value of 327+ 324. 9, Add together 3ab+be-—ca, -ab+ca, ab-2be+5ca. From the sum take 5ca+ be — ab. 3, Subtract the sum of x-y+3z and -2y—2z from the sum of 2a —-5y—3z and -3x+y+4+ 4. 4, Simplify (1) 3b —- 2b? - (2b — 30%). (2) 8a-2b —(2b +a) —(a—5d). 5, Subtract 8c?+8c-—2 from c®-1. 6, When w=3, a=2, y =4, 2 =0, find the value of Arey 9) Sata (2) 3y 7, Add together 3a?-7a+5 and 2a*+5a-3, and diminish the result by 3a7+2. (1) 2a?- Bay +4az%, 8, Subtract 267-2 from -—2b+46, and increase the result by S0— 1s 9, Find the sum of 32?-47+8, 24-33-47, and 2z?-2, and subtract the result from 62743. 10, What expression must be added to 5a?-3a+12 to produce 9a?—7? 11. Find the sum of 22, — 2°, 3x7, 2; +52, -4, 32°, ~—52?, 8; arrange the result in ascending powers of x. 12, From what expression must the sum of 5a7-2, 8a+a?, and 7 - 2a be subtracted to produce 8a?+a-—5? 18, When x=6, find the numerical value of the sum of 1-2 +2", 2a2—1, and x — x? 14, Find the value of 6ax + (2by — cz) — (2au - 8by + 4cz) — (cz + a2), nauiema= 0,0 = 1,¢=2,0=8, y=3, 25 4. 15, Subtract the sum of 4° — 3x7, 2a?-7a, 8a-—2, 5- 32°, 2a°-7 from a3+a?+a+1, aA) ALGEBRA. (CHAP. IV. 16, What expression must be taken from the sum of p*-—3p’, 2p +8, 2p7, 2p? — 3p4, in order to produce 477-3? 17. What is the result when -—32°+2z2-1lx2+5 is subtracted from zero. 18, By how much does b+¢ exceed b-c? 19, Find the algebraic sum of three times the square of x, twice the cube of x, —2?+a -22?, and 23-a—a?+1. 20, Take p?-—q? from 3pq—4q", and add the remainder to the sum of 4pq — p? — 3q? and 2p* + 6q". 91, Subtract 3b°+2b6?-8 from zero, and add the result to b4 — 2b? + 30. 99. By how much does the sum of —~m?+2m-—1, m?-3m, 2m? —2m?+5, 3m? + 4m?2+ 5m +8, fall short of 1lm* —8m?+3m? 93, Find the sum of 82° -4a3y?, Taty—axy*, 3a°y?+ 2a°y3 + dary}, y—4ayt+a%y?, 2° —y+a3y24+ vyt- ay? +3aty, and arrange the result in descending powers of x. 24, To what expression must 3x -4a3+7a7+4 be added so as to make zero? Give the answer in ascending powers of x. 95, Subtract 7a?-32-6 from unity, and x - 52? from zero, and add the results. 96, When a=4, b=3, c=2, d=0, find the value of (1) 8a? —2be —ad-+ 3b%ed. jp ee ; Ja 97, Find the sum of a, —3a?, 4a, —5a, 7, -18a, 4a7, —6, and arrange the result in descending powers of a. 98, Add-together 44+327+23, 3-22-11, x?-—22?+7, and sub- tract 22°+2?—7 from the result. 29, Ii a=5x-3y+2z, b= -2x+y-32z, c=x- 5y+6z, find the value of a+b-—-c. 30, If «=2a?-5a+3, y= -8a?+a+8, z=5a?—6a-—5, find the value of «-(y+2). CHAPTER V. MULTIPLICATION. 34, Mu.rreiicaTion in its primary sense signifies repeated addition. Thus 3x5=3 taken 5 times one Sone Here the multiplier contains 5 units, and the number of times we take 3 is the same as the number of units in 5 Again axb=a taken b times =a+a+ta+a+..., the number of terms being 0. Also 3x5=5 x8; and so long as a and 6 denote positive whole numbers it is easy to shew that ax b=-b = at ; that is, the index of a in the product is the sum of the indices of a in the factors of the product. Again, 5ba*=5aa, and 7a°=7aaa ; ba? x 7a=b x X aadaa= 35a", 22 ALGEBRA. [ciaP. When the expressions to be multiplied together contain powers of different letters, a similar method is used. Example. 5a®b? x 8a*bx? = 5aaabb x Saabuxx = 40082) Note. The beginner must be careful to observe that in this pro- cess of multiplication the indices of one letter cannot combine in any way with those of ancther. Thus the expression 40a°b’z? admits of no further simplification. 37. Rule. Zo multiply two simple expressions together, multiply the coefficients together and prefix their product to the product of the different letters, giving to each letter an index equal to the sum of the indices that letter has in the separate factors. The rule may be extended to cases where more than two expressions are to be multiplied together. Hxample 1. Find the product of 2”, x, and 2°. ‘The product)= a7 xa SG20 =e xe ee, The product of three or more expressions is called the con- tinued product. Hxample 2. Find the continued product of 5x?y?, 8y°2, and 3x24. The product 2 Hay? Sy poe nl our ay ee. 38, By definition, (a+b)m=m+m+me-+ ... takena+b times =(m+m+m-+... taken a times), together with (m+m+m-+... taken b times) =am+bm. Also (a—b)m=m+m+m-+ ... taken a—b times =(m+m+m-+... taken a times), diminished by’ (m+m+m+... taken b times) =am—bm. Similarly (a-—b+c)m=am—bm+em. Thus it appears that the product of a compound expression by a single factor ts the algebraic sum of the partial products of each term of the compound expression by that factor. Hxamples. 3(2a+3b- 4c) = 6a+9b —12c. (4a? — Ty — 82?) x Bay? = 12a3y? — Qlay® — Way, v.1 MULTIPLICATION. 29 EXAMPLES V. a. Find the value of BL oa a BER Go Ar eee one GEOG a Od face < 7c 6, 9y? x 5y®. 7, 3m? xbm*, 8, 4a® x Gat. Q, 3xx 4y. 10. ba x 60: ieee oa; TO apt og". Domocaxdbax. 14, 3qrx4or. 15, abxab. 16, 3ac x 5ad. eee ate, 19. 3x°y*x4y2 19, abe xab. 90, atx da°b* Ql, a? x ab x 5ab4. OO Rie x Op rx [pr 93, Ga®y x ay x 9aty?. OA 1G? <3b" x De, DD. Gay? x Tyz x az. 296, 3abcd x 5bca? x 4cabd. Multiply 27, ab-—ac by are. 98, x*y—a®2+4y2 by a3y2?. 99, 5a? - 3b? by 8ab?ct. 30, ab -—5ab+6a by 3a*d. Sigea-— 20? by 327. 32, 2ax?-b®y+3 by aray. 83. Tp*a-part+l by 2p. 384, m?+5mn-3n? by 4m?n. 35, zy? -3x?z-2 by 3yz. S00 oa Dy Zara. 39, Since (a —b)m=am— bm, [Art. 38. | by putting c—d in the place of m, we have (a—b)(c -d)=a(c— d) —b(c—d) =(e—d)a—(ce—d)b =(ace—ad)-— (be— bd) =uac—ad — be+bd. If we consider each term on the right-hand side of this result, and the way in which it arises, we find that (+a)x (+e) =+ae. (—b)x(—d) = +d. (—6b)x(+c) =—be. (+a) x(-d)=-— ad, These results enable us to state what is known as the Rule of Signs in multiplication. Rule of Signs. Zhe product of two terms with like signs ts positive ; the product of two terms with unlike signs ts negative. 24 ALGEBRA, [cHaP, 40, The rule of signs, and especially the use of the negative multiplier, will probably present some difficulty to the beginner. Perhaps the following numerical instances may be useful in illus- trating the interpretation that may be given to multiplication by a negative quantity. To multiply 3 by —4 we must do to 3 what is done to unity to obtain — 4. Now —4 means that unity is taken 4 times and the result made negative: therefore 3x (—4) implies that 3 is to be taken 4 times and the product made negative. But 3 taken 4 times gives +12; oie 4) Similarly —3x —4 indicates that —3 is to be taken 4 times, and the sign changed ; the first operation gives —12, and the second +12. Thus (-—3)x(-4)=+4+12. Hence, multiplication by a negative quantity indicates that we are to proceed just as if the multiplier were posite, and then change the sign of the product. Hxample 1. Multiply 4a by —3b. By the rule of signs the product is negative ; also 4a x 3b = 12ab; *. da x (- 3b) = —12ab. Example 2. Multiply -5ab®x by - ab?a. Here the absolute value of the product is 5a7)°:*, and by the rule of signs the product is positive ; “. (—Sabix) x(—ab?x)= 5a7d's?. Example 3. Find the continued product of 3a7b, -2a3b?, — abt, This result, however, may be a2 WD 3h =o Aap hos i ie ae b m= oes written down at once : for Pe Obi )” x ia a = OL . F > 3a7b x 2a30? x abt = 65’, Thus ee complete pro- and by the rule of signs the re- duct is 6a°b’. quired product is positive. EHxample 4. Multiply 6a? - 5a2b -4al? by - 3ab>. The product is the algebraical sum of the partial products formed according to the rule enunciated in Art. 37 ; thus (Ga? - 5a*b — 4ab?) x ( - 3ab?) = — 18a4h? + 15a7b3 + 12074, v.] MULTIPLICATION, 25 EXAMPLES V. b. Multiply together eo = 2, 9. —3, 42. 3, —2*, -x. 4, —5m, 3m? 5, —4¢q, 3q”. 6, —4y°, —4y°. 7, —3m3, 3m3. 8, 424, — 424, 9, —3x, —4y. 10, —5a7,4~. 11, —3p?,-49°. 12, 3ab, —4abd. Popeoa, — 0°, Zab.» 14, —a, -—0,' =c?. 15, 3a, — 2b, —4c?, —d. 16, —3ab, —4ac,3be. 17, -—2a?, -3a7b,-6. 18, —2p, -3¢q, 4s, -¢. Multiply 19, -ab+ac-—be by -ab. 20, —38a*?-4ax+5z? by — ax’, 91, ac—ac?+c* by —ae. 22, —2ab+cd -efby —32°y%, 4], To further illustrate the use of the rule of signs, we add a few examples in substitution where some of the symbols denote negative quantities. Example 1. Tfia= —4, find the value of a’, Here a® = (-—4)? =(-4)x(-4)x(-4) = — 64. By repeated applications of the rule of signs it may easily be shown that any odd power of a negative quantity is negative, and any even power of a negative quantity is posztive. Example 2. Iia=-1, b=38,c= —2, find the value of — 3a‘bc3. Here -3atbe? = -3x(-1)*x 3x (- 2)? We write. down at = —~3x(+1)x38x(-8) once, (—1)*= +1, and = 72, (-2)? = -8. RENEE V..¢. ‘al 3 “a WV faa —1,9b=0,°¢= 2,07 =, res find the value of rot: OF 0G. 3, an. 4, (-a)?. b, = 3c. 6, (-q)t tase 8, -ac. 9, ad. 10, —acn. TES 3a", 12. 4(—¥- 18, 2abe?. Thee Se. 15, -(a)* 16, -8a7q. 17, —a'n?, 18, ac’. 19, —a%c?, 20, ?g. 26 ALGEBRA. [cHAP. li ea 3 esha) fk 5; y = —], find the value of 91, 3a-2y+4k. 99, -—4c—3x4+2y. 93, -4a+5y-x. 94, ac-—ds3cy- yk. 95, 2ay -ka+4k?. 96, a? -2c?4+3y?. O17, -a-ayt3y". 98, ax-—yx-cy. 29, @-y-+y% 80, a-2?-2y. Bl, c*y2-2act+ch% 82, acy—y*t+2a?. Multiplication of Compound Expressions. 49. To find the product of a+b and e+d. From Art. 38, (¢1+b)m=am+bm ; replacing m by e+d, we have (a+b)(e+d)=a(e+d)+b(ce+d) =(c+d)a+(ce+d)b =ac+ad+be+ bd. Similarly it may be shown that (a—b\e+d)=ac+ad—be—bd; (a+b\e—d)=ac—ad+be—bd; (a—b)(e-d) =ac—ad — bc + bd. 43, When one or both of the expressions to be multiplied together contain more than two terms a similar method may be used. For instance (a—b+ cm =am—bm+em ; replacing m by x—y, we have (a—b-+o\(e—y)=ale—y)—W(e-y) + (w—y) =(ax — ay) - (bx — by)+(cx —cy) =axn—ay — bx+by+cer—cy. 44, The preceding results enable us to state the general rule for multiplying together any two compound expressions. Rule. Multiply each term of the first expression by each term of the second. When the terms multiplied together have like signs, prefix to the product the sign +, when unlike prefie — 3; the algebraical sum of the partial products so formed gives the com- plete product. 45, It should be noticed that the product of a+b and x-y is briefly expressed by (a+6)(#—-y), in which the brackets indicate that the expression a+0 taken as a whole is to be multiplied by the expression v—y taken as a whole. By the v.] MULTIPLICATION. 27 above rule, the value of the product is the algebraical sum of the partial products +ar, +bx, —ay, —by; the sign of each product being determined by the rule of signs. Hxample 1. The product Multiply «+8 by x+7. =(e-S)e4 7) =a 8a 72406 = “7+ 152+56. The operation is more conveniently arranged as follows : aie. We begin on the left and work e+ 7 to the right, placing the second e+ 8x result one place to the right, so 4+ Fa+56 that like terms may stand in the ae Fag Sree same vertical column. by addition, w7+15%+56. Hxample 2. Multiply 2a-3y by 4a-—7y. 20 = DY 4a —Ty 8a? — 12xy — l4ay + 21y? by addition, 82? — 26ay + 21y?. EXAMPLES V. d. Find the product of tp as 7, a5. Q, 2-3, +4, 3, a-6,a-7. 4, y-4, yt+4. 5h, 2+9, 2-8. 6, c-8, ct+8. Tak), k— 5. 8, m-—9, m+12. 9, w2-12,2+11. 10.6 @=14, +1. 11 ep 10s p10. a Wine fe eer aro ae 13, 2-4, —x+4. 14, -y+3, -y-3 15, -a+4,-a45. 16, «2-10, -v7+8 17, -A+4, -k-7. 18, -y-7,-y-7. 19, 2a-5, 8a42. 90, «2-7, 2a+5. Ole 3a 4 Oe 8: 92, 3y-5, y+7. 23, 5m-4,7m-3. 94, Tp-2, 2p+7. 95, x-—3a, 2a+3a. 26, 3a-2b, 2a+38b. 27, 5c44d,5c—4d. 98, a-2x, 3at+2x. 29, Th+c, 7b-2e. 30, 2a-5c, 2a+4 5c. Ol.. 3a—by, 4a+y. 82,. 2y*-3z, 2y+3z. 388, xy+2b, xy—2b. 04, 2u-3a,2x+3b. 85, 3x-4y, 2a+3b. 86, mn-—p, 2xy+3z. 28 46, KHxample 1. 3x7 — 27 —5 2x — 5 62? — 4x”?-—10x — 15x”?+ 10% +25 Ga 31942 +25. ALGEBRA. [ CHAP. We shall now give a few examples of greater difficulty. Find the product of 3a?-2a2-5 and 22-5, ~ Each term of the first expression is multiplied by 2x, the first term of the second expression; then each term of the first expression is multiplied by —5; like terms are placed in the same columns and the results added. Hxample 2. Multiply a-b+38c by a+ 20. aa b+ 3¢ & 2p : a?- ab+3ac 2ab = 20° +6be a = ab + 38ac — 2b? + Bie AT, If the expressions are not arranged according to powers, ascending or descending, of some common letter, a rearrange- ment will be found convenient. Hxample. 2a?— 3ab +4b? = Ha?+ Sab £40? — 10a4 + 15a*b -- 20a7b? + 6a°b-— 9a?b?+ 12ab? 8a2b? — 12ab? + 1604 — 10a4+ 21a) — 21a7b? + 1664. Find the product of 2a7+ 4b? -3ab and 3ab —- 5a? + 4b". The re-arrangement is not necessary, but convenient, because it makes the collec- tion of like terms more easy. : EXAMPLES V. e. Multiply together 1, 22-382-2, Qn-1. 3. 2y7%-sy+1, 3y—1. Hh, 2a?-3a-6, a-2. 7. 327-2447], 20-7. Q, xw?+x-2, w*-x4+2. ll, 2a?- 3a-6, a?-a+2. 18, at+b-c, a-b+e. 4a?-a -2, 2a+3. 3a7+4e4+5, 4a —-5. 5b? -2b4+3, -—2b-3. 5c? -4c+8, -—2ce4+1. w= 24 +5, 27-2745. 2k? —-3k=1, 3h? -—k=1. a-2b-—38c, a—2b+3c. v.] MULTIPLICATION, 29 15, 2-xyty, w+ayt+y’. 16, @-2a%+227, a?+2aa +227. 17, @-0?-38c?, -a?-b?-3c2. 18, w-3a?-a, 2?-3r+1. 19, a®-6a+5, a’?+6a—-5. 90, 2yt-4y?+1, 2y4-4y?-1. 91, 5m?+3-4m, 5-4m+3m. 22, 8a>-2a2-3a, 3a?+1- 5a. 98, 2x2+2a?- 327, 87+24+2e% 94, a? +? -a7l?, a*b?- a+b. 95, a +2°+8a274+8ara, a? + 3ax?- x? - 38a7n. 96, 5Spt-p?+4p?-2p+3, p?-2p43. 97, m®—2m*+3m3—4m?, 4m - 3m? + 2m. 98, at+1+6a?-4a?-4da, a®-1+4+38a-3a?. 99, a®+b?+c?+ab+ac—be, a-b-e. 80, 214+ Gary? + y* — 4a%y—4ay®, — at — y* — Cay? — dary? — 4a°y. 48, Although the result of multiplying together two binomial factors, such as +8 and w-7, can always be obtained by the methods already explained, it is of the utmost importance that the student should soon learn to write down the product rapidly by inspection. This is done by observing in what way the coefficients of the terms in the product arise, and noticing that they result from the combination of the numerical coefficients in the two bi- nomials which are multiplied together ; thus (e7+8)(@+7)=27+80+7x24+ 56 =7+154+56. - (@ —8)(@—7) =a? — 8x —7# +56 =x? —152-+56. (7+ 8)(4@—-7)=274+ 8x2 —Tx—56 ) =2'+x—56. (7 —8)'7+7)=x2°-—8x+ Tx —56 =? -x2- 56. In each of these results we notice that ; 1, The product consists of three terms. 2. ‘The first term is the product of the first terms of the two binomial expressions. 3. The third term is the product of the second terms of the two binomial expressions. 4, The middle term has for its coefficient the sum of the numerical quantities (taken with their proper signs) in the second terms of the two binomial expressions. 30 ALGEBRA. [oHar. v. The intermediate step in the work may be omitted, and the products written down at once, as in the following examples : (w7+2)\(v7+3)=2° + 5r+6. (v—3)(e+4)=a27+4-12. (v7+6)(@—9)=2?— 3x — 54. (a —4y)(x2 — 10y) =x? —14ay + 40y”. (7 —6y)(a@+ 4y) =x? — Qay — 24y?. By an easy extension of these principles we may write down the product of any two binomials. Thus (2% + 3y)(% — y) = 2a" + 3.2xy — 2Qxey — 3y? = 20° + xy — dy". (8a — 4y)(2a + y) =6x? — 8xy + Bry — 4y? = 6x? — Bay —4y’. (v7+4)(~—4)=n7 +427 —4x- 16 =27°—16. (2 + 5y) (2x — 5y) = 40? + 10xy — 10xy — 257? = 44? — 25. EXAMPLES V. f. Write down the values of the following products : 1, (a+3)(a—-2). 9, (a-7)(a—-6). 8, (w—-4)(~%+5). 4, (b-6)(6+4). 5 Y-Hiy—D- 6, (a-1)(a-9). 7, (c-5)(¢+4). 8. (x-9)(a— 3). OS (yA) ye7) 10, (@-3)(a+8). ll, (%-5)(x— 8). 12, (a+7)(a—-7). 13, (4-6)(/-6). 14, (a-5)(a+5) 15, (¢+7)(¢+7) 16, (p+9)(p- 10). 17. (+5)(2=8). 18, (%-9)(%+9). 19, (w-8a)(v+2a). 20, (a-2b)(a+2b). QI, (a-—4y)(x~—-4y). 22, (a+4c)(a+4c). 23, (c-5d)(c-5d). 24, (p—2q)(p+2¢). 95, (2e-8)(82+2). 26, (Ba-1)(2Q4+1). 97, (5x—2)(52+4+2). 28, (8a+2a)(8a-2a), 29, (6u+a)(6x—-2a). 80, (7x+3y)(7x-y). CHAPTER VI. DIVISION. 49, Tur object of division is to find out the quantity, called the quotient, by which the divisor must be multiplied so as to produce the dividend. Division is thus the inverse of multiplication. The above statement may be briefly written quotient x divisor = dividend, or dividend ~+ divisor = quotient. It is sometimes better to express this last result as a frac- tion ; thus dividend = quotient divisor : Example 1. Since the product of 4 and 2 is 42, it follows that when 4x is divided by x the quotient is 4, or otherwise, 4u—a = 4, Example 2, Divide 27a5 by 9a*. ma : 9705 2Taaaan We remove from the divisor The quotient = bt Oaae and dividend the factors com- mon to both, just as in arith- = 3a = 3a" metic. Therefore 27a’ + 9a? = 3a? Example 3. Divide 35a*b?c? by 7ab?c?. 3d5aaa . bb. ce 7a@. 6b. ce In each of these cases it should be noticed that the index of any letter in the quotient is the difference of the indices of that letter in the dividend and divisor. Cc = 5aa.c= 5a. The quotient = 32 ALGEBRA. [CHAP, 50. It is easy to prove that the rule of signs holds for division. Thus QOS eer a a a : a : —a —a Hence in division as well as multiplication like signs produce +, unlike signs produce —, Rule. To divide one simple expression by another : The index of each letter in the quotient vs obtained by subtracting the index of that letter in the divisor from that in the dividend. To the result so obtained prefix with its proper sign the quotient of the coefficient of the dividend by that of the divisor. Hxample 1. Divide 84a°x? by —- 12a4x. The quotient = ( — 7) x a'-4z3-1 Or at once mentally, = -— Tax? 84a°x? + (-12a4x) = — Jaz. Example 2. —45a%b?a*+(~ 9a8bx?) = 5a®ba?. Note. If we apply the rule to divide any power of a letter by the. same power of the letter we are led to a curious conclusion. Thus, by the rule Go? a? 9 = a; 3 but also roa J, a read be This result will appear somewhat strange to the beginner, but its full significance is explained in the Theory of Indices. [See Hlementary Algebra, Chap. xxxt.] Rule. Zo divide a compound expression by a single factor, divide each term separately by that fuctor, and take the algebraic sum of the partial quotients so obtained. This follows at once from Art. 38. Examples. (9x -12y+8z)+(-—3) = -—8x%+4y -2z. (86a°L? — 24a7b° — 20a4b?) + 4a*b = 9ab — 6b* — 5a°b, VI. ] DIVISION. 33 EXAMPLES VI. a. Divide Peace by x. 2. Ga’ by 3a. 3, oa’ by at. 4, 210 by 76°. 5, xy? by —axy. 6, —3axy*® by 3y. 7, 4p?q? by -2pq. 8, 1l5m®n by -—5m. 9, —2m? by —ln. a0 tse" by —6a°. J], 35242 by -72. 19, ~7a*> by \—7. lo. —28p'¢ by 28p*, 14, —Ta* by —2’. 15/ 2dxy2? by -— 322, 16, —12b%< by Gb?e® 17, -9k" by -k4. 18, 2h by —hi. 19, —45a*b%cl’ by 9a7b?c!, 90) Saryte? by —ax%yz?. 91. —168a7b?ca? by — Taba, 99, —35a%bSa? by — 7a*bta’. 23. 327-2e by x. 94, 5a®b- Tab? by ab. 25, 48p*q —24pq? by 8pq. 26, —152°+252* by —- 52°. O17, w-xy-—xz by -x. 98, 10a?-5a*b+a by —a. 99, 403+36ax2—16«% by —4za. 30, 38a?—9a2b —Cab? by —3a. When the Divisor is a Compound Expression. 51, Rule. 1. Arrange divisor and dividend in ascending or descending powers of some common letter. 2. Divide the term on the left of the dividend by the term on the left of the divisor, and put the result in the quotient. 3. Multiply the WHOLE divisor by this quotient, and put the product under the dividend. 4, Subtract and bring down from the dividend as many terms as may be necessary. Repeat these operations till all the terms from the dividend are brought down. Example 1. Divide x?+11%+30 by «+6. Arrange the work thus : +6) 22+ 11x% +30 ( divide x", the first term of the dividend, by x, the first term of the divisor ; the quotient is x. Multiply the whole divisor by x, and put the product «7+ 6x under the dividend. We then have x+6)a*+1le+30(2# o+ 6x by subtraction, Ba +30 On repeating the process above explained, we find that the next term in the quotient is +5. HA; Cc 34 ALGEBRA. [CHAP. The entire operation is more compactly written as follows : x+6)x72+1l2+380(42+5 a+ 6x ox + 30 5x +30 The reason for the rule is this: the dividend is separated into as many parts as may be convenient, and the complete quotient is found by taking the sum of all the partial quotients. By the above process #?+112+30 is separated into two parts, namely 27+ 6x, and 57+30, and each of these is divided by «+6; thus we obtain the partial quotients + and +5. Hxample 2. Divide 24a?- 65xy+2ly* by 8x -3y. Divide 2427 by 8x; this » — 297) D472 — B5y PTa2( 3x -7 8a — dy) 24a" — Oday + 2ly"( 3x - Ty gives 32, the first term of the me ES) quotient. Multiply the whole - 56xy+21y? divisor by 38x, and place the — 56ay + 21y? result under the dividend. By subtraction we obtain —56xy+2ly?. Divide the first term of this by 8z, and so obtain —Ty, the second term of the quotient. Hxample 3. Divide 16a*—46a?+39a-9 by Sa-3. 8a —3 ) 16a? — 46a? + 39a — 9 ( 2a? -5a+3 l6a°— 6a? — 40a?+ 39a — 40a? + 15a 94a —9 Y%4a—9 Thus the quotient is 2a? -5a+3. EXAMPLES VI. b. Divide ], a+2a+1 by a+. Q, U?+3b4+2 by 6+2. 3, 2?+42+3 by «+1. 4, y?+5y+6 by y+3. 5, 27?+5x2-6 by x-l. 6, 2°+2x2-8 by 2-2. 7, p?+38p-40 by pt8. 8. g@-4q-82 by q+4. 9, a?+5a-50 by a+10. 10, m*+7m-78 by m-—-S. ll, #*+ax-30a* by x+6a. 12, a*+9ab—-36b? by a+120. vI.] DIVISION. 35 Divide 18, -—2?+18x%-45 by x-15. 14, «* 4274441 by x--21. 15, 2au?-182-24 by 2443. 16, 5x74+16%+3 by «+3. 17, 6%7+52-21 by 22-3. 18, 12a*+au-6xz? by 3a- 22. 19, —52?+ay+6y? by ~x-y. 20, 6a? ac—35c? by 2a—-5e. 21, 12p?--'749q+12q? by 2p-12¢. 92, 4m?-49n? by 2m+7n. 238, 12a7--3lab+20b? by 4a—-5bd. 24, —2527+49y? by -5a+7y. 25, 21p?+1lpq—40q? by 3p+5q. 26, 82°+82?+4e+1 by 27+]. 27, —2x3+1322-17x+10 by —x+5. 98, x? +ax?-3a°x-—6u? by w-2a. 29, Ga®y—a*y?-Txy?+12y* by 2a+4+3y. 30, 8x?-122?-14%+21 by 2a-3., 52. The process of Art. 51 is applicable to cases in which the divisor consists of more than two terms. Hxample 1. Divide a*t-—2a*-—7a?+8u+12 by a?-a-6. a*—-a-6)a*—2a?—7Ja?+8a+12(a2?-a-2 a*—a?— 6a? -a-—a*+8a —a?+a"+6a — 2a7+2a+12 — 2a7+2a+12 Hxample 2. Divide 4x?-5x?+6x°-15-at-x by 3+22%--2, lirst arrange each of the expressions in descending powers of 2. 2x7 -—2+3)62°— 244+ 403-5? -a2- 15 ( 822+ a?-2a-5 6x? — 324+ 9x3 24 —5a— 5a? se a ode —493— 827- x —4a3+ 2o7?- 6a *~ 1027+ 54-15 —10x7+5x-15 36 ALGEBRA. [CHAP. Hxample 3. Divide 23a7-2a4-4a3412+2>-3lx by x2?-7x+5. x? — Te +5 ) a° — Qat — 4a3 4+ 23a? — 31a +12 ( a? - 24+3 xe” —TJao+ 5a ~ 2x4 + 3a? + 18x? - 31a — 2x4 +14a?—-10x 323+ 47-2174 12 32° —21x%4+15 hege® ae Now 42? is not divisible by x’, so that the division cannot be carried on any further ; thus the quotient is «?-2x+3, and there is a remainder 4a7— 3. In all cases where the division is not exact, the work should be carried on until the highest power in the remainder is lower than that in the divisor. 53. Occasionally it may be found convenient to arrange the expressions in ascending powers of some common letter. Example. Divide 2a?+10-16a-39a?+15a4 by 2-4a—-5a?. 2-—4a-5a?)10-16a-39a?+ 2a°+ 15a4( 5+ 2a - 8a? 10 — 20a — 25a? 4a—14a?+ 2a° 4a— 8a*-~10a° — 6a?+12a?+15a4 — 6a?+12a*?+15a4 EXAMPLES VI. c. Divide 1, w-6a?+lla-6 by a?-4a+3. 9, x -40?4+24+6 by x?-x-2. 8. yty?-9yt12 by ¥°-3y+4+3. 4, 2lm?-m?+m-1 by 7m?+2m+1. 5, 6a*?-5a*-9a-2 by 2a7-3a-1. 6, 66-k-14k4+3 by 3h°+4k-1. 7, 6x?+1la?—-392-65 by 3x74 134413. 8, 122° — 8ax? — 27a7x%+ 18a? by 62? —13ax + 6a? vi.] DIVISION. 37 Divide Q, 162°+14a?y —129xy? - 15y? by 8x7 + 27xy + 3y?. 10, 2c? — 5c2d — 3ced? - 2d? by 7c?+3cd+d?. 1], 324-1003 +12¢?-1l2+6 by 32°-2x+3. 12, 30a*+1la’ —- 82a7-12a+48 by 8a7+2a-4. 13, v°--%?-8x-13 by 274+32+4+3. 14, a+38a?+6-10a? by a?-4a+3. 15, 21m?-—27m-26m?+20 by 38m+7m?-4. 16, 18x?+24a3 — 40a?x ~- 9ax? by 927+ 7a? - 18ax. 17, 3y'-4y?+10y?+3y-2 by y-y7+3y+2. 18, 5a?+1+10at- 4a? by 5a8-2a+1. 19, l2a4*+5x3-332?-32+16 by 4a°-x-5. 90, pt-6p?+18p?-10p+7 by p?-3p4+2. D1, 28x*+69x"+2— 71x? - 35x? by 40746 —- 13a. 99, 5a°—7a4— 9a? —-1la?-388a+40 by -—5a?+17a—-10. 93, «x°>-8a® by 2?+2axr+ 4a. 24, yi+9y?+81 by y?-3y4+9. 95, «vt+4y* by x2 + 2ay + 2y". 96, 9at-—4a7+4 by 3a?-4a+2. D7, a®+64 by at-—4a?+8. 98, 1624+ 362°4+81 by 4%7+62+9. 99, 4m? — 29m -36 + 8:n?-—7m?+6m* by m? —2m?+3m -4. BO, 15a4 +22 — 320° - 302 +502? by 3-—4x + 52°. Bl, 3a?+8ab+4b?+10ac+8bce4+ 8c? by a+2b4+ 3c. 62, 9u?-4y?+4yz2-2% by -—38x+2y—-z. 63, 4c7-12c-—d?+9 by 2c+d-3. 84, 9p? -16q?+380p +25 by -38p-—4q-5. O06 Ce y tery —e+2°—y by -x-y. 86, xv +a2ty—2%y?+23-Qay? by 22 +ay—-y?. 37, a°b?+ab—9 —b4+ 3b? + 3b - at -— 3a? - 3a by 3-b+a’°, 88, z4+1 by v4+a?+e4+1. 39, 2a°+2 by a?+2a?+2a+1. 40, 2°-—6a4-82°-1 by 2? -2e-1. CHAPTER VII. REMOVAL AND INSERTION OF BRACKETS. 54, Quantities are sometimes enclosed within brackets to indicate that they must all be operated upon in the same way. Thus in the expression 2a--3b—(4a—2b) the brackets indicate that the expression 4a—26 treated as a whole has to be sub- tracted from 2a — 30. It will be convenient here to quote the rules for removing brackets which have already been given in Arts. 24 and 25. When an expression within brackets is preceded by the sign +, the brackets can be removed without making any change in the CXPTeSSON. When an expression within brackets is preceded by the sign —, the brackets may be removed if the sign of every term within the brackets be changed. Example. Simplify, by removing brackets, the expression (2a — 3b) — (3a + 4b) — (b - 2a). The expression = 2a — 3b —- 38a-4b-—b+2a = a-8b, by collecting like terms. 55, Sometimes it is convenient to enclose within brackets part of an expression already enclosed within brackets. For this purpose it is usual to employ brackets of different forms. The brackets in common use are ( ), { }, [ ]. 56, When there are two or more pairs of brackets to be removed, it is generally best to begin with the innermost pair. In dealing with each pair in succession we apply the rules quoted above. Hxample. Simplify, by removing brackets, the expression a — 2b — [4a — 6b - {3a -c + (2a — 4b +¢)}]. Removing the brackets one by one, the expression = a ~ 2b —- [4a - 6b — {3a —c+2a -—4b+c}] = a—2b—[4a - 6b-38a+c¢-2a+4b -c] =a—-2b-4a+6b+38a—c+2a-4b+¢ =2a, by collecting like terms. Note. At first the beginner will find it best not to collect terms until all the brackets have been removed. CHAP. VII.] REMOVAL OF BRACKETS. 39 EXAMPLES VII. a. Simplify by removing brackets and collecting like terms : 1, @+2b+ (2a - 3b). 9, a+2b-(2a-3b). 8, 2a-—3b—-(2a+25). 4, a-2-(4-8a). 5, (x—3y) + (2u—-4y) —(x-8y). 6, @+2b-38¢-(b-a-4e). 7, («-8y+2z) -(2-4y+4+2z). 8, 4a -(2y +22) — (38x - dy). 9. 2a+(b—3a) — (4a — 8b) — (6b — 5a). 10, m-(n-p)—(2m-2p+28n) -(n-m+2p). 1], a-b+c-(a+c-—b)-(a+b+c) -(b+e-a). 12, 5a—-(Ty+3x) — (2y+7x) — (8a + 8y). 13, (p-49)-(¢-2p)+(2p — 9g) — (p - 29). 14, 2a? - (3y?— 22) — (a? 4y?). 15, (m?— 2n?) — (2n? — 3m?) — (3m? - 4n?), 16, (#—-2a) —(x% — 2b) —{2a—a%-(2b+2)}. 17, (a+3b) -(b—3a) —{a+2b -(2a-3)}. bee 9 (9° + 2p") —{p" £39? — (2p? — 9’)}. 19, x-[y+{e-(y-x)}} 20, (a—b)-{a-b- (a+b) -(a—D)}. 21, p-[p-(¢+p)-{p—(2p-4)}]. 92, 8e-y-[w-(2y-2)-{2e—(y—2)}} 93, 38a? —[6a? — {8b? — (9c? — 2a7)}]. Q4, [Ba — {2a —-(a—b)}] - [4a — {3a — (2a — b)}}. 57, *xat=a7*4=aq%=1. [See Note; Art: 50.) ab or ae pCa Coes el = in 5 Now clear of fractions by multiplying by 5 x 7.x 4 or 140; 72a — 108 + 45a +405 = 280x — 2800 ; 2800 — 108 + 405 = 280a — 72a — 45a ; 3097 = 1632 ; Zhe: 81, Tosolve equations whose coefficients are decimals, we may express the decimals as common fractions, and proceed as before; but it is often found more simple to work entirely in decimals. Example. Solve ‘3752 ~1°875 = '12%+1-185. Transposing, *375a - 12% =1'185+41°875 ; collecting terms, (°375 — *12)~ =3°06 ; that is, 255x = 3°06 ; = 3°06 "255 = 12, EXAMPLES IX, b, Solve the equations : 1, (%+15)(a% - 3) —(x%-3)?=30-15(x—-1). Q. 15 -3a =(2x2+1)(2x-1)— (2x —1)(2x2 +3). 2 Q1—a(2a+1)+2(a—-4)(a +2) =0. 4, 3(2+5)—3(2a —1)=32—4(x —5)?+ 422, 5, 3a? -- Ta — (2 +2)(%—2)=(2+1)(%—1) + (a2 -3)(2 +3). 6, (x-6)(2e-9) - (11 -2x)(7 - x) =5a —4—7(a -2). x-1l, «x-9_ CAS - Lh Oe tere SEP ger ane “+8 5,x-6 62-2 , 3x+5_1 iT wctetear ek Ook Gish ie 58 ALGEBRA. ; [CHAP, IX. Solve the equations : 1a 13. 15. iy, 19. 21. 22, 20. 24. 25. 26. 27, 28, 29. 30, dl. 33, 30, 37, 38. ee oer 2. 12, 2+84+%27=7420. 2-0 -e=-de>> & Cee al yaar ta NU) P(g aa +5 ae2+1 _2+3 16 1i-62° 9-72 _5(¢%-1) 6 9 Wir ; 5 Oot ewe O 47 — 6x tan ed) 4—52 1-2 -13 5 en Oe see 18, 6 Fea 3¢-1 w-1_ 2xa- 31 1 Be SH OF Sa ek ees nay PD pees 5) | 4 “Ww 2 3 hs Satay ver 3 5 8 5 P(x = —6) +—. $(@-1) ~2(a—4) = 3(e-6) += = hs ] = =(2ar — -(x-1)-2 S (w@- 4) ~ 3(2n-9)=3(e -1) = (+4) — 5 (w—8) = s(8x ~ 5)-3(« ~ 6) - H(e-2) ] ] 1 ] (3 87} eee Oe 1 (8 8a) — 51 x) + 5 = 62) 1 1 Li 1 2 4 20 — a) =— (5a — 1)-_(5a%- 18) +8. Ca )- 7 x) is! ae ) go 3)+8 e+1 52+9 2+6 x-12 2 ones ae ~ l0x+1 x _l3e+4 S(x- 4) 5 - = hn, te 18 4 sd ae 6 24+5 a SOS F "i == 1 =e ae (2-3) - + 0) + 7 0 a+4 1] a 1 ESS e217) a9 M6 Sbay Ss lees 0 Be a pee Le sy eet Li oe Speed | x Se = eee ebe hes |. na AG a al 5) “Ta — 3°35 = 6°4 -— 32a. 33, *Da4+- 254 °141°25= 4a, 2 20e— lore 94 Ltbx. 34, ‘2x- 0lx+°005z7=11°7. Be — 6x = "75x —11. ‘ 96, -4r—'83x2=°7-°3. Find the value of x which makes the two expressions (8% —-1)(4a-11) and 6(22-1)(@-38) equal. What value of « will make the expression 77x — 3(2a — 1)(4a — 2) equal to 337 — 8(38%—-1)(@+1)? CHAPTER X. SYMBOLICAL EXPRESSION. 62, In solving algebraical problems the chief difficulty of the beginner is to express the conditions of the question by means of symbols. A question proposed in algebraical symbols will frequently be found puzzling, when a similar arithmetical question would present no difficulty. Thus, the answer to the question “find a number greater than 7 by a” may not be self- evident to the beginner, who would of course readily answer an analogous arithmetical question, “find a number greater than 50 by 6.” The process of addition which gives the answer in the second case supplies the necessary hint ; and, just as the number which is greater than 50 by 6 is 50+6, so the number which is greater than 7 by ais xv+a, 88. The following examples will perhaps be the best intro- duction to the subject of this chapter. After the first we leave to the student the choice of arithmetical instances, should he find them necessary. Example 1. By how much does x exceed 17? Take a numerical instance ; ‘‘ by how much does 27 exceed 17?” The answer obviously is 10, which is equal to 27 ~ 17. Hence the excess of x over 17 is x- 17. Similarly the defect of x from 17 is 17-2. Hxample 2. If x is one part of 45 the other part is 45 — x, Hxample 3. If x is one factor of 45 the other factor is 49 6; Example 4. How far can a’man walk in @ hours at the rate of 4 miles an hour? In 1 hour he walks 4 miles, In a hours he walks a times as far, that is, 4a miles, 60 ALGEBRA. [cHAP. Example 5. If $20 is divided equally among y persons, the share of each is the total sum divided by the number of persons, or ¢ 20. y Example 6. Out of a purse containing $x and y half-dollars a man spends z quarters ; express in cents the sum left. $x—4e quarters, and y half-dollars=2y quarters; .. the sum left=(4”%+2y—z) quarters, =25(4%+2y—z) cents. EXAMPLES X, a, By how much does xz exceed 5? By how much is y less than 15? What must be added to a to make 7 ? What must be added to 6 to make b? By what must 5 be multiplied to make a? What is the quotient when 3 is divided by a? By what must 6x be divided to get 2? By how much does 6% exceed 2a ? . The sum of two numbers is 2 and one of the numbers is 10; what is the other ? 10, The sum of three numbers is 100; if one of them is 25 and another is x, what is the third ? 11, The product of two factors is 4x; if one of the factors is 4, what is the other ? 12, The product of two numbers is p, and one of them is m3 what is the other ? 13, How many times is x contained in 2y ? 14, The difference of two numbers is 8, and the greater of them is @ ; what is the other ? . 15, The difference of two numbers is x, and the less of them is 6; what is the other ? 16, What number is less than 30 by y? 17, The sum of 12 equal numbers is 48x; what is the value of each number ? 18, How many numbers each equal to y must be taken to make ldxy? 19, If there are x numbers each equal to 2a, what is their sum ? 20, If there are 5 numbers each equal to x, what is their product? . COABDO POOH x SYMBOLICAL EXPRESSION. 61 91, Ifthere are x numbers each equal to p, what is their product ? 99, If there are n books each worth y dollars, what is the total cost ? 93, If mn books of equal value cost x dollars, what does each cost ? 94, How many books each worth two dollars can be bought for y dollars ? 95, If apples are sold at x for a dime, what will be the cost in cents of y apples ? 96, What is the price in cents of m oranges at six cents a score ? 97, If I spend n dimes out of a sum of $5, how many dimes have I left? 98, What is the daily wage in dimes of a man who earns $12 in p weeks, working 6 days a week ? 99, How many days must a man work in order to earn $6 at the rate of y dimes a day ? 80, If persons combine to pay a bill of $y, what is the share of each in dimes ? 31, How many dimes must a man pay out of a sum of $p so as to have left 50x cents ? 382, How many persons must contribute equally to a fund con- sisting of $x, so that the subscription of each may equal y quarters ? 33. How many hours will it take to travel x miles at 10 miles an hour ? 34, How far can I walk in p~ hours at the rate of g miles an hour ? 35, If I can walk m miles in n days, what is my rate per day ? 36. How many days will it take to travel y miles at x miles a day? . 84, We subjoin a few harder examples worked out in full. Example 1. What is (1) the sum, (2) the product of three con- secutive numbers of which the least is »? The two numbers consecutive to n aren+1 andn+2; .. the sum=n+(n+4+1)+(n+2) =3n+4+3. And the product =n(n+1)(n+2). Example 2. A boy is x years old, and five years hence his age will be half that of his father: how old is the father now? In five years the boy will be +5 years old ; therefore his father will then be 2(z+5), or 2x+10 years old; his present age must therefore be 27+ 10-5 or 2%+5 years. 62 ALGEBRA. [onar. KHxample 3. A and B are playing for money; A begins with $p and B with g dimes. B wins $a; express by an equation the fact that A has now 3 times as much as B. What B has won A has lost ; .. A has p—« dollars, that is 10(p—a) dimes, B has q dimes + dollars, that is g+ 10x dimes. Thus the required equation is 10(p—x) =8(q+10z). Hxample 4. A man travels a miles by coach and b miles by train; if the coach goes at the rate of 7 miles an hour, and the train at the rate of 25 miles per hour, how long does the journey take? The coach travels 7 miles in 1 hour ; 1 es eae SA errs Be creea 5 hour ; 2 a UAT Ieee ah eee Ch We an: hours. Similarly the train travels b miles in = hours. 5 a “, the whole time occupied is 7+ © hours. 25 Example 5. Wow many men will be required to do in p hours what g men do in np hours? np hours is the time occupied by q men ; ne Lk HOUT Aa eas etee eee qx np men; thatiis,-7 HOULSs 6), deeeeee ceanes = £2 een. P Therefore the required number of men is qn. EXAMPLES X. b. 1. Write down three consecutive numbers of which a is the least. 9, Write down four consecutive numbers of which 0 is the greatest. 3. Write down five consecutive numbers of which c is the middle one. 4, What is the next odd number after 2n-1? 5, What is the even number next before 2n ? 6. Write down the product of three odd numbers of which the middle one is 2a7+1. 7, How old is a man who will be x years old in 15 years ? 8, How old was a man 2 years ago if his present age is m years? 9, In 2x yearsa man will be y years old, what is his present age ? x. ] SYMBOLICAL EXPRESSION. 63 10, How old is a man who in # years will be twice as old as his son now aged 20 years ? 11. In 5 years a boy will be x years old; what is the present age of his father if he is twice as old as his son ? 12, A has $m and B has n dimes; after A has won 3 dimes from B, each has the same amount. Express this in algebraical symbols. 13, A has 25 dollars and B has 138 dollars; after B has won x dollars he then has four times as much as A. Express this in algebraical symbols. 14, How many miles can a man walk in 30 minutes if he walks 1 mile in 2 minutes ? 15, How many miles can a man walk in 50 minutes if he walks x miles in y minutes ? 16, How long will it take a man to walk p miles if he walks 15 miles in q hours? 17, How far can a pigeon fly in x hours at the rate of 2 miles in 7 minutes ? 18, A man travels x miles by boat and y miles by train, how long will the journey take if the train goes 30 miles and the boat 10 miles an hour? 19, If « men do a work in 5z hours, how many men will be required to do the same work in y hours ? 20, How long will it take p men to mow g acres of corn, if each man mows 7 acres a day ? 91, Write down a number which, when divided by a, gives a quotient ) and remainder c. 99. What is the remainder if x divided by y gives a quotient z? 93, What is the quotient if when m is divided by n there is a remainder r? 94, If a bill is shared equally among n persons, and each pays 75 cents, how many dollars does the bill amount to ? 95, A man has $x in his purse, he pays away 25 dimes, and receives y cents ; express in dimes the sum he has left. 96, How many dollars does a man save in a year, if he.earns $« a week and spends y quarters a calendar month ? 97, What is the total cost of 6% nuts and 4% plums, when x plums cost a dime and plums are three times as expensive as nuts ? bs 98. If on an average there are x words in a line, and y lines in a page, how many pages will be required for a book which contains z words ? CHAPTER XI. PROBLEMS LEADING TO SIMPLE EQUATIONS. 85, THe principles of the last chapter may now be employed to solve various problems. The method of procedure is as follows : Represent the unknown quantity by a symbol, as 2, and express in symbolical language the conditions of the question ; we thus obtain a simple equation which can be solved by the methods already given in Chapter LX. Example I, Find two numbers whose sum is 28, and whose difference is 4. Let x be the smaller number, then x +4 is the greater. Their sum is x+(x%+4), which is to be equal to 28, Hence xet+u+4=28; Ue = 243 w= 12, and x+4= 16, so that the numbers are 12 and 16. The beginner is advised to test his solution by finding whether it satisfies the conditions of the question or not. Example II. Divide $47 between A, B, C, so that A may have $10 more than B, and B $8 more than C. Let « represent the number of dollars that C has ; then B has x+8 dollars,;and A has «+8+10 dollars. Hence e+ (#+8) + (#+8+10) =47 ; ¢+%4+84+2748410=47, 34221 3 Bs | ; so that C has $7, B $15, A $25, CHAP, XI.] PROBLEMS LEADING TO SIMPLE EQUATIONS. 65 EXAMPLES XI, a, ], Six times a number increased by 11 is equal to 65; find it. 2, Find a number which when imultiplied by 11 and then diminished by 18 is equal to 15. 3. If 3 be added to a number, and the sum multiplied by 12, the result is 84; find the number. “ One number exceeds another by 3, and their sum is 27 ; find them. 5, Find two numbers whose sum is 30, and such that one of them is greater than the other by 8. 6, Find two numbers which differ by 10, so that one is three times the other. 7, Find two numbers whose sum is 19, such that one shall exceed twice the other by 1. 8, Find two numbers whose sum shall be 26 and their differ- ence 8. 9, Divide $100 between A and B so that B may have $30 more than A. 10, Divide $66 between A, B, and C so that B may have $8 more than A, and C $14 more than B. 11. A, B, and C have $72 among them; C has twice as much as B, and B has $4 less than A ; find the share of each. 12, How must a sum of 78 dollars be divided among A, B, and C,so that B may have 8 dollars less than A and 4 dollars more than C’? Example III, Divide 60 into two parts, so that three times the greater may exceed 100 by as much as 8 times the less falls short of 200. Let x be the greater part, then 60 —-~z is the less. Three times the greater part is 3x, and its excess over 100 is 32 — 100. Eight times the less is 8(60- x), and its defect from 200 is 200 — 8(60 - z). Whence the symbolical statement of the question is 3x — 100 = 200 — 8(60 — x) ; 3x — 100 = 200 — 480+ 82, 480 -— 100 — 200 = 8a - 32, Da = 180% x = 36, the greater part, and 60 — ax = 24, the less. H.A. E 66 ALGEBRA. [omar Example IV, A is 4 years older than B, and half A’s age exceeds one-sixth of 5’s age by 8 years; find their ages, Let x be the number of years in B’s age, then A’s age is 2 +4 years. One-half of A’s age is represented by $(~+4) years, and one-sixth of B’s age by ta years. Hence 3(xv+4)-4r=8; multiplying by 6 3x+12-—x%=48; Vi oe hoped Noe Thus B’s age is 18 years, and A’s age is 22 years. 13, Divide 75 into two parts, so that three times one part may be double of the other. 14, Divide 122 into two parts, such that one may be as much above 72 as twice the other is below 60. 15, A certain number is doubled and then increased by 5, and the result is less by 1 than three times the number ; find it. 16, How much must be added to 28 so that the resulting number may be 8 times the added part ? 17, Find the number whose double exceeds its half by 9. 18, What is the number whose seventh part exceeds its eighth part by 1? 19, Divide 48 into two parts, so that one part may be three-fifths of the other. 90. If A, B, and C have $76 between them, and A’s money is double of B’s and C’s one-sixth of B’s, what is the share of each? 91, Divide $511 between A, B, and C, so that B’s share shall be one-third of A’s, and C’s share three-fourths of A’s and B’s together. 29, B is 16 years younger than A, and one-half 5’s age is equal to one-third of A’s ; how old are they ? 93, A is 8 years younger than J, and 24 years older than C; one-sixth of A’s age, one-half of 5’s, and one-third of C’s together amount to 38 years ; find their ages. 94, Find two consecutive numbers whose product exceeds the square of the smaller by 7. [See Art. 84, Ex. 1.] 95, The difference between the squares of two consecutive numbers is 31 ; find the numbers. 86. We shall now give examples of somewhat greater difficulty. Example I. A has $6, and B has six dimes; after B has won from A acertain sum, A has then five-sixths of what B has; how much did B win ? XI.]} PROBLEMS LEADING TO SIMPLE EQUATIONS. 67 Suppose that B wins # dimes, A has then 60—«a dimes, and B has 6+” dimes. Hence 60—»=2(6+4+2); 360-—6x—= 30+ 52, Lins sa0% 7=380. Therefore B wins 30 dimes, or $3. Example II. A is twice as old as B, ten years ago he was four times as old ; what are their present ages ? Let B’s age be x years, then A’s age is 2% years. Ten years ago their ages were respectively «—10 and 27%—10 years ; thus we have 2x—10=4(x%—10) ; 2x—10=4a—40, 2x%—= 30 ; a 15, - so that Bis 15 years old, A 380 years. EXAMPLES XI. b. 1, A has $12 and Bhas $8; after B has lost a certain sum to A his money is only three-sevenths of A’s; how much did A win ? 9, Aand B begin to play each with $15; if they play till B’s money is four-elevenths of A’s, what does B lose ? 8, Aand B have $28 between them; A gives $3 to B and then finds he has six times as much money as B; how much had each at first ? 4, Ahad three times as much money as B; after giving $3 to B he had only twice as much; what had each at first ? 5, A father is four times as old as his son; in 16 years he will only be twice as old ; find their ages. 6. A is 20 years older than B, and 5 years ago A was twice as old as B; find their ages. 7, How old is a man whose age 10 years ago was three-eighths of what it will be in 15 years ? 8, Ais twice as old as‘’B; 5 years ago he was three times as old ; what are their present ages ? 9, A father is 24 years older than his son; in 7 years the son’s age will be two-fifths of his father’s age; what are their present ages ? 68 ALGEBRA. [cuap. Example III. A person spent $56.40 in buying geese and ducks ; if each goose cost 7 dimes, and each duck 3 dimes, and if the total number of birds bought was 108, how many of each did he buy ? In questions of this kind it is of essential importance to have all quantities expressed in the same denomination; in the present instance it will be convenient to express the money in dimes. Let « be the number of geese, then 108—~ is the number of ducks, Since each goose costs 7 dimes, x geese cost 7x dimes. And since each duck costs 3 dimes, 108—w ducks cost 8(108—~) dimes, Therefore the amount spent is 7x+8(108—«a) dimes. But the question states that the amount is also $56.40, that is 564 dimes. Hence 72+8(108—2) =564 ; 74+ 824 —3x=564, 4%=240, *, =60, the number of geese, and 108—x=48, the number of ducks. Note. In all these examples it should be noticed that the un- known quantity x represents a number of dollars, ducks, years, etc. ; and the student must be careful to avoid beginning a solution with a supposition of the kind, ‘‘let x=A’s share”’’ or ‘‘let x=the ducks,’’ or any statement so vague and inexact. It will sometimes be found easier not to put x equal to the quantity directly required, but to some other quantity involved in the question; by this means the equation is often simplified. Example IV. A woman spends $1 in buying eggs, and finds that 9 of them cost as much over 25 cents as 16 cost under 75 cents ; how many eggs did she buy ? Let x be the price of an egg in cents; then 9 eggs cost 9% cents, and 16 eggs cost 16x cents ; oe 9x%—25=—75—162, ’ 25x%= 100 5 eA Thus the price of an egg is 4 cents, and the number of eggs =100+4=25. 10. A sum of $30 is divided between 50 men and women, the men each receiving 75 cents, and the women 50 cents; find the number of each sex. XI.] PROBLEMS LEADING TO SIMPLE EQUATIONS... 69 11, The price of 15 yards of cloth is as much less than $10 as the price of 27 yards exceeds $20; find the price per yard. 12, A hundredweight of tea, worth $68, is made up of two sorts, part worth 80 cents a pound and the rest worth 50 cents a pound ; how much is there of each sort ? 138, A man is hired for 60 days on condition that for each day he works he shall receive $2, but for each day that he is idle he shall pay $1 for his board: at the end he received $90; how many days had he worked ? 14, A sum of $6 is made up of 50 coins, which are either quar- ters or dimes ; how many are there of each ? 15, A sum of $11.45 was paid in half-dollars, quarters, and dimes ; the number of half-dollars used was four times the number of quarters and ten times the number of dimes; how many were there of each ? 16, A person buys coffee and tea at 40 cents and 80 cents a pound respectively ; he spends $15.10, and in all gets 24 lbs. ; how much of each did he buy ? 17, A man sold a horse fora sum of money which was greater by $68 than half the price he paid for it, and gained thereby $18; what did he pay for the horse ? 18, Two boys have 240 marbles between them; one arranges his in heaps of 6 each, the other in heaps of 9 each. There are 36 heaps altogether ; how many marbles has each ? 19, A man’s age is four times the combined ages of his two sons, one of whom is three times as old as the other; in 24 years their combined ages will be 12 years less than their father’s age ; find their respective ages. 990, Asum of money is divided between three persons, A, B, and C, in such a way that A and B have $42 between them, B and C have $45, and C and A have $53; what is the share of each ? 91, A person bought a number of oranges for $3, and finds that 12 of them cost as much over 24 cents as 16 of them cost under 60 cents ; how many oranges were bought ? 99, By buying eggs at 15 for a quarter and selling them at a dozen for 15 cents a man lost $1.50; find the number of eggs. 93, I bought a certain number of apples at four for a cent, and three-fifths of that number at three for acent; by selling them at sixteen for five cents I gained 4 cents ; how many apples did I buy ? 94, If 8 lbs. of tea and 24 lbs. of sugar cost $7.20, and if 3 lbs. of tea cost as much as 45 lbs. of sugar, find the price of each per pound. 70 ALGEBRA. [CHAP. XI. 95, Four dozen of port and three dozen of sherry cost $89 ; if a bottle of port costs 25 cents more than a bottle of sherry, find the price of each per dozen. 96, Aman sells 50 acres more than the fourth part of his farm and has remaining 10 acres less than the third; find the number of acres in the farm. 97, Find a number such that if we divide it by 10 and then divide 10 by the number and add the quotients, we obtain a result which is equal to the quotient of the number increased by 20 when divided by 10. 98, A sum of money is divided between three persons, A, B, and CO, in such a way that A receives $10 more than one-half of the entire amount, B receives $10 more than one-third, and C the remainder, which is $10; find the amounts received by A and B. 99, The difference between two numbers is 15, and the quotient arising from dividing the greater by the less is 4; find the numbers. 380, A person in buying silk found that if he should pay $3.50 per yard he would lack $15 of having money enough to pay for it ; he therefore purchased an inferior quality at $2.50 per yard and had $25 left ; how many yards did he buy ? 381, Find two numbers which are to each other as 2 to 3, and whose sum is 100. 382, A man’s age is twice the combined ages of his three sons, the eldest of whom is 3 times as old as the youngest and 3 times as old as the second son; in 10 years their combined ages will be 4 years less than their father’s age ; find their respective ages. 338, The sum of $34.50 was given to some men, women, and children, each man receiving $2, each woman $1, and each child 50 eents. The number of men was 4 less than twice the number of women, and the number of children was 1 more than twice the number of women ; find the total number of persons. 34, A man bought a number of apples at the rate of 5 for 3 cents. He sold four-fifths of them at 4 for 8 cents and the remainder at 2 for a cent, gaining 10 cents; how many did he buy ? 35, Example 2. The highest common factor of a?b4, a*b°c?, atb’c is a*b*; for a? is the highest power of a that will divide a’, a?, at; b+ is the highest power of b that will divide b+, b°, b7; and c is not a common factor. 90, If the expressions have numerical coefficients, find by Arithmetic their greatest common measure, and prefix it as a coefficient to the algebraical highest common factor. Hxample. The highest common factor of 2latz*y, 35a7xty, 28a%ay is Ja*xy ; for it consists of the product of (1) the greatest common measure of the numerical coefficients ; (2) the highest power of each letter which divides every one of the given expressions. EXAMPLES XII. a. Find the highest common factor of eeoceD aur, Doc Ary, ,. 0c’, 50°C. PT ns hea Verne Dewar cn auc.) Ga.eaco,Oabe. {, 6a*ytz, Qay. 8, lby®, Sry fe, 9, 12a%bc?, 18ab?c, Oe iecyr a: Zl aae. 1]. 8ax, 6a*y, 10ab?x?, Lee oes OF cary”, 18, 14bc?, 63ba?, 56b7c. 14, lary, 60a5y72*, 25022", 15, Ixy’, Slayz, 34a°yz «16, Tia°bic?, B3a%%c%, ab ct, 72 ALGEBRA. [CHAP. Lowest Common Multiple of Simple Expressions, 91, Derrryirion. The lowest common multiple of two or more algebraical expressions is the expression of lowest dimen- sions which is divisible by each of them without remainder. The abbreviation L.C.M. is sometimes used instead of the words lowest common multiple. 92. In the case of simple expressions the lowest common multiple can be written down by inspection. Hxample 1. The lowest common multiple of a4, a3, a?, a® is a®, Example 2. The lowest common multiple of ab‘, ab®, a?b7 is a®b’; for a? is the lowest power of a that is divisible by each of the quantities a®, a, a7; and b7 is the lowest power of b that is divisible by each of the quantities b*, b°, b’. 93, If the expressions have numerical coefficients, find by Arithmetic their least common multiple, and prefix it as a co- efficient to the algebraical lowest common multiple. Example. The lowest common multiple of 2latz*y, 35a°aty, 28a°®xy is 420ata4y ; for it consists of the product of (1) the least common multiple of the numerical coefficients ; (2) the lowest power of each letter which is divisible by every power of that letter occurring in the given expressions. EXAMPLES XII. b, Find the lowest common multiple of Reo, oY, 2, abt abe, Bee ory er. A 4a, Baber. 5, 4a4bc?, 5ab?. 6. 2ab, 4ay. 7, mn, ni, lm. OP yy oor eee: 9, xy, 8yz, 4zx. LOmee oT Ag pt Pg: ell ploa ys barye. 1357 9ab?, 2107e: 13. 27a?; 816%, 18076": [Ae har Gey ait 6 Lo mewod 0, 2007 yo0z. 16. 72p79°r4, 108p%97r. Find both the highest common factor and the lowest common multiple of 17, 2ab?, 3a2b3, 4a4b. 18, 1523 y?, 5a?yz°. 19, 2a4, 8a2h3c7, OD eS ia coe. 21, 32a*b®c, 48a7bc°. 99, 5lim®p?, pn, 34mnp'. 93, 49a4, 56b4c, 2lac’., 94, 66a°*b%cat, 55ab’ayrz, 12a? yz’. XII] ELEMENTARY FRACTIONS. 73 Elementary Fractions, 94, Derrixition. If a quantity w be divided into b equal parts, and a@ of these parts be taken, the result is called the Jraction ; of x. If « be the unit, the fraction of w is called simply “the fraction @ ”+. go that the fraction ply b b represents a equal parts, b of which make up the unit. 95, In this chapter we propose to deal only with the easier kinds of fractions, where the numerator and denominator are simple expressions. Their reduction and simplification will be performed by the usual arithmetical rules. For the proofs of these rules the reader is referred to the Elementary Algebra for WSchools, Chapter xv. : Rule, Zo reduce a fraction to rts lowest terms: divide numerator and denominator by every factor which is common to them both, that 1s by their highest common factor. Dividing numerator and denominator of a fraction by a com- mon factor is called cancelling that factor. ESOS ae Lee 35a°b%c _ 5atb _ 4 Tab-c 1 oe EXAMPLES XII, c, Reduce to lowest terms : 2.3 oe 3 2203 3/2 af 232 d. “Tea ; 10. eye H. yee La. iba a Beaty Eee ag Ses ag Be 74 ALGEBRA. [CHAP. Multiplication and Division of Fractions. 96. Rule, Zo multiply algebraical fractions: as in Arith- metic, multiply together all the numerators for a new numerator, and all the denominators for a new denonunator. Hxample | 2a x 5a" v, BD? (QO Dae oUe ae é . 3b 2a7b on ~ 3b x 902 x on oq,” by cancelling like factors in numerator and denominator. 8a°b a Do Oras 0. all the factors cancelling each other. 97, Rule, Zo divide one fraction by another: invert the divisor and proceed as in multiplication. Ta? Boke, 28a%e? 4a3y? 5ab? 15b%xry? = Ta? x Beta ——— 15b*xy? 42°y2" Bab? 28a? Hxample 2. Example. all the other factors cancelling each other, EXAMPLES XII. d. Simplify the following expressions : ay, a°b? ab . 4c*d FLEE lie ANE 1. ab ‘ye 2. Qed abe 3: 3y°z * date Gara? 140% 3ab? 15b7c* Tee eee = ; SK pees fic ® : e Tab? c l2ax 5. 5b’c— 9a2b* 6, Bbc? ‘ l4be 7 am ,, 2cd? Dmy, g 4a*b , 3p7q? . PY ' By *3ab 4am? Yay * Babb? a2 ee 9 20% ,, 100? «te i 10 Uae iW Oe 34/3 ye ' Bat de®” 308" "gad ged” oe by Sore") Sax? xy? 15b2_ 14d? , 81d? x . cr ° Ll. aby * Bate * by 12. Fc * abe BIA Reduction to a Common Denominator, ‘98, In order to find the sum or difference of any fractions, we must, as in Arithmetic, first reduce them to a common denominator ; and it is most convenient to take the lowest com- mon multiple of the denominators of the given fractions. XI. J ELEMENTARY FRACTIONS. 75 Hxample. Express with lowest common denominator the fractions age anak Say’ 6xyz Qyz The lowest common multiple of the denominators is 6ayz. Multi- plying the numerator of each fraction by the factor which is required to make its denominator 6xyz, we have the equivalent fractions 20 eee 3cx 6xyz 6xyz Bayz Note. The same result would clearly be obtained by dividing the lowest common denominator by each of the denominators in turn, and multiplying the corresponding numerators by the respective quotients. EXAMPLES XII, e. Express as equivalent fractions with common denominator : Tee oer Be a 4, = Bay ae 6. ay He lh @ = 8, a = 9. ? & 105); Qa. Lge ae - Bowe Bowe wWoaet Addition and Subtraction of Fractions. 99, Rule. To add or subtract fractions: express all the fractions with their lowest common denominator ; form the algebrat- cal sum of the numerators, and retain the common denominator. 5x 712 . Simplify 27 4?x-/™, Example 1. Simplify 5 sie 7 The least common denominator is 12. 202+92—l4a_15x”_ 5x The expression = 12 =i puepie » ae 3a ab ab Example 2. Simplify Pe Fy 76 ALGEBRA. [CHAP. XII. ey azc2 3ca® . axe — C2 : Wek : The expression = ee and admits of no further simplification. CG: Ce Example 3. Simplify Note. The beginner must be careful to distinguish between erasing equal terms with different signs, as in Example 2, and cancelling equal factors in the course of multiplication, or in reducing fractions to lowest terms. Moreover, in simplifying frac- tions he must remember that a factor can only be removed from numerator and denominator when it divides each taken as a whole. 6ax —cy cae and not the whole numerator. Similarly a cannot be cancelled because it only divides 6ax% and not the whole numerator. The fraction is therefore in its simplest form. Thus in c cannot be cancelled because it only divides cy When no denominator is expressed the denominator 1 may be understood. te OL Ooty a0 Example 4. ee ee ee Be xcample 3x dy elias vy, If a fraction is not in its lowest terms it should be simplified before combining it with other fractions. vi : at ay _ax x _sav—2Qe xample 5 5 aay a 5 EXAMPLES XII. f. Simplify the following expressions : ee ee ee oe 5 5-5 DD Gey TB rap Bh Bom 9, ate 10. os 11, aoe 12, re Ry Sa ee 14, Pees 15, 18-2 Be ee uae ae 19.6 2 20 en Bes Sheen 93, oe dx, xy OA, a= a? ae ~ dy t By 3a ab? 6be CHAPTER] XLT, SIMULTANEOUS EQUATIONS, 100, ConstpEr the equation 22+ 5y=28, which contains two unknown quantities. By transposition we get By =23 — 2x ; that is, a= abs nd, RR (1). From this it appears that for every value we choose to give to # there will be one corresponding value of y. Thus we shall be able to find as many pairs of values as we please which satisfy the given equation, For instance, if v=1, then from (1) we obtain 2 Again, if = —2, then ya ; and so on. But if also we have a second equation containing the same un- known quantities, such as 8x+4y=24, we have from this y 7 Ree (2). If now we seek values of w and y which satisfy both equa- tions, the values of y in (1) and (2) must be identical. Therefore pe ee era, 5 4 Multiplying across 92-—8r=120—-152; (LZ 5 xv=A4, Substituting this value in the first equation, we have 8 + 5y = 23 ; w. SY=15; con ah and c=4, Thus, if both equations are to be satisfied by the same values of # and y, there is only one solution possible. 78 ALGEBRA. [CHAP. 101, Derrryition. When two or more equations are satisfied by the same values of the unknown quantities they are called simultaneous equations, We proceed to explain the different methods for solving simul- taneous equations. In the present chapter we shall contine our attention to the simpler cases In which the unknown quantities are involved in the first degree. 102, In the example already worked we have used the method of solution which best illustrates the meaning of the term svmultaneous equation ; but in practice it will be found that this is rarely the readiest. mode of solution. It must be borne in mind that since the two equations are simultaneously true, any equation formed by combining them will be satisfied by the values of x and y which satisfy the original equations. Our object will always be to obtain an equation which involves one only of the unknown quantities. 1038. The process by which we cause either of the un- known quantities to disappear is called elimination. We shall consider two methods. Elimination by Addition or Subtraction. Example 1. Solve BOA TY =O) senna ote shee eee (1); BiH 2y = 16.20 oes ue oe ee ee ee (2). To eliminate x we multiply (1) by 5 and (2) by 8, so as to make the coefficients of 2 in both equations equal. This gives 15x + 35y = 135, . ldx+ by = 48 ; subtracting, 297 = 877; i MN oa; To find x, substitute this value of y in ether of the given equations. Thus from (1), Noe Spd Ri Ear meas and. y = tf Note. When one of the unknowns has been found, it is immaterial which of the equations we use to complete the solution. Thus, in the present example, if we substitute 3 for y in (2), we have 5a+6=16; a = 2, as before. XIII. ] SIMULTANEOUS EQUATIONS. 79 Example 2. Solve PRS ick LIE ech ESE ee eee (1), aoe ELEM Ae agile pe cor Here it will be more convenient to eliminate y. Multiplying (1) by 2, l4x+4y = 94, and from (2) 5a —4y=13 adding, 19% = 95 ; Goa Substitute this value in (1), 30 + 2y = 47 ; . y=6, Note. Add when the coefficients of one unknown are equal and unlike in sign ; subtract when the coefficients are equal and like in sign. Elimination by Substitution. Example 3. Solve Lat NO ah cise aeantiaate ks soe nite a asst ie Ag QE OY aieniians wie NE he cote en stens (2). Here we can eliminate x by substituting in (2) its value obtained from (1). Thus 24 — T5y +1) = By; 48 — 35y —7 = by ;5x 4] =41y; noe fit and from (1) La. 104, Any one of the methods given above will be found sufficient ; but there are certain arithmetical artifices which will rnorincn shorten the work. Hxample. Solve DSA cO UY eat anon Sar Enki eigs ei tst ete reeset ‘ae Noticing that 28 and 63 contain a common factor 7, we shall make the coefficients of x in the two equations equal to the least common multiple of 28 and 63 if we multiply (1) by 9 and (2) by 4. Thus 252ax —207y = 198, 252% —-220y = 68; subtracting, 13y = 130; that is, y = 10, and therefore from (1), x= 9, [CHAP. 80 ALGEBRA. EXAMPLES XIII. a. Solve the equations: l, x«+y=19, 9. ety = 23, oO ety=ail; See TE Li Yeo. ey = —9, 4, x+y = 24, he, ey, G6, w—y= 25, x-y= 0. et 8/0. cry = 13, 7, 3a2+5y = 50, 8, x+5y= 18, 9, 4%+ y=10, 44+ 3y = 41. 3a+2y = 41. 5u+Ty = 47. 10, 7 -6y = 25, ll, 5e+4y=7, 12, 3¢—Ty=1 5a+4y = 51. 4x + 5y = 2. 4x+ y= 53. 18, 7x+5y = 45, 14, 4¢+5y =4, Lijelig= fy at433 Qu - 3y = 4. 5x — 3y = 79. 2x — 3y = 13. 16, 4x-3y=0, 17, -2x+3y = 22, 18, 72+3y = 65, 7a —4y = 36. 5a +2y = 0. iz —8y = 32. 19, 1l3e-y=14, 20. 9ux-8y = 14, Q1, 14%+13y = 35, 2a —7Ty = 9. 15x - 14y = 20. 21x+19y = 56. 0D. bf ay 21, Doaoot = sor, 04, bdea7y= it, 21x —9y = 75. 10x” = Ty — 15. etn a bap 95, 138x-9y=46, 26, 6x-5y=11, 27, lly-llx=66, llz-12y=17. 28a+21y = 7. 7x +8y = 3. 98, 6y-5x =11, 29, 32+10=5y, 80, 4y =47+4+32, 4a = Ty — 22. Ty = 4¢+13. 5a = 30- ldy. ol, 12+138y 57, Cols lly ieee Glee 13x4+ lly = 17, 292 — 39y = 17. 4lz+37y = 17. 105, We add a few cases in which, before proceeding to solve, it will be necessary to simplify the equations. Example, Solve 65(%+2y)—(38a+ ly) =14 crrcceccccceccceneeees (1); Ja Oy Sie — dy) a 38 ase ee (2), From (1), 52+10y - 3x -lly=14; Dae aap ct We ee vargas eaenecne toes (3). From (2), 7x —9y -3x+12y = 38; Ape 38 Ure, ee (4). From (3), 6a — 3y = 42, By addition, 10z = 80; whence x = 8, From (8) we obtain y = 2. xu. ] SIMULTANEOUS EQUATIONS. 81 106, Sometimes the value of the second unknown is more easily found by elimination than by substituting the value of the unknown already found. - Example. Solve 3a —4 ie = =e Lek hy a eee Cy i By +4 — lig. _ aie a 5 3(2e Weare Ieee cartes veer ecnecises es Cay Clear of fractions. Thus from (1), 42a —2y +10 = 287-21; VA Ga Le NO has cnt casiceaee ate arse (3). From (2), 9y +12 --10%+25 = ldy ; LOS UG Spencer yes clea castes steeds ev (4). Eliminating y from (38) and (4), we find that oe 13 Eliminating x from (3) and (4), we find that y 2 20! 26 107. Simultaneous equations may often be conveniently ee 1 uh ee solved by considering — and ~ as the unknown quantities. we o Example. Solve 2 se Sat LO come nat teestie Sin aennces Rese tee ce ‘id Ws Ae ey EOE TER else en: Ne ee OP (2) te Ms Multiply (1) by 2 and (2) by 3; thus 16_18_y oY 30 | 18 = 9A > og wes adding, a0 8 ; % multiplying up, 46 = 232 ; fig A and by substituting in(1), 4 =3. HLA. F 82 ALGEBRA. [cuar. EXAMPLES XIII. b. Solve the equations : 1, 2x-y=4, 9, 4e-y=1, 3. £+2y = 13, Bed Ep eed ee US Did en ne 35 HA a eye Bx a 4, eae 5, 2 =O, 6, eae G %+3y = 2 4x -3y=1 4x + 5y = 0. Ty bare Ay, 8, «x-y=9, Q, x+y=-2, 4x By _ Bd 08 e1Y_o i ae ye any 46 1 3 ] ] Ee =I, é Sop oD) = 20); ‘ Rig 20) = 10; 5(v +3) 0 11 pe ~ 2y 0 12 ate al 0 sey = 4h s(y-+8) = 2. Sm = Dy 13. 3(v@-y)+2%a+y)=15, 38(at+y)+2(x-y) = 25. 14, 3(@+y-5)=22(y—-x), 3(v-—y-7)+2(r4+y-— 2) =0. 15, 4(2x-y-6)=38(8%-2y-5), 2x-y+1)+4x = 3y+4. 16. 7(2a—y)+5(8y—4x7)+30=0, 5(y-—2+3) = 6(y— 22). +4 y-4_ oe le ya lis ot oy 17. yma ee 18. 25 peak 90, 24° =39, Ole eae cy “x sy eae ats bigs 708 Big SoEE aay ay “x y 108, In order to solve simultaneous equations which contain two unknown quantities we have seen that we must have two equations. Rule. be solved by the refes already given. Similarly we find that in .erder to solve simul- taneous equations which contain three unknown quantities we must have three equations. Eliminate one of the unknowns from any pair of the equations, and then eliminate the same unknown from another pair. Two equations involving two unknowns are thus obtained, which may The remaining unknown is then found by substituting in any one of the given equations. XIII. | SIMULTANEOUS EQUATIONS. 83 Example. Solve PRO Orees etme coe Pee bess saa seecdaceeerts (1), Miri as Vice go ycame 1 uae ae fo ane eee Retry (2), DOE EA Ber SN es ys as bog teeceaiy orsniehaanes (8). Choose y as the unknown to be eliminated. Multiply (2) by 5, 20%+10y-15z=0; Multiply (1) by 2, 14%+10y-—142= - 16; by subtraction, O75 —@ 1 On see testare sees keoe' beuvce, (E)e Multiply (2) by 2, 8x+4y-62=0; from (3), 5a — 4y + 42 = 35 5 by addition, 13x — 2z = 35. Multiply (4) by 2, 122% —2z = 32; by subtraction, Sr Ds From (4) we find a= 2, and from (2), y=-3. 109. Some modification of the foregoing rule may often be used with advantage. Example. Sol ees ba tO. xample olve 5 re id ja teen as 379 2 From the equation 5 Sle a+ 1, we have a 2) deride eral oA ee Peccceeeseae eomeeh (1). Also from the equation 57 is +2, we have rh Mio) ah re aE Nec tence an odds Aras avosalloas ieee (2). And from the equation at5e 13, we have Ri aa eae) ee ee Tatar ceestbycaetes tas (3). Eliminating z from (2) and (3), we have 21x +4y = 282 ; and from (1} 12x -4y = 48 ; whence #=10, y=18. Also by substitution in (2) we obtain z= 14, 84 ALGEBRA. EXAMPLES XIII. Solve the equations : Ub, co Lt, 12, 13, 14. 15, 16, 3x —2y+2= 4, Dh. 2x + 3y —2= 3, Lt Y+tZa S, xv + 2y +32 = 32, 4, da — dy + 6z.=.27, 7x +8y —9z = 14, 7«—4y-3z= 0, 6, 5x — 3y + 22 = 12, 34+2y-5z= 0. 3y — 62 —5a = 4, 8, 22 -3dx- y=8, x —2y+22+2=0. 10, dar (zy) = 11, Qe sey) =2, Haty) = 78-2), : 3 (CHAP. XIII. ox + 4y — 6z = 16, 4eu+ y-— z= 24, x —d3y—225 LNs Rp Sema 4¥, 62 + dy + 2z = 84, da + 4y — 5z = 13. 4x+3y- z= 9, 9G = Omer} wAYy — 32 =) 2. by +22 +52 = 21, 82 — 32 9 = 8, 22 +22 -3y = 39. (z-4x) =y. w= 4y +32. GCHAPTER: XLV; PROBLEMS LEADING TO SIMULTANEOUS EQUATIONS. 1190, Ivy the Examples discussed in the last chapter we have seen that it is essential to have as many equations as there are unknown quantities to determine. Consequently the statement of problems which give rise to simultaneous equations must contain as many Independent conditions, or different relations between the unknown quantities, as there are quantities to be determined. Example 1. Find two numbers whose difference is 11, and one- fifth of whose sum is 9. Let x be the greater number, y the less ; then yt en atta saree a ae eres § Gh): Also oe =o, or GY = 40 S232 head eee toe (2). By addition 22=56 ; and by subtraction 2y=34. The numbers are therefore 28 and 17. Example 2. If 15 lbs. of tea and 10 lbs. of coffee together cost $15.50, and 25 lbs. of tea and 13 lbs. of coffee together cost $24.55 ; find the price of each per pound. Suppose a pound. of tea to cost « cents, a eet COMLCE aie iae Ys aaa ele Then from the question we have 1g Oy = la bW aes aes weNierds ables bia 8 Ly, PAVERS GD ERO 115 Diya eer ey ee (2). Multiplying (1) by 5 and (2) by 38, we have 752+ 50y=7750, 752+ 89y =7365. Subtracting, weLiv— ooo. Y= 35. And from (1), 15% +850 = 1550 ; whence tba 1200 5 MzaS0): .*. the cost of a pound of teais 80 cents, . and the cost of a pound of coffee is 35 cents. 86 ALGEBRA. [CHAP. Hxample 3. In a bag containing black and white balls, half the number of white is equal to a third of the number of black ; and twice the whole number of balls exceeds three times the number of black balls by four. How many balls did the bag contain ? Let x be the number of white balls, and y the number of black — balls; then the bag contains 7 +y balls. We have the following equations : pag AI HO Neh RAM, l 2 3 es. ( ) 2A Y) =H OY EE vacenceatessesessescescststeaae (2). Substituting from (1) in 2, we obtain AY 4 y= By +4; whence Uai2: and from {1), x= 8. Thus there are 8 white and 12 black balls. 111, In a problem involving the digits of a number the student should carefully notice the way in which the value of a number is algebraically expressed in terms of its digits. Consider a number of three digits such as 435; its value is 4x100+3x10+5. Similarly a number whose digits beginning from the left are x, y, z =x hundreds+y tens+z units = 100% + 10y +2. Hxample. A certain number of two digits is three times the sum of its digits, and if 45 be added to it the digits will be reversed ; tind the number. Let x be the digit in the tens’ place, y the digit in the units’ place ; then the number will be represented by 10x+y, and the number formed by reversing the digits will be represented by 10y+2. Hence we have the two equations LODE 83 (20) ee coetans cout oe eee (1), _and 10% dD = LOG 12s as ssi vacks ees creek wares (2). From (1), Ley from (2), yee), From these equations we obtain x = 2, y = 7. Thus the number is 27. XIv.] PROBLEMS LEADING TO SIMULTANEOUS EQUATIONS. 87 EXAMPLES XIV. 1, Find two numbers whose sum is 54, and whose difference is 12 foal 9, Thesum of two numbers is 97 and their difference is 51; find the numbers. 8. One-fifth of the difference of two numbers is 3, and one-third of their sum is 17; find the numbers. 4, One-sixth of the sum of two numbers is 14, and half their difference is 13; find the numbers. 5, ‘Four sheep and seven cows are worth $131, while three cows and five sheep are worth $66. What is the value of each animal ? 6, A farmer bought 7 horses and 9 cows for $330. He could have bought 10 horses and 5 cows for the same money; find the price of each animal. : 7, Twice A’s age exceeds three times B’s age by 2 years ; if the sum of their ages is 61 years, how old are they ? 8, Half of A’s age exceeds a quarter of B’s age by 1 year, and three-quarters of B’s age exceeds A’s by 11 years; find the age of each ? . 9, Ineight hours C walks 3 miles more than D does in six hours, and in seven hours D walks 9 miles more than C does in six hours ; how many miles dees each walk per hour ? 10. In 9 hours a coach travels one mile more than a train does in 2 hours, but in 3 hours the train travels 2 miles more than the coach does in 13 hours ; find the rate of each per hour. 11, A bill of $15 is paid with half-dollars and quarters, and three times the number of half-dollars exceeds twice the number of quarters by 6; how many of each are used ? 12, A bill of $8.70 is paid with quarters and dimes, and five times the number of dimes exceeds seven times the number of quar- ters by 6; how many of each are used ? 18, Forty-six tons of goods are to be carried in carts and wag- ons, and it is found that this will require 10 wagons and 14 carts, or else 18 wagons and 9 carts ; how many tons can each wagon and each cart carry ? 14, A sum of $14.50 is given to 17 boys and 15 girls; the same amount could have been given to 13 boys and 20 girls; find how much each boy and each girl receives. 15, A certain number of two digits is seven times the sum of the digits, and if 36 be taken from the number the digits will be reversed ; find the number, 88 ALGEBRA. [ CHAP, XIV. 16, A certain number of two digits is four times the sum of the ; digits, and if 27 be added to the number the digits will be reversed ; find the number. 17, . 31, (—2m2n’)6, 32, (—a3y2V’, mG) me CBR on CH we To Square a Binomial. 115, By multiplication we have (a+6)?=(a+b)(a+b) 07 200 Bs on. 5 ats ate eeoeee (1); (a—b)?=(a—b)(a—b) = P= 200 4-07. aa sche steals ae Xv. ] INVOLUTION. 91 These formulz may be enunciated verbally as follows : Rule 1. The square of the sum of two quantities ts equal to the sum of their squares increased by twice their product. Rule 2. The square of the difference of two quantities is equal to the sum of their squares diminished by twice their product. Hxample 1. (2+ 2y)? = 274+2.%.2y+(2y)? = x7 + 4ay +4y*. Example 2. (2a3—8b?)?= (2a3)?—2 . 2a3 . 3b?+ (8b?)2 = 4q°— 12a3b?+ 964. To Square a Multinomial. 116, By the preceding article (a+b+c)?={(a+b)+cP =(a+b)?+2(a+b)e4+c? = a*+ b?+ 7+ 2ab+2ac +4 2be. In the same way we may prove (a—b+cP=a?+b?+ 0? —2ab+ 2ac— be. (a+b+c4+d)?=040+C4+d*+2ab4+2ac+ 2ad+ 2be+ 2bd + 2cd. In each of these instances we observe that the square con- sists of (1) the sum of the squares of the several terms of the given expression ; (2) twice the sum of the products two and two of the several terms, taken with their proper signs; that is, in each product the sign is + or — according as the quantities composing it have like or unlike signs. Note. The square terms are always positive. The same laws hold whatever be the number of terms in the expression to be squared. Rule. Zo find the square of any multinomial: to the sum of the squares of the several terms add twice the product (with the proper sign) of each term into each of the terms that follow tt. Ez. 1. («¢-—2y—382z)?=x? + 4y? + 92? --2.%.2y—-2.4.3824+2.2y.32 = 207 + 4y? + 927 — day — 6az+ 12yz. Hix. 2. (1420 -—32?)?=1 4+ 40?+4 9at+2.1.27-2.1.382?-2.27.32? =1+427+ 924+ 4a - 6x? — 1223 =]+4a — 2x? — 124° 4 9x4, by collecting like terms and rearranging. 92 ALGEBRA. [CHAP. xv. EXAMPLES XV. b. Write down the square of each of the following expressions : apa. 27; Dae = 2y: 3, a+30. 4, 2a-3b. 5, 38a+b. 6, x-—5dy. 7. 2n+7n. 8, 9-2. 9, 2-ab. 10, abc+l. ll, ab-cd. 12, 2ab+ay. Tope = 22 14, 34+2pq. 15, x?-3a. 16, 2a-+ab. 17, atb-e. 18, a-b-e. 19, 2a+b+¢. 20, 2u-y-2 91, «x+3y—-2z. 90, we ea +1, 93, 3x+2p-q. 94, 1-22-32. 95, 2-8u+22 26, xty+a—). 27, m—n+p—4q. 28, 2a+8b+%—2y. To Cube a Binomial. 117. By actual multiplication, we have (a+b) =(a+b)a+b)(a+6) = 03 +3a°b + 3ab? + b*. Also (a — 6b)? =a? —3a7b + 3ab? — Bb’, By observing the law of formation of the terms in these results we can write down the cube of any binomial. Example 1. (2a +y)3 = (2x)? + 3(2x)?y + 3(2ax)y? + y? = 8a? + 122°y + 6xy? + y?. Example 2. (8% - 2a?) = (38x) — 3(3x)?(2a*) + 3(8a)(2a?)? — (2a?)3 = 272° — 542°a? + 36xa4 - 8a, EXAMPLES XV. c. Write down the cube of each of the following expressions : Le 2, m-n. 3, a- 2b. 4, 2c+d. 5, x+3y. 6, «x+y. Ley: 8. 5442. Re rd be 10, 22?-+y7. ll, 2a%-30% 12, 4y?-3 CHAPTER XVI. EVOLUTION. 118, Derinirion. The root of any proposed expression is that quantity which being multiplied by itself the requisite number of times produces the given expression. The operation of finding the root is called Evolution: it is the reverse of Involution. 119, By the Rule of Signs we see that (1) any even root of a positive quantity may be either positive or negative ; (2) no negative quantity can have an even root ; (3) every odd root of a quantity has the same sign as the quantity itself. Note. It is especially worthy of remark that every positive quantity has two square roots equal in magnitude, but opposite in sign. Hxample. J9a?x® = +3a2°, In the present chapter, however, we shall confine our attention to the positive root. Examples. ab! = 032, because (a%b2)? = a4. / — x9 = — x, because (- 2°)® = - 2, Nc = ct, because (c4)> = c%, A/81a!2 = 3x3, because (323)! = S1x!2, 120. From the foregoing examples we may deduce a general rule for extracting any proposed root of a simple expression : Rule. (1) Find the root of the coefficient by Arithmetic, and prefix the proper sign. (2) Divide the exponent of every factor of the expression by the index of the proposed root. Hxamples. —642° = — 427, / 16a8 = 2a2, /81a1° = 9x° 25c8 Bc? 94 ALGEBRA. [oHAP. EXAMPLES XVI, a, Write down the square root of each of the following expressions : {bap ese 2.6 20a Us 3, 49c7d%, A abe De Shaw ek 6, 162%. fis 0 avian Seopa 4x6 ae isa. 144 9, 16a4 10, 36° ag “O5 iby a2 Write down the cube root of each of the following expressions : 13, x*y®. 14, -a%®, 15, 82%, 16) 2/24 pe bss 8a%h? 1250727) 64a77b? ive. 27° 18, “ys 19, ky [ae 20, ae nee Write down the value of each of the following expressions ; Dinan ey. DDN arta O38 N= ete 24, N6ta®. 25, Na?tb™, 26, Npq”. 97, N= x35y56, 298. N8laty®?. 29, §/320>bMe%, 121. By the formule in Art. 115 we are able to write down the square of any binomial. Thus (20 + 3y)? =4u? + 12xy +97". Conversely, by observing the form of the terms of an expres- sion, it may sometimes be recognised as a complete square, and its square root written down at once. Example 1. Find the square root of 25a? — 40xzy + 16y’. The expression = (5a)?- 2. 20xy+ (4y)? = (5x)? — 2(5a)(4y) + (4y)? = (5x - 4y)?. Thus the required square root is 5x - 4y. 64a? 32a Example 2, Find the square root of OOF UL Rae The expression = 3 i + (2)? + 2( 27 ) Thus the required square root is ee +2, XVI. ] EVOLUTION. 95 122. When the square root cannot be easily determined by inspection we must have recourse to the rule explained in the next article, which is quite general, and applicable to all cases. But the student is advised, here and elsewhere, to employ methods of inspection in preference to rules. To Find the Square Root of a Compound Expression. 123, Since the square of a+b is a?+2ab+b%, we have to dis- cover a process by which a and 0, the terms of the root, can be found when a?+2ab+0? is given. The first term, a, is the square root of a?. Arrange the terms according to powers of one letter a. The first term is a’, and its square root is a. Set this down as the first term of the required root. Subtract a? from the given expression and the remainder is 2ab +? or (2a+b) x 6. Now the first term 2ab of the remainder is the product of 2a and b. Thus to obtain 6 we divide the first term of the remainder by the double of the term already found ; if we add this new term to 2a we obtain the complete divisor 2a+. The work may be arranged as follows : a’ +2ab+b? (a+b ae Qat+b 2ab +b? 2ab+ 6? Example Find the square root of 9x? — 42xy + 49y. 9x? — 42xy + 49y? ( 8a - Ty 9x? 6xu — Ty | —42xy + 49y? © | — 42xy +49y? Explanation. The square root of 92? is 3%, and this is the first term of the root. By doubling this we optain 6x, which is the first term of the divisor. Divide —42zy, the first term of the remainder, by 6x and we get —7y, the new term in the root, which has to be annexed both to the root and divisor. Next multiply the complete divisor by — 7y and subtract the result from the first remainder. There is now no remainder and the root has been found. 96 ALGEBRA. [cHAP. 124, The rule can be extended so as to find the square root of any multinomial. The first two terms of the root will be obtained as before. When we have brought down the second remainder, the first part of the new divisor is obtained by doubling the terms of the root already found. We then divide the first term of the remainder by the first term of the new divisor, and set down the result as the next term in the root and in the divisor. We next multiply the complete divisor by the last term of the root and subtract the product from the last remainder. If there is now no remainder the root has been found ; if there is a remainder we continue the process. Example. Find the square root of 2527a? — 1220? + 1624 + 4a4 — 24270, Rearrange in descending powers of x. 16a4 — 2443 + 25x70? — 12x03 + 4a4 ( 4a? - 8xa + 2a? 16x4 82? — 3x0 — 2420, + 25270? —24e2a+ 9x02 82? — 6za + 2a? 1622a? — 12xa? + 4a4 16z7a? — 1220? + 4a4 Explanation. When we have obtained two terms in the root, 4x? — 32a, we have a remainder 1627a? - 12203 + 4a’. Double the terms of the root already found and place the result, 8x2 —6xa, as the first part of the divisor. Divide 16xz°a?, the first term of the remainder, by 8x, the first term of the divisor; we get + 2a? which we annex both to the root and divisor. Now multiply the complete divisor by 2a? and subtract. There is no remainder and the root is found. 125, Sometimes the following method may be used. Example. Find by inspection the square root of 4a? +b? +c¢?+4ab — 4ac — 2be. Arrange the terms in descending powers of a, and let the other letters be arranged alphabetically ; then the expression = 4a? + 4ab —4ac + b?- 2b¢4+ c? = 4a?+4a(b-c)+(b-c)? = (2a)7+2. Qa(b-c)+(b-c)?; whence tke square root is 2a+(b-c). [Art. 121.) XVI. ] EVOLUTION. 97 EXAMPLES XVI. b. By inspection or otherwise, find the square root of each of the following expressions : 1, @-8a+16. Q, «?+142+49. 8, 6444824 922. 4. 25-30m+4 9m?. 5, 36n4-84n?+49. 6, Sl+144y?+64y'%, T, 2° — 62% ytz4 + 9y8z8, 8, 4a2b4 — 12ab2c 4+ 919, ee ons 6 9a? , 24ac , 16c? 9, ra 3xy? + 9y?. 10; yet bd +—a9- 9a? 2562 16a? , 49y4 Daa 0 all) a achat bd 2x4 Nl. psp 2+ Gar 12. Foye" Tea oY 18, 1624-3223 +242?- 82+]. 14, 25 -30a+ 29a? - 12a°+ 4a4. 15, 9a®-12a° - 2a4+4a?+1. 16, 25p*- 30p? + 121 — 101p?+ 66p. 17, 8v?+14+4a4- 42. 18, 201a?- 108a? + 100 + 36a4 — 180a, 19, a?+0?+c¢?+2ab —2ac — 2be. 90, yet t+ 22x? + Py? — Qa?yz + Qary?z — Qarye. 307 1 4— q+ = Q1, at—2ab+ > - 5 +i5 93, Imi+ 4 8 OO mtn 94, 9at+ 144074 12ax?+ 4a? — 720° - 4802. 95, x8 —4ax7 4+ dao + 62° — 1404+ 40° + 9a? -6r4+1. 96, a?+9b?+c?- 6ab+6be —2ac. 27, Mp crepe 47 +4, n* n? n? 9a? b? 6a 2b. —-§4+—-— pe ze oF az 0b H.A. G 98 ALGEBRA. [OnAr. { If preferred, the remainder of this chapter may be postponed and taken at a later stage. ] To Find the Cube Root of a Compound Expression. 126. Since the cube of a+b is a?+3a%b+3ab?+b3, we have to discover a process by which a and b, the terms of the root, can be found when a?+3a%)+3ab?+ 03 is given. The first term a is the cube root of a’. Arrange the terms according to powers of one letter a; then the first term is a, and its cube root a. Set this down as the first term of the required root. Subtract a’ from the given expression and the remainder is 3a°b + 3ab? + D3 or (3a°?+3ab +b") x b. Now the first term of the remainder is the product of 3a? and 6. Thus to obtain b we divide the first term of the re- mainder by three times the square of the term already found. Having found b we can complete the divisor, which consists of the following three terms : 1. Three times the square of a, the term of the root already found. 2. Three times the product of this first term a, and the new term 0. 3. The square of 6. The work may be arranged as follows : a +3ab+ 3ab?+b? (a+b he 3(aPr = 3a? 3a°b + 3ab? +3 3xaxb= +4+3ab Qe = +0? 3a7+3ab+b? 3a°b + 8ab?+ 63 Lxample 1. Find the cube root of 8x? — 36z°y + 54ary? — 27y?, 8x3 — 36x?y + B4ay? — 27? ( Qa — 3y 823 B23)" = 109° 3x 2x x (- 3y) = — 1ory (—3y)' = + 9y" 1227 - 18xy + 9y? — — 86x7y + 54ay? — 27y? — 36x%y + 54ay? — 277 va *PUNOF SI JOOL oY} PUL JopUIvUIeI OT sI rey} { ovagqns pue g— Aq JostAtp ozopdur0s oy Ajdynut Mon “g— jo orenbs oy4 ose puv “g— pue LF — LZ Jo qyoupoad oy} sow, g oye} 9M AOSTATP OT} aya{duroo OJ, *4OOL oy} JO W104 MoU B E— SOATS SIYy £ IOSLATP oY} Jo W199 4s1y oY) LTT Aq ‘AopuTVUTEL OY} Jo W194 4SIG OU} ‘,79E— SPLAT _“AOSTATP Mou Ot} jo qied si 0} Se “rept USF — VST “4[Nsert yy sovjd puv punos Apvarye joor oxy jo orenbs oy4 sourry g exe], *L]— XSOT — e206 — eX PFT 1 pXOE — JopuTeUTeI B OAT] OM “Zp — XZ “JOOI OY UT SUITE} OM} POUTeAGO OAvT OAL UOT] AA "wounUDnjidxy Z 1G — LOT — -@O6 — eLFFI + X98 - 6 +298 + LOE + PSP — ASI eS 6+ =(¢—) RB LOE + UST - =(¢—) x (xp p#Z) x8 iS LZ — X8OI — X06 — LEFT +4298 — LOb + COP — eTL = (ZH — eB) XE & = XF9 —pUOG + —USP—-| XII + ek FS— LVI erg + ‘ =(xp—) eXFG — = (xp—) x (2B) XE eX0S +5L09 + 6XSF - PXSL = o(zlG) XE 9S 2 -— XP— XLS) LZ — CSOT — X06 — eVOS8 + 5X09 t+ oT SP — 978 ‘1G — L8OL — e706 — eX08 + 4009 + oXSPF — 9B JO JOor oqno oy} PUNT °% apdumagy XVI. ] 100 ALGEBRA. [CHAP. XVL EXAMPLES XVI. c. Find the cube root of each of the following expressions : ], @+12a?+4+ 48a + 64. 9, 82? +120?+6xr+1. 9 6423 — 1442+ 108a — 27. 4, 8p*-36p!+54p? - 27. 5, m?—18m?+ 108m - 216. 6, x°+ 6xty? + 12a°y4 + 8y®. 7, 1-3c+6c? — 7c? + 6c4 — 3c° + c%, 8; 8+36m + 66m? + 63m? + 33m4 + 9m +m, Q, 216 -108k+ 342k? — 109K? + 171K4 — 27K + 27K8. 10, 48y°+ 108y + 6044 — 90y? - 27 + 8y® — 80z°. 11, 64+192k4 + 240k? + 160K? + 60K4 + 1245 + h°, 12, x? —62°y — 3272+ 12xy? + lQxyz + 3x27 — 8y3 — 12yz - Byz? — 2. [For additional examples see Hlementary Algebra. | 127, The ordinary rules for extracting square and cube roots in Arithmetic are based upon the algebraical methods explained in the present chapter. The following example is given to illustrate the arithmetical process. Hxample. Find the cube root of 614125. Since 614125 lies between 512000 and 729000, that is between (80)? and (90)°, its cube root lies between 80 and 90 and therefore consists of two figures. at+b 614125 ( 80+5 = 85 512000 3a7 =.3 K (80) = 19200) 1102125 3xaxb=3x80x5= 1200 A= Bx Os 25 20425 In Arithmetic the ciphers are usually omitted, and there are other modifications of the algebraical rules. CHAPTER XVII. RESOLUTION INTO FACTORS. 128. Derritiox. When an algebraical expression is the product of two or more expressions each of these latter quanti- ties is called a factor of it, and the determination of these quantities is called the resolution of the expression into its factors. In this chapter we shall explain the principal rules by which the resolution of expressions into their component factors may be effected. Expressions in which Each Term is divisible by a Common Factor. 129. Such expressions may be simplified by dividing each term separately by this factor, and enclosing the quotient within brackets ; the common factor being placed outside as a coefficient. Example 1. The terms of the expression 3a? — 6ab have a common factor 3a; 3a? -- 6ab = 8a(a — 2b). Example 2. 5a*ba? — 15aba? — 20032? = 5bx?(a*a — 3a — 407). EXAMPLES XVII. a. Resolve into factors : ], 2+ax. Oy 262 =. a0. Ce Os. 4, ai--a?b. 5, 3m2-6mn. 6, p?+2p?q. T, 2-52, 8, yxy. Q, 5a? -—25a7b. 10, 12%+4827y. ll, 10c? — 25c4d. 12, 27-162z. 13, x°y?+82y. 14, 17x2-5la. 15, 2a?-a?+a. 16, 32°+6a?x? - 3a°x. 17, 7p?-7Tp? + 14p%. 18, 40°+6a7b? — 262. 19, xy? — 2?y?+ 2Qary. 920, 26a°b°+39a%b2, 102 ALGEBRA. [onar. Expressions in which the Terms can be so grouped as to contain a Compound Factor that is Common. 180, The method is shown in the following examples. Example 1. Resolve into factors «2—ax+bxe—ab. Since the first two terms contain a common factor x, and the last two terms a common factor b, we have x? ~— ax +ba—ab = (x?—ax)-+(ba-ab) a(x ~a)+b(~-a) (a —a) taken w times plus (w-—a) taken 0 times = (x-—a) taken (x+b) times = (~-—a)(~+b). II I | EHxample 2. Resolve into factors 6a? — 9ax + 4ba - Gab. 6x? — 9ax+4bx - bab = (6x? — 9ax) + (4ba — 6ab) = 3a(2a —- 3a) + 2b(2x - 3a) = (24 — 3a)(3a” + 2b). Example 3. Resolve into factors 12a? + ba? — 4ab — 3ax. 12a? + ba? — 4ab — 38ax? = (12a? — 4ab) — (Bax? — bx?) ¢ = 4a(3a —b)- x°(8a —- b) = (3a —b)(4a - x). EXAMPLES XVII. b. Resolve into factors : 1, x?+ayt+ue+y% 2, w—xz+auy-y2. 8, a?+2a+ab+2b. 4, a®?+ac+4a+4e. By Qa+2xe+an+x2, 6, 3q-3p+pq-p. 7, am—bm-an+bn. 8, ab—by-ay+y’. 9, pataqr-pr-7. 10, 2ma+na+2Qmy + ny. ll, awv-2ay - ba +2by. 12, 2a?+3ab —2ac — 3le. 18, ac?+b+bc?+a. 14, ac*—2a -bc?4+ 2b. 15, a@-a?+a-l1. 16, 22°+3+4+2a+4 32. 17. ax—aby+2ax—2by. 18. axy+bexy—az—bez. Trinomial Expressions. 181, In Chap. v. Art. 48 attention has been drawn to the way in which, in forming the product of two binomials, the coefficients of the different terms combine so as to give a trino- mial result. XVII. ] RESOLUTION INTO FACTORS. 103 Thus (BED GAS) Sa FBG ALD, ets scse veces (1), (a —5)\ae— et HO Arty Le cain cea ina ies cath pae (2), CoD BN OD cciaces voden se Catees eee CoD Reto a 2b LO ans cein ew saddens (4). We now propose to consider the converse problem: namely, the resolution of a trinomial expression, similar to those which occur on the right-hand side of the above identities, into its component binomial factors. By examining the above results, we notice that : 1. The first term of both the factors is 2. 2. The product of the second terms of the two factors is equal to the third term of the trinomial; e.g. in (2) above we see that 15 is the product of —5 and —3; ‘and in (3) we see that —15 is the product of +5 and —-3. - 3. The algebraic sum of the second terms of the two factors is equal to the coefficient of # in the trinomial ; e.g. in (4) the sum of —5 and +3 gives —2, the coefficient of w in the tri- nomial. The application of these laws will be easily understood from the following examples. Example 1. Resolve into factors v2+ lla +24. The second terms of the factors must be such that their product is +24, and their sum +11. It is clear that they must be +8 and +3. x? + 1la+24 =(x+8)(~+4+3). Example 2. Resolve into factors x? - 10x” +24, The second terms of the factors must be such that their product is +24, and theirsum —10. Hence they must both be negative, and it is easy to see that they must be -6 and —4. x? —10x”+24 = (a -—6)(a - 4). EHxample 3. x?-18x%+81 = (x-9)(x-9) = (x - 9). Example 4. + + 10a? + 25 = (2? + 5)(y?+5) ea (ro Ve Example 5. Resolve into factors x? - llaa+10a?. The second terms of the factors must be such that their product is +10a7, and their sum —lla. Hence they must be —-10a and —a. —llax+10a? = (x - 10a)(x — a). Note. In eed: of this kind the student should always verify his results, by forming the product (mentally, as explained in Chap. v.) of the factors he has chosen. 104 ALGEBRA. [CHAP. EXAMPLES XVIL ec. Resolve into factors: ik +3e+ 2. 2. y+5y+6. 3. x tiy+12. 4. a?—3a+2. 5. a?—6a+8. 6, B-55+6. 7, +135+42. 8. 6-135+40. 9, =-132+36. 10, 2°-15r+56. ll, 2?-157454. 12, 24+152+44. 13, 126436. 14, a@84+15a+56. 15, a?—120+27. 16, 2°+92+20. 17, z2-10r+9. 18, 2°-162+64. 19, »°-23y+102. 90, y®-24y+95. 21. 2+ 54y +729. 99. a-+10ab+21F%. 23, a? +12ab+11F. 24, a? - 23064 132%. 95, m*+8m?+7. 26. 429m*n?+14n*. 27, 6-S5rix. 98, 54-l5a+a?. 29. isp 30, 216-35a+a*. 1382. Next consider a case where the third term of the tri- nomial is negative. Example 1. Resolve into factors z*+2z—- 35. The second terms of the factors must be such that their product is —35, and their algebraical sum +2. Hence they must have opposite signs, and the greater of them must be positive in order to give its sign to their sum. The required terms are therefore +7 and —5. > x? + 2a —35 = (x+7)(x-5). Example 2. Resolve into factors z* — 3z - 54. The second terms of the factors must be such that their product is —54, and their algebraical sum —3. Hence they must have opposite signs, and the greater of them must be negative in order to give its sign to their sum. The required terms are therefore —9 and +6. x? -3z2-54 =(x-9)x+6). Remembering that in these cases the numerical quantities must have opposite signs, if preferred, the following method may be adopted. Example 3. Resolve into factors z*y* + 23zy — 420. Find two numbers whose product is 420, and whose difference is 23. These are 35 and 12; hence inserting the signs so that the positive may predominate, we have ry"? + 23zy — 420 = (zy + 35)(zy — 12). xviIL] RESOLUTION INTO FACTORS. 105 EXAMPLES XVIL @ Resolve into factors: 1, 242-2 2. -z-6. 3, 2-2-2. 4. y+4y-12 5. y+4y-21. 6. y°-5y-36. 7, @+Sa-3. 8. 2-14-30. 9, wia- 132. 10. --—12%-4. lL &+145-51. 12. F+10-39. 13, m*-—m—56. 14, m-im-S. 15, minm- 5. 16. ~-—Sp-6&. 17. #+3p-108. 18. #+p-110. 19, 2+27-4. 90, = -—7z-120. 91. =-2z-132 92. xt+139¢-48. 23. t4dzry-S9. 24. ¥+izy-Ke. % aia FT 96. a+ab-—240F. o7, 14-i-&. 28. 3-%-¥. 29. 9645-1. 90, 72+5-F 133. We proceed now to the resolution into factors of tri- nomial expressions when the coefiident of the highest power is not unity. Again, referring to Chap. v. Art. 48, we may write down the following results - se et Oe A, (1), — 2) x—4}=32°— 1448.0... (2), ons (x—4)=322— eoige hee eee, (3), —9\(x+-4)=32"+410e—8........---------- (4) The converse bea fe presents more ORS than the cases we have yet considered. Before endeavouring to give a general method of procedure it will be worth while to examime im detail two of the identities given above. Consider the result 32° — 14r+8=(3r—2yYxr—4)} The first term 32° is the product of 3r and x. The third term +8....................- —land -4+4 The middle term —14r is the result of adding together the two products 3r x —4 and xx —2 Again, consider the result 3x.°*—10r—8=(3r+2\2-—4) The first term 3 is the product of 3r and x. The third term —8..................... +2and -+ The middle term — 1 ts the result of adding together the two products 3rx —4and 2x2; and its sign is negative because the greater of these two products is negative. 106 ALGEBRA. (CHAP. 134, The beginner will frequently find that it is not easy to select the proper factors at the first trial. Practice alone will enable him to detect at a glance whether any pair he has chosen will combine so as to give the correct coefficients of the expres- sion to be resolved. Example. Resolve into factors 7x? - 19x - 6. Write down (7z 38)(@ 2) for a first trial, noticing that 3 and 2 must have opposite signs. These factors give 7x? and —6 for the first and third terms. But since 7 x2—3x1=11, the combination fails to give the correct coefficient of the middle term. Next try (7x 2)(a 3). Since 7x38-2x1=19, these factors will be correct if we insert the signs so that the negative shall predominate. Thus 7a? — 192-6 = (7x +2)(x — 8). [Verify by mental multiplication. ] 1385, In actual work it will not be necessary to put down all these steps at length. The student will soon find that the different cases may be rapidly reviewed, and the unsuitable combinations rejected at once. It is especially important to pay attention to the two follow- ing hints : 1. If the third term of the trinomial is positive, then the second terms of its factors have both the same sign, and this sign is the same as that of the middle term of the trinomial. 2. If the third term of the trinomial is negative, then the second terms of its factors have opposite signs. Example 1. Resolve into factors 14x%?+ 29% -15 ....... ee, (1) ]474— 203-15 se eee (2). In each case we may write down (72 ~3)(2x% 45) as a first trial, noticing that 3 and 5 must have opposite signs. And since 7 x 5-3 x 2=29, we have only now to insert the proper signs in each factor. In (1) the positive sign must predominate, Ati A ZY UNS NEGA UY Cr...:-cas+s ccs aecrest aeee eye Therefore 1427+ 29a —15 = (7a — 3)(2a +5). 7 XVII. J RESOLUTION INTO FACTORS. 107 Example 2. Resolve into factors 547+ 172 +6 .....cccccceccseeee eee (1); arid PEA Oe tan tis eae coo (2). In (1) we notice that the factors which give 6 are both positive. EDC ys Sis oan sen ie a aC Oe a asc negative. And therefore for (1) we may write (5v+ )(x+ ). And, since 5x 3+1x2=17, we see that 5a? +172+6 = (5a+2)(x+3). 52? — 17246 = (5a —2)(x- 3). Note. In each expression the third term 6 also admits of factors 6 and 1; but this is one of the cases referred to above which the student would reject at once as unsuitable. EXAMPLES XVII. e. Resolve into factors : 1, 2a7+3a+1. 9, 3a7+4a+4+1. 3, 407+5a4+1. 4, 2a?+5a4+2. 5, 3a7+10a+3. 6, 2a?+7a+3. 7, 5a7+7a+4+2. 8, 2a7+9a+10. Q, 2a?+7a+6. 10, 2%7+92%+4+4. 11, 2%?+52 - 3. 12, 327+52-2. 13, 39?+y-2. 14, 3y?-7y-6. 15, 2y?+9y-5. 16, 207-5b-3. 17, 60?+7b-3. 18, 26°+b-15. 19, 4m?+5m-6. 90, 4m?-4m-38. Q1, 6m?-7m-3. 99, 427-8xry—-—5y*. 98, 6x?-Tayt+2y*, 94, 6a?-1382y+2y?. 95, 12a?-17ab+6b?. 26, Ga?-5ab-6b7. 97, Ga?+35ab— 607. 28, 2-3y- 2y*. 29, 3+23y -8y?. 80, 8+18y—5y2. 9], 44172-1522. 39. 6-13a+6a%. 33, 28-31b—502 When an Expression is the Difference of Two Squares. 186. By multiplying a+b by a—b we obtain the identity (a+b)(a—b) =a?—}?, a result which may be verbally expressed as follows: The product of the sum and the difference of any two quantities is equal to the difference of their squares. Conversely, the difference of the squares of any two quantities is equal to the product of the sum and the difference of the two quantities. Thus any expression which is the difference of two squares may at once be resolved into factors. 108 ALGEBRA. [CHAP, Example. Resolve into factors 25x? - 16y?. 25a? — 16y? = (5x)? — (4y)?. Therefore the first factor is the sum of 5% and 4y, and the second factor is the difference of 52 and 4y. 2527 — 16y? = (5a + 4y)(5ax — 4y). The intermediate steps may usually be omitted. Example. 1 — 49c® = (1+ 7c3)(1 — 7°). The difference of the squares of two numerical quantities is sometimes conveniently found by the aid of the formula a? —b?=(a+b)(a—b). Example. (329)? — (171)? = (329 + 171)(329 - 171) = 500 x 158 = 79000. EXAMPLES XVII. f. Resolve into factors : 1, a’-9. 9, a? -49. 3, a’-8l. 4, a’?-100. Boa = 25. 6, «144. 7. 64=:2. 8, 81-42". Q, 4y?-1. 10, y?-9a% Il, 4y°-25. 12, 9y?-49z%. 18, 4m?-81. 14, 36a?-1. 15, k?-640?. 16, 9a?- 250". L]7, (121 1697, 18, 121-3627. LON, 25 =e 90, a?b? — a7y". 91, 49a*— 1006. 992, 64a? - 492. 93, 4p?q?-81. OA rat hea a D 95, 2x®- 4a, 06, x7 = 2027. 1 0 = po. 98, 16a!6- 90%. 99, 2xa’-4, 380, a%b8c4 — 92. Find by factors the value of Ql, (89)? — (31). 82, (51)? - (49). 33, (1001)?-1. 34, (82)-(18)?. 35, (275)?- (225), 36, (936)? — (64). When an Expression is the Sum or Difference of Two Cubes. 187. If we divide a?+b3 by a+b the quotient is a?—ab+6?; and if we divide a?—b3 by a—b the quotient is a?+ab+b”. We have therefore the following identities : a®+b3=(a+b)(a?—ab+b?) ; a®—b3=(a—b)(a?+ab+b?). These results enable us to resolve into factors any expression which can be written as the sum or the difference of two cubes. XVII. | RESOLUTION INTO FACTORS. 109 Hxample 1. 8a? — 27y? = (2x)? - (By)3 = (2a — 3y)(4u? + Gay + 9y?). Note. The middle term 6zy is the product of 2a and 3y. Example 2. 64a? +1 = (4a)? +(1)3 = (4a + 1)(16a2 — 4a +1). We may usually omit the intermediate step and write down the factors at once. Examples. 343a® - 272° = (7a? — 3x)(49a4 + 21a? + 92"). 8x9 +729 = (223+ 9)(4a% — 1823 +81). EXAMPLES XVII. g. Resolve into factors : ec = U2, 9. a +b°. Bala: 4, 1-7. 5, S8a?+1. 6, x? — 823. Le 0 270°. 8, 2-1. 9, 1-8a?. 10, 08-8. A ant bee 12, 64-p*. 13, 12503+1. 14, 216-23 15, xy? +343. 16, 1000a°+1. ie Silas 7. 18, a*b°®c? — 27. 19, Sx? - 343. 90, x? +2167". 91, x8 -— 272. 92, m—1000n°. 93, a? —729b%, 94, 125a%+512b%. 138, We shall now give some harder applications of the foregoing rules, followed by a miscellaneous exercise in which all the processes of this chapter will be illustrated. Example 1. Resolve into factors (a + 2b)? — 162°. The sum of a+ 20) and 4x is a+2b+ 4a, and their difference is a+2b-—- 4a. (a +2b)? — 16x? = (a+2b +4x)(a+2b—4zx). If the factors contain like terms they should be collected so as to give the result in its simplest form. Example 2. (8”+7y)? — (2x -3y)? = {(32+ Ty) + (2a - 8y)} {(3a-+Ty) - (2x — 3y)} = (8a¢+7Ty + 2x - 3y)(8x+ Ty — 2a + 3y) = (5¢+4y)(~+10y), 110 ALGEBRA. [ CHAP. 139. By suitably grouping together the terms, compound expressions can often be expressed as the difference of two squares, and so be resolved into factors. Hxample 1. Resolve into factors 9a? — c?+4cu - 4a”, 9a? — c? + 4ca — 4a? = 9a? — (c? — 4ca + 477) = (3a)? — (c — 22)? = (8a+c¢ — 2x)(3a —-c+2z2). Example 2. Resolve into factors 2bd — a? — c? + b?+ d?+2ac. Here the terms 2d and 2ac suggest. the proper preliminary arrangement of the expression. Thus 2bd — a? —c? + b? + d*+2ac = b?+ 2bd + d? — a? + 2ac — c? = b?+ 2bd + d? — (a? — Qac +c?) = (b+d)?-(a-c)? =(b+d+a-c)(b+d-atec). 140. The following case is important. Example. Resolve into factors 24+ 22y? + y4. x4 ee al ab y4 = (a*+ Dy224 2 +y') ms raf = (22+ 92)? = (ay)? = (2? -+y? + ay)(a?+y? — xy) = (x? +ay+y?)(a?-xy+y?). 141, Sometimes an expression may be resolved into more than two factors. gf £1 1 Example 1. Resolve into factors 16a‘ -— 814. 16a4 — 81b4 = (402+ 9b2)(4a2 — 9b?) = (4a? + 9b*)(2a + 3b)(2a — 3b). Example 2. Resolve into factors 2° — y°, af — yf = (a? + y°)(28 — y?) = (e+ y)(a?— ay + yu —y)(a?+ay ty), Note. When an expression can be arranged either as the dif- ference of two squares, or as the difference of two cubes, each of the methods explained in Arts. 136, 137 will be applicable. It will, however, be found simplest to first use the rule for resolving into factors the difference of two squares. XVII. | RESOLUTION INTO FACTORS. 111 142, In all cases where an expression to be resolved contains a simple factor common to each of its terms, this should be first taken outside a bracket as explained in Art. 129. Huample. Resolve into factors 28x4y + 64a°y — 60x7y. 28u4y + 642%y — 60x?y = 4x?y(7x? + 16x — 15) = 4a°y(7x — 5)(@ +3). EXAMPLES XVII. h. Resolve into two or more factors : 1, (w+y)-2. 2, (w-y)-2. 4, (a+3c)?-1. 5, (2¢-1)?-a?. 7, 4a?-(b-1)?. 8, 9-(a4+2). 10, (18%+y)?-(17"-y). its 12, 4a? - (2a - 3b). 13. 14, (x+y)? (m-n)?. 15. 16, a@-2ax%+a?- 4b. live 18, 1-a?—2ab-—b?. 19. 90, c?-a?-b?+2ab. O11. 99, x*+y*—2t-at4 Qary? —2Qa72?, 93, 94, at+a?+l. 95, a‘b*—16. 27, 16a4b?- 6%. 98, 64m7 — mn, 80, «@7b>-81a?d. 81, 400a?x — x, 83, 2160° + ad, 34, 25027 +2. 386, aa? — ax? -240ax. 37, 88, mt+4m?n2p? + 4ntp+. 39, 40. Gay? + 1l5a°y? - 36ay". 41, 42, 9824 — Tay? — yy? 48. 44, 2? -2u7-4%+4+2. 45, 46, a’2x? — 8a2y? — 4b7x3 + 32077". 47, 2p-3q+4p? -9q°. 48, 49, 24a7b? — 30ab? — 3604. 50, HI, 2*+427+4+ 16. bo: 58, a*t—18a7b? + b+. 54, 8, (a+2b)? -—c?. 6, @-(b+c). Q, (2a-—3b)?-c3. (6a + 3)? — (5a — 4)”. x — (2b -- 8c)". (3x + 2y)? — (2a - 3y)?. x? + a? + Qaa — 22, 12ay +25 — 4x7 - 9y?. x? — Qa +1 —m?—-4mn —4n2. (m+n+p)?—-(m-n+p). 96, 25624 -81y%. 99, at—aty?. ou. L—=129y*. 95, 1029—3x% aca? + bex — adx — bd. 82243 — 20°. . 2m8n*t — Tmin’ — 4n8, a°b? — a? — 6? +1. (a@+bpP+1. 119+ 10m — m?. 240? + aSy/4 — a7l0y8, at + y+ — Ta2y?, 8 +o4+1. [For additional examples see Elementary Algebra. ] 112 ALGEBRA. [crraP. Converse Use of Factors. 148, The actual processes of multiplication and division can often be partially or wholly avoided by a skilful use of factors. It should be observed that the formulze which the student has seen exemplified in this chapter are just as useful in their converse as in their direct application. Thus the formula for resolving into factors the difference of two squares is equally useful as enabling us to write down at once the product of the sum and the difference of two quantities. Hxample 1. Multiply 2a+3b-c¢ by 2a-3d+¢. These expressions may be arranged thus: 2a+(8b-—c) and 2a-—(3b-c), Hence the product = {2a + (3b —c)} {2a — (3b —c)} = (2a)? - (8b -c)? = 4a? — (9b? — 6bc +c?) = 4a? — 9b? + 6be - c?. EHxample 2. Find the product of x+2, 2-2, w?-Qa+4, u?+2r+4. Taking the first factor with the third, and the second with the fourth, the product = {(a + 2)(x? — 2a +4)} {(a - 2)(u?+2x%+4)} = (a + 8)(a — 8) = 26 — 64. Example 3. Divide the product of 2u?+a”-6 and 62?-5x+1 by 3x7+5x-2. Denoting the division by means of a fraction, (2a?+ x2 — 6)(6x? —5x”+1) 327 +52 —2 (2a — 8)(a +2)(3x —1)(2a—1) (8% —1)(%+2) = (2% —3)(2x- 1), by cancelling factors which are common to numerator and denomin- ator. the required quotient = XVII. ] CONVERSE USE OF FACTORS. TS Example 4. Prove the identity 17(5x + 3a)? — 2(40a + 27a)(5x + 3a) = 25x? - 9a. Since each term of the first expression contains the factor 5x + 3a, the first side = (5% + 3a){17(5x + 8a) — 2(40x + 27a)} = (5x + 8a)(85x + 5la — 80x — 54a) = (5% + 3a)(5x — 3a) = 252? — 9a?. EXAMPLES XVII. k., Employ factors to obtain the product of 1, a-b+c, a-b-e. 2, 2w—yt2, W+y+z 8, 14+2x-x7, 1-2x-2?, 4, ¢c?+3c4+2, c?-3c-2. 5, a+b-—c+d, at+b+c-d. Go Po rey, p-g— ary. 7, a’®—4a2d + 8ab? —-8b°, a*+4a°b + 8ab? + 8b? Find the continued product of 8, (a—b), (a+b), (a? +b?) 9, (icy, (L+2)°, (ia? )". 10, a-4a+3, a?-a-2, a?+5a+6. ll, 3-y, 3+y, 9-3y+y’, 94+3y+y’. 12, l+ct+e?, l-c+e, 1-c?+c4 18, Divide a3(a+2)(a*-a-56) by a?+7a. 14, Divide the product of z7+a”-2 and 27+4x%+3 by 2?+5x+6. 15, Divide 32°(2+4)(z?-9) by 2?+a”-12. 16, Divide the product of 2x%7+lla-21 and 38a?-20a-7 by a” — 49. 17, Divide (2a?-a-3)(8a?-a-2) by 6a7-5a—-6. 18, Divide x6-7x?-8 by (~#+1)(u?+2u +4). Prove the following identities : 19, (a+b)?-(a-b)?(a+b) = 4ab(a+b). 90, ct-—dt—(c—d)(c+d) = 2cd(c? - d?). 91, (m-n)(m+n) — m+ nt = 2mn(m? — v7). 22, (a+y)t-8ay(a+y)? = (a+y)(a+y"). 93, 3ab(a—b)?+(a—b)* = (a —- b)(a? — b*). H.A. vee 114 ALGEBRA. (CHAP. MISCELLANEOUS EXAMPLES III. ], Find the product of 10z?-12-3% and 2%-4+432%. Qo Leal. b= 1 ea 2, 0.0, tindsthe walle of a?—b? b*-cd , cb - 0? a2+b? 2b?+cd Babe° 3, Simplify 2[4a - {2y+(2e-y)-(«+y)}]. 4, Solve the equations : (1) z—-3 3-9 12x (2) 3u — 4y = 25, 5 3 15 ie bt 2y =i. 5, Write down the square of 223-—a+5. 6, Find the H.C.F. and L.C.M. of 3a2b°c, 12a4b?c?, 15a%b°c. - 7, Divide a*+4b* by a?—2ab+26?. 8, ind in dollars the price of 5x articles at 8a cents each. 9, Find the square root of 24-82? + 24a? — 32x+16. QO, lh.a=5, b=3.¢ =]; ind the value of (a—b) , (b-c) , (ac) a+b b+e a+c¢ Ti eeccire (78 +5) -~73=13 -f(a i AI. 12, A is twice as old as B; twenty years ago he was three times as old. Find their ages. 18, Simplify (1 - 2a) -{3-(4-5x)}+{6- @7 -8z)}. 14, The product of two expressions is 6x4 + Say + 6x?y? + Bay? + 6y4, and one of them is 247+ 3ay+2y?; find the other. 15, How old is a boy who 2x years ago was half as old as his father now aged 40? 16, Find the lowest common multiple of 2a?, 3ab, 5a%bc, 6ab*c, 7a”. 17, Find the factors of (1) 2?-axy-72y*. (2) 6x?-18x%+4+6. 18, Find two numbers which differ by 11, and such that one- third of the greater exceeds one-fourth of the less by 7. XVII. ] MISCELLANEOUS EXAMPLES III. 115 ifo@al, b=—)) c= 2, d.=0, find.the value of at+b c+d, ad— be een a-b ce-d bd+ac a?+b™ - : 3 ae | ad be 90, Simplify 8x —y—{2v-ly-7-(%-4)+(2-}e)l 91, Solve the equations : (1) (3a -8)(8x% +2) -— (4% -—11)(2%+1) =(%-3)(x%4+7) 5 x — ; 25 ene (2) SU Siva 2, aty-5=ay-2). 99, + 7c4 — 20c?. 9 a ° 4m —9m?, 6m? —5m?-6m, 6m4+ 5m — 6m?. 3a42x3 — Sax? + 40723, 3a5x? - llatz? + 6a%x, Satz? + 16a%23 — 120223, 118 ALGEBRA. [CHAP. 147, The highest common factor should always be determined by inspection when possible, but it will sometimes happen that expressions cannot be readily resolved into factors. To find the highest common factor in such cases, we adopt a method analogous to that used in Arithmetic for finding the greatest common measure of two or more numbers. 148, We shall now work out examples illustrative of the aleebraical process of finding the highest common factor; for the proof of the rules the reader may consult the Hlementary Algebra, Arts. 102, 108. We may here conveniently enunciate two principles, which the student should bear in mind in reading the examples which follow. I. Jf an expression contain a certain factor, any multiple of the acpression is divisible by that factor. Il. If two expressions have a common factor, ct will divide their sum and their difference ; and also the sum and the difference of any multiples of them. Example. Yind the highest common factor of 423 — 3a?- 244-9 and 8x3 — 2x? - 53x — 39. x | 4a? — 8a? — 24a -—9 8a? — 2a? — 53a — 39 | 2 4a3 -- 5a? — 21a 823 — 62? — 48a -—18 Pee Na 272 — 32-9 2 Gaeta OA Mal? Qn — 6x 47 —. 62-18 3 32-9 x- 3 32-9 Therefore the H.C.F. is x—-3. Explanation. First arrange the given expressions according to descending or ascending powers of x. The expressions so arranged having their first terms of the same order, we take for divisor that whose highest power has the smaller coefficient. Arrange the work in parallel columns as above. When the first remainder 42? — 5a — 21 is made the divisor we put the quotient x to the left of the dividend. Again, when the second remainder 2a?-—3x-—9 is in turn made the divisor, the quotient 2 is placed to the right; and so am. As in Arithmetic, the last divisor x-3 is the highest common factor required. 149, This method is only useful to determine the compound factor of the highest common factor. Simple factors of the given expressions must be first removed from them, and the highest common factor of these, if any, must be observed and multiplied into the compound factor given by the rule. XVIII. ] HIGHEST COMMON FACTOR. 119 Hxample. Find the highest common factor of 2424 — 2a3 — 60a? - 32a” and 1824 — 62? — 39x? — 182. We have 2424 — 223 — 602? — 32a = 2a (122° — x? — 30x — 16), and 1824 — 6x3 — 39x? — 18x = 32 (623 — 2x? — 13x - 6). Also 2x and 3x have the common factor x. Removing the simple factors 2a and 3x, and reserving their common factor x, we continue as in Art. 148. 2% | 6a? — 2a? —- 1382-6 1227 -— x?-30x-16|2 622 —8x?-— 8x 122° — 4a? — 26x — 12 2 6a? -— 52-6 3u7— 4r— 4) 2 62?— 8x-—8 3x7 + Qe 38x24+2 —- 6x%- 4/-2 — 6a- 4 Therefore the H.C. F. is 7(8%+2). 150, So far the process of Arithmetic has been found exactly applicable to the algebraical expressions we have considered. But in many cases certain modifications of the arithmetical method will be found necessary. These will be more clearly understood if it is remembered that, at every stage of the work, the remainder must contain as a factor of itself the highest common factor we are seeking. [See Art. 148, L. & IL] Example 1. Find the highest common factor of 3x? — 1327+ 23% -21 and 62° +2? - 447421, 3x3 — 13274+23x2-21| 623+ x2?-44r421]2 6x? — 26x? + 46x — 42 27x? — 902 + 63 Here on making 27z?-90x+63 a divisor, we find that it is not contained in 3x3 — 132?+23x%-21 with an integral quotient. But noticing that 27x? — 90x + 63 may be written in the form 9(3z?-10x+7), and also bearing in mind that every remainder in the course of the work contains the H.C. F., we conclude that the H. C.F. we are seeking is contained in 9(3x7-10x+7). But the two original expres- sions have no simple factors, therefore their H. C. F. can have none. We may therefore reject the factor 9 and go on with divisor 32? - 10x+4+-7. 120 ALGEBRA. (CHAP. Resuming the work, we have x | 323 — 13224 234-21 3x2-10e+7 |x 32° -—10a?+ Tx Ste | ae -] — 32°+162-21 —- 3x4+7|-1 — 327+10a- 7 — 3x+7 2)6x-14 Auaias fi Therefore the highest common factor is 3x —7. The factor 2 has been removed on the same grounds as the factor 9 above. 151, Sometimes the process is more convenient when the expressions are arranged in ascending powers. Hxample. Find the highest common factor of Oo 40160? = 0gr oo ee ee (1), and AiG LOG Sa" 2. rete ere ee Oo} As the expressions stand we cannot begin to divide one by the other without using a fractional quotient. The difficulty may be obviated by introducing a suitable factor, just as in the last case we found it useful to remove a factor when we could no longer proceed with the division in the ordinary way. The given expressions have no common simple factor, hence their H.C.F. cannot be affected if we multiply either of them by any simple factor. Multiply (1) by 4 and use (2) as a divisor : 4— Ja- 19a?- 8a? 12 - 16a — 64a? - 36a? | 3 5 12 — 21a — 57a? — 24a 4|20-35a— 95a2— 40a a|5a— Ta? — 12a" 20-28a-— 48a? 5-— Ja —12a2|5 — Ja- 47a*- 40a3 5+ 5a at 12a —12a?|- 124 7a ~ 35a-+ 2350? +200a° —12a -—12a? 3da— 49a2-— 84a? 2840? | 284a? + 284a3 l+a Therefore the H.C.F. is 1+a. XVIII. ] HIGHEST COMMON FACTOR. 121 After the first division the factor a is removed as explained in Art. 150; then the factor 5 is introduced because the first term of 4—7Ta-—19a?—8a? is not divisible by the first term of 5—7a-— 12a? At the next stage a factor —5 is introduced, and finally the factor 284a? is removed. 152, From the last two examples it appears that we may multiply or divide either of the given expressions, or any of the remainders which occur in the course of the work, by any factor which does not divide both of the given expressions. EXAMPLES XVIII. b. Find the highest common factor of 243 + 322+ a+6, 22+ 4242443. 2y? — 9y?+9y-—7, y> — 5y? + 5y —4. 2a? + 8a? — 5a -—20, 6a? -4a?-1524+10. a’? +3a2—-16a4+12, a®+a?-10a+8. 62° — a? — 7a —-2, 2a? —Ta?+x2+6. g—3q+2, g@-59q?+7q-3. at+a®—20?+a-3, 5a°+3a?-17a+6. 3y4 — 3y® — 15y?-9y, 4y° — 16y4 — 44y? — 24y/?. 1524 — 152? + 1022-102, 302° + 12024 + 2027 + 8022, 2mt+7m?+10m?2+ 35m, 4m*+ 14m? — 4m? -— 6m 4 28, 3act — 9a? + 12a?-12%, 6a? -6x7-15%+6. 2a° —4a4*- 6a, a’ +a*—- 3a - 3a? x? +442 -—922—-15, x2? —2Qlx — 36. 14, 9at+2a72?+a4, 3a*- 8a%x + 5a7a? — 2a’. 15, 2-3a+5a?-2a3, 2-5a+8a?~ 3a’. 16, 3a?-52°-15a4- 40°, 6a — 7x? -— 2923 — 1224, - DOONIOARwWNH EH ee oONnNrH © [For additional examples see Hlementary Algebra. | GHA PPE Rex FRACTIONS. 153, Tue principles explained in Chapter xvi11. may now be applied to the reduction and simplification of fractions. Reduction to Lowest Terms. 154, Rule. The value of a fraction is not altered if we multi- ply or divide the numerator and denominator by the same quantity. An algebraical fraction may therefore be reduced to an equi- valent fraction by dividing numerator and denominator by any common factor; if this factor be the highest common factor, the resulting fraction is said to be in its lowest terms. 9407 ¢7a2 18a°2? — 120223 Hxample 1. Reduce to lowest terms ~ 2Aaen* 6a7x7(3a — 22) =. ac 3a — 2a The expression = 6x? — Say Jay — 12 On a5 A The expression = aA _ 2% by(3u—4y) 3y Note. The beginner should be careful not to begin cancelling until he has expressed both numerator and denominator in the most convenient form, by resolution into factors where necessary. Hxample 2. Reduce to lowest terms EXAMPLES XIX, a. Reduce to lowest terms : a 9 ae —2a_ 9 3ab +b? 622 —32y * 4a? — 8a? * 6a*b? + 2ab 1. CHAP. XIX. ] FRACTIONS. hes Reduce to lowest terms : 0 eae Baryz Fetes ox? —y" 6 222y7 — 8 * Bay + 10x72 * 62°y — Qary? * Bay + 6x i, pies Gen Gilin OaC eens et = 30 pa ie 12. * Bea? — 10c2a + 5c" * Bat + 30a” 2a +b)? a? + 63 2c? + Bcd — 3d? Pe el ee ee 4a° — ab? a? — ab — 2b? c7 + 6cd + 9d? x? —4¢ —21 a — I —15 92 +4 -3 Re ag pe ee pe es 3274+10c+3 327 — 122-15 227+ 1lla#+12 3a? — 24 4a? —"Bary? 18a? + 6a2x + 2aa? iy | ere Ab et See ele ae ee es 4a” + 4a —24 227 4+- ay — 1oy" 2 Oe 155. When the factors of the numerator and denominator cannot be determined by inspection, the fraction may be reduced to its lowest terms by dividing both numerator and denomi- nator by the highest common factor, which may be found by the rules given in Chap. XVIII. 3a? — 1327+ 23% — 21 15a? — 3827 — 2x%+21 Example. Reduce to lowest terms The u.c.¥F. of numerator and denominator is 3a — 7. Dividing numerator and denominator by 32-7, we obtain as respective quotients 2? -2x+3 and 5x7-x-38. oa? —l3at+23e—21 (82 —7\(x2-2e+3) 272-2243 15a? — 3827-22+21 (8%-—7)(5z*?7-x2-3) 522-2 -3 Thus 156, If either numerator or denominator can readily be resolved into factors we may use the following method. x? + 3x7 — 4x 72 — 1827+ 62+5 The numerator = 2(x?+ 3x - 4) =a(x%+4)(x-1). EHxample. Reduce to lowest terms Of these factors the only one which can be a common divisor is zx-—1. Hence, arranging the denominator so as to shew z-1 asa factor, a(a+4)(a—-1) Fax —1)—1lx(z—1)—d5(x—-1) ee ee le ale ts) ~ (@—-1)(7a? — lla -—5)” 7a2?-lla—5 the fraction = 124 ALGEBRA. [coHaP. EXAMPLES XIX. b. Reduce to lowest terms : 1 x? — 4? +24 —2 9 oe+at+2 © “Bat4 72242 * @—402+5a—-6 9 yey ly 3) 4 m>—m?—2Qm * 3y2+4y?+4y+1 * m—m?-—m—-2 5 a? — Qab?+ 21 b3 ; 6 Qa? — aa = 90° ‘ * a? — 4a7b —2lab? * 323 — 1l0ax? — Ja2x — 4a? 7 5a? —4a-—] : 9 +27 — 12cd? = 9d? : * 923 — 32741 * Qc? + 6c2d — 28cd? — 24d3 9 aA —21e+8_ 10 ap Oy ly = oy". * 8a4—21a? +1 * y+ Ty’ + 3y?- lly ll Rea ings 12 2 —5ax — 4a? + 323 2-a +923 4+ 4a + 922+ 423 — Bat [For additional examples see Hlementary Algebra. ] Multiplication and Division of Fractions. 157, Rule. To multiply together two or more fractions: multiply the numerators for a new numerator, and the denome- nators for a new denominator. Ce Caae Thus B* dba Sunilarly, as del sl BaF bdf’ and so for any number of fractions. In practice the application of this rule is modified by re- moving in the course of the work factors which are common to numerator and denominator. 2a? + 3a y 4a? —6a dae) 18a 15) a(2a +3) 2a(2a — 3) 4a 6(2a + 3) _2a-3 oom by cancelling those factors which are common to both numerator and denominator. EHxample. Simplify The expression = 158, Rule. Zo divide one fraction by another: invert the divisor, and proceed as in multiplication. XIX.] FRACTIONS. 125 d ad Thus sa a oe ah by di bd -e* 06 Fe Vy PC) ae Example. Simplify Cea aed pae A ASS aD er ax — az 9u7-4a? 38ax+2a? : 6x? — ax — 2a? x-a 3ax + 2a? Th Ss = : x ) e expression en ara eo a : _ (8% ~2a)(2a +a) u-% ee C(O -+ 2a) . a(x — a) (84+2a)\(8a2-2a) Weta =, since all the factors cancel each other. EXAMPLES XIX. c. Simplify 7 zea pe, 9. ee 2b — Bab® te ARG UG e724 2-12ab a*b?-4 9 2c? + 3cd , e+d 4 5y - lOyees ] -Qy * 4c? - 9d? ° 2cd - 3d? * 12y? + 6y? © Qy + y? 5 x? —4 Nate 6 b2 uy oes 16a? * 9244744 °> 424+2 1 3b=—da~ = $2295 ~ 7 Eee a 8 aU heey ey = 24 * @?+5e+4 a74+3%+4+2 ; pees y+ 6y+9° G74 2) ee 40 = 2) a*—3a—2_ 3a?-8a-3 Drees gyee gr eaee 10, x pee Ay 6? + 125 . 2O0 tn 12 38m?—m—-2 , 4m?+m—-6 * 5b?+24b-5 63—5b2+25d * 3m2?+8m+4° m+2 13 merely Os 5p +6 ee 2p-- 15 “ “p=9 p?-5p p-4 64a72—=1 4#7=49 . -w=7 14, ie eat Ee ee Sa 15 407+ 44-15 x+8 ; 2a? + 5x Piet -46 (abe eis 6 16 a? + 8ab — 9b? a? — Tab +12? a+ ab fab? , ard 27b ai? — b? a? — 3ab — 4b? 17 — 16a? 4 x*+ae—2007 , x? - pee Cat ; ae —ax—3002 ax2+ 9a2x +2003 x? + 8ax + l5a2 (a—b)?-c? a?+ab+ac . (a+b)?-¢? 18, Gee ele (a—c)?- 0? 7 (a+b+c)? [For additional examples see Llementary Algebra. ] CHAPTER XX. LOWEST CoMMON MULTIPLE. 159, Derinition. The lowest common multiple of two or more algebraical expressions is the expression of lowest di- mensions which is divisible by each of them without remainder. The lowest common multiple of compound expressions which are given as the product of factors, or which can be easily resolved into factors, can be readily found by inspection. Example 1. The lowest common multiple of 6x°(a—2)?, 8a®(a — x), and l2ax(a— 2x) is 24a%x?(a — x)°. Hor it consists of the product of (1) the L.C.M. of the numerical coefficients ; (2) the lowest power of each factor which is divisible by every power of that factor occurring in the given expressions. Haample 2. Find the lowest common multiple of 38a7+9ab, 2a*- 18ab?, a® + 6a2b + 9ab?. Resolving each expression into its factors, we have 3a? + 9ab = 8a(a+3d), 2a? — 18ab? = 2a(a+3b)(a — 3d), a? + 6a*b + Yab? = a(a + 3b)(a+4+ 3b) = a(a + 3b). Therefore the L.C.M. is 6a(a+3b)?(a— 3b). Example 3. Find the lowest common multiple of (y2—xyzy, ya — x3), x44 Qad-+ a2, Resolving each ee into its factors, we have (yo —ay2)= {yee 2) =P 2-2), yy? (202 — 203) = y?ar( 2? — x?) = wy?(z — x)(z+2), ZA 4 Dare? 4 9022? = 2( 2? + Daz + 2?) = 22(z-+-00), Therefore the L.C..M. is wy?2*(z+a)?(z-—a)%, CHAP. XX.] LOWEST COMMON MULTIPLE. 127 EXAMPLES XX, a. Find the lowest common multiple of 1, a’, a-a’. 9, x7, 2? — 323, 3. 4m?, 6m —8m?2. 4, 627, 2*+327. 5, b+), b-b. 6, «2-4, 2°+8. I, 9a?b-b, 6a? +2a. 8, B-k+1, B-1. YY. m?-5m+6, m?+5m-14. 10, y?+8y?, y?-9y’. ll. 2?-9x2+14, x?+4e-12, 12, 2+27y*, x?+xy- by’. 13, 067+9b+20, b?+6- 20. 14, c?-3cx-182?, c?-8ex+ 122%. 15, a*?-4a-5, a?-8a4+15, a?- 2a?- 3a, 16. 2a?-4ay-16y?, 2?-6ary+8y?, 3x? - 127°. 17, 3a3-12a?x, 4x7+1l6arv+16a7, 18, a’c—ac®, (a2c¢+ac?)*. 19, (a%x—2ax*)?, (Qax — 42)2, 90, (2a-a7)3, 4a?-4a?+a4, 91, 2u?-x-3, (2x -38)?, 4z7-9. 99, 2x2-Tx-4, 6x?-Tx-5, x2? - 8274+ 16x. 28, 10x°y%(x?-y*), ldy*(a—y)?, l2a%y(a — y)(a? — y?), 94, 2a? +x-6, 7x?+1lxa-6, (7x7- 32). 95, 6a? — Tax -3ax", 10a?x - llax?- 6x, 10a? -2lax — 102". 160, When the given expressions are such that their factors cannot be determined by inspection, they must be resolved by finding the highest common factor. Hxample. Find the lowest common multiple of 2a + a? — 20a? — 7x +24 and 2a4+ 3x° — 138x2-7a+15. The highest common factor is z?+2z.- 3. By division, we obtain Qa + a3 — 20a? — Ta 4+ 24 = (x? + 2a — 3) (Qa? — Ba - 8). Qart + Ba3 — 13x? — Ta +15 = (2? + Qa — 3)(22? — w — 5), Therefore the L.C.M. is (xz? + 2a — 3)(2a? - 3x” —8)(2u?- a” -5). 128 ALGEBRA. [CHAP, Xx. EXAMPLES XX. b. Find the lowest common multiple of ], «-Qa?-132-10 and x - a? 10x- 8. 2, y'+3y?-- 38y-9 and y*®+ 3y? - 8y— 24. 8, m+3m?—-m—3 and m?+6m*+1l1m+6. 4, 2Qat-2e3+274+3x2-6 and 4a4-2a°+ 3x -9, . Find the highest common factor and the lowest common multiple of (%—«?)®, (~?-a°)?, a? — a4 6, Find the lowest common multiple of (a*- ax), (a?+az)’, (aa — x7). 7, Find the highest common factor and lowest common multiple of Ga?+5¢-6 and 6a?+x%-12; and show that the product of the H.C.F. and L.C.M. is equal to the product of the two given expres- sions. 8. Find the highest common factor and the lowest common multiple of a?+5ab+6b*, a*—4b?, a? — 3ab?+2b°, 9, Find the lowest common multiple of 1-a?-a4+2° and 1422 +2? — at — 2, 10. Find the highest common factor of (a? —4ab?)?, (a?+2a°b)°, (a*x + 2abx)?. 11, Find the highest common factor and the lowest common multiple of (3a? —2ax)?, 2a7xa(9a? — 4a"), 6a? — 18a7x? + bax’. 12, Find the lowest common mrltiple of #+a°y+ay?, wy -y', . amy + ay? + xy, [For additional examples see Hiementary Algebra. ] CHAPTER XXI. ADDITION AND SUBTRACTION OF FRACTIONS. 161, To find the algebraical sum of a number of fractions we must, as in Arithmetic, first reduce them to a common denominator. For this purpose it is usually most convenient to take the dowes¢ common denominator. Rule. Zo reduce fractions to their lowest common denomin- ator: find the L.C.M. of the given denominators, and take it for the common denominator ; divide it by the denominator of the first fraction, and multiply the numerator of this fraction by the quotient so obtained; and do the same with all the other given fractions. Hxample. Express with lowest common denominator 2a(x — a) 32e(a2 — a?) The lowest common denominator is Gax(x%-a)(w+a). We must therefore multiply the numerators by 32(v+a) and 2a respectively. Hence the equivalent fractions are 15x7(a~+ a) er 8a? 6ax(«-a)\(x+a) 6ax(x—-a)(v+a) 162, We may now enunciate the rule for the addition or subtraction of fractions. Rule. Zo add or subtract fractions: reduce them to the lowest common denominator; find the algebraical sum of the numerators, and retain the conmmon denominator, 1 a,c _ad+be zou ue he and a_¢_ad—be 6d bd 130 ALGEBRA. (citar. 168. We begin with examples in further illustration of those already discussed in Chapter x11. : 2e+a 5a?-4ax Hxample 1. Find the val fee ae xample ind the value of + 2-5 The lowest common denominator is 9a. Therefore the expression = da(2x + a) + Sat — dam 9a? = 6Gax + 3a?2+4+ 52? -4ax tt 3a? + 2ax + 5a? i 9a? > 9a* x—-2y 3y-a 34-240 ee eR ae The lowest common denominator is axy. Thus the expression = COA at eae axy ac —2ay + 3xy -ax—38xy + 2ay = waxy = 0, since the terms in the numerator destroy each other. Hxample 2. Find the value of Note. To ensure accuracy the beginner is recommended to use brackets as in the first line of work above. EXAMPLES XXTI. a. Tind the value of Be-1 2+3 2e-1 1 a-2 a-l at+5 9 Sure oe eOe lpia GEE ees peor iy LO MueuNaG caret. 5 eg! 2-2 6 22-5 2-4 a2 —dAg we ye yp RS) Mt Mel eu: ener Ye ee atx at2ex x-5a. 1. ye | wet xy the 2a 3a. | 64 i 3 2 joe a PAs 9. 2a°—-5a a pee reads 10 See. Ey eee, 4x? a a" a? y oe Bry R20 30 aC, ab-be a 2a®-ab TSE Re Ope a Se tea) 13. ay-axy+4e_ 1 _ a. 14, Sa era Quy Ee a" be ca XXI.] ADDITION AND SUBTRACTION OF FRACTIONS. Tal 164, We shall now consider the addition and subtraction of fractions whose denominators are compound expressions. The lowest common multiple of the denominators should always be written down by inspection when possible. 22-30 2x-a Hxample 1. Simplify 5 : x-20 x-a The lowest common denominator is (% —2a)(z—- a). Hence, multiplying the numerators by x-a@ and x-2a respec- tively, we have the expression = es ee) (2 — 2a)(x — a) _ 2a? — 5am + 8a? — (2a — Sax + 2a?) he (a —2a)(%— a) Qu? — Fax + 3a? -— 27+ 5ax - 2a? (x —2a)(x—-a@) a > (a —2a)(x — a) Note. In finding the value of such an expression as — (2% -—a)(x - 2a), the beginner should first express the product in brackets, and then remove the brackets, as we have done. After a little practice he will be able to take both steps together. Example 2, Find the value of ae + ame The lowest common denominator is (a -- 4)(~% +4). (82% +2)(x+4) +(x —5)(x - 4) (7 -4)(~+ 4)? _ 3x7 +14e+8+2?—9x+20 (x —4)(~ +4)? _ 447+ 5a +28 (x—4)(a +4)? Hence the expression = 165, If a fraction is not in its lowest terms, it should be simplified before it is combined with other fractions. 132 ALGEBRA. [cHaP. B be Wa are v?+5ey—4y?_ xy — 3y? cample. Simplify 2 16,2 a wrap, 2 — Ay? = The expression =~ ey. ay = ED a*—l6y? (w+ 4y)(a — 3y) Ua Oly tee oe x” — 167? x+4y + bay Ay yee ay) x? — 16y? _v+5ay —4y—- xyt+4y? x? — 16y _e+4ay wetd4y) & 2 —l6y? (a+4y)(a—4y) x—4y EXAMPLES XXI, b. Find the value of ] 1 1 1 i 1 1. a—2*a-3 2. asd ae a 3. b-2 642 a b Gt be a+3 a-3 4, x-a «x-—b 5, ate 'a-@ 6. a-3 a+3 z he 3a 1 a Leta? ata x uw? C ye : : _9 Example 2. Simplify = a Hoe Here the reduction may be simply effected by multiplying the fractions above and below by 6a, which is the L.C.M. of the denominators. 18 + 2a? — 12a a?+3a-18 _2(a?-6a+9) 2(a-3) \(a¢6)(a—38) “a+6 Thus the expression = a2 +0? a? 0? Example 3. Simplify ere a0 aD TI des (Ga Oe ae 4a°b* ae numerator (a?+ B5(a2 02) = (2 ED a2 BY similarly the denominator = ne : 4a?b? dab H the fraction = ae ence the fraction (®t B\ ew) * @Lbanb a 4a7b? (a+ b)(a—b) (a@2@+e)\(a2—b)~ 4abd = 20d a2+b? Note. To ensure accuracy and neatness, when the numerator and denominator are somewhat complicated, the beginner is advised to simplify each separately as in the above example, XXII. ] MISCELLANEOUS FRACTIONS. 141 174, In the case of Continued Fractions we begin from the lowest fraction, and simplify step by step. Haample. Find the value of : ; ae eee oe The expression ek ae Eten ed inl 7 3 ceo — x) Q2-24+2% 2-2 Si iS 1 1 8-44-3432 5-a@ 2-2 2-2 2+2 62 EXAMPLES XXII, a, Find the value of 1 a 1, pry Us oa o - is 4, —. +2 b= = ee | = e Zz d a> l-a aes en pel P_y ae te NOE d 4g 5, ea 6. y a le a 1 8. ab Tr ie OL xy Bard Pa p a+? -5 y-34+2 aoe = 9, 10. . ak, ea ee eee noe a2 a y n oan p-2-—8, ee 12, ———. Bo ——— ss 14, — x-44_l% Dens BF coat 243 b+3 at abt R c+d c-d qa - 222 2+3 «+3 ae TG l-ab a 4 15. c+ad c-d 16. 1 a(a—b) 17. x-3 2-3 c-d'ct+d L—ab z—1 1 1 eek 18, 1 * tomes 19. C+ ¥ 20, a c 1+- x — 4--= a a ad 142 ALGEBRA [CHAP. Find the value of ] 21, —— 22, ——— 23, ——. ]-—= L+H lee x x y Rew D4, 2 “ ~ 95. a 26. “ ~ es l= ee eae | ie loa dae Lite, 2-y ip 22] 8a —2 i Sera 28, ae De ia Si Oe ee lee [eae x—-1 3a — 2c 175, Sometimes it is convenient to express a single fraction as a group of fractions. Day — lay loy? bay 10ry? | be Example. 102: ~ 10x%y? 10x%y* 102%? Lael ae, ee eae 2 pe aeaete 176. Since a fraction represents the quotient of the nume- rator by the denominator, we may often express a fraction in an equivalent form, partly integral and partly fractional. +7 (#+2)+5_ 44. 5 Hxample | x+2 x+2 x+2 Example 2. 3x-2 _ 3(@ +5) — 15-2 _ 3(@4 +o)— 17 17 ae LZ ate 2+5 x+5 x+5 x+5 Example 3. Shew that 2Qu*— Te—1 =22-—1= meal she © 2-3 By actual division, x -3)2x4?-TJa-—1(2xr-1 200 — 62 - x-l —- £+3 -4 Thus the quotient is 22-1, and the remainder ~4, Peas cs Therefore vast adel Me 2a—-1- Lies «“-3 x-3 XXxII.] MISCELLANEOUS FRACTIONS. 143 177. If the numerator be of lower dimensions than the denominator, we may still perform the division, and express the result in a form which is partly integral and partly fractional, Example. Prove that j —_ = 2% — 6x? + 182° - tees By division 1+322 ) 2e ( 2a — 6a? +1825 2x + 623 Fe3 — 62? — 1825 1825 182° + 54x7 — 54x47 whence the result follows. Here the division may be carried on to any number of terms in the quotient, and we can stop at any term we please by taking for our remainder the fraction whose numerator is the remainder last found, and whose denominator is the divisor, Thus, if we carried on the quotient to four terms, we should have Qa ; 16229 g PEs Ont G8 1905 <4 a aS ree ch me aia The terms in the quotient may be fractional; thus if 2? is divided p x*—a', the first four terms of the quotient are ied ae ea ae sua and the remainder is — A ei gw CO ae pe 178, The following exercise contains miscellaneous examples which illustrate most of the processes connected with fractions. EXAMPLES XXII, b, Simplify the following fractions : l 1-23 9 12a? +x2—-1 ,1+6%+92x? * 14294 2424 ge * 1-824 1622" 1622-1 ~ a+b. 4ab a+b 2a 3. i teas. = at 4. o—ab—26? a?—4o° 5 a—1 xt+e2+1 6, (w+y)?_(#- (w-y) gz-l x+e+) CV-Y “bY 144 ALGEBRA. [crrap. Simplify the following fractions : 7 aba" — aca + bay — cy 9 “I ee! )- a ax*+xy—-ax—y ‘ g\a-x% a+3x/ a+3u 2 3 5 9. 6848 bat alle +08, Oneal: Se 02+ a* 1 3a.) a Cee 10. 3a+1 “ORE H a? + 23 Sore 1 1 eee =, z—1 12, ae a 18, 92 _ 3(a? — 2ia+ 54) cd(a? + b?) + ab(c? + d?) ( x ee 1d. cd(a? — b*) + ab(c? — d2) 16. lta a a 16 ] ., 1 : —8a22+4a? 23 — aa? — 4074+ 403 2 Bae oO 2(422) 17 2 eae mW 0 ngs e = 3 * ° = 204 a’+8 lig 3 a a 19 223+ 42-32 | 5a?-8x-21 , 2x?-3x-9 * 8542+ 24a -—35° w+ 7a? -S8a ° 722+5la—-40 Cat Pr Te p= Dad ee MS IISA Cae al 1rd 21 ye ee ee ; “a-y xt+yl \x-y “t+y 2a7y + 2ry" a?—(b-c)? b?-(c-a)? , 2 -(a—b/y 22, (ctaf—B atbpie (b+c)?-a 23, (= +¥)( : )- ye a el YING) ae Pay ae x+y 94. y a1) te a*—4a+3, a? — 4 — tale a+a-6 a2 —-4a4+4 weet ae Waste at sea ~4ax 25. (55 Toa z) (sa saa) 6a*(x - 2a)(% — 3a)’ XXII. ] MISCELLANEOUS FRACTIONS. Simplify the following fractions : i] ] i 2a? 20. 6a —6 6a+6 30243 8a4+3° 97 4ah? Ae 1 a) a x 1 * 2a4+32b4 8a+16b 4a7+1602 8(2b-a) 3b7-+6 .2b=7, 26°=3b ; A Rae EB at 28. Geb t 195 28 43" 99, att ND I HI x x y y Z 2 Dga 2 LT we FES een : 20 (ger) xy 9] m m+m ee “- Pigede ie 1 ns x) \a a m4 m ab ac ] ] if 32. (sea) A aie a [ba bes ha re b Cc eli pends 6 99 Ltac = 1 -—ae Eee a) ny | sree cose aa eee) | a l+ac l-—ac 1 ] (30-- wee oe 1 34. ul to 3a Riera iy) TPR) a a 145 CHAPTER XXIIL HARDER EQUATIONS. 179. Some of the equations in this chapter will serve as a useful exercise for revision of the methods already explained ; but we also add others presenting more difficulty, the solution of which will often be facilitated by some special artifice. The following examples worked in full will sufficiently illus- trate the most useful methods. 6x-3 32-2 2+] e+5° Clearing of fractions, we have (6a — 3)(v7+5) = (8a —2)(24+7), 6x7 + 27x — 15 =62?+17x-14; Example 1. Solve 1022 1% 1 ea=—.,. 10 Note. By asimple reduction many equations can be brought to the form in which the above equation is given. When this is the case, the necessary simplification is readily completed by multiply- ing across or ‘‘ multiplying up,’’ as it is sometimes called. 87+23 5a+2 2%4+3 Example 2. Sol ie = eat; xanrple olve 50 te hae 1 Multiplying by 20, we have 8a +23 _ 20(5% + 2) cxaisd Ge = 82x + 12- 20. : a 20(52 +2) By t t BY fee ee y transposition, Nera Multiplying across, 93x7+124=20(57+2), 84= 725 ¢= 12, CHAP, XXIII. ] HARDER EQUATIONS. 147 180. When two or more fractions have the same denomin- ator, they should be taken together and simplified. 24-52 Sx-49 28 E. Lax kee OLY 6 ane = — 13. aample NGS ary tap tame eae By transposition, we have 8a — 28 — (24 — 5x), ee eect a ea Ook 4— : x-2 Gree dae 4-¢ 2-2 Multiplying across, we have 3a — 5a?-6+102 = 16-424 20x — 52? ; that is, — 32 =22; Bix _ 22 = Example 2. Solve a EN ea am “2-7 we 108 wae Boe ESO. This equation rent be solved by at once clearing of fractions, but the work would be laborious. The solution will be much simplified by proceeding as follows. The equation may be written in the form (10) Pl (e = 6) oe Ge ye (ae 9) 2 Sanyo © Eee yh een ay whence we have 2 2 2 2 i el Pa (ae ae es Seay, =9t A A 1 1 i ] h = : ‘ al la #210) eEbiao7 2-9 Transposing ! st Vis Rie a ec ae : x-10 x-7 «x-9 x-6 3 3 (a —10)(z7—7)° (%—9)(a—6) Hence, since the numerators are equal, the denominators must be equal ; that is, (a —10)(a-—7)=(a%-9)(x%-6), x? —17a+70=27-15%+543 TG es da ee “& 8. 148 ALGEBRA. EXAMPLES XXIII. a. Solve the following equations : ape! 9 ie Gabe 5e-9 42-10 © 62-17 42-13 7_3-4a 1 ee, 9 4-52 4, Fhe ees ae Bu-8_5x+14 F 8a-1_4a-3 pad “aToo °° 6x42 32-1 220-12 _ 4, 3e+7 9n- 22 38a-5_ 8a-5 4248 8. On 5 One] 8x-19 1_3e-4 10, 2#t2 1, 6e=1, 42-10 2 2x41 = 3(¢ 21) ee ook e-5 22] oe =1e 52 2 3x42 10 5 Bu-17 , 2¢-11_28_3e-7 13 -—4a 14 42 Ai 4x-—3 1-92 _4%+3 1 ald eh tee en gOd ea eee eae al xt+l “+2 %x4+3 3x2+6 Be aes 2 es zx-4 3x-18 47-16 xw-6 Liye eee ae “X+6 3a+12 2%+10 6(@+4) 2-1 #£-5_ “2-3 0-7 e-2 x-6 «x-4 “2-8 1 és ] 1 ] oO ee l7. 2 llete— 15: §x-64_ 4%-55_2x—-IT x- 6 2-18 @-14 2-6 -#=7 - [crrA. XXIII. ] HARDER EQUATIONS. 149 Solve the following equations : 99 5a+31_ 22+9_ 42-6, 2e-13 we t6 (#45 2-5 2-6 12¢+1 Bo bia 2a Sa-1 1-922 1432 ~ es | ev 9 x—3 a = hi eee 24. x*7-9 x-3 xL+3 28, [For additional examples see Hlementary Algebra. ] Literal Equations. 181, In the equations we have discussed hitherto the co- efficients have been numerical quantities. When equations involve literal coefficients, these are supposed to be known, and will appear in the solution. Haample 1. Solve (x+a)(x+6)-—c(a+c) =(x-—c)(x+c)+ab. Multiplying out, we have x*+axn+ba+ab-—ac—c?=27-c?+ab; whence ax +bx = ac, (a+b)x=ac; ac Tete : a+b 6b _a-b Hxample 2. Solve Simplifying the left side, we have a(x —b)—b(x-a)_a—b (e-a)(x—-b) 2aw-—c a 1 (w-a)(x—b) xw-c Multiplying across, x?-—cwv =2?-ax-bxtab, ax+ba—cx = ab, (at+b-c)u=ab; ab 150 ALGEBRA. Example 3. Solve the simultaneous equations : OL — DY A Cae, eset howto ea eee DEO =P Moe Eanes me ee To eliminate y, multiply (1) by g and (2) by 6; thus agqx — bay = cq, bpx + bay = br. By addition, (aq+bp)x =cq+br ; ey SE br aq +bp We might obtain y by substituting this value of x in ether of the equations (1) or (2); but y is more conveniently found by eliminat- ing x, as follows. Multiplying (1) by p and (2) by a, we have apx — bpy = cp, apx+aqy=ar. By subtraction, (aq+ bp)y = ar-cp; _ar-cp ay+ bp EXAMPLES XXIII, b, Solve the following equations : 10, x+(2-a)(x—b)+a?+b0%? =b+27-a(b-1). ae ek ae 2 ys. ps 2 2 vik Qu a_ 3x b_ 3a 8b 19. a-x b-2x a’ +b b a _ ab a—-b at ~ at oF ax—-b bxe-c_a-cx x+a-b_x+b-c 13. ¢ a a bai 14 xt+tb+e x+at+b ty a eee aN ey LO) nee q)*- pP(P a) + pa(Z 1) 0 1, av+b?=a?- ba. 9, 2*-a? = (Qa-x), 3. a@(a—x)+aba = b*(x—-b). 4, (b4+1)\(x%+a) =(b-1)(a-a) 5, a(«+b)-b?=a?-b(a-2), 6. ca Ct=ad2+c. 7, a(e—a)+b(a—b)+e(x-c) = 2(ab+be+ca). a? b? Qe) ee A es psy : = Fa eee, tbe aby Oxas i aa XXIII. ] HARDER EQUATIONS. Tol Solve the following simultaneous equations : 6. x-y=a4+), 17, ce-dy=+a, Toe exer by, y ax + by = 0. w+ y= 2c. UC, tes ab _ EN hah 19, pe ne 90. x y 0, yA a-y bv i Yn oe ¥-9, BELG 25) pon - ute a+b z_y_a,b etY _€-Y_¢, = Ss C= Ye PY So he a(a+2) = b(b-y) Dp ae 2 [on AGS 94 2a—-b_2yta_su+y 95 axtby_1_ a? i a b a+2b * betay 2 bx+ay Irrational or Surd Equations. 182. Derintrion. If the root of a quantity cannot be ex- actly obtained the indicated root is called a surd. Thus ./2, 9/5, /a3, Na?+0? are surds. A surd is sometimes called an irrational quantity; and quantities which are not surds are, for the sake of distinction, termed rational quantities, 183. Sometimes equations are proposed in which the un- known quantity appears under the radical sign. For a fuller discussion of surd equations the student may consult the e- mentary Algebra, Chap. xxxi1. Here we shall only consider a few simple cases, which can generally be solved by the follow- ing method. Bring to one side of the equation a single radical term by itself: on squaring both sides this radical will disappear. By repeating this process any remaining radicals can in turn be removed. Example 1. Solve 2 Je —n/4e-11 = 1. Transposing, 2/2 —-1=/4x—11. Square both sides; then 4xv-4,/x+1=42—11, 4,/x = 12, ee oy e='9: 152 ALGEBRA. [CHAP. XXIII. Example 2. Solve 24 Vxe—5 = 13. Transposing, Na-5=11. Here we must cube both sides; thus 2-5=1331; whence x = 1336. Hxample 3. Solve 6Je-ll_2Je+1 3,/x Multiplying across, we have (6,/%-11)(,/7+6) = Jxz+6- 3r/2(2J/a+1)3 that is, 62-11 ,/2+36/x - 66 = 6%4+38,/2, —11,/x+36,/x-3,/x = 66, Da Mace tits Vx =3; Me eee EXAMPLES XXIII, c, Solve the equations : L Neoe= 1. 0. NS — 2a = 7. 3. Ne=7 =e. 4, oJe+1=3. ieee fhe = ae 7 1-52 =3N 1-2. 8, 2N5x2-3-7/x=0. 9, N4a?— 112-7 = 2-3. 10. 38V1-7x+422 = 5-62. ll. 14+ V2°—-322+72—-ll=z, 12) Nee wee) 13, V4a+134+2,/a = 13. 14, 34+ V12z2—33 = 2A/32. 15 Je-1_J/¢-3 16 Net _ J/e+5 ie. ir ie ‘ 32-8 3\/2=7 ; ] 17 Mepey ae 19, 2v%=7-9, 1 Jxa-1 Wee Ja 14 4,/x-13 3 3 3 Sane 1 Ee poecae, ly —-3=—_——. 19. 1+474+2,/x AEs 90. Va+nNa-8 pe Q1. N4a+7-NaetrlsNe-3. 99, Nde-3-Nat+3= Ve-4. CHAPTER XXIV. HARDER PROBLEMS. 184, Is previous chapters we have given collections of problems which lead to simple equations. We add here a few examples of somewhat greater difficulty. Example 1. If the numerator of a fraction is increased by 2 and the denominator by 1, it becomes equal to 8; and if the numerator and denominator are each diminished by 1, it becomes equal to 4: find the fraction. Let a be the numerator of the fraction, y the denominator ; then - . xv She fraction is =. From the first supposition, H+2_5 yt ae) peor reerer eee eeeoeroee eeerreeoesses (Ly from the second, rd alt — = Fever sevescnes Peorereese ee evcoen 2, s ae (2) From the first equation, 8x#-d5y=-11, and from the second, 22—- y=1; whence x= 8, y=15. Thus the fraction is = Hxample 2. At what time between 4 and 5 o’clock will the minute-hand of a watch be 13 minutes in advance of the hour-hand ? Let x denote the required number of minutes after 4 o’clock ; then, as the minute-hand travels twelve times as fast as the hour- hand, the hour-hand will move over 5 minute-divisions in 2 minutes, a At 4 o’clock the minute-hand is 20 divisions behind the hour-hand, and finally the minute-hand is 13 divisions in advance; therefore the minute-hand moves over 20+13, or 33 divisions more than the hour- hand. 154 ALGEBRA. [crar. a Hence soSe + 33, 11 —— i 33 ; 12" Viera. Thus the time is 36 minutes past 4. If the question be asked as follows: ‘‘ At what times between 4 and 5 o’clock will there be 13 minutes between the two hands?” we must also take into consideration the case when the minute-hand is 13 divisions behind the hour-band. In this case the minute-hand gains 20 - 13, or 7 divisions. Hence x = wat 7; which gives Sehr i , Therefore the times are i past 4, and 36’ past 4. Example 3. A grocer buys 15 Ibs. of figs and 28 Ibs. of currants for $2.60; by selling the figs at a loss of 10 per cent., and the cur- rants at a gain of 30 per cent., he clears 30 cents on his outlay ; how much per pound did he pay for each ? Let 2, y denote the number of cents in the price of a pound of figs and currants respectively ; then the outlay is 154%-+28y cents. Therefore LODE 28 Y= 200 Fe eeitn se lawienaeatenes tee (3; The loss upon the figs is “6 x 15a cents, and the gain upon the currants is x 28y cents ; therefore the total gain is a) Te cents ; 5 2 pp eey 2h Tay EAD ; that is, BY = Daa 00g ag ete h en eee een (2). From (1) and (2) we find that «=8, and y=5; that is, the figs cost 8 cents a pound, and the currants cost 5 cents a pound. Example 4. Two persons A and B start simultaneously from two places, ¢ miles apart, and walk in the same direction. A travels at the rate of p miles an hour, and B at the rate of g miles ; how far will A have walked before he overtakes B ? XXIV, ] HARDER PROBLEMS. 155 Suppose A has walked x miles, then B has walked a —c miles. A walking at the rate of p miles an hour will travel 2 miles in x—-—C * hours; and B will travel a—c miles in hours; these two ) times being equal, we have a _m-¢ fo) ee Gx = px—pe 5 whence peed San Ped Therefore A has travelled ae miles. 9) coat Example 5. A train travelled a certain distance at a uniform rate. Had the speed been 6 miles an hour more, the journey would have occupied 4 hours less ; and had the speed been 6 miles an hour less, the journey would have occupied 6 hours more. Find the distance. Let the speed of the train be x miles per hour, and let the time occupied be y hours ; then the distance traversed will be represented by zy miles. On the first supposition the speed per hour is 7 +6 miles, and the time taken is y—4 hours. In this case the distance traversed will be represented by (x+6)(y- 4) miles. On the second supposition the distance traversed will be repre- sented by (x—6)(y+6) miles. All these expressions for the distance must be equal ; xy = («+6)(y —4) = (w@-6)(y+6). From these equations we have xy = xy + by — 4x - 24, or Oran aerlee aes itn. gnaw an tanhihrare eee: Cl ys and xy = xy — by + 6x — 36, or Goes Ope VOM ate ae seein en'n.s 8016 Riss anions (2). From (1) and (2) we obtain x = 30, y = 24. Hence the distance is 720 miles. EXAMPLES XXIV. 1, If the numerator of a fraction is increased by 5 it reduces to 4, and if the denominator is increased by 9 it reduces to 4: find the fraction. 156 ALGEBRA. [cHAP., 9, Find a fraction such that it reduces to 2 if 7 be subtracted from its denominator, and reduces to 2 on subtracting 3 from its numerator. 3. If unity is taken from the denominator of a fraction it reduces to 4; if 3 is added to the numerator it reduces to +: required the fraction. 4, Find a fraction which becomes # on adding 5 to the numerator and subtracting 1 from the denominator, and reduces to 4 on sub- tracting 4 from the numerator and adding 7 to the denominator. 5, If 9 is added to the numerator a certain fraction will be increased by 3; if 6 is taken from the denominator the fraction reduces to 2: required the fraction. 6. At what time between 9 and 10 o’clock are the hands of a watch together ? 7, When are the hands of a clock 8 minutes apart between the hours of 5 and 6? 8, At what time between 10 and 11 o’clock is the hour-hand six minutes ahead of the minute-hand ? 9, At what time between 1 and 2 o’clock are the hands of a watch in the same straight line ? 10, When are the hands of a clock at right angles between the hours of 5 and 6? 11, At what times between 12 and 1 o’clock are the hands of a watch at right angles ? 12, A person buys 20 yards of cloth and 25 yards of canvas for $35. By selling the cloth at a gain of 15 per cent. and the canvas at a gain of 20 per cent. he clears $5.75 ; find the price of each per yard. 13, ) Hind: the WH, ©. ot 323 —1le?+a%+4+15 and 5a4— 7x3 — 20a? - lla -3. 23, Express in the simplest form See he at 3 C1y ee eee e eee ee LEY Ee e-l1 «+1 x + y x 24, — u4+427’ (2) Bote! 1, 2+3 ates pa: 98, What value of a will make the product of 3-8a and 8a+4 equal to the product of G6a+11 and 3-4a? XV] MISCELLANEOUS EXAMPLES IV. 161 99, Find the L.C.M. of 2? -—2?-3x-—9 and 2° - 2z?-5x2—- 12. 30, A certain number of two digits is equal to seven times the sum of its digits: if the digit in the units’ place be decreased by two and that in the tens’ place by one, and if the number thus formed be divided by the sum of its digits, the quotient is 10. Find the number, 81, Find the value of 6a? — Bay — Gy? . 3x? — xy — 4 2, 9a" — Gay — Sy" 2x? +ay—y? Qa? — Bayt dy?” La —Bayt+y? 82. Resolve each of the following expressions into four factors : (1) 4a4—17a?b? + 404; (2) a8 —256y8. 383, Find the expression of highest dimensions which will divide 24a4b — 2a%b? — 9ab* and 18a° + a4b? —- 6a°b*? without remainder. 34, Find the square root of (I) w(a+1)(e+2)(4+3)+1; (2) (2a?+13a+15)(a?+ 4a —5)(2a2+ a - 3). 35, Simplify 36. A quantity of land, partly pasture and partly arable, is sold at the rate of $60 per acre for the pasture and $40 per acre for the arable, and the whole sum obtained is $10000. If the average price per acre were $50, the sum obtained would be 10 per cent. higher: find how much of the land is pasture, and how much arable. CHAPTER XXV. QUADRATIC EQUATIONS. 185, Derinirion. An equation which contains the square of the unknown quantity, but no higher power, is called a quad- ratic equation, or an equation of the second degree. If the equation contains both the square and the first power of the unknown it is called an affected quadratic; if it contains only the square of the unknown it is said to be a pure quadratic. Thus 2x?7—5xz=3 is an affected quadratic, and 5x2=20 is a pure quadratic. Pure Quadratic Equations. 186, A pure quadratic may be considered as a simple equa- tion in which the square of the unknown quantity is to be found. Example. Solve ——— Hiatal — ei. v?—-27 2-11 Multiplying across, 9x?—99=25x?— 675 ; Hak Veni Jaa yh ios ob 5 and taking the square root of these equals, we have L= +6. [In regard to the double sign see Art. 119. ] 187. In extracting the square root of the two sides of the equation z?=36, it might seem that we ought to prefix the double sign to the quantities on both sides, and write +z2=+6. But an examination of the various cases shows this to be un- necessary. For +x=+6 gives the four cases: t+z4=+4+6, +x=—-6, —x7=+6, —x=—6, and these are all included in the two already given, namely x=+6,2=—6. Hence, when we extract the square root of the two sides of an equation, it is sufficient to put the double sign before the square root of one side. CHAP, XXV. | QUADRATIC EQUATIONS. 163 Affected Quadratic Equations. 188. The equation z?=36 is an instance of the simplest form of quadratic equations. The equation (#—3)?=25 may be solved in a similar way; for taking the square root of both sides, we have two simple equations, Bao 5, Taking the upper sign, #2—3=+5, whence 7=8; taking the lower sign, x—3=—5, whence r= — 2. .. the solution is = 6,06 2. Now the given equation (7—3)?=25 may be written x* — 6x +(3)?=25, or x? —62=16. Hence, by retracing our steps, we learn that the equation w= 6216 can be solved by first adding (3)? or 9 to each side, and then extracting the square root; and the reason why we add 9 to each side is that this quantity added to the left side makes it a perfect square. Now whatever the quantity a may be, a+ 2ar+a°=(¢@+a)’, and uv —2ax+a?=(«4-ay; so that if a trinomial is a perfect square, and zits highest power, x*, has unity for its coefficient, we must always have the term without w equal to the square of half the coefficient of w. If, therefore, the terms in xz? and z are given, the square may be completed by adding the square of half the coefficient of vr. Example. Solve 2?+14x = 32. The square of half 14 is (7)?. that is, (fF 7 j= 81 ; 189, When an expression is a perfect square, the square terms are always positive. Hence, before completing the square the coefficient of x” should be made cqual to +1, 164 ALGEBRA. [orar. Hxamplel. Solve Fx =2?-8. Transpose so as to have the terms involving x on one side, and the square term positive. Thus ele = 8. Completing the square, 2? -—7a+ & = oe : that is, Ga ; 2 + 4 bis iin 2 pee), odio > Ze $ _ 8x45 38a+1 3x2+1- Example 2. Solve 4- Clearing of fractions, 12%+4-8 = 327+5; bringing the terms involving 2 to one side, we obtain 3x7 — 12% =-9. Divide throughout by 3; then x?-4e=-3; x? -444+(2)?=4-3; that is, (w= Que e-2=+1; Gis. OL ad EXAMPLES XXV. a. Solve the equations : Ll.) W(=7) = 627. 2. (24 8)(e—-8)=17. Be (i eal aye eee oe eerie ee an” eae Drea: 8, 2?+6a = 40. = Baa gale 4a 10) 927 £25. Lie? 1G =a: Loe Leia 18, 27+47=32. 14, 974+36=27. 15, 2«7+152-34=0. XXV.] QUADRATIC EQUATIONS. 165 Solve the equations : ] a9 J eb) 2 2 9 tA _ “+37 16, ee ee) Ne ae +3). Lis ane ei +3, %-2_ 5 Deaar Le near ge 18. e2+2 “2-3 (#+2)(x—3) 19. Qa +2 eri 2 Se eee 90, 5 aay x+2, 190. We have shown that the square may readily be com- pleted when the coefficient of #2 is unity. All cases may be reduced to this by dividing the equation throughout by the coefficient of 2°. Example 1. Solve 32-3827=10z. Transposing, 3x? + 10% = 32. Divide throughout by 3, so as to make the coefficient of x nity. oy ANU Sy TI Co . lus 3 op 3 5\2 OF Completing the square, 27 + Put 3 a4 ; ae oy 4h. that is, (« oe rg 5) 11 C+—= *-— 3 oS e a3 =2, or —5% Example 2. Solve 527+1llx=i2. Dividing by 5, eo AZ aae Completing thesquare, 4 1 Gar = + “i - that is, (2+ i = ae ; ete 166 ALGEBRA. [cIAP. 191. We see then that the following steps are required for solving an affected quadratic equation : (1) If necessary, simplify the equation so that the terms in x“ and x are on one side of the equation, and the term without x on the other. (2) Make the coefficient of x* unity and positive by dividing throughout by the coefficient of x”. (3) Add to each side of the equation the square of half the coefficient of x. (4) Take the square root of each side. (5) Solve the resulting simple equations. 192. When the coefficients are hteral the same method may be used. Hxample. Solve 7(#+2a)*+ 8a? = 5a(7x + 23a). Simplifying, 1x7 + 28ax + 28a? + 8a? = 35ax + 11da? ; that is, 7x? — Tax =84a?, or Tae Ce OO. Completing the square, 2*-ax+ OF = 12a + ; 2 2 that is, (« oe B = ae 3 C— x = +14 R 2 2 x=4a, or —3a 193, In all the instances considered hitherto the quadratic equations have had two roots. Sometimes, however, there is only one solution. Thus if #?—-227+1=0, then (7—1)’=0, whence w=1 is the only solution. Nevertheless, in this and similar cases we find it convenient to say that the quadratic has two equal roots. EXAMPLES XXV., b, Solve the equations : 1, 3a2+2a = 21. 9, 5a2=S8a42]. 8 6a2-x-1=0. 43-1 le =e"; He 2a = Dares. 6, 10+232+4 122? = 0. 7, 152?-6x = 9. §, 4a0°-17a = 15. Q, 8a?-197-15=0. XXV. J QUADRATIC EQUATIONS, 167 Solve the equations : HH) 1077 oe = 1. Pee Ta = 12: 12, 2027—2-—1=0. toe pean loa. 14° 29*—S8at= lbax. 15, 3a*=k(2k— 5a). nGseloee 20e ans 17, 9a" — 14h = box, 18. 2e72* = ar). 19, (w7-3)(~—-2) = 2(a7-4). 90, 5(%+1)(8%+4+5) = 3(82?+ lla +10). D1, 322+134+ (%-1)(2%4+1) = 2x(2x 4+ 3). Tx-3_ 3x 2 2-] 3a-1_2a-9 22, a O° 23, 375577 24, eto ei 4 62-5 2 x-4 1 11 == Oo. See Shed oot Se ec 25. z+5 3 26, je+1 2 2(3+ 22) 27, 3(2a+ 3)? + 2(2x% +3)(2- a) = (w- 2)", 98, (38a —7)? — (2a - 3)? = (w—4)(8a+1). [For additional examples see Elementary Algebra. | 194, Solution by Formula, From the preceding examples it appears that after suitable reduction and_ transposition every quadratic equation can be written in the form an +br+c=0, where a, 6, c may have any numerical values whatever. If therefore we can solve this quadratic we can solve any. Transposing, ax’ +be=—c¢; anon b c dividing by a, Cg, a a Complete the square by adding to each side cal ; thus 20 faa nee bee, a 2a Aa? a’ : b\2 6b%—4ae that is (« ey = s ; oe 4a? extracting the square root, + Ge 2a 2a bie Ot (Ot 4ac) Qa ; 168 ALGEBRA. [CHAP. —b+ ,/(b’ — 4ac) 2a 4 it must be remembered that the expression ,/()?—4ac) is the square root of the compound quantity b? —4ac, taken as a whole. We cannot sumphfy the solution unless we know the numerical values of a, b,c. It may sometimes happen that these values do not make b?—4ac a perfect square. In such a case the exact numerical solution of the equation cannot be determined. 195. Inthe result w= Hxample. Solve 52?-13x-11=0. Here a=5, b=-13, c=~—11; therefore by the formula we have (—13) + \(S19P—4 7) B11) oa 2.5 _13 + V169 +22) 10 _ 13+ /389 10 Since 389 has not an exact square root this result cannot be simplified ; thus the two roots are 13+ ,/389 13— ,/389 102s 10 196. Solution by Factors. There is still one method of obtaining the solution of a quadratic which will sometimes be found shorter than either of the methods already given. Consider the equation 2+ la = Clearing of fractions, $3274 72-60 assent -ceecesnessennaee CLs by resolving the left-hand side into factors we have (3x2 — 2)(a+3)=0. Now if either of the factors 32 —2, «+3 be zero, their product is zero. Hence the quadratic equation is satisfied by either of the suppositions a7 —-2=0, or r+3=0. > Ef) It appears from this that when a quadratic equation has been simplified and brought to the form of equation (1), its solution can always be readily obtained if the expression on the left-hand Thus the roots are XXv.] QUADRATIC EQUATIONS. 169 side can be resolved into factors. Hach of these factors equated to zero gives a simple equation, and a corresponding root of the quadratic. Example 1. Solve 2x°-ax+2bx = ab. Transposing, so as to have all the terms on one side of the equation, we have 2x7 —ax+2bxe—-ab=0, Now 22:7 — ax + 2ba -—ab = x(2% -a)+b(2x%-a) = (2%-a)(x+b). Therefore (Qe-a)\(v+l)=0; whence 22-—-a=0, or 7+b=0, ane, or —b. Y Example 2. Solve 2(x?-6) = 3(a”—-4). We have 2e2 \2 = oe = 12 > that is, CA laa Woe ee RAPE APT tee NORA (1). Transposing, 20% Bu =), x(2%—-3)=0 v= O,or 27 = 3 =0 Thus the roots are 0, 2 Note. In equation (1) above we might have divided both sides by x a] 3 . > oO . F, and obtained the simple equation 2a = 3, whence x = 5, which is one of the solutions of the given equation. But the student must be particularly careful to notice that whenever an x is removed by division from every term of an equation it must not be neglected, since the equation is satisfied by x=0, which is therefore one of the roots. 197, Formation of Equations with given roots, It is now easy to form an equation whose roots are known. Example 1. Form the equation whose roots are 4 and —3, Here : v= Ay or asia) * G=4'3Q, Of T+ a= 0 3 both of these statements are included in (x —4)(x+3) =0, or a*-x2-12=0, which is the required equation, 170 ALGEBRA. [CHAP. Example 2. Form the equation whose roots are a and - : : Here the equation is that is, or 320? (x —a)(8a+ 6) = 0, —8axr+ba-—ab=0. EXAMPLES XXV. c. Solve by formula the equations : a oi Uk —24-1=0. 3. x7 ore: Ae Oar =e eel 5, Qa?-9x = 4. 6. 327+7x2=6. 7 4a2—14 = 82. 8. 6a2-3-—Ta=0. 9, 1222410 = 23m. Solve by resolution into factors : 10.30 —92 =90; Teale] 1625 19) ar 8a = 1 13, 2_ 37 = 2, 14) 3224+52-+42=0"" 15, 427° ite 16.6 Ser Sle 2 0) 17 a2 — 0.0; 182° 2? = Tax =8e% 19, 122? — 2362 + 100? = 0. 90, 3ax?+2ba = Tax. 91, 2422+ 22cx = 21c?. 09. a — 22 44h = 2he, Solve the equations : 93, 2a(a+9) =(e%4+1)(5--z). 94, (2x-1)?-11 = 5x+(x-3)2. 95, G(x -—2)?+138(1 - x)(a - 2) + 6a? = 6(2x -1). = 4 3 2 2. aan 9, EASA = 26, x-6 x-5 27. Sx-1l x+1 2 10 oe +3 4(x~6)_ a a ese se ea Oo ae ie ] Gon Wines eee sat as a Sseeeig S13 Mead} Tee OG hs TE) e415 Sakon teal apres ee ao. 32, io ales wre Xxv.] QUADRATIC EQUATIONS. 171 Solve the equations : eee et" oot 34, Gry Vel Oa” 35. Bar’ — 4 aap Ce eee — 3 Pel D) u ei? Hi og oT el eA 3, SEM a G2 99, (p— abe + 2 = Wn +9. b P 5b c\? 2 aN a 40, Spans 41. i 1) (: i) ® [For additional examples see Hiementary Algebra. | 198. Simultaneous Quadratic Equations. If from either of two equations which involve w and y the value of one of the unknowns can be expressed in terms of the other, then by sub- stitution in the second equation we obtain a quadratic which may be solved by any one of the methods explained in this chapter. Example. Solve the simultaneous equations 5xt+Ty=1, 447+3axy —2y?=10. From the first equation, x = a , and therefore by substitution in the second equation, we have a(S ures ye 272=10; 25 whence 4 — 56y + 196y? + 15y — 105y? — 50y? = 250 ; that is, 4ly*—4ly —246=0 ; | y’-y-6=0; (y—8)(y+2)=05 y=3, or —2. From the first equation, we see that if y=3, then x= —4, and if y=—2, then x=. Homogeneous Equations of the Same Degree. 199, The most convenient method of solution is to substitute y=me in each of the given equations. By division we eliminate x and obtain a quadratic to determine the values of m. ve ALGEBRA. (CHAP. XXV. Example. Solve the simultaneous equations 527 + 3y?= 32, “x? -a2y +2y?=16. Put y= mez and substitute in each equation. Thus roms MaRS (Ed beciea ors Pare, a Pe SM eT A, oe | (qj and col 22) = LG eae eee ee 2). By division, Maweisil ayia 2 l1-m+2m* 16 that is, m* -2m-3=0; (m—3)(m+1)=0; = 3,.0F «2.1, (1) Take m=3 and Syustitnte in either (1) or (2). From (1), 32x? = 32: whence x= + 1. WA = a a es (2) Take m= —1 and substitute in (1). Thus 827=32; whence v= + 2. Y=MNE= Lf. 2. EXAMPLES XXV. d. Solve the simultaneous equations : 1, w+3y=9, 9, 38x-—4y=2, oe 2r+y=5, vy bs iy =e De — wy = 2. ee Te aaa 5 ot er vans, 6, 28-215 x? + 4y? = 29. 3xy — y= 9. mete Lb co on ae ies) ai pot hs aa 1; 8, Batry fs ’ 9, x oe xy = 24 10zy=1 xy —yr=4 eel eat Pees be ieee . — A Lie Dates itm 10.5 5 Le ong ae ; 3.Y ee) seein Prv=] peas a | — = 8, 22 ee ey oe 10; 327-79? 255... 14, léry—-Se*= 77, 15. a2 ey a 2c? ey OU. Tay + 3y? = 110. Say + 3y7 = 195. 16, 27+ 2ay + 2y? = 17, 17, 2la?+3ay - y? =37)1, 3a? — Oxy — y= 119. 5a? + day + Sy? = 265. CHAPTER XXXVI. PROBLEMS LEADING TO QUADRATIC EQUATIONS. 200. Wr shall now discuss some problems which give rise to quadratic equations. Hxample 1. A train travels 300 miles at a uniform rate; if the speed had been 5 miles an hour more, the journey would have taken two hours less: find the rate of the train. Suppose the train travels at the rate of x miles per hour, then the : See ay time occupied is —— hours. i ry On the other supposition the time is a5 hours ; 5 300 _300_o. Se Ce ee ’ ce oe ie whence e+ 5a —750 = 0, or (7+ 30)(% — 25) = 0, * oa = 25,06) — 30. Hence the train travels 25 miles per hour, the negative value being inadmissible, {For an explanation of the meaning of the negative value see Elementary Algebra. | EHxrample 2. Aman buys a number of articles for $2.40, and sells for $2.52 all but two at 2 cents apiece more than they cost; how many did he buy ? Let « be the number of articles bought; then the cost price of each is ae cents, and the sale price is 5 cents. HY Spas 252 240_.. fo ee f that is, Oe me Eas 174 ALGEBRA. [cHAP. After simplification, 6x +240 = a? - 2a, or x? —8x2-240=0; that is, (7 —20)(4+12)=0; Lia UP Orie, Thus the number required is 20. Example 3. A cistern can be filled by two pipes in 334 minutes ; if the larger pipe takes 15 minutes less than the smaller to fill the cistern, find in what time it will be filled by each pipe singly. Suppose that the two pipes running singly would fill the cistern in x and «-15 minutes; then they will fill ba i yet 5 of the cistern 2 a respectively in one minute, and therefore when running together they will fill (+ i of the cistern in one minute. But they fill a , or a of the cistern in one minute. 1 1 3 H = eee niet a fen anOG 100(2% - 15) = 38x(x—- 15), 3a? — 245x + 1500 =0, (a —75)(3a —20)=0; x ='75, or 62. Thus the smaller pipe takes 75 minutes, the larger 60 minutes, The other solution 62 is inadmissible. 901. Sometimes it will be found convenient to use more than one unknown. Example. Nine times the side of one square exceeds the peri- meter of a second square by one foot, and six times the area of the second square exceeds twenty-nine times the area of the first by one square foot; find the length of a side of each square. Let x feet and y feet represent the sides of the two squares ; then the perimeter of the second square is 4y feet ; thus 9x. 4y-= 1, The areas of the two squares are x” and y” square feet; thus 6y? - 2927 = 1, XXVI.] PROBLEMS LEADING TO QUADRATIC EQUATIONS. 175 From the first equation, Ye a _ By substitution in the second equation, Se Woe oon 1 : 8 that is, llz?-—547%-5=0, or (2—d5)(llz+1)=0; whence x =5, the negative value being inadmissible. Also, y = a L STi Thus the lengths are 5 ft. and 11 ft. EXAMPLES XXVI, ], Find a number which is less than its square by 72. 9. Divide 16 into two parts such that the sum of their squares is 130, 3, Find two numbers differing by 5 such that the sum of their squares is equal to 233. 4, Find a number which when increased by 13 is 68 times the reciprocal of the number. 5, Find two numbers differing by 7 such that their product is 330. 6, The breadth of a rectangle is five yards shorter than the length, and the area is 374 square yards: find the sides. 7, One side of a rectangle is 7 yards longer than the other, and its diagonal is 13 yards ; find the area. 8, Find two consecutive numbers the difference of whose reci- procals is sz. 9, Find two consecutive even numbers the difference of whose reciprocals is <3. 10. The difference of the reciprocals of two consecutive odd numbers is 725: find them. 11, A farmer bought a certain number of sheep for $315; through disease he lost 10, but by selling the remainder at 75 cents each more than he gave for them, he gained $75: how many did he buy ? 12, By walking three-quarters of a mile more than his ordinary pace per hour, a man finds that he takes 1} hours less than usual to walk 293 miles; what is the ordinary rate ? 176 ALGEBRA. (CHAP. XXVI. 13, + 16p*- 33p°+14p? by p?+7p. 3, Find the sum of a-2(b-3c), 3{a-2(b+c)}, 2{b -2(a —2b)}. 4, Simplify by removing brackets 7[8a-4{a-—b+3(a+b)}}. 5, Solve the equations : x+4,«a-4 (1) agentes =4; 6. A is three times as old as B; two years ago he was five times as old as B was four years ago : what is A’s age ? “i 7. Find the product of 2a -3b-(a-—2b-—c) and b-2c-—(a-c). Gide] b= 0, ¢=-11) do =2, ei 2) find the value of +4+0+E+4+0R—-8+a24+l2+C°+d?-e. 9, Remove brackets from the expressions : (1) a -[5b-{a—(3c -3b)+2c- (a - eel (2) Qa - 3{b-4(c - d)}]-[a—4fb - 6(c--d)}]. 10, If the price of 5 acres of land is $a, what is the price of x acres ? and how many acres can be bought for $b ? 11, Divide at—4 by a?—2a42. 12, There are 150 coins in a bag which are either half-dollars or quarters. If the value of the coins is $58.50, find the number of each kind. H.A. M 178 ALGEBRA. 13. Add together a—{b+c-(a+b)!+c, 2(3a+2b) -4(b+2a)-c, and 3(2b — a) -2(3b-a)+e. 14, Find what value of x will make the product of +3 and 2%+3 exceed the product of x+1 and 2x+1 by 14. 15, Divide b?+8-125c?+30bc by b-5c+2. 16, *Simplify (1) 13ab*c? ‘ 87c°d? , 26bed , 29c4d ~ 2a2bt * 4ab3 ’ 202 ee a®bx , atb%a4 ) “b2 \2Qab 2c? Ga2b — Qakb4x3/° 17, How old will a man be in m years who n years ago was p times as old as his son then aged 2 years ? (2) 18, I bought a certain number of pears at three for a cent, and two-thirds of that number at four for a cent; by selling them at twenty-five for 12 cents, I gained 18 cents, How many pears did I buy ? 19, Solve the equations: a a _ 4%42 74+14 (1) — alee e =5—6%+ 9) X+Y 8z—5Y _o ada bate a ey az : cAlce 90. Divide a2xa§+ (2ac—b?)x#+c? by axt+c—b2?. 91, Ifa horses are worth b cows, and c cows are worth d sheep, find the value of a horse when a sheep is worth $2. 99, Find the highest common factor of 8a7b3c, 12a%bc?, 15a2b5 ; and the lowest common multiple of 4ab2c3, 12a%b, 18ac?. aah oy IPOS MUS 2ab2c 362 b2d =~ 6. a2? 93, A gentleman divided $49 amongst 150 children. Each girl had 50 cents, and each boy 25 cents. How many boys were there ? 94, Ii V=5a+4b—-6c, X=—8a—9b+7c, Y=20a+7b—85c, Z=18a—5b+4+9c, calculate the value of V—(X+ Y)+Z. Also find the value o MISCELLANEOUS EXAMPLES V, 179 95, Solve the equations : l 3(6 — 5x) 63a _ 3a _ 3G () a) a KE Oy Rae 2) plety)=e+1, Hy-n)=20-1. 96, Find the factors of (1) a?-a-182; (2) 8a?+13x-6. Oi 4 eye AN ey, ye the value of Ni{5(y? — 22) — ah + V/Bfa(a? — 2?) — 1h. 98, The product of two expressions is («+ 2y)?+ (3%+z)°, and one of them is 4%+2y+2; find the other. 99, When 4 and B sit down to play, B has two-thirds as much money as A ; after a time A wins $15, and then he has twice as much money as 6. How much had each at first? 80, Find the square root of 16a°+4a+4 -16a?+a?-8at. | 81, Find the value of | 1 ae Teed pte 2a {2 p(e3)} {a —3le42)} (e-F(2-5)}, and subtract the result from (a+ 2)(%—38)(~+4). 382, Find the square root of a4 293 lla? 9 Gs. PAG eG BO ela ciieiaD =.c= es = 1, find the value of arc _ nt Jatd — Vaid + o me a 34, Separate into he simplest factors : (1) 2?—xy—-6y’; (2) 2° —4xy?- ay +4y%. 85, Solve the equations : (1) (a-—1)(a%-2)(%-6) = (w- 3); ~ ome: (2) 2 5 3 2a! By =0, ms ys 13. 36, A farmer ere to one person 9 horses and 7 cows for $375, and to another 6 horses and 13 cows at the same prices and for the same sum: what was the price of each ? 180 ALGEBRA. 37, When a=3, b=2, c=—7, find the value of (1) 3Bb%c | Bbc? _ Ga? ; a a-c b+2c¢ (2) 4c+{c-(3¢-—2b) + 2b}. 88, Solve the equations : 1,1 1 (2) 542+3y = 120, 10.c = Sy + 90. 39, Find the highest common factor of 5a°+2z?-—15a2-—6 and 7x? — 447 — 21a +12. 40, A coach travels between two places in 5 hours; if its speed were increased by 3 miles an hour, it would take 34 hours for the journey : what is the distance between the places ? 41, From 2(a+a-b)(c-—a+b) take (%-a)(x%—-b)(vt+a+tDd). 42, Find the value of 2a*+5a~3 3 3a? —102+3, 62? -—52+1 a3 — 9x 2+ 3042 © Bae +7a%+2a 43, Divide 2°+y?+3xy-1 by x+y-—-1, and extract the square root of «4 - 3.° +5 4 204% 44, Aman can walk from A to B and back in a certain time at the rate of 4 miles an hour. If he walks at the rate of 3 miles an hour from A to B, and at the rate of 5 miles an hour from B to A, he requires 10 minutes longer for the double journey. What is the distance from A to B? 45, Find the highest common factor of 7x4 — l0ax® + 8a72? - 4da®x+4at, 8rt— 13az3+ 5a2x? - 3a%a + 3a'. 46. Solve the equations : (2) w-2y4+z2=0,° 9%-8y+32=0, 22+3y+5z2 = 36. 47, Find the lowest common multiple of 6a?-a2-1, 3a7+7x4+2, 227+32—-2. MISCELLANEOUS EXAMPLES V., 181 48, The expression ax+3b is equal to 30 when 2 is 3, and to 42 when 2 is 7: what is its value when 2 is 1; and for what value of x is it equal to zero? 49, Find the lowest common multiple of 4(a?+ab), 12(ab?-b*), 18(a?—b?). 50, Extract the square root of 4a? _ 128, on | 24a , 16a? da 8G: x Figs 51, Reduce to lowest terms 1224 + 423 — 23x? — 9a — 9 8x4 — 1427-9 52. Solve the equations : OU CY ett OY Qu —Ty 3a —7 1 a et se Fh a ee PP ames | Vireo : Gy 2— Fer em eg y 6 ial ae (2) 2e-y+32=1, 4¢+3y-2z2=13, 6x-4y+2=20. 53. Simplify (1) GHD 2a 8 obo b a+b ab—b?’ ee 1 24+ 8a+15 x?+1le+30 54, The sum of the two digits of a number is 9 ; if the digits are reversed the new number is four-sevenths of what it was before. Find the number. (2) 55, Solve the equations : (1) 4ar- (5y-4)=1, es 1-1, (2) Sx+dy-11l=0, 5y—-6z=-—8, 7z2-8x%-—13=0. 56, Find the value of : 2 ] 3x a atx a-x x -a@ (a+a)* 57, Resolve into factors : A) 28 -2a* a; (2) af+at—ai—l1. 182 ALGEBRA. 58, Two persons started at the same time to go from A to B. One rode at the rate of 74 miles per hour and arrived half an hour later than the other who travelled by train at the rate of 30 miles per hour. What is the distance between A and B? 59, Find the square root of Ae Loge ays 4 Gay 16a? 9y2 2 l5yz 1622 52? 2527" 60. Find the factor of highest dimensions which will exactly divide each of the expressions 2c4+ c8d — c2d? --7cd?—4d4, 3c4+ c8d - 2c?d* — 9cd? — 5d4. : : 2 3 — Qn? Simplif Tye et ee ee G1, Simplify (1) —;| Fail Mee DACiie aler) 62, Find the highest common factor of 624 — 2a3+927+9"—4 and 924+ 80x? — 9. What value of x will make both these expressions vanish ? 63. Solve the equations : 9 293 (1) x-3 x-6 2¢ Se 1 ba 15> yi ~ SAN (2) rE NE eS x=2 e 9 6x? — Say —6y? — 15x? + 8ay — 121? 4x? — Bay +3y? 35a?+47xy + by” 64, Simplify i 65, Find the value of x-2a_x2+2a Il6ab Shane e+2b 2-2b 462-2? .. 4ab a+b 66, An egg-dealer bought a certain number of eggs at 16 cents per score, and five times the number at 75 cents per hundred; he sold the whole at 10 cents per dozen, gaining $3.24 by the transac- tion. How many eggs did he buy ? MISCELLANEOUS EXAMPLES V. 183 67, If a=-1, b=-2, c=-3, d=—4, find the value of 2a? +ab?—8abe_a-b+c-d_a ” Ze a? — b?-—abc be — 2ad brad 68, Solve the simultaneous equations : Dee TY ee al eZ <=) > 2 14 8 4 ey YO 5(y +1) MAA ite aE sau 710 Pe ay 69, Simplify the fractions: (iy cae ey) x-%Z x-Yy (%-x)(y- 2) wig ut el 5 B a b a eee) ee b a/\b a a2 hb? ab 70, From a certain sum of money one-third part was taken and From the sum thus increased one-fourth part $50 put in its stead. If the amount was now $120, was taken and $70 put in its stead. find the original sum. 71, Find the lowest common multiple of (a4 —a%c?)?, 4a®-Sate?+4a2ct, a+ 3a3c + 3a%c?+ ac’. 72, Solve the equations: ; ‘135% —°225_ 36 -09%-°18 , (1) *l5a+ 03 es, z= 9 ; ll - 5a, da —-33 6 l0x+4, 7-20? _ (2) SAY EVER S as 73, Simplify x?—44¢—-21 y C+ Ga? -247a , x? - 202491 ae era e+17 x? —a2-12 74, What must be the value of x in order that (a+2a)? a? + 70ax + 3.x? may be equal to 14 when a is equal to 67 ? 184 ALGEBRA. 75, Find the highest common factor of 1674+36a?+81 and 8x? +27; and find the lowest common multiple of 82°+27, 16a*+ 3627+81, and 62?-5x-6. 76, Solve the equations : (1) a(a—a)-—b(a-—b)=(a+b)(x-a-—b). (2) (a+b)z--ay = a?, (a? + b*)x -— aby = a’. 77. A farmer bought a certain number of sheep for $30. He sold all but five of them for $27, and made a profit of 20 per cent. on those he sold: find how many he bought. 78, Find the value of ey vty 2a-8b 3b (1) oe eA ee eth gh (2) 2Qa—6b 2a a?—y? atty?? ~" 2a=3b 3b ory? 72 — 2a 2a —6b 79, Find the H.C.F. of 212? - 26a?+ 8a and 6a723 — a7a? — 2a°x. Also the L.C.M. of z?-2, ax?+2a2—-3a, x?-—72?+6z2. 80, Simplify the expression (a+b+c)(a-b+c)-{(atcP—-b— (a+b +c")}. 81, Solve the equations : T+u QWw-y 5Sy-7 , 4a-3 < ee nD : sis oF; (1) 5 p 3y — 9, 5 “b 6 Ov (2) Se hie anh) 3u—-2Z x2+1 2243 82, Extract the square root of aes 3003 ey: #4 V2 +4) 4+9(2 +4) ~4(Z 42) 45. Ope ee pe 3 ae JP 83, Find the value of 4a+6b | 6a-4b _4a°+65? , 4b?- 6a? , 200! a+b a—b a? — b? a+b? at—bt 84, A bag contains 180 gold and silver coins of the value alto- gether of $144. Each gold coin is worth as many cents as there are silver coins, and each silver coin as many cents as there are gold coins. low many coins are there of each kind ? MISCELLANEOUS EXAMPLES V. 185 85. Solve the equations: ibe ob Soaiom ke glo a+4 247? (2) V22+6-Nx—1=2. 86, Find the factors of (1) 20a?+2lab-27b?; (2) a3-3z?--9x+27. 87, If the length of a field were diminished and its breadth increased by 12 yards, it would be square. If its length were increased and its breadth diminished by 12 yards, its area would be 15049 square yards. Find the area of the field. 68, Simplify the expression {(a+b)(at+b+c)+c7{(a+b)? - 7} {(a+bP-C}{at+b+ey 89, Find the square root of (2x + 1)(2a + 3)(2% + 5)(2%+7) + 16. 90, Simplify 10a? paket fe 2 (1+a?)(l-4a2) 1-2a 1+a?™ 91, Solve the equations: liye PER SNE ae ine ain me oe Oh ye mae ee 99, Resolve into factors: (1) 622+52-6. (2) 9at-82ax°y?+9y4. a4 — 154? 4+- 28a - 12 Red ea Ema Be to its lowest terms. 94, Simplify the fraction (a+b)? a+ 2b+a , (a+ b)z ak (a-a)(xtat+b) Wwe-a) w#+be-a?-ah 2 186 ALGEBRA. 95, A person being asked his age replied, ‘‘Ten years ago I was five times as old as my son, but twenty years hence he will be half my age.” What is his age? 96, Find the value of { 2a, Lee ee eee (a-a)(ate) a-a wtasy az Oy. When ia yoo ee g d=-1, find the value of a’ — 03 —(a—b)> - 11(80 +20)(2e-F). a“ 98, Find the square root of a* + b4 — a®b — ab? + 9a" 99, Solve the equations: (1) 3x 24 = =92- 9 2 = 493. a OeeT 4 (2) 327+ 227 = 493 LOO DEY Negi onan ined 101, What value of x will make the sum of pear and Hane equal to 2? 102, Aman drives toacertain place at the rate of 8 miles an hour ; returning by a road 3 miles longer at the rate of 9 miles an hour he takes 74 minutes longer than in going: how long is each road ? 103, Find the product of (1) 32°?-4ay+ Ty’, 3a*+4ay+ Ty? ; (2) #?-2y7, 2?-Qay+2Qy?, x?+2y?, 2? +2ry+2y?. 104, Extract the square root of 309 1-50? 4-208 + Sat - S054 a MISCELLANEOUS EXAMPLES V. 187 105, Find the highest factor common to x(62? — 8y?) — y(3a?-4y?) and Qay(2y —x) +423 — 2y3, 106, The sum of the digits of a number is 9, and if five times the digit in the tens’ place be added to twice the digit in the units’ place, the number will be inverted. What is the number? 2 the yeeaaes (a+1) = 3, prove that a® neo. (eR? 108, Solve the equations : (1) (x+7)(y-3)+2-(y+8)(x-1) =5x-1ly+35=0; (2) pastes) = 3 64-32 3(x-2) 109, If x=b+c, y=c-a, z=a-b), find the value of x+y? +27 — Qay — Quea+ Qyz. 110, Express in the simplest form en Leer Pee tee . a 02+ 3e+1 6a2+bet+1] 1222+7x24+1 2022+9xe41 11]. Resolve into factors : (1) 4a7b? — (a2 +b? -c?)*; (2) ab(m?+1)+m(a? +b”). 112, Simplify the fractions : ] a-x ——— at Nee (1) —: @) — ete CH= the 14.21 ie 3-2 x 1138, Solve the equations : 75-2 , 802+21_ 23 1 : = 55 (1) 3(e41) 5(8a42) £41 (2) Ve+12— Ja =6. 188 ALGEBRA. 114, Ten minutes after the departure of an express train a slow train is started, travelling on an average 20 miles less per hour, which reaches a station 250 miles distant 34 hours after the arrival of the express. Find the rate at which each train travels. 115, Simplify a’ — 64a , [ 2a27+5a4+2. ee a*—4 °2a?+9a+4° \a?+4a° a?+a—2 ; 116. Resolve into four factors 4(ab + cd)? — (a? +b? - c? - d’)?. LL When Oa 4 OH ec :, d=-1, find the numerical value 4c? — ala — 2b — d) —‘Vb4e + 11b3a?. 118, Find the value of 119, Solve the equations : 120, A has 19 miles to walk. At the end of a quarter of an hour he is overtaken by B who walks half a mile per hour faster ; by walking at the same rate as & for the remainder of the journey he arrives half an hour sooner than he expected. Find how long the journey occupied each man. IN PREPARATION. AMERICAN EDITION OF ALGEBRA FOR BEGINNERS. By H. S. HALL, M.A., and S. R. KNIGHT. 10mo. Cloth. 60 cents. REVISED BY FRANK L. SEVENOAK, A.M., M.D., Assistant Principal, and Professor of Mathematics and Natural Sciences, in the Stevens School, Hoboken, N.J. The Algebras by Messrs. Hall and Knight have been introduced in many Colleges and Schools, from among which may be mentioned: Brown University. State Normal School, Ypsilanti, Stanford University. Mich. Northwestern University. Stevens Institute of Technology. Dalhousie University. Cascadilla School, Ithaca, N.Y. Vassar College. Thayer Academy, Braintree, Mass. Illinois College. St. Paul’s School, Concord, N.H. U. S. Naval Academy, Annapolis. Etc., etc., etc. MACMIELE ANS -CO., 66 FIFTH AVENUE, NEW YORK. I MATHEMATICAL TEXT-BOOKS SUITABLE FOR USE IN PREPARATORY SCHOOLS. SELECTED FROM THE LISTS OF MACMILLAN & CQO., Publishers. ARITHMETIC FOR SCHOOLS. By J. B. LOCK, Author of “ Trigonometry for Beginners,” “ Elementary Trigonometry,” etc Edited and Arranged for American Schools By CHARLOTTE ANGAS SCOTT, D.SC., Head of Math. Dept., Bryn Mawr College, Pa 16mo. Cloth. 75 cents. “* Evidently the work of a thoroughly good teacher. The elementary truth, that arithmetic is common sense, is the principle which pervades the whole book, and no process, however simple, is deemed unworthy of clear explanation. Where it seems advantageous, a rule is given after the explanation. ... Mr. Lock’s admirable ‘Trigonometry ’ and the present work are, to our mind, models of what mathematical school books should be.” — The Literary World. FOR MORE ADVANCED CLASSES. ARITHMETIC. By CHARLES SMITH, M.A., Author of “ Elementary Algebra,” “A Treatise on Algebra, AND CHARLES L. HARRINGTON, M.A., Head Master of Dr. $. Sach’s School for Boys, New York. 16mo. Cloth. 90 cents. A thorough and comprehensive High School Arithmetic, containing many good examples and clear, well-arranged explanations. There are chapters on Stocks and Bonds, and on Exchange, which are of more than ordinary value, and there is also a useful collection of miscellaneous examples. bby MACMILLAN & CO., 66-FIP TH VAVEN UE. NEW sy ORK. INTRODUCTORY MODERN GEOMETRY OF THE POINT, RAY, AND CIRCLE. By WILLIAM B. SMITH, Ph.D., Professor of Mathematics in the Tulane University of New Orleans, La. Cloth. $1.10. ““To the many of my fellow-teachers in America who have questioned me in regard to the Non-Euclidean Geometry, I would now wish to say publicly that Dr. Smith’s conception of that profound advance in pure science is entirely sound. . ._ Dr, Smith has given us a book of which our country can be proud. think it the duty of every teacher of geometry to examine it carefully.’’— From Prof. xEORGE Bruce HALstTED, Ph. D. (Johns Hopkins), Se ‘of Mathematics, University of Texas. “‘T cannot see any cogent reason for not introducing the methods of Modern Geometry in text-books intended for first years of a college course. How useful! and instructive these methods are, is clearly brought to view in Dr. Smith’s admi- rable treatise. This treatise is in the right direction, and is one step in advancing a doctrine which is destined to reconstruct in great measure the whole edifice of Geometry. I shall make provision for it in the advanced class in this school next term.” — From Principal Joun M. Cotaw, A.M., Monterey, Va. MODERN PLANE GEOMETRY. Being the Proofs of the Theorems in the Syllabus of Modern Plane Geometry issued by the Association for the Improvement of Geometrical Teaching. By G. RICHARDSON, M.A., and A. S. RAMSAY, M.A. Cloth. $1.00. ‘‘Tntended to be an Introduction to the subject of Modern Plane Geometry and to the more advanced books of Cremona and others. It has a twofold object: to serve, in the first place, as a sequel to Euclid . . .; and, secondly, as a systematic means of procedure from Euclidean Geometry to the higher descriptive Geometry of Conics and of imaginary points.”’ MACMILLAN & CO. 667P TE TH eAVENU NEW ) YORK, TEXT-BOOK OF EUCLID’S ELEMENTS. Including Alternative Proofs, together with Additional Theorems and Exercises, classified and arranged By H. S. HALL and F. H. STEVENS. Books I.-VI. and XI. $1.10. Also sold separately as follows - Book J. 3% Moai 30 cents. | Books III.-VI. . . . 175 cents. Books I.andIIl.. . 50 cents. | Books V., VI., and XI. 70 cents. BooksI.-IV. .. . 75 cents. | Book XI. .... . 380cents. ‘The chief peculiarity of Messrs. Hall and Stevens’ edition is the extent and variety of the additions. After each important proposition a large number of exer- cises are given, and at the end of each book additional exercises, theorems, notes, ete.; ete., well selected, often ingenious and interesting. ... There are a ereat number of minute details about the construction of this edition and its mechanical execution which we have no space to mention, but all showing the care, the patience, and the labor which have been bestowed upon it. On the whole, we think it the most usable edition of Euclid that has yet appeared.’’ — The Nation. THE ELEMENTS OF SOLID GEOMETRY. By ROBERT BALDWIN HAYWARD, M.A., F.R.S., Senior Mathematical Master in Harrow School; Late President of the Association for the Improvement of Geometrical Teaching. 16mo, Cloth. 75 cents. ‘*A modification and extension of the first twenty-one propositions of the eleventh book of Euclid, developed out of a Syllabus of Solid Geometry submitted by the author to a Committee of the Association for the Improvement of Geometricai Prenerre S and reported upon by that Committee with a considerable degree or avor.”’ MACMILLAN & CO., 66. -PIPITH AVENUE NE Way Ornk: ELEMENTS OF SYNTHETIC SOLID GEOMETRY. By NATHAN F. DUPUIS, M.A., F.R.S.C., Professor of Pure Mathematics in the University of Queen’s College, Kingston, Canada. 16mo. pp. 239. $1.60. FROM THE AUTHOR’S PREFACE. ‘*T have been induced to present the work to the public, partly by receiving from a number of Educationists inquiries as to what work on Solid Geometry I would recommend as a sequel to my Plane Geometry, and partly from the high estimate that I have formed of the value of the study of synthetic solid geometry as a means of mental discipline... . ‘‘In this work the subject is carried somewhat farther than is customary in those works in which the subject of solid geometry is appended to that of plane geometry, but the extensions thus made are fairly within the scope of an elemen- tary work and are highly interesting and important in themselves as forming valu- able aids to the right understanding of the more transcendental methods.” Introductory to the Above. ELEMENTARY SYNTHETIC GEOMETRY OF THE Point, Line, and Circle in the Plane. 16mo. $1.10. ““To this valuable work we previously directed special attention. The whole in- tention of the work has been to prepare the student to take up successfully the - modern works on analytical geometry. It is safe to say that a student will learn more of the science from this book in one year than he can learn from the old- fashioned translations of a certain ancient Greek treatise in two years. Every mathematical master should study this book in order to learn the logical method of presenting the subject to beginners.’’ — Canada Educational Journal. MACMILLAN & CO.,, S66 pRI RTE CANE NUE NEV YORK: N In filling orders for the American Edition of Smith's Ele- wane Mentary Algebra, the Briefer Edition will be sent wherever the Complete Edition is not distinctly ordered. A pamphiet con- taining the answers will be supplied free, but only upon the written order of the teacher for whose classes they are required. AMERICAN EDITION OF Charles Smith's Elementary Algebra. FOR THE USE OF PREPARATORY SCHOOLS, HIGH SCHOOLS, ACADEMIES, SEMINARIES, Etc. BY IRVING STRINGHAM, PH.D., PROFESSOR OF MATHEMATICS, AND DEAN OF THE COLLEGE FACULTIES IN THE UNIVERSITY OF CALIFORNIA. BRIEFER EDITION (408 pages) : 4 C - - $1.10 This edition is the same as Oe I.-XXVI. of the COMPLETE EDITION (584 pages) . A 4 : ‘ 5 5 ole “T have always liked Charles Smith’s Alge- bra, and the new edition contains a good many improvements, and seems to me an excellent work. The use of the book in schools prepar- ing for Harvard College would be satisfactory to our Mathematical Department. I have already privately recommended it to teachers who have consulted me.’’ — Prof. W. E. BYERLY, Harvard College. MACMLEEAWN Sa eCOz, 66 FIFTH AVENUE - - - - - NEW YORK. PelneotloleON ALGEBRA: By CHARLES SMITH, M.A. Cloth. $1.90. No stronger commendation of this work is needed than the fact that it is the text used in a large number, if not in the majority, of the leading colleges of the country, among which may be mentioned Harvard University, Cornell University, University of Ohio, of Pennsylvania, of Michigan, of Wisconsin, of Kansas, of California, of Missouri, Stanford University, etc., ete. ‘““Those acquainted with Mr. Smith’s text-books on conic sections and solid geometry will form a high expectation of this work, and we do not think they will be disappointed. Its style is clear and neat, it gives alternative proofs of most of the fundamental theorems, and abounds in practical hints, among which we may notice those on the resolution of expressions into factors and the recognition of a series as a binominal expansion.’ — Oxford Review. HIGHER ALGEBRA FOR SCHOOLS. By H. 8. HALL, B.A., and S: R. KNIGHT, B.A. Cloth. $1,90. ‘“The ‘Elementary Algebra,’ by the same authors, which has already reached a sixth edition, is a work of such exceptional merit that those acquainted with it will form high expectations of the sequel to it nowissued. Nor will they be disappointed. Of the authors’ ‘ Higher Algebra,’ as of their ‘ Elementary Algebra,’ we un- hesitatingly assert that it is by far the best work of the kind with which we are acquainted. It supplies a want much felt by teachers.’’— The Athenwum. MACMILLAN & CO., GORE PELE Asay ENG Ey ra VVeey OR IK, ELEMENTARY TRIGONOMETRY BY H. S. HALL, B.A., and S. R. KNIGHT, B.A. ” Authors of ‘Algebra for Beginners,’ “Elementary Algebra for Schools,” ete. Cloth. $1.10. “I consider the work as a remarkably clean and clear presentation of the principles of Plane Trigonometry. For the beginner, it is a book that will lead him step by step to grasp its subject matter in a most satisfactory manner.” — E. MILLER, Unzversity of Kansas. “The book is an excellent one. The treatment of the fundamental relations of angles and their functions is clear and easy, the arrangement of the topics such as cannot but commend itself to the experienced teacher. It is, more than any other work on the subject that I just now recall, one which should, I think, give pleasure to the student.” — JoHN J. SCHOBINGER, 7he Harvard Schood. WORKS BY REV. J. B. LOCK. TRIGONOMETRY FOR BEGINNERS. AS FAR AS THE SOLUTION OF TRIANGLES, 16mo. 75cents. ‘(A very concise and complete little treatise on this somewhat difficult subject for boys; not too childishly simple in its explanations; an incentive to thinking, not a substitute for it. The schoolboy is encouraged, not insulted. The illustrations are clear. Abundant examples are given at every stage, with answers at the end of the book, the general correctness of which we have taken pains to prove. The definitions are good, the arrangement of the work clear and easy, the book itself well printed.” — Fournal of Education. ELEMENTARY TRIGONOMETRY. 6th edition. (In this edition the chapter on Logarithms has been carefully revised.) NGMO.) 51.10. ‘“‘ The work contains a very large collection of good (and not too hard) examples. Mr. Lock is to be congratulated, when so many Trigonometries are in the field, on having produced so good a book; for he has not merely availed himself of the labors of his predecessors, but by the treatment of a well-worn subject has invested the study of it with interest.” — ature. MACMILLAN & CO., OGL EPI TH AV ENO aE NView Oli uy ea ) Paw! vale sb bey ‘By ’ Lh om ; , ‘i Oy, iit Vi Na , RA Ab AY i i A wy) Ath} mf ay i lal VPA ae : i a VNN Fe ny ee Ware: Pay Th i | Nar thi }) Ai a? 4 ie ie rh wy oon |