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Les diagrammes suivants illustrent la mAthode. 1 2 3 4 5 6 L/i> AD y ALGEBRA FOR BEGINNERS. BY JAMES LOUDON, M.A. PKOFESSOR OF MATHKMATICS AND NATURAL PIULOSOPHV, l'NIVF.R.SITY COI.LKUF,, TOUONTO. TOROXTO : COPP, CLARK & CO., FRONT STREET. 1 8 7 (5 . 4^CQ PEE FACE V Be,.nners, for who„. this book is intended, are recon.- mended to work according to the rules printed in itahcs, and to om.t, on a first reading, the articles in sn.all tvpe. Those ^sho desire only a practical acquaintance with the method. of working, may also omit Chapters X and XXI. The author will feel obliged to tearhnrc f,..- o / ^'tot-ti lo leaciieis for any suggestion. they may have lo communicate SKPlKAIlJiiK. 1S75. CONTENTS. PACiK. I. Introduction r II. Addition 8 III. Subtraction i6 IV. The use of Double Signs and Brackets 20 V. Multiplicatio' 23 VI. Division 30 VII. Examples involving the application of the first four Rules 39 VI II. Simple Equations 42 IX. Problems 49 X. Particular results in Multiplication and Division 55 XI. Involution and Evolution 61 XII. The Highest Common Measure 72 XIII. The Lowest Common Multiple 84 XIV. Fractions 88 XV. Simple Equations, continued 105 XVI. Problems, continued 108 XVI I. Quadratic Equations 113 XVI 1 1. Problems , 119 XIX. Simultaneous Equations 123 XX. Problems 129 XXI. Exponential Notation 1 33 Answers 141 I ■I 1 I I 8 ALGEBRA. ■•o*- CHAPTER I. f I INTRODUCTION. 1. The operations of addition and subtraction, which in Arithmetic are stated in words, are denoted in Algebra by the signs + and — , respectively. Thus, add together 12, 6, and 6, is expressed 12 + 5 + 6 ; from 15 tahe 9, is expressed 15 — 9 ; add together 3, h, 5^ and from the sum lake i and 3^, is expressed 3 + 2 + 5^—^—31; and so on. 2. The sign + is called the plus sign, and the sign - the minus sign. Thus, 24 + 2 - 15, is read 24 plus 2 minus 15. Exercise I. Employ plus and minus signs to express the following ope ;ations in Algebraical language : — (1.) Add together 56, 10 and 15; 10, 12 and |; 2, land f ; 6> 10* 12> «*"^ 3« (2.) From 29 take 15 ; from 2^ take 1^ ; from 2*5 take 1*6. (3.) From the sum of 11, 35 and 6, take 17; from the sum of 81 and 75 take 69 and 42. 9 INTRODUCTION. (4.) To the difference between 15 and 7 add the sum of 8 and 9. (5.) To the difToi'cnce between 28 and 16 add the difference between 10 and 4. 3. The signs + and — are also used to denote that the quantities l)efore which they are written are, respectively, io he added to and subtracted from some quantity not neces- sarily expressed. Thus, +4 denotes a number 4 to he added to some number; —5 denotes a number 5 to be subtracted from some number. 4. Quantities to be added are called positive quantities, and quantities to be subtracted negative quantities. Thus, +2, +7, +^, are positive quantities; and —3, — 1, — I, are negative quantities. 5. As an illustration of positive and negative quantities, we may take the following examples : — (i.) A man has a certain amount of cash in hand ; he owes $150 to one man, and $280 to another ; and there are owing to him the several sums of $100, $210, and $120. Now, in order to determine what the man is worth, these several sums are to be considered in connection with the cash in hand ; $150 and $280 are evidently to he subtracted, and $100, $210, and $120, to be added. The former, therefore, may be de?ioted by -150, -280, and the latter by +100, +210, + 120. In other words, in the process of finding out how a man's business stands, the sign + may be used to denote his assets, and the sign — his liabilities. (ii.) The mercury in a thermometer rises and falls in con- sequence of changes in the temperature, and the amounts of these variations are expressed in degrees. In order to deter- mine the reading of the thermometer, some of these variations must be added to, and others subtracted from, the reading before the variations took place. Suppose, for example, that during the day there take place a rise of 5°, a fall of 7°, a INTRODUCTION. I rise of 12°, and a fall of 8° ; 5° and 12° are to lie added to, and 7° and 8° subtracted from, the reading of the morning. The forjner, therefore, may be denoted by +5°, + 12°, and the latter by —7°, —8°. If the degrees are measured from a zero point, distances above may therefore be denoted by + , and distances below by — . Thus +20° means 20° above zero, and —5° means 6° below zero. 6. In like manner the signs + and — may generally bo employed to denote the two relations of contrariety which magnitudes of the same kind may bear to one another in some defined respect. Thus, if money received be denoted by + , money paid away will be denoted by -- ; if distances walked in one direction be denoted by + , distances walked in the contrary direction will be denoted by — ; if + denote games won, — will denote games lost ; and so on. Exercise II. (1.) A owes B $60, B owes C $30, and C owes A $20; employ the signs + and — to denote the assets and liabilities of A, B, C. (2.) B and C owe A $20 each, C and A owe B $30 each, and A and B owe C $40 each ; express in Algebraical lan- guage the assets and liabilities of A, B, C. (3.) A pays B $10, B pays C $7, and C pays A $4 ; express Algebraically the amounts paid and received by A, B, C. (4.) Denote the following variations in the thermometer ; a fall of 2°, a rise of 5°, a fall of 3°. (5.) Denote that the thermometer rises and falls alternately 1° per hour for 5 hours. (6.) Denote that the thermometer £Eillsand rises alternately 2° i)er hour for 4 hours. (7.) Denote the following readings: 25° above zero, T below zero. 7. In Algebra the numerical values of quantities are denoted B INTRODUCTION. by (1) figures, as in Arithmetic; (2) tho letters of the alphabet, either alone or iu combination. Thus, if there lie three points. A, B, C, situated in that order on a right line, the distances between A and B, and between B and C, if unknown or variable, may be denoted by o and h feet, respectively ; and, consequently, the distance between A and C will be denoted by a + 6 feet. If lie between A and B, and a and h denote, as before, the lengths AB, BC, the length AC will be denoted by a—h feet. Again, if x denote tho length, and y the breadth, in feet, of a room, the dimensions of a room 8 feet longer and 5 feet narrower will be a? +3 and y— 5 feet, respectively. 8. In Arithmetical operations which are performed with figure symbols alone, all mention of the unit is suppressed, the results of these operations being true, whatever be the unit in view. Thus, 10 and 5 added together make 15, whether the suppressed unit be a pound, a gallon, or an inch. Algebraical symbols have a still greater generality. Not only is the unit suppressed, but the number of units is not assigned, as in Arithmetic. Thus a may represent any number referred to any unit. Operations performed in Algebraical symbols will, therefore, give results which are true for any numerical values which may be assigned to the symbols. 9. Tho product of symbols which denote numbers, is repre- sented by writing them down in a horizontal line one after another, in any order, with or without the multiplication sign X , or dot . , between them. Thus a6, 6a, «•&, b'aaxb,bx a, all denote the product of a and b; ubc, a-b-c, axbxc, the product of a, h, c; and so on. Figure symbols are written first in order ; thus, da, 6ab, 6abc, f«. When there are two figure symbols, the sign x only must be used between them. Thus, the product of 5 and 6a is denoted by 6 x 6a, and not by 5-6a, or 56a, whose values avejive decimal six times o, waA. fifty-six times a, respectively. .«f. INTRODUCTION. \ When thero arc throe or more figure symbols, either the sign X or • must bo used. Thus the product of 2, 5 and 7 is denoted by 2-5 -7, or 2x5x7. 10. The symbol a^ stands for im ; a^ for naa ; a* for aaaa ; and so on. «^ is read a s(/Hared, or a to the power of 2 ; u^ is read a cnbnl^ or a to the power of ^\ a* is read « to the power oj 4 ; and so on. 11. Tlio power to wliich a letter is raised is called the iiiikx or exponent of that power. Thus, 1 is the index or exponent of a ; 2 of tt^ j 3 of a*, 12. The quotient of one quantity, n, divided by another, &, is denoted by either of the forms, « -f- i or ^ . Thus the quotient of 2a divided by 36c is denoted by 2a-r-36cor by|;i. 13. Any object or result of an Algebraical operation is called an Algebraical quantity or eapression. 14. It is to be observed that the numerical value of an ex- pression which is not fractional in form may be a fraction, and the numerical value of an expression which is fractional in form may be an integer. Thus, if to a we assign the value 2, and to 6 the value ^, tho value of =- will be 2 -f- 2, or 4 ; whilst if the values of a and 6 be ^ and i, respectively, the value of ah will be i x i, or i, and the value of ^ will be J -r i, or 2. 15. The symbol = stands for is or are equal to, and is written between the quantities whose equality it is desired to express. Thus a = 3 denotes that the value of a is 3. INTRODUCTJON. Exercise III. If rt = 1, ft = 2, c = 3, ^ = 4, e = 5, a; = 0, find the values of the following expressions : — (1.) a + 2i + 8c + 4'i. (2.) 13a-4/> + 5c. (3.) ah\de. (4.) ah-^'k—cx. (5.) 4a&c, SictZ, Gfifca;. (6.) a2 + 6-+c2. (7.) ^.d'^-W-cK (8.) 5rt(^+4/>3-cU a t7 (9.) h''~e (^^•) ^f + f- X, I V ah ad bx ^ de be ce' a' 3 2b Sd'e (^^•) :x^-Q,.2w + yZ;^""3c2(^'^8a^6** (13.) If h = c=d=a, find what each of the following be- comes in terms of a : 2bca,3a%2a^ + dahc-\-4:b^cd. 16. When two or more quantities are multiplied together, each is said to be a factor of the product. Thus, a and 6 are factors of ah ; 3 and a^ are factors of 3a'^; and 6, h, ar d c are factors of 5&c. 17. One factor of a quantity is said to be a coefficient of the remaining factor, and is said to be a litend or a cmtrical coefficient according as it involves letters or not. Thus, in 3^ and fa''' the numerical coefficients of x and a^ are 3 and *, respectively ; in ax^ and 3cd the literal coefficients of x'^ and d are a and 3c, respectively. Also, since a; = 1 x x, the coefficient of x in the quantity x is 1. The sign + or — when it precedes a quantity is also a sign of the coefficient. Thus, the coefficient of a; in +803 is +3, of x^ in — 5aas* is —5a, and of dz^ in —^cdz^ is — fc. i INTRODUCTION. I f I In the case of fractional numerical coefficients the letter symbols are sometimes written with the numerator ; thus, 18. Quantities are said to be like or unlike according as they involve the same or different combiuatiors of letters. Thus, +5a, —7a are like quantities ; and so also are +Qx-7/, —bx^y; —2a, — a^ are unlike quantities. Exercise IV. (1.) Name the coefficient of x in 2aj, ^ax, fx, lex, 4ta%x, -^(XX, (2.) Name the coefficient of a in —a, +3a, —fa, +2ab, — 5ax2. (3.) Name the coefficient of x^ in -\-x^, -x^, -3ax\ -\-^dx\ (4) Name the coefficient of xym ^xy, H-Sa'^a;?/, — faxyz. (5.) Name the numerical coefficients in ^, ^, -|, +?^' 5.15 (6.) Name the niunerical coefficients in a:, — «/, —2x2, -.3^^^ (7.) Name the like quantities among 2x, ax, x, 3ca;l (8.) Name the like quantities among — a^, +2a2a;, +|a-', — a'flj. (9.) Name the like quantities among —Za^x, ax\ ahx^. I C 8 ) CHAPTER II. ADDITION. I. Figure Symbols. 19. In order to explain the meaning of Algebraical addition, we shall, in the first instance, suppose the numerical values of the quantities to be represented by figure symbols, as in Arithmetic. We shall consider in order the addition of (i.) Positive Quantities. (ii.) Negative Quantities. (iii.) Positive and Negative Quantities. 20. (i.) Positive Quantities. We have seen that a positive quantity, as + 5, represents a quantity 5 to be added to some number which may or may not be known or expressed. The sum of any numher of positive quantities is denoted hy writing them in a row with their signs between them, or by a positive quantity ivhose numerical value is their Arithmetical sum. Thus the sum of +4, +2, and +10 is +4+2 + 10,or +16; that is to say, the addition of 4 and 2 and 10 to a number is equivalent to the addition of 16 to that number. In like manner, the sum of +2, +f, +-*- and +5 is + 2+1 + 1 + 5 =+8/^. The Algebraical statement +5 + 6 + H3^ = +14f ADDITION. may therefore be read the addition of^^^,\ and 3^ to a nurnber is equivalent to the addition of 14| to that number, 21. (ii.) Negative Quantities. Since —4 denotes a quantity 4 to he subtracted from some number, when there are several negative quantities, as —4, —7, —8, denoting that they are all to be subtracted from some number, the operation may be denoted by writing them in a row, thus —4—7—8, or by a negative quantity, —19, whose numerical value is liheir Arithmetical sum ; that is to say, -4_7«8 = -19. In Algebra —4— 7— 8, or —19, is called the sum of —4:, —7, and —8. Thus the sum of —1, —10, — |, and — f is -1-10-t-f = -12ii. It follows therefore that the Algebraical sum of any number of negative quantities is a negative quantity whose numerical value is their Arithmetical sum. It will be observed that, instead of saying to subtract 12, we may say in Algebra to add —12. The Algebraical statement -2-7-10 =-19 may therefore be read, in Arithmetical language, the subtrac' tion of 2, 7, and 10 from a number is equivalent to the subtrac- tion of Id ; or, in Algebraical language, the addition of —2,-7, and —10 ^0 a number is equivalent to the addition of —19. As an illustration of the foregoing phraseology we may take the following example : Suppose a man's gains to be denoted by +, and his losses by — ; then the statement -200-60-500 = -760 may be read, if a dollar is the unit understood, the sum of a loss of 200 dollars^ a loss of 60 dollars, and a loss of 500 dollars is equivalent to a loss of 760 dollars. It may also be read, the subtraction of a gain of 200 dollars, a gain of 60 dollars, and a gain of 600 dollars is equivalent to the subtraction of a gain of 7QQ dcUars. J, ,1 lO ADDITION. 22. (iii.) Positive and Negative Quantities. Since +5 denotes a number 5 to he added, and —2a number 2 to be subtracted, the performance of both these operations may be denoted by +5—2; and since the performance of these two operations is equivalent to addir.g 3, we may write + 5-2 = +3. In Algebra +5—2, or +3, is called the sum of +5 and -2. Thus, +7-5, or +2, is the sum of +7 and — 5 ; +f — i, or +i, is the sum of +? and — i. Again, since to add 2 to a number and then subtract 5 is equivalent to subtracting 3, we may write + 2-5= -3; that is, in Algebraical language, to add +2 and —5 to a number is equivalent to adding —3. The statement +7-5+2= +4 may therefore be read, in Arithmetical language, to add 7 to, then subtract 5 frort\ and finally add 2 to a number is equi- valent to adding 4 ; or, in Algebraical language, the sum of + 7, —5, and +2 is +4. So also the statement -3+10-15= -8 may be read, in Arithmetical language, to subtract 3 from, then add 10 to, and finally subtract 15 from a number is equi- valent to subtracting 8 ; or, in Algebraical language, the sum of —3, +10, and —15 is equal to —8. As an illustration of the foregoing phraseology, we may again take the case of a man's gains and losses. Thus the statement + 10-8= +2 may be read, if a dollar is the unit understood, a gain of 10 dollars and a loss of 8 dollars are equivalent to a gain of 2 dollars. So also the statement ADDITION. II -25 + 20= -5 may be read a loss of 25 dollars and a gain of 20 dollars are equivalent to a loss of 5 dollars. 23. From the preceding cases we can deduce the follow- ing rule for finding the sum of any positive and negative numbers. (i.) When the signs are all alike— i^'mc^ their Arithmetical sum, and 'prefix the common sign, (ii.) When the signs are different— i^«i(Z the numericcd difference between the Arithmetical sum of the positives and the Arithmetical sum of the negatives, and prefix the sign of the numerically greater sum. Examples. (1.) The sum of +4, +3, +h, and +7 is +4+3 + 1 + 7= +14^. (2.) The sum of -5, -12, -f, and -3 is -5-12-^-3= -20|. (3.) The sum of +4, -2, -3, +5, and +7 is +4-2-3 + 5 + 7= +4+5+7-2-3 = +16-5 = +11. Here +16 is the sum of the positives, and -5 of the negatives ; 11 is the numerical difference between these sums, and has the sign of the numerically greater + 16. (4.) The sum of +i, -2, -^, +1, and -f is + i-2-^ + l-|= +i + l-.2-|-| — 4- 5 _ l_3 = -2. Here +f is the sum of the positives, and —^-^ of the negatives ; 2 is the numerical difference between these sums, and has the sign of the numerically greater — ^^. 12 ADDITION. Exercise V. Find the sum of (1.) + 2, + 5, +18. (2.) +i +3, +f. (3.) _ 8, -13, - 7. (4) -i -1, -|. (5.) -rl8, -13. (6.) -26, +20. (7.) +i -1. (8.) -t, +1. (9.) +2-5, -3-2. (10.) +2, -3, +12. (11.) -3, +4, -6, +7. (12.) +5,-8, -12, +3. (13.) +^, -1, +1, -f (14.) ~|, -1, +2, -1 +i-. (15.) +2-58, -3-26, +1-089, -0-067. II. Letter Symbols. 24. (i.) Like Quantities. We shall now show how to find the sum of quantities whose numerical values are represented by letters. When these quantities are like quantities, their sum is obtained by the rule — The sum of any numher of like quantities is a like quantity vjhose coefficient is the sum of the several coefficients. Examples. (1.) The sum of + 2r<, + 5a, and + 10a is + 2a + 5a + 10a = + 17a. Hero + 17 is the sum of +2, +5, and + 10. (2.) The sum of -3c, -10c, and -12c ie _3c;_l0c-12c=-25c. Here —25 is the sum of —3, —10, and —12. (3.) The sum of + 8a;, — 12a5, and + Ix is + 8a;-12x + 7.x=+3.x. Here +3 is the sum of +8, —12, and +7. (4.) The sum of —10a?, +12a5, and —6a; is — 10a; + 12a;— 5a= — 3a;. Here —3 fs the sum of —10, +12, and —5. Fine (1.) (4.) (6.) (9.) (11-: (13. (15.: 25. ( The writing I between I Thus — 5y is +3&, ai I —2a; a: -3c. 26. I + or - Thufi ADDITION. 13 Exercise VI. Find the sum of (1.) +a, +2a. (2.) +3a, +6a, +7a. (3.) -a, -4a (4) -2a, -6a, -5a. (5.) +5x2, ^3^2^ (6.) -a^ +4a^ (7.) +4a, -7a. (8.) -2c, +5c, +7c. (9.) -10c, +8c, -3c. (10.) +a;^ -7a;2, +3a;2, -a;'^. (11.) -2a&, +lla&, +a6, -3a&. (12.) -ia, +|<i. (13.) +|a2, -fa2. (14) _2a, +^a, -a. (15.) -^-a,—^, +|<i, —2a. 25. (ii.) Unlike Quantities. Tlie sum of any number of unlike quantities is denoted hy writing them in a row in any order, with their proper signs between them ; and each quantity is called a term of the sum. Thus the sum of +2a and +36 is +2a+36; of +Sx and -5y is +3x-52/; of -W md -26^ is -ia2_2&2; of +2a, + 36, and — 5c is + 2-* + 36 — 5c. The terms of — 2ic + 6y are — 2a3 and +5y ,• and of — 6a+76— 3c are —6a, +76, and —3c. 26. If a quantity contains no parts connected by the sign + or — it is called a mononomial. Thus +2x, — 3a6, +6x^2/, are mononomials. 27. When a quantity consists of two terms it is called a binomial expression; when it consists of three terms it is called a trinomial expression ; and generally when it consists of several terms it is called a polynomial, or multinomial expression. Thus +2a— 36 is a binomial, and +a— 26 + 3c a trinomial expression. 28. The sign of a mononomial, or of the first term of a I polynomial, if it is positive, is generally omitted. Thus 2x^ stands for +2a;2, and a— 6 + c for +a— 6 + c. 29. Like terms when they occur must be added together. I The operation may be conducted by arranging the several 14 ADDITION. quantities in rows under eiich other, so that like terms shall stand in the same column. Exam'phs. (1.) Find the sum of 2a+3i and 5a— 26. 2a + 3& 5r^26 Here la is the sum of %x and 5a, and +& of +36 and —26. (2.) I'ind the sum of xy—^x and x—^xy. —^£-\-xy x—'^xy ^bx^lxy. Here — 5rc is the sum of —6^- and x, and —Ixy of -^-xy and —^xy. (3.) Find the sum of 3 + a; + a7?/-8.T^ -3a;?/ + 2- 6a;, and 4!r?/+fl32-|.l. 3+ a; + cry -8x2 2— 6.r— 3a;y 1 + 4^xy + re'' (d-Zx^^xy-lx^. Here the sum of 1, 2, 3 in the iirst column is 6; of +ir, -6x in the second is -5a:; of ■\-xy, -3.r?/, +4a?/ in the third is ^2xy ; and of -Sa;^, +a;2 in the fourth is -7x^ (4.) Add together ha-^b-ic, ^J + ic + ^a, Ac-ia-i6. — ^-T^ + k Here -j-^a is the sum of ^a, ia, and -ia ; - JL6 and + ^^c are the sums of the quantities in the second and third columns, respectively. ADDITION. 15 Exercise VII. Find the sum of (1.) 2a, -3&. (2.) -oj, +3y. (3.) -2aj, -3y, -2. (4.) 3a, 2.x, -57/. (5.) 4, -.r, %j. (6.) ««, -.52, i. (7.) a, -' \ 3c, -c?. (8.) -a:, 2?/, -2, 1. (9.) a-3&,8a + &. (10.) -2a'''+&c, 3a2-56c. (11.) 3a-5?^ -4a + 2&, 5a-6&. (12.) a-.3&, 26-5c-, 4c-3a. (13.) 4x— 3?/+22, — 3;c + ?/— 42, a;— 4y + 2. (14.) a-a5 + 3, 5a + 2a5-5, -2a + 7. (15.) 2«-7, 5a+4, -6:c + 3. (16.) a-2&+3c, &-2c+3a, c-2a+3?). (17.) a;-2y+33-l,2ir + 3-42, 52/-2 + 7^. (18.) a;2 + 2aa; + a^2a;2-2a^£c2-2aa; + a2 (19.) 3aS+a''^&-2a&2+&3^ 3a&2-2a26+a3, a%^al^^W. (20.)a-?> + |, a + &-|, &-a + |. '^"■'•^2 3^4' 2 3^1' 2 2^4' (22.) a+56-c, 2a-4&-c, |-a+|. ( i6 ) CHAPTER III. SUBTRACTION. fM ■■ h-}> 30. The Alyehraical difference between one quantity and anotlier is the quantity wliicli added Algebraically to the latter will produce the former. Thus the difference between 2 and —4 is the quantity whicli added to —4 will produce 2; the difference between —5a and 2a is the quantity which added to 2a will produce —6a ; and the difference between 3.r and 2a;— 5 is the quan- tity which added to 2x—6 will produce '6x. 31. The quantity to bo diminished is called the minuend, and the quantity to be subtracted the subtrahend. 82. The difference between two quantities is found by the rule — Add the first quantity to the second with its sign or signs changed. The reason for tliis rule will appear from the following Examples. (1.) From 5 take 3. Here the difference is the sum of 5 and —3=5—3=2, because 2 added to 3 makes 5. (2.) From 7 take -4. Here the difference is 7 -r 4=11, because the sum of 11 and -4=11-4=7. SUBTRACTION, 17 i (3.) From -6 take -4 Here the diflference = —6+4=— 2, because the sum of -2 and -4= -2-4= -6. (4.) From -8 take 5. The difference =— 8— 5=— 13, because the sum of —13 and5=-13 + 5=-8. (5.) From 2a take —3a. The difference =2a+3a=5a, because the sum of 5a and — 3rt = 5a — 3a =2a. (6.) From —hx take 4. The difference =— 5a;— 4, because the sum of —5a;— 4 and 4 is —5a;. (7.) From 2a take -3a + 2&. The difference = the sum of 2a and 3a— 2&=2a+3a— 2&=5a— 2&, Iv^cause the sum of 5a— 2& and —3a + 26 is 2a. The operation of changing signs and adding may be per- formed mentally, and the difference exhibited as in the fol- lowing examples, in which the minuend and subtrahend are written in rows with the difference underneath. In this arrangement the first row is equal to the sum of the second and third rows. (8.) From Zx^-^y take a;^— 5?/. 3x2+2/ 2x2 +6y Here Sx^ is the sum of 3x2 ^nd ^^-^ and +6y of +2/ and (9.) From 5a+3&— c take a— 6+ 3c. 5a + 3?>— c a - &+3c 4a+46— 4c ]8 SUBTRACTION. Hero 4a is tho sum of 5a and —a; +4Jof +36 and +&; and — 4c of — c and —3c. 33. From tho foregoing examples it appears that, in Alge- braical language, to subtract a positive ijuitntity is eciuivalcnt to adding a negdiive ; and to sidUract a negative (juantity is equivalent to adding a positive. This phraseology may bo illustrated by taking the case of a man's gains and lasses to he denoted by + and — , resi)ee- tively. Thus, to subtract a gain of 10 dollars is equivalent to adding a loss of 10 dollars ; and to subtract a loss of 25 dollars is equivalent to adding a gain of 25 dollars. Moreover, if a man gains a dollars and loses h dollars, we say, in Arithmetical language, either that he gains a—h dollare, if a is greater than b, or loses b—a dollars, if b is greater than n. Either of these phrases may be employed indifferently if we agree that a gain of —c dollars means a loss of c dollars, and that a loss of —c dollars means a gain ofc dollars. Thus if a man gains 10 dollars and loses 5 dollars, we may cither say that ho gained 10—6, or 5 dollars, or that he lost 5— 10, or —5 dollars. Again, if ho gains 8 dollars and loses 12 dollars, we may either say that he gained 8—12, or —4 dollars, or that he lost 12—8, or 4 dollars. Exercise YIII. (1.) From 1 take -3. (2.) From 1 take -1. (3.) From —5 take 4. (4.) From 12 take 15. (5.) From -3 take -8. (6.) From -8 take -5. (7.) From 4-56 take -6*04. (8.) From -I'Oi take 235. (9.) From -432 take -2-16. (10.) From -1089 take -0*123. SUBTRACTION, 19 (11.) From 2a take 8a. (12.) From —5a; take 2aj. (13.) From Sa^ take -2a^ (14.) From —3c take —5c. (15.) From 2a take h (16.) From -a' take 2a». (17.) From bx^ take -Gail (18.) From la take 4a-&. (19.) From a + a; take a— a;. (20.) From 5a-2a;+3 take 2a-a;-l. (21.) From 3a2-4a& + 62 ^ake SaHaJ-i*. (22.) From aa;— 4&y + 3c2 take 2ax-^hy—cz, (23.) From -4a+6-l take 2a-a; + 3. (24.) From 12a;2-5a; + l take 7a;3-16a;2+l. (26.) From^+3/+| take |+|-;3. (26.) From|o+6-^ take a+|6-^. o2 ( 20 ) CHAPTER IV. TEE USE OF DOUBLE SIGNS AND OF BRACKETS. 34. The operations of addition and subtraction of positive and negative quantities may also be denoted by the use of + for the former operation and — for the latter. Thus instead of saying add together 2a and —Sh, we may employ the notation 2a -\ — 'Sh, the equivalent of which is of course 2a— db. So also tJw sum of —5a^, +Sb, and —2c may be written — 5cr+ +3iH 2c, which is equivalent to — 5a'-^ + 3&— 2c; and the difference between 6a; and —7y may be expressed hx ly, the equivalent of which is hx-\-ly. Accordmg to this notation, therefore, 2a'^H — 3&+ +2c means the sum of2a^, — 3&, and +2c; —60;+ +83/ H 1 the sum of —5a?, +Sy, and —1; 7a \-2b the difference between la and +2'>; —2x -5 the difference between —2x and —5; 4a'^ 7 the difference between 4a- and —7. 35. When any of the quantities before which the double signs are to be used contains more terms than one, it must be enclosed in a bracket; thus + (2a— 3&), + (—a; + 4), - (.r-5), -(-2 + aHa;). Thus a+(3i— c) denotes the sum of a and 36— c ; 2x + (4y-2) + (-2^ + 1) the sum of 2x, 4y-2, and -22 + 1; 2a2_j_ (3a2+4) the difference between 2a^-b and Sa-+4:; and 2a+ {b—1) — (c+4) the sum of 2ct and &— 1 IckSs c+4. Exercise IX. Retaining the given quantities, denote by using the double signs and brackets (when necessary) in the following opera- tions : — i USE OF DOUBLE SIGNS AND OF BRACKETS. 21 (1.) The sum of 2x2, _i. 3^^^ __^y^ __^.^ 2(t, -3&,+4c. (2.) From 2c. take —6a. (3.) From -Stake +5^. (4.) From the siim of 2a and — 3& take +7. (5.) From the sum of 5 and +a; take —3a. (6.) The sum of 5a and &— 4. (7.) The sum of —a and —6+5. (8.) The sum of a— 4 and 2&— c. (9.) The sum of x^ and 2?/ + 5 less 2?. (10.) The sum of a— 1 and 3& + 5 less —3c. (11.) The sum of x, 2*2-1, and-3a:2-8. (12.) The diflference between da^ and V'—c. (13.) The diflference between a^^^ and — 2&+3. (14.) The diflference between 2a— 5 and a2>_2a+3. (15.) From the sum of a+&+c and a—h—c take — ct +26-3c. 86. Double signs may be equivalently replaced by single ones by the rule : — Like signs produce + , and unlike signs — ; that is + + = -- = +, Thus a++5=a + 5; 2x a=2x+a; 3+— 4g=3— 4c; c— +2a=c— 2a. 37. Expressions may be cleared of brackets by the rule : — The sign + he/ore a bracket does not change the signs ivithin^ whilst the sign — changes every sign within. Thus 4 + (&-c)=4 + 6-c, 2a+(— x + c— 2(f)=2a— cc+c— 26^, 4a2_i_(26' + c)=4a2-l-262-c, 3a;-(-4?/ + 5)=3«;+42/-5, 03— (?/— 2)-f4— (— 3?/ + a;)=£c— y + «+4 + 3?/— .r. 22 USE OF DOUBLE SIGNS AND OF BRACKETS. EXEBGISE X. Eeplace the double signs \y single ones in the expres- sions:— (1.) 2a++3&+-c. (3.) a;2+-4a;+-l. Clear of brackets : — (5.) 8a^.-(J+c). (7.) 8a-(-2&+30. (9.) (»+5-(2-42/)+8. (11.) 3a;2-l-(-aj+4)+2iB-(sc2-.6). (2.) ah—+hc c. (4.) 5x'-\--Sx' ?a;-+8. (6.) 8a-(5-c). (8.) 2a-l + (&-6)+c. (10.) a + (&-c)-(a-c). ( 23 ) CHAPTER V. MULTIPLICATION, 38. When it is desired to denote the operation of multiply- ing several expressions together so as to exhibit the various factors, we enclose each in a bracket and write them together in a row in any order. Thus ( + 2a) (— 3&) denotes the product of -I- 2a and — 3& ; (2a -1) l-h) the product of 2a--l and -&; (x'-d) (2a; + o) the product of 05^—3 and 2x + 5; and («— 1) (a:; + 2) (2x—b) the product of a;— 1, a; + 2, and 2ic— 5. 39. Each of the quantities so enclosed in brackets is called a, factor of the product. Thus —2a, a^— 1, and 2a— 3 are the factors of (—2a) («"— 1) (2a-3). 40. When a factor is mononomial, it is called a simple factor ; otherwise a compound factor. Thus in the preceding example —2a is a simple factor, and a^—1 and 2a— 3 compound factors. 41. In the case of a simple factor the bracket may be omitted (i.) if the simple factor is written in the first place ; (ii.) if the sign of the simple factor is not expressed. Thus we generally write —2a (a3— 1), a:(a;''— 2), (a— &)a;, (a— Z>) (c^d)x. If the sign of a simple factor is expressed, the bracket may be replaced by a multiplication sign, x . 24 MUL TIPLICA TION. Thus -2^ X +3A = -2r(( + 36), 3ic X — 5?/ = 8.t(— 5y), (a-1) X -2tr = (rt-l) (-2^2). JC?/^^. Exercise XI. Express in Algebraic<al language, retaining the given factors and using brackets wlion necessary : — (1.) The products of a-1, 2a2-3; -2+a, -3-a2; x-h, -2x + 7. (2.) The products of -%.i^, 6^-1; a^-l, -3a; 5x, —a;- + 3. (3.) The products of \, jk-I ; \, 2i»-3; -f, x^-S. (4.) The products of — oo?, a:— 1, ^7 + 2; a;-— 4, +5x, 2a + 3. (5.) The products of +8a5, —hyjXy—\\ —la, ab—'d, +86. 42. The mode of performing the operation of multiplication whereby products are expressed as mononomials or poly- nomials will now be explained. It is convenient to make three cases : I. The Multiplication of Sijiple Factors. II. The Multiplication of a Simple and a Compound Factor. III. The Multiplication of Compound Factors. 43. I. The product of two simple factors is obtained by the following rule : — (i.) The siffn of the 'product is obtained by the rule of signs : like signs produce -\- , and unlike signs —. Thus the signs of +2a ( + 36), -2a (-3&), +2a (-3&), —2a ( + 36) are, respectively, +, +, — , — . (ii.) The numerical coefficient of the product is the product of the numerical coefficients of the factors. Thus the numerical value of the coefficient of the product of ~2a and +36c is 2 x 3 - 6. (iii.) The literal part of the product is the prodmt of the literal parts of the factors (9). MUL riPLICA TION, 25 Thus tho literal part of the product of — 3aj and ^ifz is %yh. (iv.) The 'p oditct of two powers of the same letter is a power whose index is the sum of the ii. dices of the factors. Thus 1 1 Examples, (1.) The product of +2a and -36= + -2x3a6=-6a&. (2.) The product of -5a and 2d=-5x2ad=-10ad. (3.) The product of -^a and +3&'-=- + ^ x3a62=-.|a6='. (4.) -a?<-3c) = 3a&c= + 3a&c. (5.) -ia^( + ^Z>)=- + ^ • ia2&=— ia^ft. (6.) -2x(-f3a;-)=-+2x3x. a;2=:-6a;3. (7.) - 6ftX- Sit «) = 6x3 a3. «*= + 18al (8.) +2a2&c3(-5«6V)= + -2 x 5 a • a^ . j . j2 . ^ . c*- Exercise XII. Find the product of (1.) +3^, -26; -«, +5c; -2a2, -36; 5x, -6y. (2.) 2a6, -7c2; -4a^ +56c; -2iB, S?/^; -6, -8a. (3.) -ix, 3^; |a, -26; -hx, -\y; 2a\ -^. (4.) 2^2/^ —3a;-?/ ; —ax^y, —Zxy^ ; |-a6^c, — ^a^6c^ ^-Ks a __2a^ ah a-W _%c,ij^ 3x^2/' . _jxx^ _J)Oiy^ ^^ 2' "3 * 5' IT ' "5"' "8" ' T' 3 • 44. II. The product of a simple and a compound factor is the Algebraical sum of the products of the simple factor and the several terms of the compound factor. 26 MULTIPLICA TION. Thus the product of 2a and 3ft— 5c+<^ is the sum of the products of 2a, 3&; 2a, —5c; and 2a, +c?; and is therefore equal to 6a6— 10ac+2a(^. The work may be arranged as in the following Examples^ (1.) Find the product of —803 a;ttd 2?/'*— 403^—5, -3a; — 6rt2/^ + 12a3'^y + 16a;. (2.) Find the product of 2x'^—^-\-\ and ^^^xy, 2x2-ia; + i — fa;y Fs'^y* (3.) Clear of brackets the expression —2a (Sa"— 5a+l). As this expression denotes the product of —2a and Sa^— 5a + l, it is equivalent to — GaHlOa^— 2a. Exercise XIII. Find the product of (1.) 4a-36 + c, -2x; 3a;2-2x + l, 4?/; 2a&-3c, -d (2.) a;2«2x--5, 3^; 2rt2-3a + 7, -a»; a;'''-aa; + 2a2, -4aa;. (3.) 2x-\-y--^z,%iyz; la^-ah^ + ah, -4ub. (4.) a + ^ft-fc, 2a5; 2a2-3a+4, -|a; ^x^''ax-{-ia^,^ax. ^rs 4a 2ah ^ ,^ 2.k ^ , 3aa; ,. W-) g — -g- +1,-15; _- 1+__,A| /X'» Clear of brackets : — (G.) 5(^2-a^-f-4); -2(a-a& + 3); -a(a2- 2aa; + l). (7.) (a-a;)a;; (a^h + c)c; (-!-{■ ab-Sa^)bab. (8.) |'(6a''-9a + 12);- |'(-10 + 2x-15x«). 45. in. The product of two compound factors is the Alge- hraiml sum of the products of one factor and the several terms of the other. MUL TIP Lie A TION, 2^ I Thus the product of 2a— 36 and 4c +5^/ is the sum of the products 2a— 3?', 4c and 2a— 36, +5<?, and is, therefore, equal to 8ac-126c + 10a(^-156d The work may be conveniently arranged as in the following Examphs. (1.) Multiply 2a;2-3a;+5 by 4a;-7. 2ic2-3a!+6 4a; -7 8a;»-12a;H20a; -14a;H21a;-35 8a;»-26xH4ia;-35. Here ^-llx^^-'ildx is the product of 4(r and Ix^—^x-^-^-^ *-14icH21a5— 35 is the product of —7 and 2ar^— 3a;+5; and the sum of these two partial products, which are arranged so that like terms stand in the same column, is the product required. It will be observed that in the foregoing process we work from left to right, and not from right to left, as in the cor- responding Arithmetical operation. (2.) Multiply 2a-6 by c-M, 2a-6 2ac-^hc "Qad-^-Shd 2ac •^hc— Gad + dbd. In this case, as there are no like terms in the partial pro- ducts, the second is placed entirely to the right of the first. (3.) Multiply l-2x+3x- by 4(r-5a;2+2. It will be found most convenient in this and similar examples to arrange the given factors according to ascending or descending powers of x ; that is, so that the exponents of tlie successive terms shall continually increase or decrease, in the former arrangement the numeral stands first, in the latter last. I.. 28 MULTJPLICA TION. iii m ■i In the present case let them be arranged in the order l-2£c + 3x=^ 2+4x-5»2 2-4r+6x2 + 4:03-8** +12^3 -5a;- + lQx-"-15a;* 2 -7ar^ + 22x3- 15a;*. (4.) Multiply 2a2-a& + Z>2 ^^y aHa&-36^ 2a2-a& + 62 2a4-a3i+^r2 -6a'6H3a63-354 2aHa'6-6a''^6H4aft«-36^ Here the factors are arranged according to descending powers of a and ascending powers of h. Exercise XIV. Multiply together (1.) 2*-3, .a-{-4; 4aH-5, -.r + l; 2-3x, l+cc. (2.) a;^-2,2.r-l; 1-a-, 2 + 3^2; \^x\^-x^, (3.) 2a--ti + 4, 3a~2; !-« + «=, 1 + a; l + a + a^^ 1-a. (4.) « + i, 1)1— n ; a-{-l)'-^c,7n-\-2n ; 2))i—ii, 2m -{-n, (5.) fi;?/ + u.^a;y/-a'2; x + o:y-y\ ar-'ly. (6.) 2«-^-5« + l,aH3^i-4; a + 26-3c, a-26 + 3c. (7.) 3a + 2i-c,o^~2/^ + 8c,- 3xH2x2/ + 2/', «'-2a'y + 8?/2. (8.) a;2-i,.r2-.>; « + ! ^-i ; 2a-^, a + i. (9.) «;2_^x + l,2^— i; 3*--|:« + i, 3x— ^ aO.) ^+|-2,|-3y; ^'-2x + i3^-|. a^ 2a a- 2a Ui.^g— 3- + ^' 2 ^'3~'"-^' MULTIPLICA TION, 29 (12.) c»H2/^-a;y+a'+2/-l»«'+2/-l- (13.) aH&Hc''— &c-ca— a&, a+&+c. 46. The <product of three expressions is found hy muUijplying thejproduct of two of them by the third. Examples. (1) Here (2.) Multiply together 2a;, -Sx% ^ixy^z. 2a;(-3x22/)=-6x' Multiply together c^y ; and —Qx^y{'-ixyh)= +|a;y2;. X -L aj-2, a;-3. X X -I -2 ay^—x ~2ig+2 a^-dx+2 {B8-3x2+2a; -3a;'^+9a;~6 ar'-6x2 + lla;-6. EXEBCISE XV. Multiply together (1.) -3a2, +2a% -6ab^; Ix, -^x', +|(b*; -S:x^y, —\y\ ^yz. (2.) -'2x,Zxy,^x'-hy; 2a&, -^^a^, 2a2-3a& + l. (3.) 2a;-3,4a; + l, a;-2. (4.) a;2_a; + l, a;-l, x + 1. (5.) a;2+2a£C+a2, a;2-2aa;+a2, a;H2a2a;2+a*. " > ( 30 ) CHArTEB VI. DIVISION. MJi 47. Division being the inverse of multiplication, it follows that, when two factors are nmltiplied together, either factor will be the quotient of the product divided by the other. Thus, since +a(— 26) =—2ab, +a is the quotient of — 2a6 divided by —28, and — 2& is the quotient of — 2a6 divided by Again, since •~2x^ (x^'-4a}-\-d)=^2x^-\-Sx^—6x^, it follows that 05^—403 + 3 is the quotient of —2x^-\-Sx^—Qx'^ divided by —203^, and — 2a;^ is the quotient of — 223^+80^—6052 divided by oj^— 403 + 3. 48. "When it is desired to denote that one quantity is to be divided by another, they are enclosed in brackets and written in a row with the sign ~ between them ; or the second quantity is written below the first with a line between them. Thus (-2a;2)-f-(+3!r), or -2ar^ +Sx' denotes that — 203^* is to be divided by +dx; (3a:2-2a5+6)-f-(aJ-4), or ?^-^-±_5. 03—4 that Sas^— 2a3+5 is to be divided by as— 4. 49. The first or upper quantity is called the dividend, and the second or lower the divisor. 50. The bracket is generally omitted in the case of a mononomial. Thus (-203) -r (-32^) is written -2a;-7--32/; (2a;2-l) -r(4x) is written (2x2-1) ~ 4a; ; (-3x3)-;- (2a- 1) |g written -3a!3-r(2!»-l). DIVISION. 3« EZBBOISE XVI. Betaining the given quantities and employing brackets only when necessary, express in Algebraical language — (1.) Divide 2a-5 by -3a; da^-Sa + l by 3a-4. (2.) Divide 2a by -36; -a;^ by +2a;; 3a; by 2a. (3.) Divide4a;'^by 2a5— 5; —aa;'* by oc- a. 51. The mode of performing the operation of division whereby quotients are expressed as mononomials or polyno- mials will now be explained. We shall consider in order three cases : I. When the Dividend and Divisor are Mononomials. II. When the Divisor only is a Mononomial. III. When the Dividend and Divisor are Polynomials. 52. I. The quotient of one mononomial divided by another is obtained by the following rule : (i.) The sign of the quotient is obtained by the rule of signs : like signs produce +, and unlike signs — . Thus the signs of -^2ah-^Scd, Sx^-7'-2x, 4a2-i--2a, —bx^y-r- ■\-xyy are, respectively, +, +, — , — . This rule follows from the rule of signs in multiplication; thus, since +a (+i) = +a&, it follows that + a6. + a' So also from the equivalent forms — a ( + &) = — o&, — a (— &) = + a6, +a (— &) =— a6, we deduce — a& , 7 +a6 , —at- , —a —a +a (ii.) The numerical coefficient^ without regard to sign^ of the quotient is obtained by dividing the numerical coefficient of the dividend by the numerical coefficient of the divisor. Thus the numerical coefficient of 12a^-^Sa is 12-7-3=4, of 2a^-T-Sx is f, and of la^^fa is I -~|=|. (iii.) The literal part of the quotient is obtained by dividing the literal part of the dividend by the literal part of the divisor (12). •= + b. If I'll 1: 32 DIVISION. Thus the literal part of the quotient of 2a2 -f-Jc is ~. In be applying this rule the fractional form of expressing the quotient should always bo used. (iv.) The (jHothnt of two poirem of the same letter ia a power inJaise index in the di'jference 0/ the indices of dividend and diiHHor. The reason for this rule will he evident from the following examples : — From (10) a^ aaa a a a" aaaaa ■ aa =a2— 0^3-1 . =iaa=a^=a'~^; a" aaa of aaaaaaa =aaa = a^ssa^"* : a aaaa ir .i So likewise a" ±=a^-^=a\ a^ 53. Since the quotient of any quantity divided by itself ia 1 , this rule can be applied when the indices of the dividend and divisor are equal, if the zero power of a letter is considered ' a^ " d~ eiual to unity. Thus -n=l ; find, by the rule, --zra^-s—fjO. a^ a^ therefore a°=l. Whenever, therefore, by applying the pre- ceding rule, we get such symbols as x^, y^, c", we must replace each of them by 1. Examples. (1.) -.i2-~+6=-V=-2; -15~-10=+i^=+f; 1 4.IJ: #— _=* — -_2 f (2.) -6aS-3/. = -«. ?=-^*. u (1- (2. (3 DIVISION. n ^=:rt"^ (3.) -2a6-r--5c = +f^^=+|^. c 6c (5.) -4a6-f--7a2= + |a°-!^=+-fa3. (6.) 2tt263c» -r 3a6c2 = ^a'-'-i ft^-i cO-2 ^ 1^52^5^ Exercise XVII. Divide (1.) -16 by +4 ; 20 by -4 ; -f by -|; +5 by -f, (2.) -a by +2a;; Sa^ by -26; -Gajy by -3a. (3.) ^ by -I; -' ^ oy g, ^ oy -^. 'A " 3 (4.) 2a» by a^\ 3a* by 3a; 8iB« by 2a;*. (5.) -4a26 by 2a& ; lOa^JSc* by 2a6c. (6.) a^y^ by — 3aa;2/; —ZQ^xy^ by -"Baxy. 54. II. T/je quotient of a polynomial divided hy a mono- tiomial is the Algebraical sum of the quotients of the several terms of the former divided by the latter. Thus the quotient of Sx^—Qx^-\-Sx divided by —2x is the sum of the quotients 3x'-^ — 2x, —Gx'^-i — 2x, +8x-z — 2a5 ; and is therefore equal to — fa;"+3a3— 4. The work may be arranged as in the following Examples. (1.) Divide Hx^-^x^+^x by -2x. -•2x)S x^-4:X^+2 x -4xH2iB-l. (2.) Divide 6a»6-10a26='+2a& by 5ab. 5a6)5aV-10a26H2a6 a2- 2db^ + 1 (3.) Collect coefficients of x in 2aa;— S*. "it II 34 DIVISION, As this means that 2aa;— Sec is to be expressed as the pro- duct of two factors one of which shall be x, we have merely to divide the given quantity by x to get the other factor. Thus 2aa5-3x=(2a-3)a5. (4.) Collect coefficients of cc^ in ^ax^—bx^+x^. By dividing the given quantity by x'^ we get ba3?—hx^ + flj2=(5a— 6 + 1) x^. «t Exercise XVIII. Divide (1.) 10a-15&+20by -5; -4aa! + 12--8a2by -4. (2.) 4a2a;-3aHa'by-a2; 12x3-6!B2+9a; by 3x. (3.) 3a''-12aH15a2by -S^^. 2a;32/-6xV+8iC2/* by 2a;«^. (4.) 2a^6c— 3a6^c+a&c2 by a&c. (5.) 20a26c2-15a&V+5afeby -5a6. (6.) Collect coefficients of x in ax—hx, 2ax—cx+x. (7.) Collect coefficients of as?/ in 4ixy—axy, Sx^y—xy^. 55. Since by (54) ^^ x-1 X X it follows that likewise g .=^-i and by (44) Ka'-l)-^-^, and K^~l) ^J*® equivalent forms. So 2a;-3 i(2aj-3), 56. III. When the dividend and divisor are polynomials, the quotient is obtained by the following rule : — (i.) Arrange both dividend and divisor according to ascending or descending powers of some letter. (ii.) The first term of the quotient is found by dividing the leading term of the dividend by the leading term of the divisor. The product of the divisor and the first term of the quotient is subtracted from the dividend, giving the first difference. ro- ely DIVISION. 35 (iii.) Tlie second term of the quotient is found hy dividing the leading term of the first difference hy the leading term of the divisor. The product of the divisor and the second term of the quotient is subtracted from the first difference, giving the second difference in the process. And so on until the last difference is zero. y- Examples. (1.) Divide Qx^-hx'-'^x^-^ by 2f3!r. 3ic + 2) 6a;3-5.x2-3x + 2(2x2-3a;+l 605^ + 4tx^ -9jr;^-"-3x + 2 — 9£c^— 6a3 " 3;rT2 3x+2 — 1. a* So iais, iing J the isor. nt i& Here the divisor is first arranged, like the dividend, ac- cording to descending powers of x. The first term of the quotient 2x^ is obtained by dividing 605^, the leading term of the dividend, by 'dx, the leading term of the divisor. The product of the divisor 3«+2 and 2x'^, which is 6a5^+4a5^, is then subtracted from the dividend, giving — Qx'^— 3£c+2, the first difference. The second term of the quotient —^x is obtained by dividing —9*^, the leading term of the first difference, by 3a5, the leading term of the divisor. The pro- duct of the divisor .ind —^x, which is — 9a5^— 605, is then subtracted from the first difference, giving 3x+2, the second difference. The third term of the quotient +1 is obtained by dividing 3x, the leading term of the second difference, by 3cc, the leading term of the divisor. The product of the divisor and +1, which is 3a3 + 2, is then subtracted from the second difference, giving the last difference zero ; and thus the process ends. The latter terms of the differences need not be expressed until the corresponding like terms in the par^i il products are to be subtracted from them; thus in the following ex- D 2 36 DIVISION. I % !ir If It i ample — ITx + G is not expressed in the first difference, nor + 6 in the second. (2.) Divide 6x4^5a;3 + 6x--17^ + 6 by 2.c-l. 2x-l)6xH5x3 + 6.z;'^-17u; + 6C3x3 + 4x2 + 5x-6 -12aj + 6 (3.) Divide l-2x3+&;'' by l-2ic+a;2. l-2x + x2)l-2ie3-ha:«(l + 2a; + 3x2 + 2x3 + a;* l-2x+u;^ 2a;-x2-2:^3 2ic— 4a;^ + 2x* 3x2— 4a;^ 3.c2_5a;=' + 3x* 2x3-3x4 2x3-4.r* + 2xS a;'* — ■ia'-'' 4- x* Here +x' is not expressed until we reach the last dif- ference. (4.) Divide 2a5-6a3& + 13a262-6a63-3 1<& by 2a-3&. Here wo shall arrange the dividend and divisor according to descending powers of a. 2a-36)2a5-3a*6-6a36 + 18a262-6a&3(a4-3a26+2a6''^ 2a°-3a*6 -6a36 + 13a2&2 -6a36+ 9a262 • ^ 4a=*62-.6a6' 4a262-6a63 DIVISION. 37 Exercise XIX. Divide (1.) a;2-7a! + 12 by a;-3; and 3x2 + 7a; + 2 by a; + 2. (2.) a;2-4x-5 by a;-5; and 4x2-9 by 2x + 3. (3.) 6x2-5x-6 by 2x-3 ; and 9x'- 18x2 + 26^-24 by 3x-4. (4.) x3-4x2 + 6x-2 by x2-3x+2; and cc^ + x^+l by x'^ + X + 1. (5.) 4x4-15x3 + 14x2-6x + l by 4x2-3x + l. (6.) X*— 1 by X— 1; and x^ + 1 by x + 1. (7.) x2— 2x?/ + 2/'^ by x—y ; and x^-^-if by x + y. (8.) aHa'&H&^bya2-a& + &2^ (9.) 20x2 + 9x?/-12x- 182/2 + 92/ by 4x-32/. (10.) x6-2ax* + 3a2x3-3a3x2 + 2a*x-a5 ijy x3-ax2+a2x'-a8, (11.) x^-^x^y—xy^-\-y^ by x2 + x2/ + ?/2. (12.) |a*X-\fa3x2 +|a2^3 +_3_o^a;4_^5 l^y |.f^3__A,j2^^1a;3^ 57. When the division is not exact, the last difiference is called the remainder. In this case the product of the quotient and divisor added to the remainder will be equal to the dividend. Example. Find the quotient and remainder in dividing 10x3+ 7x'* -8x-2by 2x + 3. 2x + 3)10x3+7x2~8x-2(5x2-4x+2 10x3 + 15x2 -8x2-8x -8x2- 12x 4x-2 4x+6 -8 Quotient =5x2— 4x + 2. Remainder = — 8. I i 38 DIVISION. Exercise XX. Find the quotient and remainder in dividing (1.) 4x3-4a;H8a;+2by2a5+l. (4.) 2a5«-2x* + 9x3 by 2a;' + a? + 1. (6.) 2a;*x2x*+5a;3 by a;»+a^+»+l. I H ( 39 ) CHAPTEE VII. EXAMPLES INVOLVING THE APPLICATION OF THE FIEST FOUB BULE8. 58. In the following examples some of the given quantities are expressed by letter symbols, and the object of the exercises is to express in like manner other quantities which by the conditions of the question are related to the former. When a doubt exists as to the manner of solving a question, it will be well to substitute numbers for letters in order to see what operations ought to be performed in the given symbols. 59. The sign .*. will be used to mean hence, or therefore, and the sign '.* siuce^ or because. Exumples, (1.) I buy goods for 2a+Sb—c dollars, and sell themfo^r 4a— 6 + 2c dollars ; what do I gain ? 4a— 6+ 2c 2a-H36-c 2a-4&+3c .*. the gain, which is the selling price less the cost price, is 2a— 4&+ 3c dollars. (2.) A man has 3*^^ + Tas + 2 dollars and spends aj +2 of them per day ; how long will his money last ? a;+2)3xH7a;+2(3a;+l »+2 »+2 40 EXAMPLES OF THE FIRST FOUR RULES. .*. the required number of days, which is equal to the num- ber of times the amount of his daily expenses is contained in the amount he possesses, is Sx- + 1. (3.) A man walks x miles in y hours : at what rate per hour does he walk ; how far will he walk in 5 hours ; and how long "will he be in walking 12 miles ? *.* he walks x miles in y hours, „ ^ „ „ 1 hour, and 1 mile in ^ hours. • • » X Again, X he walks - miles in 1 hour, y » >j 5x y ,, 5 hours; and he takes ^ hours to walk 1 mile, X if i> 12y X » » i> 12 miles. or OuC The answers are, therefore, - miles, — miles, and y y » 12^/ hours. Exercise XXI. (1.) A man walks x, +a, and x—2a miles in tb'^ same direction ; how far does he walk altogether? (2.) A man has 100 dollars, and owes 50— x dollars; what is he worth ? (3.) A man walks a-\-h miles and returns a—h miles; how far is he from the starting point ? (4.) What is the area of a room a? +2/ feet long and x—y feet broad ? (5.) A man walks x miles at a miles an hour ; how long is he on the road ? (6.) A has X dollars, B 50?/ cents, and C 76z cents ; how many dollars have A, B, and C together ? EXAMPLES OF THE FIRST FOUR RULES. 41 (7.) A has X pounds, B has y shillings, and C z pence ; how many pounds have A, B, and C together ? (8.) A spends a dollars in x days ; in how many days will he spend 10 dollars? (9.) How many square yards in a floor which is a feet by X feet ? (10.) What is the cost in dollars of painting a floor aj-hy feet by 05— y ^66* at x'^-k-y^ cents per square foot? (11.) A owns a acres, B 6 acres, and C 5 acres less than one- half of A's and one-third of C's together; what is the whole amount possessed by A, B, and C ? (12.) A owns a+ 6 acres, B a— & acres, C half as much as A, and D half as much as B ; how much more do A and C own than B and D? (13.) A walks a miles in t hours, and B half as far again in the same time ; how far will B walk in 10 hours? (14.) A walks 10 miles in x houis ; how long will he be in walking a miles? (15.) A spends at the rate of x dollars a day for a days, a dollar a day less for twice that time, and a dollar a day more for three times that time; how much does he spend altogether? c 42 ) CHAPTER VIII. SIMPLE EQUATIONS. I I i; V 60. An equation is the statement of the equality of diiferent quantities, and these quantities are called the equation's members or sides. Thus 205+3=7 is an equation whose sides are 2a? + 3 and 7, and a;*— 5a; +6=0 is an equation whose sides are a;'''— 5a+6 andO. 61. An identity is the statement of the equality of two like or different forms of the same quantity. Thus 2a + &=2a + 6, 2a; + 3a;=5a;, x^-bx-\-6=(x-'2X«:-S), are identities. 62. In the case of an identity the equality holds for all values of the quantities involved, whereas in an equation the equality does not exist except for a limited number of values of the quantities involved. Thus the statement a;^+2a;+l= (a; + l)^ holds no matter what X is; but 5a;— 3=7 holds only when a;=2,and a;^+6=5a; only when a;=2, or a;=3. 63. A symbol to which a particular value or values must be assigned in order that the statement contained in an equa- tion may be true is called an unknown quantity. Thus the unknown quantity in 6a;— 6=9 is x, and in 22/2-2/=8 is y. The letters a, &, c, Z, m, ?^, ^, q, r are generally used to denote quantities which are supposed to be known, and x,y, z those which are for the time unknown. SIMPLE EQUATIONS. 43 -2/,» &i:. Quantities which on being substituted for the unknown reduce the equation to an identity are said to satisfy the equation, and are called its roots. Thus 5 is a root of 2ac— 3=7, because 5 when substituted for X reduces the equation to the identity 10—3=7. So 2 and 3 are the roots of a;'' + 6=5a;, because when either is sub- stituted for X the equation is satisfied. 65. The determination of the root or roots is called the solution of the equation. 66. An equation is said to be reduced to its simplest form when its members consist of a series of mononomials involving positive integral powers only of the unknown. Thus5a;-8=0, a;2-5a; + 8=0, 2a;H6a;=7, a;^- 6*2= 7a; -8 are in their simplest forms. 67. Equations when reduced to their simplest forms are classed according to their order or degree. 68. Simple equations, or those < f the first degree, are those in which the highest power of the unknown quantity is the first; as, for example, 2£c=5, 5a5— 8=0, 305—7=0. 69. Quadratic equations, or those of the second degree, are those in which the highest power of the unknown quantity is the second; as, for example, a;^— 2xH-3=0, 2a;2=9, 4a;2_3=10a;. 70. Equations of the third and fourth degrees arc called cubic and biquadratic equations, respectively ; thus r' + 2a;=10 is a cubic, and a;^— 2a;^=10a;— 5 a biquadratic. 71. It is proved in works on the Theory of Equations that the number of the roots of an equation is equal to its degree ; so that a simple equation has one root, a quadratic two roots, a cubic three roots, and so on. 72. In order to solve an equation it is generally necessary to reduce it by one or both of the following processes : — I. Transposition of Terms. II. Clearing op Fractions. 44 SIMPLE EQUATIONS. %\ These operations will be illustrated by applying them in order to the solution of simple equations. I. Transposition of Terms. 73. If an equation contains no fractions it may be solved by transposition of termSf which consists in taking the unknown quantities to one side of the equation and the known to the otheVy the signs of the quantities which are so transposed being changed. Thus, if the equation is 4a; +5=10, by subtracting 6 from each side we get 4a; +5-5=10-5, or 4a;=5; and so any quantity may be transposed from one side to the other by changing its sign. Examples. (1.) Solve 5a;+15=25. Transposing + 15 we get 5a;=25-15=10. The value of x is then found by dividing both sides by its coefficient 5. /. x=2. (2.) Solve 8a;-4=2a; + 20. Transposing —4, 8a!;=2a5 + 20 + 4. Transposing 2x, 8a; — 205 =20 + 4 ; 6a;=24. :.x =4. (3.) Solve 10 + 2(6a;-l) =32-3(a;-4). Clearing of brackets, ^; 10 + 12X— 2=32-3a; + 12. Transposing 10, —2, —3a;, i 12a; + 3a;=32+12-10+2; [ 15x=36. • «._ 3.6 —i)2. "^ SIMPLE EQUATIONS. 45 (4.) Solve 3(a;*+2x) + 13 = 3a;2-7+4(3a;-l). Clearing of brackets, 3a;H6a; + 13 = 3a;2-7+12aj-4. Transposing +13, 3a;'», +12a?, 3a!2-3a;2+6a;-12a; = -13-7-4; Dividing by— 6, - 6x = -24. 03 = 4. When, as in this case, the same quantity is common to t)oth sides, it may be struck out without actually transposing ; thus 5a;2-6a; + 7=8a; + 5!»2_10 becomes — 6a3 + 7=8a;— 10. (5.) Solve ax-\-'b=c. Transposing +&, aa5=c— 6. Dividing by a, a;=£rL. a (6.) Solve ax-\-'b—cx-\-d. Transposing +&, ex, ax—cx = d'-'b. Collecting coeflBcients of oj, (a— c).r = c?— 5. Dividing by a— c, a—c (7.) Solve a(a;— &) = &(a; + a)— c. Clearing of brackets, ax—ab = hx-\-ab—c. Transposing — a&, hx, ax—bx = ab-\-ab—c. Collecting coefficients of a;, («— &)a; = 2a&— c. Dividing by a— &, a;= 2a6— c 4« SIMPLE EQUATIONS. Exercise XXII. (1.) 3-|-a;=6. (2.) «-6 = 4. (3.) a; + 5 = 12. (4.) a; + 9=4. (5.) 2x-l=3. (6.) 5a;+4 = 29. (7.) 4-3a;-5. (8.) l-aj=6. (9.) 3=6-2». (10.) 2a; + 3 = .'r + 5. (11.) 5i»-2-2x + 7. (12.) .'c+4 = 18-4x-4. (13.) 2r+3 = 3x-4. (14.) 16-2«=46-5a;. (15.) 3(a:-l)+4 = 4(4-aj), (16.) 5-3(4-2a;)+4(3-4a!)=0. (17.) a!-l-2(a;-2) + 3(£c-3) = 6. (18.) 6(c»-.5)+2(a;-3)-(»>-l)=9. (19.) 2(x-.2)-3(«-3) + 4(c«-4)~5(a;~5) = 0. (20.) 4(a;-ll)-7(x-12) = 6-(a;-8). (21.) «=2a-». (22.) 2a-3a;=8a-5a;. (23.) a;-2&=2a~a;. (24.) a + o:-& = a + 6. (25.) aas— a&— ac=0. '26.) ax—a = b-—bx. (27.) ax-a^=hx^b\ (28.) «(«-&) = c(a3-a). I II. Clearing of Fractions. 74. If an equation contains fractions, it may be reduced to a form capable of solution by transposition, by multiplying both sides of the equatio7i by the L. C M. of all the denominators of the fractions. In the following examples numerical denominators only will be considered. (1.) Solve XXX 2 3 5 Multiplying by 30, the l.c.m. of 2, 3, 5, 15x — lOx = 6x — 30. Transposing, 15a; -- lO.r — 6x = — 30 ; -a; =-30. .*. x = 30. (2.) Solve ^^^^&« + l?^ A o o Multiplying by 24, the l.c.m. of 2, 3, 8, 12(a!-l) + 8(2a; + 3)=:3(6.K+19); SIMPLE EQUATIONS, 47 whence on clearing of brackets and transposing we get It must be observed that ^=i(a;-l), ?^= i(2a; + 3), 3 6x4-19 fl-'^d y =J(6a? + 19); and therefore the brackets must be supplied in the first step since the numerators become bino- mial factors. (3.) Solve 8-^:=l + ^+2^(j 2 • 3 Multiplying by 6, the l.o.m. of 2 and 3, ■18~3(a;-l) + 2(aj + 2) = 0; whence on clearing of brackets and transposing we get a5=25. m X X Exercise XXIII. X , X (l.)i-i=3. (2.) ^+^=7. (3.) 1-1+1=10. (4.) ^+^=10. 2 3 '4' (5.) ^+1=20-^9 (6.) ^+^=4-^. (7.) ^ + 2=^-K«^ + l). (8.) 2(a)-l)-K2a;-9) = K17-2x). (9.) §^=4-^^-12 (10.) |(x-4) + K«'-6)=2x-15. (11.) 3-|cc=l-J^(7a;-18). (12.) 6(x-l)-l=f(5-2x) + .3(x + l)+4a;. (13.) x+^^—^^-9h=0. (14.) 4.+2,V+^-^^-^-^=^+M. /iK\ * + 4 3a;--2 , 1 x—1 \^'^') — o -ITT- +4 = 12 3 • ..n. S-2x 6x 5_3(2x + 6) 2x ^,10.; -^-+7—7 14- -y 48 SIMPLE EQUATIONS. ,.^.?>x-\ 13-;r 7.T_^ll.x-|-33 n (1 i .) —^ >.- —-rr+ ^. =^- 5 6 ^^'•^2 3'T 5- + " 9"-^* (20.) |(x-.8)-'^--*-g=0. ih •I ( 49 ) 11 CHAPTER IX. PROBLEMS, 75. When a question is assigned for solution the unknown quantity, or number, is generally involved in the various conditions which are proposed for its determination. The expression of these conditions in Algebraical language leads to an equation, the solution of which will be the solution of the question. 76. In some cases, although there are more unknowns than one, they are related to each other in such a manner that when one is determined the others become immediately known. In such cases the unknowns can be expressed in terms of one unknown. Thus, if the sum of two unknowns is equal to 8, we may denote one of them by x and the other by 8— jc, or one of them by 4: + a3 and the other by 4— a?,- if the greater of two unknowns exceeds the less by 3, the former may be denoted by X, and the latter by jc— 3 ; if there be two numbers of which one exceeds 4 times the other by 7, the former may be denoted by 4a3 + 7, and the latter by x. In like manner, if there be three unknowns, of which the first exceeds the second by 3, and the second exceeds the third by 5, the first may be denoted by a?, the second by 03—3, and the third by as— 8. 77. The following examples will illustrate the method of solving problems by means of simple equations : — (1.) What number exceeds its fifth part by 20 ? Let X be the required number. E 50 PROBLEMS. 1^' m Then its fifth part =. | ; and by the condition of the question 5 /. x=25. (2.) The Slim of two numbers is 71, and their difference 43 : find them. Let x be the greater number. Then .a— 43 is the less : and since their sum is 71, we have a; + a;-43=71. /. a; =57, the greater; and a; -43 = 14, the less. This question may also be solved as follows: Let X be one number, the greater suppose. Then 71 -a; is the less i and since their difference is 43, we have „^ ^ Ao a;-(71-x)=43. .*. a; =57, and71-aj=14. (3 ) A boy is one-third the age of his father, and has a brother one-sixth of his own age; the ages of all three amount to 50 years. Find the oge of each. Let the boy's age z=x years. Then the father's age —'6x years, And the brother's age =| years. % \ And by the condition of the question o /. .x=12, 3a;=36, ^-2. Fractions may be avoided by supposing the ages of boy, father, and brother to be 6a?, 18a7, x years, respectively. PROBLEMS. 5' 5: re we IS a iree (4.) A and B start from two places, 90 miles apart, at the same moment, A walking i miles per hour, and B 5 ; when will they meet, and how far will each have walked ? Let the time of meeting be x hours after starting. Then A will have walked 4a5 miles, and B 5x miles ; and since the sum of these two distances is 90 miles, 4a; + 5x=90. /. aj=10. /. 405=40, and 5x=50, are the distances in miles walked by A and B, respectively. (5.) How much tea at 90 cents per lb. must be mixed with 50 lbs. at $1"20, that the mixture may be sold at $1*10 ? Let X = the number of lbs. at 90 cents, the value of which will be '900; dollars. Then, since there will be a? + 50 lbs. in the mixture, its value will be 1*10 (cc + 50) dollars ; and since the value of the 50 lbs. at $1'20 is 60 dollars, we have •90x + 60=l-10(£c + 50). Multiplying by 100, 90a; + 6000=110(x- + 50). .*. 3J=25. boy, Exercise XXIV. (1.) Divide 25 into two such parts that 6 times the greater exceeds twice the less by 70. (2.) Divide 135 into two parts such that one shall be |- the other. (3.) The sum of two numbers is 37 and their difference 3 : find them. (4.) A fish weighed 71bs. and half its weight : how much did it weigh ? (5.) At a meeting 43 members were present, and the motion was carried by 9 : how many voted on each side ? E 2 5« PROBLEMS. Xh (6.) Divide 326 into two parts, such that f of the one shall be equal to the other diminished by 7. (7.) What is the number whose 4:th and 5th parts added together make 2i ? (8.) Forty-two years hence a boy will be 7 times as old as he was 6 years ago : how old is he ? (9.) A father is 57 years old, his son 13 : when will the father be 3 times as old as his son ? (10.) I have made 164 runs at cricket this season in 12 innings: how many must I make in my next innings to average 14? (11.) My grandfather told me 10 years ago that he was 7 times as old as myself; I am now 18: how old is my grandfather ? (12.) If in a theatre f of the seats are in the pit, ^ in the lower gallery, \ in the upper, and there are 50 reserved seats, how many are there altogether ? (13.) After losing \ of our men by sickness, and 210 killed and wounded, the regiment was reduced by \ : how many men did the regiment originally contain ? (14.) In a certain examination f of a boy's marks were gained by translation, ^ by mathematics, and -^ by Latin prose : he also obtained 1 mark for French. How many marks did he obtain for each subject ? (15.) Two men receive the same sum ; but if one were to receive 15 shillings more, and the other 9 shillings less, the one would receive 3 times as much as the other. What sum did they receive ? (16.) A and B begin trade, A with 3 times as much stock as B. They each gain £50, and then 3 times A's stock is exactly equal to 7 times B's. What were their original stocks? (17.) One-tenth of a rod is coloured red, one-twentieth orange, one-thirtieth yellow, one-fortieth green, one-fiftieth PROBLEMS, 53 •e to the sum blue, oue-sixtieth indigo, and the remainder, which is 302 inches long, white : what is its length ? (18.) Find three numbers whose sum is 37, such that the greater exceeds the second by 7, and the second exceeds the third by 8 (19.) Find a number such that if 5, 11, jmd 17 be suc- cessively subtracted from it, the sum of the third, fourth, and sixth parts of the respective results shall be equal to 19. (20.) How much wheat at 44s. a quarter must be mixed with 120 quarters at 60s. that the mixture may be sold for 50s. a quarter ? (21.) How many lbs. of tea at 2s. 6d. per lb. must be mixed with 18 lbs. at 5s. per lb. that the mixture may be sold for 4s. per lb. '? (22.) How much sugar at 4id. per lb. must be mixed with 50 lbs. at 6^d. per lb., that the mixture may be worth 5d. per lb. ? (23.) A bag contains a certain number of sovereigns, twice as many shillings, and three times as many pence ; and the whole sum is £267 ; find the number of sovereigns, shillings, and pence. (24.) I wish to divide £5 4s. into the same number of crowns, florins, and shillings ; how many coins must I have of each sort ? (25.) A person gets an income of £550 a year from a capital of £13,000, part of which produces 5 per cent, and part 4 per cent. : what are the amounts producing 5 and 4 per cent., respectively ? (26.) I invest £800, partly at 4| per cent., and partly at 5^ per cent. ; my income is £39 10s. : what are the sums invested at 4 J and 62 per cent., respectively ? (27.) A garrison consists of 2600 men, of whom there are 9 times as many infantry and 3 times as many artillery as there are cavalry : how many men are there of each ? (28.) My grand&ther's age is 5 times my own ; if I had .1 54 PROBLEMS. U\ been born 100 years ago, I should have been born 15 years before my grandfather : how old am I ? (29.) There is a number consisting of two figures of which the figure in the unit's place is 3 times that in the ten's; if 36 be added, the sum is expressed by the digits reversed : what is the number ? (30.) A miner works for 6 weeks (exclusive of Sundays), his wages being it the rate of 24s. per week, but he is to forfeit Is. besides his pay for each day that he is absent ; at the end of the time he receives 4 guineas : how many days was he absent ? (31.) A contractor finds that if he pays his workmen 2s. 6d. per day, he will gain 10s. per day on the job ; if he pays them 3s. a day, he will lose 18s. : how many workmen are there, and what does the contractor receive per day? (32.) An officer on drawing up his men in a solid square finds he has 34 men to spare, but increasing the side by 1 man he wants 39 to make up the square : how many men had he ? (33.) If the mean velocity of a cannon-ball at effective ranges is 1430 feet per second, and that of sound 1100 feet, how far is a soldier from a fort who hears the report of a gun 1% of a second after he is hit ? (34.) An army in a defeat loses one-sixth of its number in killed and wounded, and 4O0O prisoners. It is reinforced by 3000 men ; but retreats, losing a fourth of its number in doing so. There remain 18,000 men. What was the original force ? (35.) Su])pose the distance between London and Edinburgh is 360 miles, and that one traveller starts from Edinburgh and travels at the rate of 10 miles an hour, while another starts at the same time from London and travels at the rate of 8 miles an hour : it is required to know where they will meet. (36.) There are two places 154 miles apart, from which two persons start at the same time with a design to meet ; one travels at the rate of 3 miles in 2 hours, and the other at the rate of 5 miles in 4 hours : when will they meet ? ( 55 ) CHAPTER X. PARTICULAR RESULTS IN MULTIPLICATION AND DIVISION. 78 There are several results in multiplication and division which should be committed to memory, as they enable us to dispense with the labour of performing the operations. The following cases occur most frequently. I. Since by actual multiplication (a + by=a^ + b^ + 2ah, &c. = &c. we can hence write down the square of a polynomial bv the rule: ^ The square of a polynomial is equal to the sum of the squares of the several terms and twice the sum of the products of every tivo terms. ^ Thus in the last example a\ +&2, a,c\ are, respectively, the squares of a, +&, -c; +2a& is twice the product of a and + &, -2ac IS twice the product of a and -c, and -2ic is twice the product of +& and — c. In taking the products of the terms, two and two, it will be found most convenient to take in order the products of the first term and every term that follows it, then the pro- ducts of the second term and every term that follows it, and «o on, if there be more terms than three. m^ 56 PARTICULAR RESULTS IN Examples. (1.) (a + 2ie)2=a2+4a!2+4aa;. Here +4aa3 is twice the product of a and +2a;. (2.) (2a-5£c)*=4a2 + 25a;2-20aaJ. Here +250?^ is the square of —6x, and — 20aaj is twice the product of 2a and —6x. (3.) (2ar2-3a;+4)2=4a;* + 9a;2 + 16-12a;3 + 16»2_24a; =4x^-12a;H25!»'^-24x + 16. Here — 12a5^ is twice the product of 203^ and —3a!, +16a!r' of 2x^ and 4, and —24a; of —Sx and +4. Like terms are added together and the terms are arranged according to descending powers of x. (4.) 992 =(100-1)2=10000+1-200=9801. Exercise XXV. Write down the squares of (1.) 03—1, x+a, x—5, x-\-d. (2.) 2a; + l, Sx-1, 2x+3, 3a;-2. (3.) x"-a, 2xy + l, Sx^-'2a, aa;2-46. (4.) 05— 2/ + Z, 2oj + 32/— s, x—2y^^z, 2a5— 42/ + 1. (6.) 2a2 + a + 3, 3a2-4a + l, a2-2a-4. (6.) Find the squares of 49, 98 and 995. 79. II. Since (a +6) (a— 6) =a^^b^, it follows that the pro- duct of the sum and difference of two quantities is equal to the difference of their squares. Thus (2xVSy)(2x-Sy) =^x^^Qy^; (a2+l)(a2_l) =a4_l; (50)2+42/) (5a;2-42/) ='26x*-16y^; (2ar»+a*)(2a;3-a^) =4:x^^a^; 501 X 499 =(500 + 1) (500-1) =50(^-1 =249999. MULTIPLICA T20N AND DIVISION. 57 Exercise XXVI. Write down the products of (1.) oj-l, x + 1; a+3, a-3; 2+03, 2-a;. (2.) 2j5 + 1, 2x-l; 5a+2, 5a-2; 4a3+a, 4a;-a. (3.) aHa;,a^--a5; aHl,a'-l; a'* + a;2, aO^^jZ^ (4.) 3aH26, 3a2-2&; 4a»+2a;2, 4a3-2x2; 7a*-5a3, 7a*+5a8. (5.) Find the products of 48, 52; 95, 105; 695, 705. 80. III. Since by actual multiplication Id'^W-k-ab) (a-&) =a3-63, it follows that the sum of the squares less the product of two quantities multiplied by their sum is equal to the sum of their 9ubes. In the latter identity the two quantities are a and —h ; the sum of their squares less their product is, therefore, ^2+62 ah=a'^-\-b^-\-ab; and, since the cube of —6 is —6', the sum of their cubes is a^—b^. Examples. (I.) (x^-x + l) (x + 1) =x^-\-l. Here the two quantities are x and 1. (2.) (x^+x+1) (x-l)=x^-l. Here the two quantities are x and —1. (3.) (4a;2-2x + l)(2«+l)=8x» + l. Here the two quantities are 2x and 1, the cubes of which are 8x^ and 1. (4.) {x'--a^x' + a')(x^-\-a^)=x''-\-a\ Here tlie two quantities are x^ and a^, the cubes of which are .x^' and a*^. (5.) (4:x^-\-6x^ij-\-di/) (2x^-Si/) =8x^-27 f. Here the two quantities are 2x^ and —dy, the cubes of which are Sx*' and —27y\ 58 PARTICULAR RESULTS IN \ ' EXEUCISE XXVII. Write down the products of (1.) m^"mn-\-ii^,in-^u; 2^"-^Pl-^Q^>P-'Q' (2.) m^-m + l, ?/i + l; l + q-{-r,l-fj. (3.) x^-Bx + 9,x + 3', a2+4a-l-lG,«-4. (4.) 4a2-2a + 1, 2a + 1 ; leas^ + 4ax + a-, ^x—a. (5.) ^a'-Qab + W, 2a + 36; 9xH 15x7 + 25^/2, 3a;- S^/. (6.) a;< + a;2 + l, a;2-l; (c^-aV+a*, ccHa^ 81. IV. By actual division it can be shown that the sum of any the same odd powers of two quantities is exactly divisible by the sum of the quantities. Thus, ^±i' =1, x+y t±y'=x'-xy + y\ x-\ry J-Ji-J- = 03* — xhj + x^y^ — xy^ + y^, &c.=&c. It will be observed that the signs of the quotient are alternately + and — , and that the successive powers of x are in descending whilst those of y are in ascending order. 82. V. The difference of any the same odd powers of two quantities is exactly divisible by the difference between the quantities. Thus, — X, x—y _ — ^=zx^-\-xy-\-y'^, x-y x—y ic* + x^y + x^^y"^ + xy^ + y* &c.=&c. h MULTIPLJCATJON AND DIVISION, 59 Hero tlie signs of the quotient are all + . It may be noted that this case is included in the preceding (81) by supj)osing the two quantities to be x and — y. Thus the sum of the cubes of x and — ?/ is a;^— t/^ which is exactly divisible by their sum x—y. In fact the formulas of (82) are deducible from those of (81) by substituting —yioxy in the latter. 83. YI. The dljference hetvjeen any the same even powers of tioo quantities is exactly divisible by the sum of the quantities and also by their difference. Thus (i.) y, x + y — ^ =jf-j'y-\rxy^^y^, x-{-y 03' P — lfi yi. =x'^—'ji*y + x^y^—a^y^-\-xy*--y^f (ii.) x-i-y &c. =&c. x-y -^^ =x^-{-x^y + xy^-\-y^f x-y /yjC _^ yd iZ. = x** + 05*2/ + x^y^ + xy^ + xy* + y'^, x-y &c.=&c. It will be observed that when the divisor isx—y, the signs of the quotient are all + ; and when the divisor i&x + y the signs of the quotient are alternately + and — . It should also be noted that the formulas (ii.) are deducible from (i.) by substituting —y for y in the latter. Exercise XXVIII. Write down the quotients of (1.) x^+1 and 03^ + 1 divided by a; + 1. 6o PARTICULAR RESULTS. II ! i: (2.) a;3_i and ;r'-l divided by x^l. (3.) ^2-1 and «<-! divided by « + !. (4.) x^-l and a;*-l divided by a;-l. (5.) 4a2-9i2 divided by 2a + 36. (6.) 9j^«-4a2 divided by 3.x3-2a. * (7.) ia'-a;« divided by i«Ha;3. (8.) Find what the quotient of x^ + if divided by aj+y becomes when (i.) a;=2a, y=36 ; (ii.) x=a\ y=2. (9.) Find what the quotient of x^^y^ divided by x-^y becomes when (i.) a;=3a, y=6 ; (ii.) aj=2a2, y =36. 84. expr( of pc expm^ 85. the p the p with of the Thi i; I Ii i: t ( 6l ) CHAPTER XL 3y a;+y jy x—y INVOLUTION AND EVOLUTION, 84. The process by which the powers of quantities are expressed as mononomials is called Involution, The powei-fl of polynomials when so expressed are said to be developed, or expanded. 85. We have already explained the notation for denoting the powers of a single symbol, as a, as, y. In all other cases the power of a quantity is denoted by enclosing it in brackets with the number indicating the power above and to the right of the bracket. Thus (—2a)'' denotes the third power of —2a; {a^lif the square of d^h\ {a%c^y the fourth power of a%c^\ (a— by the cube of a— 5; (x"— 2x+3/ the fifth power of a;'^— 2x+3. 86. The same notation is used for denoting powers of powers of a quantity, brackets of different shapes being em- ployed when necessary. Thus (a^y denotes the square of a'; {(— 2icy)'^p the cube of (-2«2/)2 ; {(x^-S)'}^ the fourth power of (x^-bf. Exercise XXIX. Retaining the given quantities, denote (1.) The cubes of -a, 2x, Zxy\ 'la^Wc. (2.) The squares of 2a-l, a-& + l, x^—l. (3.) The squares of (xO^ (-2a)3, {4mx)\ (3a%c*y. (4.) The cubes of (a-6)3, (jb^-I)^, {9^^3x+2y, 62 INVOLUTION AND EVOLUTION. (5.) The squares of the cubes of x, — 2vC, aP-h^ a;— a, (6.) The cubes of the squares of —a, x^, 4x— 1, as^— a'*. 87. A power of a power of a quantity is expressed as a power of that quantity according to the rule Thus, a ) —CI . 88. So also Thus, (a"'¥c'^y=ia""''hP"-ci\ &G, (abhy=a^b'V-'. 89. A rule has already been given in Art. 78 for expanding the square of any polynomial. The expansion may also be effected as in the following examples, in which the various parts are arranged in rows. Jn the first row occurs the square of the first term of the given qucDittfy ; in the second row the ^>roduct of twice the first term added to the second and the second; in the third row the product of twice the first term added to twice the second term added to the third term and the third ; and so on. Mf Examples. (1.) (a + 6)2=a2 + (2a + 6)&=&c. (2.) (&-c)^=62 + (26-f)(-c')=&c. INVOLUTION AND EVOLUTION. ^^h, jB— a, (3.) (af&+c)2=a2 + (2a + &)6 f^a\ + (2a + 26 + c)c=:&c. ised as a 1 (4.) (a-5-.c)2=a2 i + (2a-.&X~5) 1 + (2a-2&-c)(-c)=&c. 1 (5.) (a2~& + c2-(^)2=((^'2)2 1 + (2«'-&)(-&) 1 + (2a2-.2& + c2)c2 J + (2a2-2& + 2c2-t^)(-c?v.==&C. 63 panding also be Various (3 square row the and the ^st term and the Exercise XXX. Express as powers or products of powers (1.) (x')\ (2xy, (x^)3, (2x3)8, (3x2)*. (2.) {ax'f, («V)2, (ah;h/)\ (3.) {ahcj, {aV)\ (2a^6V)3. (4.) {x-ifzy, (a6V)«. Expand (5.) (x + l)2, (2^-3)2, {x^^bf, {x^-^a'f. (6.) (a;2 + 2.c + 3)2, (a;2_3x + 4)^ (2x«-x«+5)2. 90. Pligher powers of polynomials are developed by tlio Bmoraial and Multinomial Theorems, the explanation of which may be found in more advanced works. 91. The process by which the roots of quantities are deter mmed is called Evolution., 92. The nih root of a quantity is denoted by writino- the quantity under the sign V" , the line above being sometimes replaced by brackets enclosing the given quantity. Thus, a/2«. denotes the square root of 2 / ; K,.2 J> 5r«2j ^«^ „ cube ^^^^'+3 „ fourth „ xH3; ^(«2-2a + 3) denotes the lifth root of «'^-2r< + 3. !! li 1! 64 INVOLUTION AND EVOLUTION. 93. The mth root of the wth root of a quantity is denoted by writing the quantity under the sign "v ^ . Thus, V V2a denotes the cube root of V2a ; V^4^6^ „ fourth „ >^5^; V Vo;*— ir'+l „ cube » 94. The mth root of the wth root of a quantity is expressei as a root of that quantity by the rule Thus, 3 VVa = v^(i; V V23D = '>i/2«; 95. The reason of this rule will appear from the following case : — Let s/ i>Ja-s.x, Then on cubing both sides ot this equation we have Squaring a=a?«. Extracting the sixth root, Exercise XXXI. Retaining the given quantities, denote (1.) The square roots of 2aj, ax^^ 03^—1, a;^— 3a;+4. (2.) The cube roots of — a;^ 3a^ a - &, (a^— 3a+4). (3.) The square roots of the cube roots of 3aa;, a;— 1, (4.) The fourth roots of the cube roots of 2, 3a;— 1, 2a*-aH3. Express as roots of the quantities under the double sign (5.) ^/^!a, y/I^, \/V3a, \/,yi^. IN VOL VTION AND E VOL UTION. 65 (6.) \/-^x^-l, V^'-^'Jar^-O, V^^x«-6;k* + 7. 96. Since the square of a quantity is equal to the square of the same quantity with its sign or signs changed, it follows that there will be two square roots (if there be any), one being derived from the other by a change of signs. Thus, since ( + fr)^=(— a)-= + rt'-^, it follows that the square root of +«- is +a or — tt. These two roots may be repre- sented by the symbol ±« (read jilas or minus a) ; so that wo have ^a^=±a; ^x*^±x^; /^dx^=±3x. Again, since by Art. 78 &c.=&c., it follows that /^a'—Saft + &''=«—&, or — a + 6, = ±(a-b); v'CaH&^ + c^— 2t«6 + 2ac— 2&c)=:a— /) + c, or — (i + &— c = ±(a-& + c). In the following examples we shall only determine thtict square root of a polynomial whose leading term is +, the other being derivable by a mere change of signs. 97. Since \^a;'^"'-=a;'», it follows that j^ X ^^X" f \X ^^X , j^ X =: X \ where it will be observed the index of the root is one-half the index of the given power. 98. The square root of a polynomial can generally be found by the following rule. (i.) Ari'd'iuje ih*' f/iiX'n ijiiantitu uccordiHg to afictniding or tiescettdhvj jiowers 0/ f^oiiie letter. (ii.) The first ternt of the root is the square root of the leading tor lit of the given (jnuhtit^,froui wJtich <7.s sqiatre is suhtractcdy leaving the first difference. (iii,; The frst divisor is fn'ire the first term <f the root added W 66 INVOLUTION AND EVOLUTION. to the second term. The second term is the quotient of the leading term of tlie first difference divided hy the leading term of the first divisor. The product of the first divisor and the second term of the root is subtracted from the first difference, leaving the second difference. (iv.) The second divisor is twice the sum of the first and second terms of the root added to the third term. The third, term is the quotient of the leading term of the second difference divided hy the leading term of the second divisor. The product of the second divisor and the third term of the root is then sub- tracted fro7n the second difference, leaving the third difference. The process is thus continued until the difference is zero. Examples. (1.) Find the square root of 9fl5^— 12x+4:. 9a;2-12x+4(3x-2 9x^ 6a;-2)-12x+4 -12a;+4 Here the first term 3a; is the square root of the leading term of the given quantity, from which its square 9x^ is sub- tracted, leaving — 12ic+4. The leading term of the first divisor is 2 x Sx=6x. This is divided into — 12j;, the leading term of the first difference, giving —2, the second term of the root, which is also the second term of the first divisor. The product of 6x—2 and —2 is subtracted from the first difference, leaving remainder zero. The root is therefore 3a;-2. (2.) Find the square root of 4«^— 12*H5a?2 + 6x + l. 4^ *__ 4x2 -.3^)« 12x3 + 5^ij + 6x + 1 ''12>:^-\-9x' 4a;2-(k-l)-4xH 6a; + 1 -4^H6,r + l INVOLUTION AND EVOLUTION. 67 of the ig term md the ference, rst and \e third •fftrence product hen siih' fence. s zero. leading is sub- lie first lleading term of livisor. lie first lerefore The first two terms, 2x^ and — 3x, are found as in Ex. 1. The first two terms of the second divisor = 2 (2/j^ — Sx) = ^x^—6x, the leading term of which is divided into —4:x^, the leading term of the second difference, giving— 1, the third term of the root, which is also the third term of the second divisor. The product of 405^ — 6aj — 1 and — 1 is then sub- tracted from the second diff'erence, leaving remainder zero. The root is therefore 2x^-^3x—l. The latter terms of the differences need not be expressed until the corresponding terms in the partial products are to be subtracted ; thus in the foregoing example 602 + 1 might have been omitted from the first difference. (3.) Find the square root of x^^ix^y -\-10x^y^—12xy^ -\-9y\ x*-4.<€hj + lOxV- 12a.y + 92/* {x^-^^xy + 3y^ ^Ix^ — 2x2/) — 4x^2/ + lOx'^2/'^ —4x^2/+ ^^y^ 2a;2-4^^H^^^2^~6«V-12a;2/H 92/* 6a;y-12x2/^4-92/* In this example the given quantity is arranged according to descending powers of x, and the first two terms only of the first difference are expressed. 99. The reason for the rule given in the preceding Article will appear from the following method of considering the last example. The given quantity is thei*e seen to be equal to a?» that is, to ^i2x"-^2xy)(i-2xy) -ti2x^-Axy+'5y')Sy^ which by Art. 89 is equal to {x^ — 2xy-\-oy^y. Now, since a?' = (a--)-, the first term, x^, of the root is the square root of X*, the leading term of the given quantity. Also since —4tx^y — 2x\—2xy), it follows that the second term —2xy is the quotient of F 2 •:H 68 INVOLUTION AND EVOLUTION. — 4a;'y, the leading term of the first difference, divided by 2j;-, the leading term of the first divisor. Again, since 6a;-//- = 2x^(3,7^), it follows that the thii'd term ^xj^ is the quotient of 6,/;-^-, the leading term of the second difference, divided by 1x^^ the leading term of the second divisor. 100. When the process for extracting tho square root is applied to a quantity which is not an exact square, a result is obtained the square of which added to the last difference is equal to the given quantity. Example, Find three terms of the square root of 1— 2a;. x^ 1— 2a5(l—a;— o" _1_^ 2-!c)-2a; -2x+a;2 |-2a;-|]- X' -ajHccH?- — 05^ — X* : r,ti' I The dd 03"— ^ square root is, therefore, l^x—% and remainder Hence (l-.-fj-.?--^ = l-%c. EXERCISIS XXXII. Find the square roots of (1.) ^ti*h\ 25x2//6, ^UY^ (3.) a6x2-36.»-H9. X (5.) a^H^-f,^. (2.) 16^H40a; + 25. (4.) 1 + 6a; + 9x2. (6.) x2~7x + ?. INVOLUTION AND EVOLUTION. 69 2.C-, the 3/y2), it leading I of the root is - result •ence is linder (7.) 4a;2-Aa: + yi^. (8.) 4a;'-'-12x2/ + V. (9.) a;H4a;3 + 6a;2 + 4x + l. (10.) «* + 2xH3a;2+2.x + l. (11.) .x*-4a;3?/ + 6.xV-4a:2/H?/. (12.) 4,x«-4a;^-ll** + 14.«3 + 5ir-^-12.« + 4. (13.) Extract to three terms the square root of 1 + ic. 101. The method of extracting the square root of a numerical quantity is founded on the Algebraical process, as will appear by- comparing the examples giv^n below. We shall first show how the number of figures in the root is determined by dividing the given quantity into periods. Since ^\-\, ^lOO^lO, ^10000=100, ^1000000=: 1000, &c., it follows that the square root of a number between 1 and 100, that is, containing 1 or 2 figures, lies between 1 and 10, and therefore contains 1 figure ; the square root of a number between 100 and 10000, that is containing 3 or 4 figures, lies between 10 and 100, and therefore con- tains 2 figures ; so likewise the squai'e root of a number containing 5 or 6 figures contains 3 figures ; and so on. If, therefore, we divide a number into periods of 2 figures each, beginning at the units, the number of such periods, whether complete or not, will be the number of figures in the root. In Arithmetic if the root is a+64-c4- &C'> 2a is called the p'st trial-dmsorj 2a +26 „ second „ 2a + 26+ 2c „ thinl find so on. j> »> Instead of obtaining the divisors as in the previous examples, we may form them as below, where it will be observed that the sum of a divisor and its last term or digit gives the next trial-divisor. Examples. (1.) a a2+2a6+62+2ac+26c+c2(a+6+c a a^ 2a+b ) 2a6+62+2ac+26c+c2 6 2a6+62 2a+26+c ) 2flrc+26c+c2 2ac+26c+c2 )--t i 70 INVOLUTION AND EVOLUTION. 300 300 10'69'29 (300 + 20+7 9 00 00 600 + 20 20 ) 169 29 124 00 600+40+7 ) 45 29 45 29 In the numericiil example, since the given number contains 3 periods, the root will contain 3 figures. The leading figure of the root, which is also the number of hundreds, will be 3, since the given number lies between 90000 = 300^ and 160000 = 400=. The second figure of the root, which is also the number of tens, is obtained by- dividing the first remainder 16929 by the first trial-divisor 600. The third figure of the root, which is the number of units, is obtained by dividing the second remainder 4529 by the second trial-divisor 640. Omitting all unnecessary figures, we may arrange and describe the work as follows: — 3 8 2 647 10'69'29 (327 9 )T69 124 )4529 4529 li The leading figure of the root is 3, the squai-e of which is the greatest square number under 10, tiie first period ; the square of 3 is subtracted from the first period, and to the remainder is annexed the second period 69 to form the first dividend 169. The first figure of the root is doubled to give the first trial-divisor 6, the division of which into 16 indicates the second figure of the root- The second figure of the root is annexed to the trial-divisor to form the first divisor 62, which is multiplied by the second figure of the root, and the product is subtracted from 109. To the remainder is annexed the third period to form the second dividend 4529. Under the divisor 62 is written its last <ligit, and the sum forms the second trial-divisor 64. The third and last figure of the root is 7, because when annexed to the trial-divisor to form the INVOLUTION AND EVOLUTION, n 7TLT'""''' '^' ^''^'''' '^ '^' '-^ 7 is equal dividend (2.) 2 2 44 4 481 1 4825 5 48306 to 4529, the last 5'83'51'23'36 (24156 4 T83 176 751 481 27023 24125 28^836 289836 ir ( 72 ) CHAPTER XII. TEE HIGHEST COMMON MEASURE, ^.| 102. A QUANTITY is Said to be of so many di'mensions, in any letter, as are indicated by the index of the liighest power of that Ujtter involved in it. Tlids 'dx^—2x+4tis of 2 dimensions in x; 3?/ + 2?/— 5isof 4 dimensions in y ; and ax^—hx^ + cis of 3 dimensions in x. 103. A whole exvression, or quantity ^ is one which involves no fractional forms. Thus 3x^, —^xy, 2a;*— 3x4-4, are whole expressions, as are also all positive and negative integers. 104. When two or more whole expressions are multiplied together, each is said to be a measure of the product, and the product is said to be a multiple of each factor. Thus 1, 4, a, and h are measures of 4a6; 1, 3, x and cc+1 are measures of 3cc" + 3ji;; 5, a;-, x—\, and y'^-^l are measures of 5x2 (^_i) (r-1). 105. It must be carefully observed that the terms measure and multiple are to be used only in connection witli whole expressions. In order, therefore, to obtain a multiple of a quantity it must be multiplied by another whole quantity ; and to obtain a measure of a quantity it must be divided by a whole quantity, the quotient also being a whole quantity. Thus the terms measure and multiple cannot bo us- d in connection with facc^, a:;^— _ + 3, because they involve tractions; whilst 1, 3, a, a;, x^, and « — 1 are measures of 3c<x^(a;— 1), THE HIGHEST COMMON MEASURE. 73 because the quotient of the latter divided by oach of tho former is a whole quantity. lOG. The hlyhei't mean/i.re of a quantity is tho quantity divided by + 1 or —1, that divisor being taken which will make tho first term of tho quotient positive. Thus the highest measure of — 4x- is 4:x^, oti>x—7 is 5;c—7, and of —2x~+ x—S is 2.c^—x-\-3. 107. The Joivest mnUqjIe of a quantity is the quantity mul- tiplied by +1 or —1, that multiplier being taken which will make the first term of the product positive. Thus the lowest multiple of i2x— 3 is 2x— 3, and of —flc^ + aj— 5 is x'^—x + 6. 108. When one quantity is a measure of two or more others, it is said to be a common measure of those quantities. Thus 2x is a common measure of 4x^ and 2.^'-— 4x, and x—1 is a common measure of '2x—2, x^—^.*: + 1, and x^—1. 109. The highest common measure of two or more quantities is the common measure of highest dimensions and greatest numerical coefficient or coefficients. Thus the common measures of li;^ and Qxhj are 1, 2, X, 2x, 2x% of which the last is called the highest common measure (h.o.m.) ; the common measures of 4:{x'^—T) and 6(£c— 1)^ are 1, 2, x-1, 2(x-l), of which the last is the h.c.m. 110. We shall consider the process for finding the h.c.m. in the three following cases, namely — I. When one of the quantities is a mononomial. II. When the two quantities are liolynomials, neither of which has a mononomial measure. III. When the two quantities are polynomials, one or both of which have nioiionomiai measures. 111. I. (i.) When the given quantities consist of two or more mononomials, their h.c.m. is the product of the G.C.M. 74 THE HIGHEST COMMON MEASURE. of the numerical coefficients and the highest power or powers common to the several (jiven quantities. Examples. (1.) Find the h.cm. of ISa&^x^ and Iha^'W. Here the g.cm. of 18 and 15 is 3; the highest power of a common to both is a ; ft it ^ f> M " » and there is no power of x common to both ; .'. H.O.M. is 3aP. (2.) Find the H.O.M. of 12xyz\ IQxYz^ and 28x*yz\ The G.CM. of 12, 16, and 28 is 4; the highest power of x common to the three quantities is x"^ ; It i» y a it » » z it M y ; i .*. H.O.M. is ^a?yz^. 112. (ii.) The h.cm. of a mononomial and a polyn^ni •' is the H.O.M. of the mononomial and the h.cm. of the several terms of the polynomial ; and may be found by the following rule : — Express the polynomial as a product one factor of which is the H.CM. of its several terms: the H.CM. of this simple factor and the given mononomial will he the h.cm. required. Examples. (1.) Find the h.cm. of &aW and 8o:'W-VlaW. Here ^a^h^-VlaW^^a^W (2&-3a), where 4^262 is the h.cm. of U%^ and \2a%^; and the h.cm. of ^^h' and &aW is 2ai^ the H.CM. required. (2.) Find the h.cm. of 15x7/V and lQx'y'^z--l^xhfz^ +20a;Vs^ Here lOxhfz^ - Ibxhfz^ + IQxhfz^ = 6xYzX'2x - 3?/ + 4^), where 5a;2?/V jg the h.cm. of lOxYz^, I5:>y^y^z^, and 20.c:^y^z^ ; THE HIGHEST COMMON MEASURE. 7$ and the h.c.m. of bxhfz^ and li^xifr} is 6xyh'\ the h.o.m. required. Exercise XXXIII. Find the h.c.m. of (1.) 12rt&2 and IGa^fts. (2.) 15a^6 and 20ah\ (3.) da.7i/ and SQxyz. (4.) 'Jax\y^ and 15a^.r2;. (5.) 4:2a^x^y and 35a^ic'^2/'*' (^O «&^cw'y and Sa'^twy'^i^. (7.) 8a*&, 12a»6^ and IGa^i^. (8.) 30a«<2/^ 426a;32/3, and IQcxY- (9.) 4a&2 and Ua^hx-^SahY (10.) 10tt&2cand30a»6H45a2i*. (11.) 10a&2a;y and 4:2a¥cx~70h*ci/. (12.) Swv^t^; and 12uhv^-24:u^viv^+S6uHo*. 113. II. When two polynomials, neither of which contains a mononomial measure (other than unity), involve powers of a single letter, their h.c.m. can be obtained by the following rule : — (i.) Having arranged the given quantities according to descend^ ing powers^ choose that one which is not of lower dimensions than the other as divisor. (ii.) Divide this into the other multiplied hy the least number ivhich will make its leading term a mtdtiple of the leading term of the divisor. When this numher is unity, actual multiplication may he dispensed with. (iii.) Divide the first difference hy the highest mononomial measure contained in it. When this measure is +1, actual division may he dispensed with. (iv.) Bepeat the steps (i.), (ii.), (iii.), ivith respect to this last quotient (or diff'erence) and the first divisor; and so on, until there is no diff'erence. The last divisor ivill he the H.C.M. required. i 76 THE HIGHEST COMMON MEASURE. ,.* It will be observed that no fractions occur in the process, and that the leading signs of all divisors are made positive. Examples. (1.) Find the h.o.m. of 2x2-7a; + 5 and Sx^-Tac+i. 2 2x2-7x+5)6x'^-14a;+ 8(3 6a;2-21x + 15 7)7^£T X- 1 Here since the dimensions of the two given quantities are equal, either one may be made the divisor. 205^—70; + 5 being taken as divisor, Sx^— 7^ + 4 is multiplied by 2 in order to make the leading term Gx^ a multiple of the leading term 1x^ of the divisor. The first difference 7i»--7 is divided by its highest mononomial measure 7. In the next step x—1 and 2x^—lx-\-b are to be treated in the same manner as the given quantities. a;-l)2a;2_7a; + 5(2a; 2x2-2x —5) -5a? + 5 03 — 1 The leading term of ^x^—lx-\r^ is a multiple of the leading term of u:— 1, and therefore the multiplication of the former by 1 is omitted. The difference — 5» + 5 is divided by its highest mononomial measure —5. In the next step the quotient x—1 and divisor x—\ are to be treated as the previous quotient and divisor were. a;~l)a;-l(l OJ-l Th( X — oft the THE HIGHEST COMMON MEASURE. 77 process, sitive. ;ities are - 5 being order to ^erm 2a5^ d by its eated in I leading former by its 1 are to The process thus terminates and the h.c.m. is the last divisor a;— 1. Whenever as in the second step the difference is a multiple of the divisor, the division may be continued and the work of the last step avoided. Thus a;-l)2a;2-7a; + 5(2a;-5 2x2 -2x- — 5x+5 -5x + 5 le whole work may be arranged j JB follows : 3x2 -7x+ 4 2 2ic2. -7aj+5) Gx'^- 6x2. -14x+ 8(3 -21x + 15 7)7x- 7 :=a;~l« X- 1)2x2 2x2 -7x+5(2x- -2x -5 H.O.M. -5x+5 — 5x + 5 (2.) Find the h.c.m. of x2+2x— 3 and xH5x + 6. a;2+2x-3)x2 + 5x + 6(l X- + 2X-8 ~3)3x + 9 ^'+3)x2 + 2x~3(x-l X" + 3x H.C.M.=:X + 3. — X — 3 Here the multiplication of x2 + 5.r + 6 by unity is unneces- isary. The other steps are similar to Ex. 1. (3.) Find the h.c.m. or2x''-7x-2 and 2':^-x-G. 78 "//£ HIGHEST COMMON MEASURE, 2a;2-a;-6)2x3--7a'-2(a;, 1 2i«;^— x"— 6. X X ^--x-^ 2x2-2x-4: 2x2- X- ■6 -l)-a: + 2 X --2)2x2-cc-.6(2x+3 2x^—4:0; fi.o.M.=:a;— 2. 3x--6 Sic— 6 Here 2x'^— cc— 6 is used as divisor in the second step, the dimensions of the first difference being 2. The partial quotients x, 1 of 2x^— 7x— 2 and 2x^—'2x—4: divided by 2x2— cc— 6 ^j,Q separated by a comma to distinguish them from parts of an ordinary quotisnt. (4.) Find the h.o.m. of 4x2 + 3x-10 and dx^+Tx^-Sx-lS. ^^■{■3x-^10)ix'-\-7x'- 3x-15(x + l 4a;H3x-2-10x 4a;2 + 7a;— 15 4lx^-\-3x-10 Sic^)4a;H3x-10(a; + 2 H.C.M,=:4x — 5. 8.-10 8r-10 In this example there is no necessity to introduce or suppress any mononomial factors. 114. The process of the foregoing examples will frequently enable us to find the h.c.m. of polynomials involving powers of several letters, as in the following bep, the partial cled by li them 2 ice or lently kowers THE HIGHEST COMMON MEASURE. 79 Exam.'ple. Find the h.c.m. of %^-^xy~6if and 3x2—40:^+2/'^. 2 _6x2 + 3.i7/-%2 -lly)-ll.rv/ + 112/2 H.C.M. =£13 — ?/• cc — 7/) 2«;2 -^xij— 2>if- (2x + By 'dxy-^'dif Here the mononomial factor — lly is suppressed. 115. The reason for the rule in Art. 113 will appear from the follow- ing proposition and its application in the next Art. V/licn one quantdij is a mea'^ure of two others, it will measure the sum and difference of any multiples of them. Let the quantities be A, L\ C : and let A measure B and C, so that JB = inA, Cz:=7iA, where m and 7i are wliole quantities. Take any multiples pB, qC of B,C, where p and q are any whole quantities whatsoever. Then, since ^9^ =/wi^4, qC=qnA, pBdzqC = p}nA±qnA = (pm±iqn)A. . pB±.qC , A that is, A is a measure of j)BdzqC, the sum or difference of any multiples of B and C\ because the ([uotiont pia-±.qn is a whole quantity. Thus, *Ja'-, which is a measure of (j.r-' ;ind 8,r-y, will measure (U"'( - -la) - ^jrii{ - ox), (Jx^-f y.r-//, ^ix^ - 8x'-//(4-.i7/), &ic. 11(3. Sui)pose, now, that A and J! denote two ]iolynomials (as in Art. ll'i), neither of which contains a mononomial measure other than unity ; and let the dimensions ot' A be not greater than the dimensions of B. Divide A into // n)ultiplied by a mouonomiid whole quantity a, which makes its lirst term a multiple of the lirst term of ^4 ; and 8o THE HIGHEST COMMON MEASURE. divide the difference G by the highest mononomial meatiure which it contains, and lot the quotient be D. B A)aBib 64 c)C D Now, C being equal to aB—bA, or the diffbrence of two multiples of A and B, is a multiple of all the common measures of A and B, and therefore of their H.c.M. Again, every common measure of C and A is a measure of C+bA, or ctB, and therefore of B, because A has no mononomial measure. Hence the H.C.M. of A and B is the same as the H.c.M* of ^1 and C, which is the same as the n.C.M. of A and D, because A has no mono- nomial measure. The problem is thus reduced to finding the H.c.M. of A and 1), These two quantities, A and I), are then treated in precisely the same manner as A and B ; and the process is continued until it tciminates as /> " ws, when the last divisor, F (suppose), is a measure of the last divide. 1 Q. P)Q(r rP The problem is thus finally reduced to finding the n.c.M. of P and Q. This is evidently P. Hence the last divisor in the above procef^ will be the H.C.M. required. PJXERCISE XXXIV. luiid the H.c.M. of (1.) 3.r2 + 2.t;-'il ?vA .^r>;2-fl3.r-6. (t2.) 2.tH;r-auivi8:<;-.-l^i 1. (3.) a;2-5.r + 6 and .'.'^-f;:^:-! 9. (4.) i>jHi0a; + 21 and x2-2.r-15. THE HIGimST CPMMOJV MEASURE. 8i v'hich it lUltiples i i)\ and '+6^, or « A and C, lO mono- sely the until it measure f P and H.C.M. (5.) 2xHa;--15and2a;^-19a; + 35. (6.) x2-4a; + 3 and 4«3-9a;2- 15^ + 18. (7.) ct" + 10.rj + 25 and x^ + 15^;2 + 75 ,. ^ i25. (8.) ;/;3-6a-- + ll*-6 and a;3-it;^-Ux- + 24. (9.) ;t3-3x2-9x + 27 and 3,c3-a;a-27.« + 9. (10.) 3.>;2-22x + 32 and x^ -\\x" ^■Z'lx-'l^, (11.) 7a)2 - 12x + 5 and Ix? + .*- - Sa? + 5. (12.) 5.xH 2x^-15*;-. 6 and 7x^— 4cb2_21xH-12. (13.) 2x3 + 9^-2 ^ 4,^, __ 15 j^nd 4a;3 + 8^2 + 3x + 20. (14.) a;3-6x2 + lla5-6 and a;4-2x3-13a;2 + 14a; + 24. (15.) a;*-2x2 + l and a;*-4x3 + 6x2-4x + l. (16.) x^-^xy— Ykf and x^ — bxy + 6?/-. (17.) 2x2+3x2/ + ?/2and3x2+2x2/-2/'^. (18.) £c^ + a;22/ + ^2/ + 2/'^ ^Ji<i *''■~■2/^• (19.) 5*2 + 26jry + 33?/2 and 7x^ + 19i»2/ - Qtf. (20.) 3a;*-a;V-2i/^ and2x-* + 3x3?/-2xV-3a;2/». 117. III. The H.o.M. of two polynomials involving mono- nomial measures is found as follows : Express each polynomial as the product of a moiioiioimal and a polynomial luhich contains no mononomial measure. Then the H.C.M. required is the product of the H.CM. of the rnono- Qiomial factors and the H.C.M. of the polynomial factors. Example. Fin( '. the h.c.m. of '6x'^y + l%9y + ^xhj and Qx^y^—Qx^y^ — l^xy"^. Here the given quantities arc equal to 4^2^(2x2 + 3,/j + l) and ijxif(^-x^'^. The H.C.M. of 4^7/ and (Sxy^ is %ry ; and the h.c.m. of 2*^ + 3x- + 1 and x- - x - 2 is u; + 1. .*. the H.C.M. re(iuired is %icy{x + 1). 118. The H.C.M. of three polynomials is the H.o.jr. of any one and of the H.c.31. of the other two. o 82 THE h 'CHEST COMMON MEASURE. Example. Find the h.c.m. of x^-l, ccH 2^2-3, and 2x^ + 2x3 + 3a; + 3. The H.C.M. of x^—1 and a;^ + 2x^—3 is cc^— 1; and the h.c.m. of x^— 1 and 2x'* + 2;*3 + 3x + 3 is as + l, the h.c.m. required. Exercise XXXV. Find the h.c.m. of (1.) 12aa;^-27aa;2and2a2ic3+aV-3a2a;. (2.) 10x2+40^ + 30 and 4a;3-16a;2-84a;. (3.) 2x«-6x3-4x2and3x*-3x3-12x. (4) 2x2+x-3, x2-l, and x'-^^.x-h. {L) 6x2-a;-2, 21x2-17x + 2, and 15x2+5x-10. 119. When all the component factors of the given quan- tities are known or can be determined, the h.cm. may be found by the rule of Art. 111. Examples. (1.) Find the h.c.m. of 4(x-l)2(a; + 2)3 and 6(x-l)3(a;+2). The g.o.m. of 4 and 6 is 2 ; the highest power of x— 1 common to both is (x— 1)* )} X + 2 }) 03 + 2. the h.c.m. required is 2(x--l)2(ic + 2). (2.) Find the h.c.m. of By Art. 80, Sn^x(x'--l) = H(fx(x-tXuHx-hi); Pj Art. 79, 12r/,TV-l) = 12ax'-^(.t;2-l)(a;^ + l) -I2r/x2(x-l)(x + l)(x2 + l); ^;^fJ^fffffSrC0MM0J^ MEASURE. Now the O.O.M. of 8, 12, and 20 is 4 the highest power of and the other factors a. + l, ^2^'i^ ^2 IS a; a common to the three quantities X x—\ a:— 1; . +, + »'• + 1 arc not common . . tlie H.C.M. required is 4ax(x--l). Exercise XXXVI. Find the h.o.m. of (2.) 6a2(^+2Xa.-3)and8a^(«:-3)0 (3.) ax2_2aa; + a and 2a^x'^^2a'^ X +3). (4.) a;2-l,a;3+l and a?*— 1. C5.)a;+2,a;2_4,andx3+8 (60 3^3_8i,«,2^6^^9^^^^2^3__ 18a7. I 1 11 g2 li I' ( 84 ) i. 1- \ CHAPTER XIII. THE LOWEST COMMON MULTIPLE. 120. When one quantity is a mnltiplc of two or more others, it is said to be a common multiple of those quantities. Thus l%t^ is a common multiple of 2.c and 3*^; and Xhx(x—V) of 3, 5, 15.^- and a; — 1. 121. The lowest common midtiple (l.c.m.) of two or more quantities is the common multiple of lowest dimensions and least numerical coefficient or coefficients. Thus of the follo\*ing common multiples of 1x and Sx^, namely, &x\ 12^2, 18x2, 6x^ 12xS 18xS &c., the first 6x2 ^g q^\\q^ the lowest. 122. The L.C.M. of two quantities is found by the following rule : — (i.) If they contain no common measure except unity, their L.C.M. is their product. Thus, tho L.C.M. of 4x and Tab is 28a/;x. (ii.) If they contain a common measure, their L.c.m. is c^jual to one of the given qtiUiitifies multijilied by tjie quotient (f the other dlnlded liy their H.c.M, It will be generally found most convenient to expresK the L.C.M. as the product of several fuciors. THE LOWEST COMMON MULTIPLE. 85 (1.) Find the l.c.m. of ^.rhj and Sixif, The H.o.M. of these quantities is '6xy, .-. by the rule, l.c.m. J^f^- x Sixy'^X^xhf. '6xy (2.) Find the l.c.m. of 2x2-7a: + 5 and Zx^'-lx^^, The H.c.M. is found to be x—\\ and since = 2x-5, the L.C.M. will be (2a;— 6) (^x'-lx-Y^). 2a;2-7a; + 5 X' 123. The following is the proof of the rule given in the preceding Article : — Let the two quantities be denoted by A and B^ and their ii.c.M. by C; and let A — aC^ B — bC, where a and h are whole expressions which have no common measure except unity. Then, since the L.C.M. of a and b is ab, the L.C.M. of aC( = Ji) and bC( = B) IS abC= —— =-^= -^.B=-^.A. 124. The L.C.M. of three quantities is the L.C.M. of any one and the l.c.m. of the remaining two. The L.C.M. (if four quantities is the L.C.M. of any one and the L.C.M. (f the remaining three. And so on. J/Jxample. Find the l.c.m. olUx^ifz, Gxyz^, and lOx^yz^. The L.C.M. of 3x'i/'z and 6.///^!- js Gx'VV; and the l.cm. of QxYz^ and lOx'vjz^ is SOx'-y'z^ 125. When alx the component factors of the given quan- tities are known or can be found, their l.c.m. may also be obtained hy multiplyimj the l.c.m. of the numerical factors by the Jiiyhest j)oiuer or 'powers of the several factors that occur in the fjiven quantities. 86 THE LOWEST COMMON MULTIPLE. I- I f' ! .^ d Examples. (1.) Find the l.c.m. of 6x^^22^ i.x?u\ and 8xYz. The L.C.M. of 6, 4, and 8 is 24 ; the highest power of x which occurs amongst the factors of the given quantities is £c* ; and the highest powers of y and z are, respectively, y^ and z\ Therefore the l.c.m. required is 2ix* ijh^. (2.) Find the l.c.m. of 15«&(a— ft), 21a(a + bXa—h), and The l.c.m. of 15, 21, and 35 is 105; the highest powers of a,h,a—h, a + b, wliich occur amongst the given quantities, are, respectively, a, U^, a—b,a-\-b. Therefore the l.c.m. required is 105ahXa—b){a-\-b). (3.) Find the l.c.m. of x^-^l, a;^— 1, and cc^+l. Here x^-l = (x + l)(x^l); a;3-l = (a;-l)(,;:2 + aj + l); x'^+l = (x-\-lXx'-x-\-l). /. the L.O.M. = (cc + l)(x - l)(a:;2 + a; + l)(a;2-a; + 1) = lx'-lXx'+x^ + l). Exercise XXXVU. Find the l.c.m. of (1.) Sahx, 2bxy. (2.) Sa^xy, Uax% (3.) a¥, bc\ ca\ (4.) Sa%c, 12ah% 24Mhc\ (5.) Uhcu^, lQcav\ 20ab7v'-, 40a2'>V. (6.) x^-7x-\-12,x^-^x-G. (7.) 2x'-5cr-3,4a;2+4« + l. (8.) 3a;^-ll« + 6,2a;2-7a; + 3. (9.) cc^— 4aa;^ + 5a2£c— 2a^, o;'-?-^^*— 4a'. (10.) 8(a;2-l), 12(x-l)2. .0^ TJIEJ^ESrcOMj^O^ MULTIPLE. (13.) a^-{.a%, ab"b\ a^-~b'i (14.) 2x(:c^ + a; + i)^3^,3_3^^^a_^ «7 (15 /^ +r,i>'-?^i>H«3^ (16.) r -I,i>*-1,7>«-.1. (17.) (a-&) (a-,)^ (j.,.>> ^j_^^^ ^^^^^ (18.) 8a«6(a«6), 12a6(6-a), 3(a^-J2), W(j, J^,^^ IMAGE EVALUATION TEST TARGET (MT 3) <- s^^ ^ '^ 1.0 I.I ■SO ^ 1^ 1^ 12.2 1^ 2.0 m F' ¥' r < 6" ► 7 ^ ^ %> V \y^ > ^;7* Hiotographic Sciences Corporation 33 WEST MAIN STREET WEBSTER, N.Y. 14S80 (716) C '2-4503 ^^i^" ^ ^ u:^ rU 'V It ill ( 88 ) CHAPTER XIV. FBACTI0N8. hi 126. When one quantity is not exactly divisible by another, the quotient is represented by writing them in the form of a fraction. Thus, — ^ denotes the quotient of —2 divided by +3; n?- the quotient of -2a divided by -^Sx; .'^"■^ ^ the quotient of a;— 1 divided by a;'*— 3a; +4. 127. A fraction is not altered in value by multiplying or dividing the numerator and denominator by the same quantity. -2_-2x3_--6 ±6 _ +^6 X -4 _ -24 Thus, ^3 - ^3^3 - ^9; _8 " -8x -4 "" +82^ -20 _ -20-j — 5 _ +4, -4 _ -4xf __ -^f . -25- -25-r— 5~ +5' +5 + 5x| -^'■r a ac —ax ^ h ~~ he ~ ~-hx ' x-l __ (x-l)(.r + l) 2:^-3 ~ (2x-3)(x + l) a;2-l 2x2-a;-3' 128. The statement in the preceding Article depends on the two following propositions : — I. 77ie numerical value of a fraction is unaltered by multiplying or dividing its numerator and denominator by Vie sam^ quantity, (i.) Let a, 6, m, be integers. Then, since ^ denotes a of the 6 parts into which a unit is divided, it follows that FRACTIONS. 89 b ' mb a b ma . (1) . (2) . (3) Before considering the case where a, 6, m are fractional, we shall show how the operations of multiplication and division of numerical fractions must be performed. mc Let Of, 6, c, c?, m be integers. Then, since — =m, it follows that But by (3), a mo a , ma -.— =-xm=-- h c b ma _mao 1 'be , a mc_mac "^ c bo . (4) Hence, if - be multiplied by an integer in a fractional form, the b product L a fraction whose numerator is the product of the numerators and denominator the product of the denominators. If, now, we wish ft o o to find the product of - and -, where - is not equal to an integer, the *^ b d a operation of multiplication must be in accordance with this rule ; for any application of the term multiplication to cases where its primaiy meaning (which is repetition) does not apply, must not be inconsistent with the cases where it does apply. , a c _ac "VlTbd (5) Again, since diyision is the inverse of multiplication, both in its primary and extended applications, it follows from (5) that ac a bd d b __acd '~bcd __ac d ~bd' c by (3) by (5) ^i; i \ I '9' M'l 90 FRACTIONS. Also ac ,a hdrV _c 'd _abc abd bd a by (3) by (5) * /I Hence it follows that the quotient of the fraction - divided by b -. is equal to the product of - and — : that is, d b a.c a d_ad 6 d _a _ b c he . (6) (ii.) Now, let a, 6, m be fractional, which will include the case where some of them may be integers. Let Then y u d X z yz z u And mh' c c X ex cz by (5) d u du __cxdu czdy by (6) =^ by (3) yz ••T .Tna mb' II. If, in the fraction -, a and 6 denote positive and negative quantities, the sign of the fraction depends on the signs of the nu- merator and denominator. These signs will be either like or unlike, and on multiplying or dividing by a positive or negative quantity they will still be like or unlike, and the sign of the fraction will therefore remain unchanged. FRACTIONS. 91 ivided by the case negative the uu- r unlike, tity they therefore For example, multiplying by — 1, I 2 — 2 2 — „ = — - = +- , a positive fraction ; + — o o _i_3 _3 q — - = — _ = - a negative fraction. — 4 +4 4 129. A fraction which involves fractional coefficients in the terms of the numerator and denominator can always be reduced to one whose numerator and denominator are whole expressions by multiplying the numerator and denominator by the l.o.m. of the several denominators. Thus, n^ii_3j ■j ^ is reduced by multiplying nxmierator and denominator by 12, the l.c.m. of 2, 4, and 6, to the . . _ 12x-6 equivalent from 24^2^:y^^2- 130. A fraction is said to be in lowest terms when its numerator and denominator contain no common measure, except unity. Hence a fraction is reduced to lowest terms hy dividing its num>erator and denominator hy their H.C.M. Examples. (1.) Eeduce " lOa^S'-^y/ *^ lo^^st terms. The H.O.M. of SaWx and Ua^b^y is 4a^h\ Sa^b^x _%x •'• UaWy "" 3ay* ■x^—1 (2.) Reduce -3-r-i to lowest terms. The H.O.M. = 03 + 1. x^—\ x—\ ^2^~^Tl* ceHl ~ as' 2^2 I g^ 2 (3.) Eeduce 03,2^53; 7 9 to lowest terms. The H.O.M. = 2x— 1. 2^3^-2 _ a^-f^ •*• 2a;2-5;r + 2 "■ x-2' ! i, ' 92 FRACTIONS. m !-M I, I I. 'I Exercise XXXVIII. Reduce to lowest terms — (5.) (8.) (11.) (13.) 6a2Z>2 (3.) Qx^yT? Sxy'^^ O" a»-l ^-7«J-10 3 + l%+3a;2 S-\-Sx-3x^' (6.) OJ + l C7) -"^ (9.) x^—y^ a^—y^' (10.) jc"— .4 a;-3 (12.) a;2-4x + 3' a;H7a; + 12 a;2-ic-20' n4N g;'^-3ag-70 'Ti^ cc^— 6a;— 9 ^ '^ a;* + 3x3-9a;-9' (16.) 12x^-15x+3 6*3_6x2+2x-2* (17.) a;^— 4a3V x3-6a;'^^ + 12ay-82/3 nS) «'^— 3a;'^y + 3a;;/''— y ^ x^—x^y-'xy^-^y^ 131. Fractions ate said to be like or unlike, according as thej have the same or diflferent denominators. 2a Thus -, - are hke fractions, as are also SiC X X a;2--r a;=^-l' '3o 2a jr- are unlike fractions. Sx 132. Unlike fractions are reduced to like fractions by mul- tiplying the numerator a:>d denominator of eadi fraction by the quotient of the L.O.M. of the several denominators divided by lis deno7ninator. The common denominator will accordingly be the l.c.m. of the several denominators. Examples. (1.) Reduce p^, -^ to like fractions. 00 4» The L.C.M. of 3& and U k i%d. 1 6ah 4x+3* 3 -2* ig as thej SiC 2a^ ^-1' 3»/ s 6y mul- lon by the ivided by ;he L.c.M. FRACTIONS. 93 The multiplier for the first fraction is -w. -^l\ •* Wh~md' The multiplier for thti second fraction is -j-^- = 36 ; . 3c__ %c " 4a5 126c^' (2.) Eeduce a—b ' 6— a to like fractions. The L.o.M. of a—b and b—a is a--6, the quotient cf which divided by 6— a is — 1 ; 6— a a— 6* (3.) Eeduce 1 x-l xj-2^ x-r x^+x+v x^-i to like fractions. The L.c.M. of the denominators is x^—l, the quotients of which divided by x-1, jrHas + l, jk»-1 are, respectively, tr/^ + cc + l, x—1, 1. a;-l __ a;'^-2a; + l aHaJ + l a^'-l ' as— 2»'^_£c— 2ar £C^— 1 os^— 1' EXEECISB XXXIX. Eeduce to like fractions : — 1 2 (1.) a ^^•^ i^ ' ~ab' (3.) '^ , -5^ , cc xy xyz (5.) i , J- . A. a —a ax- — xy ««;«/ (4.)". ^ a 4 la (6.) - - , -^-y . (7.) - x-\-l u;+3 W --r, 3— a; x -r i-x- (9.) a •a a—b^ b—a m 94 FRACTIONS. 1.1 (10.) (12.) (13.) (14.) 4 3^ 1_ 2 x^\ 3aJ (^^•) a^^-x + l'a^ + l'x^+l* 3 1 (a: + 2) (x-lj ' (1-u^) (2-x) ' (a— 2) (x + 2) 1 _ __ 1. _ 1 (6_a) (c-ii) ' («-/>) (c-&) ' (a-c) (6-c)' 1^ 1 1 a((0—b) (.c— a) ' h(h—aj (x—h) ' a&x'* i ^t Addition and Subtraction. 133. The operations of addition and subtraction of frac tions may be denoted by the signs + and — respectively. (c-2 a -2& ;2-3 -2b Thus, the sum of , -^ , ajid may be denoted by _a , — 'J<> , a?— 2. x-l x^-3 dx-^' and ^Ac diference between -_,^ , a?^c? v^— * by Sx"^—! oa;--4 4a5 a;-l 3a;''^— i 5a;— 4* Thus also +? denotes the fraction ? ^o be added to, and — - the fraction - to be subtracted from some quantity not expressed. 184. Sums and differences of fractions when expressed as single fractions are said to be simplified, the operation being IDerformed according to the three following rules : — (i.) The sum of any number of UJce fractions is a like fraction ivhose numerator is the sum of their numerators. Thus — i- + ~^ + ^'^— ^ = 4--a; + 2a;— 3_a? + 1 i»— 1 x—1 OS— 1 a;— 1 cc— 1' (ii.) The difference between two like fractions is a like fraction whose numerator is the difference between their numerators. FRACTIONS, 3a; af frac- vely. loted by to, and ity not ssed as 11 being ''raction raction 95 (1.) (2.) Examples, 5a;~4 _ 6 _ 5a;-4- 6_ 5a;~10 2x-3 2a;-3 2a;-3' 2aj-3* 4ag --2|K_4a: + 2x_ 6aj 03' It will be observed that the — before the second fraction changes the sign of — 2ic. ^ '^ " X+l X^ + 1 X^ + 1 _Sx^-x-]-6-x^-{-Qx + S _.2x'^ + 5x + 13 a^ + 1 " In this case the — before the second fraction changes the signs of all the terms of a;^— 6a3— 8. It appears from the preceding rules that _ a—b _ — a + 6 c c and conversely ; in other words, the suhtraction of a fraction is equivalent to the addition (fa like fraction, whose numerator is the numerator of the former icith its sif/n or signs changed. Thus, a;2-2a: + 3 , -a-2 + 2x-3 + x^-l -2^5 X^—1 _ 5 3a;2-7* 2^-b (iii.) Addition and subtraction of unlike fractions are per- formed by reducing the unlike to like fractions and proceeding as above (i.), (ii.). Examples. (1.) Simplify .- + - + be ca ab >h III ||.' )|i*'" |t|H i;. ■#' 96 FRACTIONS. einco the l.o.m. of he, ca, ah is ahc, this sum is equal to a abc ubc ^ abc "^ nh,.— a^\■h■\^c ubc 2r* + i-^. - l_a2 (2.) Simplify 1^^ . i_^^ Tlie L.c.M. of tho denominators is l--a\ 1 2« •'• 1 + a "^ 1-a "l-a^ - l-a2 + i_^2 -JZ:^ 1— « 1+a 2a 1— a + l + a— 2a - l-a" _2--2a ~ l-a2 2 ~ 1+a (3.) Simplify ^^—y - ^^::;^ - (^3^- The L.O.M. of the denominators is (x-\-yy (x—yY . 1 1 1_ " (x+yf y'-x^ (x-yy __ (j^—yf y'^-^x^ (?i-\-yf - (a;2-2/2)2 - (^2_2/2)2 - (052-2^2)2 _ x'^—^xy^y'^—y^-\-x^ ~'X^'-^lxy'^i^ x^—Axy—y'^^ ~~ (x^—y^y Simplify — EXEBCISE XL. (^•) 2a"'"(/ X /ox- 1 2 FR^CT/ONS, 97 1 2 2 8a 1 Wx~3^' (S-^^+^^-i^'* <6-) a;— 1 ^—8 2a5 ~ a" ' 2a-3fe 3a--26 a-36 <.y-; a~6 a + 6 a+6 a-& (8-) ^3^6+^6- (10.) i:i^.+ii:^- (^^•-^ a'*"a + 6 + a*+a6 ' ^^^'J a"2x-l""4x2-l * ^^^•''a;2 + a; + l+a;2-a;+I ^^*'^ 2"(»-l)"2(a; + l)"a;«' (■"•^0 ^c^iy'-^^'iy ■*■ x^^^p ■ .... 2 1_ a;+3 . 2y' . ^ -^ a:;=' + a;?/+2/^^a;'*— as*/ + «/'*'»* +33 V+!/ /iQ\__^""l JL_ g ^ + 1 1 2 2 1 (^^•) (¥:^"XiT2)~5^)(a;-2)''"(x-2)(a;+2)' ah ac (^0-) (a-6)(6-c),'''(a-cXc-6)' V^J-O (a;-6XaJ-c)'*"(a;-cXiC-a)+(a;-a)(x-6) (ic— a) (as— ft) (as— c) * OS— a cc— 6 as- c (22.) (fc-aXc-«)"*"(a~&Xc-«»)'''(«^0(^-^c)' (^^•) ("^:^(^:::^+(6-«x^-c)+((;-6Xc-a) ' 135. A mixed quantity is the sum of a whole expression and a fraction; as, for example, 6 „ 1 3a; _ as— 1 FRACTIONS. 136. A mixed quantity may he expressed as a fraction by considering a whole expression as a fraction whose denominator is unity ; and conversely, a fraction may be expressed as a mixed quantity when part of its numerator is a multiple of its denominator. liii ii i [11 Examples, ,- . _ 1 2a; 1 2aa; + l (1.) 2x+-=-j- + -=— - . , 1 a;— 1 1 a^ (3.) p-+i-»"+i=,;=-TT — r=x^+i- a;'— 1 It will be observed here that —a^ + 1 = —(a;''— 1)= — -| — (4.) Express ^.^ as a mixed quantity. On dividing 2a5*— 3a;+4: by ac+1 we get quotient 2a;— 6 and remainder 9. x-\rl x + 1 (5.) Express ^±^^^^ as a mixed quantity. In this case the quotient is a? +2, and remainder — aj+1. . a;8 + a;2_2a;+3_^ . o. -fc + 3^ *, 5 x-r^-t-A — i . a;''— a; + l US'*— a; + l = a; + 2- a;-l a;'^— a; + l ' Here the + before the fraction is changed to — by changinp: at the same time the signs of all the terms in the numerator -a;+l. iion by minator ed a8 a )h of its —5 and FRACTIONS. 99 Simplify — (1.) 1 + 5. (4) ^+a-l X EXEBCISE XII. X (2.) ^,-1. (5.) ??-o+l. X (3.) 2-?. a (6.) 2+?-i. (7)8- 03* 05 — 1 (12.) 4a; -1- (10.)-^, -a. (11.) 2- a + 6' 2a; + l a; + 3 Express as mixed quantities — (13.) xHiry + t/'^-^"^'. (14.) (17.) (19.) (22.) a 2a;2-aj + 3 X a;' + 3a; + 4 a;«-3 a;H2 (15.) ^-i. a; (18.) (20.) ^^-6 (16.)-^ 4-3:^ + 6a:'' 2-3a; 3x ' 'Sx (21 >i ^ + ''^'' (23.) 6a:'— 4a; + 5 2a;=^-a; + r' (24.) jc*— a; + 5 ^\ B + 1. langmg nerator Multiplication. 137. To denote that two or more fractions, or a fraction and a mononomial whole quantity, are to be multiplied together, they are written in a row with the multiplication sign x , or • (dot), between them. Thus ? X -,, or V * 4 denotes that ~ is to be multiplied by d a b c, x-l 20^-3 jej^oteg the product of '^"- and ?^'"?. d' a' x^-l X--1 ^^^ X — 3aaj denotes the product of ^. - and — 3«a:. b h2 100 FRACTIONS, 1^ K j Hi ^ lit * ■S v- Sums and differences when they are factors must bo enclosed in brackets. Thus, _„A_/a'^— aaj+cc^ denotes the product of -3^ and a^— ic^ a^— as' a^— ax+a;2; _^(a;+^) denotes the product of -^ — and cc+aX XI flj + a 0; + ^; -^(^+?-i\ denotes the product of 4^:? and X 05-^— 1\2 X a2V cc*— 1 138. Multiplication of fractions is performed according to the following rule : — The product cf any number of fractions is a fraction whose numerator is the product of their numerators and denominator the product of their denominators. A whole quantity is to be considered as a fraction whose denominator is u iiity ; and sums and differences of fractions and mixed quantities must first be reduced to fractions. (1)^ x^—a Examples. ax _(x^—a~)ax_x—a 2ax x + a 2ax(x-^a) 2 Factors common to any numerator and any denominator may be struck out. Thus ax which is common to the first denominator and second numerator, and x+a which is com- mon to the first numerator and second denominator, may be struck out before multiplying, and the result will be in lowest terms. When possible, therefore, the component factors of the several numerators and denominators should be obtained. (^\ 4a^— 4aa;^ hc-\-hx_4ia(a+x)(a-^x') ^ &(c + cc) ^ "^ Sbc^-Sbx^ ' o^^ac" 35(c + a;) (c-a;) * a(a-x) __A(a+x) 3(c-a;)' FRACTIONS., lOI Here a,a^Xy b, and c+a; are struck out, being common to the numerators and denominators. (3.) X (a; + 6)(a; + 7) (a;2-49) = X (x + 6)(a; + 7) x(x-7) (x + 7)(x^7) \ x^—i)x^ + l x^—1 x^ + 1 xHl x-l (x-tl)(x-l) JcHl 1 x+r ExEBCiSE XLU. Simplify — yo\ b^c <?a o^}> KP'J ~~~-\ • — o • — 5« 2/2S 2i» ^32/ ^2 a c?-W (5.) . — ^. .ON a- <>- c- oc ca ao (4.) '^ y (6.) 1—03 ' l + ai* (7.) (9.) ^S + p • ^2^2,2 1 ac + l a rl ^°'^ -(7-6)3 --^3 • 05 + 1 0)2 + 1 a;*+l (11.) /I+I+IW \x y zj a3.)(..:.«i)(«-g. (17.) (t,— ■^-+i)^ \x* xy y^Jx (10.) x^—l x^—\ a;*— 1 x^y^ (12.) -^(^-«\ . (14.) 5;+* f(-5 1-). ae.) ^-(&+^Ui— ^V 6x\ <* / \ a + xj (18.) (l±'+^\4dt' . I- • * 1 i ■ w it : 102 FRACTIONS. Division. 139. To denote that one fraction is to be divided by another, they are written in a row with the sign -f- between them. The same notation is employed when the dividend or divisor is a mononomial whole quantity Sums and differences when they are the object of division must be enclosed in brackets. Thus, --f-- denotes that - is to be divided by - ; 4a a |4-(2a-l) 4a ,2 a »> ii X y it X' n ii a Ii it a ii Ii 3c; 2a-l; 3a; X 2a X .2__: a The same thing may also be denoted by writing the quantities which are the objects of division in the form of a fraction. 2^ a Thus, i==27j-^a* 2^n=6-^-(2a-l); a a. — •*' X 140. When the product of two quantities is unity, each is said to be the reciprocal of the other. Thus, since axl=l , ? . ^ = 1 , ^^ . ?£r§_i :. r^, lows that f'R ACTIONS. 103 1 .. a a h 2x-3 is the reciprocal of a , and ft )} tt »i b a' 2x-S x-l » ■5 » 1. a a . a " b' 2.r-3 x-1 a of b n x-1 " 2a:-3' 141. Division of fractions is performed according to the following rule : — The quotient of one fraction divided hy another is the product of the former and the reciprocal if the latter, A whole quantity is to be considered as a fraction whose denominator is unity ; and sums and differences of fractions and mixed quantities must first be reduced to fractions. Examples, ^ ''' bxy'' ' luxhj 5xy^ ' 2ab^~ 'ET' (2.) a' ^y 5xy^ a^ + ax _ a^ by a^-}-ax + x^ a^—a^ ' a^^ax+x^ a-'—x^ ' a^ + ax . a^ (a— a;) (a^ + a.x + a;^) a a^—x^' a^-\-ax-\-x^ a(ci-\-xj Wh) ' \a b) ab" ' ~ab _bx-\-ay ab ab bx—ay _bx-\-ay bx—ay _y^—xy-\-x^ x^y"^ ix+y)(x^-xy-\-y^) ■M 1 1 * I I aV(a;+2/)' ini i •■■■■'• II : I' W': 104 FRACTIONS. EXBBOISE XUn. Simplify— (3.) — , -^ -ir- ^ ' cp + l • as— 1 (9.) a; y (11.) ^-r^+y. 05—2^ as+y ^ - ^ - i+6 S"*'G a a (6.) (4.) -J^-:--^-. a+cc • a— as ... /I 1\ . 1 .g. ia^h a^^\a^ 6 a+ac a— a; /in\ a—x'_a±x (12.) ij^ 1 (B 9? i9 «? (14.) 8~10 10~li} S+io 10+12 ( I05 ) CHAPTER XV. SIMPLE EQUATIONS (continued). I ,• 142. We shall give in this chapter some examples of equa- tions involving fractions with literal denominators. Such equations may be cleared of fractions by the rule already given in Art. 74. In some cases before applying this rule it will be found more advantageous to simplify parts of the equation separately. (1.) Solve Examples, x—a x~-b b ~ a Multiplying by ah, the l.c.m. of the denominators, we get (x—a)a = (x—h)b. Clearing of brackets and transposing, ax—bx^a^-^hK Collecting coefficients of x, (a-'b)x = a^—b\ Dividing by a—h, (2.) Solve x = r- = a+&. a—b 3a;-l 4a;-2 Multiplying by the l.c.m. 6 (2x— 1) (3c5— 2), 6(3a;-lX3x-2) . 6(4a;-2X2a!-l)=(2x-l)(3a!-2). io6 SIMPLE EQUATIONS. m w n-T I Clearing of brackets, 54a;2-54:a;H-12-48a;2+48x-12=6x2-7x+2. Transposing, 54x2-48x2-6a;2-54a;+48a;+7a;=2-12+12, .*. a;=2. a;— l__a;— 2_a7— 4__aj— 5 03—2 as— 3 as— 5 as— 6* (3.) Solve Simplifying the sides separately, (sc-1) (a;-3)-(a;-2y_ (a;-4)((g-6)-(a;-5)a (a;_2)(a;-3) (!B-6)(a;-6) Clearing the numerators of brackets, -1_ ^ -1 (a;-.2Ki»-3) (ir-5)(a;-6)* Multiplying by the l.o.m. (as— 2) (ai— 3) (as— 5) (as— 6), clearing of brackets, and solving, as =4. Solve- EXEBOISE XLIV. a\ 12 1 _ij •-^¥"^12^-^*' (2.) (3.) (5.) (7.) 16 27 3aj-4 5a;-6* a;-l_7a;-21 a;_2 7a;-26' _^+2 = -^. a; + l as + 2 (4.) .- 42 _ 35 a;-2 a;-3* 45 57 ^QN 5as— 3 __ 2as + 3 _j^ (ll.)^=:i+«^~^ 2a; + 3 4a;-5* C6 "i 2as— 6 _ 2 as— 5 '^ '^ 3x-8 3^^* xo N 2as— 3 , 2as— 1_Q ^^'^ 2HHhl + 2^+3~''' /I n\ Gas + 13 2as_ 8.r + 5 2a;-5 (12.) g!-14 _ 2a;-29 _ 1 a? 2aj-20 2!b' (13.) (2aj+3)(a!-5) (3x-2) (jb-11) ' SIMPLE EQUATIONS. 107 X -^ = 0. /15) JP a;--l_g;— 3_ a;— 4- ''^ a-l a;-2 a;-4 a;-5' (16.) zr"' 6— a? 6— a; 4— a;~ Uo-; — T~ + = r + -. a, a (20.)^-^+^=!. (22.) ^— ^ -L^'"^ - ^"^^ I ^jj. (17.) 2-5= c. a b (19.) 5+»=5+*. (21.) A-.«=j2-a2. ( io8 ) -it-- 1' I'm'* i ■■ I':. *^ CHAPTEE XVI. PROBLEMS (continued). 143. We shall give in this chapter some examples whicl are more difficult than those in Chapter IX. Examjples, (1.) If A can perform a given work in 60 days, and B in 4C days, in how mauy days will A and B, working together, be able to perform it ? Let w denote the work to bo done, and x the required number of days. Then w amount of work done by A in one day=gQ» B AandB $$ » li W io M f9 t> X w w w 1 1 Divide by «;,- = gQ+|^- Multiply by 120a;, 120=2x + 3a;. .*. a:!=2-l. (2.) At what time between 5 and 6 is the minute hand of a watch 5 minute divisions behind the hour hand ? PROBLEMS, 109 les whicl id B in 4C ;ether, be required Lot X =5= the number of minute divisions between the hour hand and 5; then 5— a; = the number of minute divisions between the minute hand and 5. But the number of minuto divisions between 12 and 5 is 25 ; therefore the number of minute divisions between 12 and the minute hand is 25— (5-a;)=20 + a;. The hour hand thus moves over x minute divisions while the minute hand moves over 20 + a;; and since the latL.r moves 12 times faster than the former, it follows that 20 + a:=12a;. • T — l-fi- • > •*' — "'■ 1 T* Hence tho required time is 12!r=21^Y minutes past 5. (3.) A grocer bought 200 lbs. of tea and 1000 lbs. of sugar, the price of the sugar being \ of that of the tea. He sold the tea at a profit of 40 per cent., and the sugar at a loss of 2i per cent., gaining on the whole $4550. "What were his buy- ing and selling prices ? Let the cost price of the sugar per lb. =x dollars. •*• a ii tea „ =033 „ Then „ „ 1000 lbs. of sugar = 1000a; dollars. „ „ 200 „ tea =1200a! „ 40 The profit on the tea-TQ^* 1200a; =480a; 2J The loss „ sugar = jQQ- 1000x = 25a; /. 480a;~25x=45-50 a;=-10 .". the buying price of sugar is 10 cents, and the selling price 91 cents per lb. ; the buying price of tea is 60 cents, and the selling price 84 centa per lb. ( ■■ ; te hand of Exercise XLV. (1.) I arrange 1024 men 8 deep in a hollow square : how many men will there be in each outer face ? no PROBLEMS. If !!l- Ik (2.) A regiment containing 700 men is formed into a hollow square 5 ranks deep : how many men are there in the front rank ? (3.) A man has a number of cents which he tries to arrange in the form of a square ; on the first attempt he has 180 over ; when he increases the side of the square by 3 cents he has only 31 over. How many cents has he ? (4.) On a side of cricket consisting of 11 men, one-third more were bowled than run out, and 3 times as many run out as stumped ; two were caught out. How many were bowled and run out, respectively ? (5.) Water expands 10 per cent, when it turns to ice. How much per cent, does ice contract when it turns to water ? (6.) A manufacturer adds to the cost price of goods 20 per cent, of it to give the selling price ; afterwards, to eflfect a rapid sale, he deducts from the selling price of each article a discount of 10 per cent., and then obtains on each article a profit of 8 shillings. What was the cost price of each article ? (7.) A person invests £14,970 in the purchase of 3 per cents, at 90 and 3r per cents, at 97. His total income being £500, how much of each stock did he buy ? (8.) A and B join capital for a commercial enterprise, B contributing £250 more than A. If their profits amount to 10 per cent, on their joint capital, B's share of them is 12 per cent, on A's capital. How much does each contribute ? *(9.) In a concert room 800 persons are seated on benches of equal length. If there were 20 fewer benches, it would be necessary that two persons more should sit on each bench. Find the number of benches. *(10.) A man travelled 105 miles, and then found that if he had not travelled so fast by 2 miles an hour, he would have been 6 hours longer in performing the journey. Determine his rate of travelling. (11.) An express train running from liondon to "Wake- field (a distance of 180 miles) travels half as fast again as an These questions belong to Exercise XLVIII. PROBLEMS. Ill a hollow he front tries to )t ho has ^ 3 cents rac-third J run out e bowled ice. How ,tcr? ds 20 per ) efifect a article a article a article ? of 3 per ne beincr rprise, B nount to is 12 per e? benches ivould be h bench. hat if he uld have etermine o Wake- tiin AS an ordinary train, and performs the distance in two hours less time ; find the rates of travelling. (12.) A can do half as much work as B, B can do half as much as C, and together they can complete a piece of work in 24 days; in what time could each alone complete the work ? (13.) Three persons can together complete a piece of work in GO days ; and it is found that the first does ;I of what the second does, and the second % of what the third does : in what time could each alone complete the work ? (14.) What is the first time after 7 o'clock when the hour and minute hands of a watch are exactly opposite ? (15.) The hour is between 2 and 3 o'clock, and the minute hand is in advance of the hour hand by 14i minute spaces of the dial. What o'clock is it ? (16.) At what time between 3 and 4 o'clock is one hand of a watch exactly in the direction of the other hand produced ? (17.) The hands of a watch are at right angles to each other at 3 o'clock : when are they next at right angles ? (18.) How much water must be mixed with 60 gallons of spirit which cost £1 per gallon, that on selling the mixture at 22s. per gallon a gain of £17 may be made ? (19.) How much water must be mixed with 80 gallons of spirit bought at 15s. per gallon, so that on selling the mix- ture at 12s. per gallon there may be a profit of 10 per cent, on the outlay ? (20.) If 16 oz. of sea-water contain 0'8 oz. of salt, how much pure water must be added that 16 oz. of the mixture may contain only 0*1 oz. of salt ? (21.) I have a bar of metal containing 80 per cent, pure gold, which weighs 30 grains : how much must I add to this of metal containing 90 per cent, pure gold, in order that the mixture may contain 87 per cent. ? (22.) How much silver must I add to 2 lbs. 6 oz. of an 1 ■. l!, 112 PROBLEMS. iii> oUoy of silver and gold containing 91*7 per cent, of pure gold, in order that the mixture may contain 84 per cent, of gold? (23.) A person started at a certain pace to walk to a rail- way station 3 miles off, intending to arrive at a certain time ; but, after walking a mile, ho was detained 10 minutes, and was in consequence obliged to walk the rest of the way a milo an hour faster. At what pace did he start ? (24.) A person started at the rate of 3 miles an hour to walk to a railway station in order to catch a train, but after he had walked \ of the distance he was detained 15 minutes, and was obliged in consequence to walk the rest of the way at the rate of 4 miles an hour. How far off was the station ? (25.) A wins the 200 yard race in 28i seconds, B the con- solation stakes (same distance) in 30 seconds: how many yards ought A to give B in a handicap ? (26.) A wins a mile race with B in 5' 19". B runs at a uniform pace all the way ; A runs at \^ of B*s pace for the greater part of the distance, and then doubles his pace, win- ning by a second : how far did A run before changing his pace ? (27.) A boy swam half a mile down a stream in 10 minutes ; without the aid of the stream it would have taken him a quarter of an hour. What was the rate of the stream per hour ; and how long would it take him to return against it ? (28. A contractor undertook to build a house in 21 days, and engaged 15 men to do the work. But after 10 days he found it necessary to engage 10 men more, and then he accom- plished the work one day too soon. How many days behind- hand would he have been if he had not engaged the 10 additional men ? (29.) Two crews row a match over a four-mile course; one pulls 42 strokes a minute, the other 38, and the latter does the distance in 25 minutes; supposing both crews to row uniformly, and 40 strokes of the former to be equivalent to 36 of the latter, find the position of the losing boat at the end of the race. ( "3 ) CHAPTEB XVn. QUADRATIC EQUATIONS. 144. Wb have already defined Quadratic Equations in Art. 69. They are further called adfeded or pure, according as the term involving the first power of the unknown quan- tity does, or does not, appear. Thus, 4«;2-5x + 7=0, «;2+6x~3=0, a:^-3x=0,areadfected quadratics; 3a;2-8=0, 0^24-6=0, are pure quadratics. 145. I. Pure Quadratics are solved bi/ transposition of terms and extraction of tlie square root. Examples. (1.) Solve jb2-4=0. Transposing, a?=L Since the square root of a positive quantity is either + or — , we have, extracting the square root, a:=±2. Thus the two roots are +2, —2. (2.) Solve 0^+6= Var'-ie. Clearing of fractions, 3a;H15=10ay»-48. Transposing and dividing by —7, X ^=9. .-. X =±3. Thus the roots are +3 and -3. I 114 QUADRATIC EQUATIONS. if'i:i; ! !!'■« lil,; *il n Li's' 146. n. Adfected Quadratics may be solved by one of the following three rules : — (i.) When tlie equation is in the form of the product of two factors, each containing the unknown, equated to zero, the solu- tion is effected by equating to zero each factor in turn. Examples. (1.) Solve (x-i-l) (2a;-3)=0. Since, when the product of two factors vanishes, one or other must be zero, we have either a; +1=0, and /. a;=— 1; or 2a5— 3=0, and .*. a;=f. Thus the two roots are —1 and f. (2.) Solve a;2-5aj=0. Factoring, x{x—b)=0. * :. either a;=0, or cc— 5=0, and /. £c=5. Thus the roots are and 6. Whenever, as in this case, the terms of an equation are divisible by the unknown x, we can infer that one root is zero. (3.) Solve (2x-5) (ax-43}) =0. Here either 2x— 5=0, and .*. a;=f ; 46 or aaj— 4&=0, and /. a;= a ib Thus the roots are f and -- ■* a Q UADRA TIC EQUA TIONS. "5 •y one of \ct of two the solu- one or ;ion are root is Exercise XL VI. (1.) a)2„36=o. (2.) 5a;2=45. (4.) 2(a;2-7)+3(x2-ll)=33. (5)- K«^'H4) + Ka^H3) = a;2 + l. (3.) I =27. ,(.. _6 7_ (7.) x'=Sx. (8.) 2+6^^=0. (^') i(x-'-4x)=6x. (11.) 4x2+1=0. (10.) x^-'~=0. (12.) a;2— ^=2a:2+iK. (13.) (x-S)(x-5)=0. (14.) (a5 + 5)(a;-7)=0. (15.) (a; + l)(a; + 3)=0. (16.) (2i»-l)(3x-4)=0. (17.) (3x-5)(2a; + 7)=0. (18.) (5u; + 6)(6a; + 7)=0. (Id.) (ax^b){cx-{-d)=0. (20.) a;^— aaj=a£c--a5'^. 147. When tlie quadratic is not in a lorm adapted for applying rule (i.), it may be solved by either of the following rules : — (ii.) Having transposed the unknowns separately to one side, make the coefficient ofs? unity by division (if necessary). Then add the square of^ the coefficient of x, and the solution ?s effected hy the extraction of the sqtcare root of both sides. 148. (iii.) Having cleared the equation of fractions (if neces- sary), and transposed the unknoums separately to one side, mid- tiply both sides by 4 times the coefficient of x^, and add the square of the coefficient of x. The solution is then effected by the extraction of the .iquare root of both sides. Examples. (1.) Solve a;2- 12a; + 35=0. By rule (ii.), transposing, a;2-12r=-35. i2 - p. 4 kiu* '*!' I I I [^'■1' ! ii6 QUADRATIC EQUATIONS. Adding the square of one-lialf 12, a;2-.12^ + 6"=:36-35=l. Extracting the square root, a;-6=±l; that is, a:— 6=1, and /. a; =7, or 05— 6= —1, and ,*. a;=5. Thus the roots are 7 and 5. (2.) Solve 2a;2+5;«-3=0. By rule (ii.), transposing and dividing by 2, Adding the square of one-half -|, «.2 1 5^ I /5\2 — 2 5 _1_3— 4.9. X -\r-^x-t Vi^ --^6^^2--l6• Extracting the 6quar(3 root. »+^=±I. • • flj— -5±7 4 =¥ or -3. Thus the roots are \ and— 3. (3.) Solve a;2+i)a;+g'=0. By (ii.), transposing, 03^+^03=— g'. Adding the square of one-half^, Extracting the square root, QUADRATIC EQUATIONS. Thus the roots are 2 2 (4.) Solve 2a;2 + 5a;=3 by rule (iii.). Multiplying by 4x2=8, 16x2+40x=24. Adding the square of 5, Vox' + 40a; + 5^=25 + 24=49. Extracting the square root, 4ir+5=-fc7. /. a;=2, or — 3. (5.) Solve aic2+6ic+c=0. By (iii.), transposing, aa)2+5x=— c. Multiplying by 4a and adding W, Extracting the square root, 2aa; + &=±V62— 4ac. j^^-6±V6''-4ac 2a Thus the roots are " 2a" and — &-V&^— 4ac 2tt 117 i Exercise XLVn. (1.) (B2-6ir+8 = 0. (3.) a;2+4a;-21==0. (5.) l-a:^=|. (2.) x2«4a,«6=o. (4.) 2*2_5^^2=0. (6.) ^=x^+-^. ^ -i ili f 1^' lis QUADRATIC EQUATIONS. (7.) 4a;=^-4a;=15. (9.) |'-a.=12. 1-a; 2- 03 (8.) 6a;2-lla;+4=0. (10.) a;+l=" (14.) £C a; + 60 3x-5* (17.) a;=5— i-. a? — o (19.) ?±i_5fe±|)=8. (20.) (21.) (a;-3)2-3(x-2)(a;-7)=21. (22.) »2-(a+6)a;+a6=0. y-ip s 2a7— 3 a;— 3 _t '^ "^--^ 2^+l"3^T2-^ (18.) ^-2xz:^=f . 2a;-5 3x--2 a; + 2 4— a; 1,# ( "9 ) CHAPTER XVm. PROBLEMS. 149. When the Algebraical statement of a problem leads to a quadratic equation, the unknown quantity will be one of the roots. In some cases either root may be taken, but it will generally be found that one of the roots must be rejected as being inconsistent with the conditions of the particular question proposed. Examples. (1.) A person laid out a certain sum of money in goods which he sold again for $24, and lost as much per cent, as he laid out. Find out how much he laid out. Let X = number of dollars laid out. .'.05—24:= „ „ lost. But the loss is also x per cent, of x =— x»= ^ 100 100' X" ^x-IL " 100 a;2-100a;=-2400. 03=40 or 60. The amount laid out was, therefore, $40 or $60. Thus both roots satisfy the conditions of the problem. (2.) A person buys a certain number of shares for as many dollars per share as he buys shares; after they have risen as many cents per share as he has shares, he sells and gains $100. How many shares did he buy ? I20 PROBLEMS, Hi** #; Let X = the number of sLares bought, the price of which at X dollars per share is x^ dollars. The rise being x cents or ^ dollars, the price for which he afterwards sells the x shares at a^+^Tw^ dollars per share is (ic+=^W But the gain is $100. a;2=10000. /. X = + 100, or -100. As the negative root would not answer the conditions of the problem, it must be rejected. The answer is, therefore, 100. Exercise XL VIII. (1.) A rectangular room which contains 1800 square feet is twice as long as it is broad : find its dimensions. (2.) Divide 20 into two parts whose product shall be 91. (3.) Find a number whose square increased by 20 is 12 times as great as the number itself. (4.) Divide 15 into two parts such that their product shall be 4 times their difference. (5.) By what number must I divide 24 in order that the sum of the divisor and quotient may be 10 ? (6.) Find three consecutive numbers such that the square of the greater shall be equal to the sum of the squares of the other two. (7.) A ladder 34 feet long just reached a window of a house, when placed in such a position that the height of the window above the ground exceeded the distance of the foot of the ladder from the wall by 14 feet. Find the height of the window. PROBLEMS. 121 (8.) A horse is sold for £24, and the immber expressing the profit per cent, also expresses the cost price of the horse : what did he cost ? (9.) An article is sold for £9 at a loss of as much per cent, as it is worth. Find its value. (10.) A and B start together for a walk of 10 miles; A walks 1-2 miles an hour faster than B, and arrives 1^ hours sooner than he does : at wl)at rate did each walk ? (11.) After selling a part of an estate, and the same part of the remainder, I find I have left nine- tenths of the part first sold : what part did I sell at first ? (12.) An uncle leaves 14,000 dollars among his nephews and nieces, but 3 of them having died in his lifetime, the others received 600 dollars apiece more : how many nephews and nieces were there originally ? (13.) A number is composed of two digits, the first of which exceeds the second by unity, and the number itself falls short of the sum of the squares of its digits by 26. What is the number ? (14.) The sides of a rectangle are 12 and 20 feet : what is the breadth of the border which must be added all round that the whole area may be 384 square feet ? (15.) One hundred and ten bushels of coals arc distributed among a certain number of poor persons; if each had received one bushel more, then he would have received as many bushels as there were persons. How many persons were there ? (16.) A sum of £23 is divided among a certain number of persons ; if each one had received 3 shillings more, he would have received as mary shillings as there were persons. How many persons were tnere ? (17.) A company at an inn had £7 4s. to pay, but before the bill was settled 3 of them left the room, and then those who remained had 4s. apiece more to pay than before; of how many did the company consist ? ■ I '' i 122 PROBLEMS. , ii (18.) A person rents a certain nuu* jer of acres of pasture land for £70; he keeps 8 acres in his own possession, and sublets the remainder at 5s. an acre more than he gave, and thus covers his rent and has £2 over. How many acres were there? (19.) An oflBcer can form the men in his battalion into a solid square, and also into a hollow square 12 deep ; if the front in the latter formation exceed the front in the former by 3^ find tho number of men in the battalion. t ,, li'^ii of pasture ission, and gave, and acres were ( 123 ) lion into a 3p ; if the ihe former CHAPTER XIX. SIMULTANEOUS EQUATIONS, 150. If two unknowns are to be determined, there must be two independent equations. These equations are called simultaneous equations, because the same values of the un- knowns X and y must be substituted in both equations. Thus if 2x^y = 9, the only values which satisfy both these equations at the same time are x=7,y=6. 151. It must be borne in mind that there is an infinite number of values which will satisfy either equation sepa- rately. Thus, in the equation 203— 2/ =9, ifa;= 1, 2--2/=9, and .'. 2/=- 7; if 05= 2, 4— y=9, and /. 2/=— 5; if a:=10, 20-2/=9, and .*. y= 11 ; and so on. 152. If three unknowns are to be determined, there must bo three independent equations ; and generally the number of unknowns must be the same as tho number of independent equations connecting them. 153. The solution of simultaneous equations is effected by deducing from them other equations, each of which involves one unknown. This process is called elimination, and may be conducted according to one of the following methods :~ I m 124 SIMULTANEOUS EQUATIONS. I. Substitution. II. Comparison. III. Cross Multiplication. I. Method of Substitution. 154. Tins method consists in finding from one equation the value of one unknown in terms of the other, and substituting the value so found in the second equation, which is thereby reduced to a simple equation in one unknown. For convenience of reference the given equations and others which arise in the process of solution are numbered (1), (2), (3), &c. Example. Solve x-\'y=d .... (1), 2aj+2/ = 4 .... (2). From (1) we find 2/ = 3— a; .... (3). Substituting this value of y in (2), 2a; + 3-a;=4:. Substituting this value of x in (3), 2/=3-l=2. Thus the solution is x~l. y=2. II. Method of Comparison. 155. This method consists in finding from each of the pro- posed equations the value of one and the same unknown i7i terms of the other f and equating the values so found. Solve Example. 7a:~32/ = 19 .... (1), 4a;+7y=37 .... (2). lation the tuting the 'educed to Qd others ^ (1), (2), ne pro" n terms SIMULTANEOUS EQUATIONS. 125 From (1) we find y __7i»-19 • . (3); and from (2) y 7- .... (4). Equating these values of ?/, 7aj-19_37-4a; 3 T~' ,\ a;=4. Substituting this value of x in (3) Thus the solution is a;=4, y=6. III. Method of Cross Multiplication. 156. T/u-s ^^e^7,of^ co«,s/sf.s in Tnulfiplyhifj the qiven equations {reduced to the form ax + by=c) Inj such quantities as will render the coefficients of the same nuhioum mtmerically equal By adding or suhtracting the equations so found, ive obtain a simple equation in one unknoivn. Examples. (1.) Solve lx-2y= 5 . . . 13«+4?/-30 . . . Multiplying (1) by 4 and (2) by 9, 28x-36v/= 20 . . . 117a^ + 36;y=270 . . . Adding (3) and (4), 145ir = 290, Again, multiplying (1) by 13 and (2) by 7, 91X-117?/- 65 .... (5), 91as-+ 28^=210 .... (6). (1), (2). (3), c- \^ ■ f! 126 SIMULTANEOUS EQUATIONS. Subtracting (5) from (6), 145?/ = 145, /. 2/ = l. Thus the solution is x = %y = \. \ (2.) Solve 8a; + 25?/ = 9 . . . . (1). ^ |, 12a;-10iy = 4 .... (2). ,!|"' Multiplying (1) by 2 and (2) by 5, [ 16x + 502/ = 18 . . . . (3), ;Li. 60a;-50?/ = 20 . . . .(4). Adding (3) and (4), 76a; = 38, 1, * rf — 1 1 ..»«'— a • Again, multiplying (1) by 3 and (2) by 2, 1 i 24a; + 75?/ = 27 . . . . (5), j ; 24a;-20//= 8 . . . .(6). Subtracting (6) from (')> ! ; 95?/ = 19, .1 V • ?y-i ! • • 2/ - 6' -i^ Thus tho solution is x=\, ij=^. I*'^- m Exercise XLIX. (1.) 4x + 2/=ll, a; + 4?/=14. (2.) 2x + 32,'=21, 3a; + 5,?/=34. (3.) 3x=:23-2?/, 10 + 2.t=5?/. ^^•^ 2 3' 32 ^^■^5^6~2^'^' 3 *'10 4* (6.) 3a;-22/ = 3(6-a;), 3(4a;-3?/) = 72/. (7.) 7(a)-l)=3(2/+8), ^+?=5^. SIMULTANEOUS EQUATIONS. 127 ^ON 2a;—;/ 3 3y « (SO "1-^-2 = -4 -^-2, a7+y = 8. (10.) 2a;-2^3=5fr2 g.,_^~5_7?/-7 5 2 ' *^ 3 2~' (ll.)T^(aj + ll)+Ky-4)=:a;-7,-Ka; + 5)-.K2/-7) = 3y-a:. (12.) a;-24 = |+16, Ka'+y)+x=3(2y-a;) + 105. (13.) K3^-72/)=K2a;+2/ + l), 8-K«'-2/) = 6. (14.) ^^±^=l + 2(2a;-6?/ + l) a;_,, (15.) a; 4- 2/ = a, a; -2/ = 6. (16.) aa; + a2/ = a2+6^ a;=a. (17.)^ + |=l,c.+y=c. 157. When there are three simultaneous equations containing three unhnoivns, the solution is effected hy eliminating one of the unknoivns hetiveen the first and second equations, and also he- twecn the first and third, or second and third. Tu-o equations are thus obtained involving two unknoivns, ivhich may he found hy the methods already exi>lained. The value of the third un- knoivn may then he found hy substitution. Example, Soke 2.r-32/+ 2= 1 ... . (1), 305-52/ + 4^= 3 . . . . (2), 4x + 22/-32=13 .... (3). Multiplying (1) by 3 and (2) by 2, 6a;- 92/ + 32=3 .... (4), 6a;-102/ + 82=6 .... (5). >}' 128 SIMULTANEOUS EQUATIONS. if :■■ Subtracting (5) fr^^m (4), 2/--5,i=— 3 .... (6). Thus a- is eliminated from (1) and (2). Again, multiplying (1) by 2, 4a— •62/ + 22= 2 . . . . (7), and 4a; + 2y-32=13 .... (3). Subtracting (7) from (3), 8?/-5^= 11 ... . (8). And y—5z=-'3 .... (6). From (6) and (8) we find y=2, 2=1. Substituting these values of y and z in (1), (2), or (3) we get a; =3. Thus the solution is x=3, y=% 2=1. n. EXEKCISE L. (1.) x + By-^2z=ll, 2x + y-i-3z=U, Sx -{-^y + z=ll. (2.) a; + 2^ + 37;=13, 2.b + 3?/ + 2=13, 3x-{-y + 2z=10. (3.) 2x-{-Sy-'4:Z=10, 3x-4:y + 2z=6, 4:X-9aj+Sz=21. (4.) lOx-2?/ +42=10, 3a; + 52/+32=20, a;+32/-2.2=21. (5.) 3^ + 2^=13, 3?/+22=8, 32+2x=9. (6.) ^-f 1+1=3, 4:x + 5y-{-6z=77,z+x=2y. 4 5 D (7.) 3a;-2.y=6, 3y-22=5, 32-2a;=-2." (8.) |^(«-l)-2/=35, %-5z=43, x+y-{-z=60. (9.) 2/— 2 + 3=0, z;--iC=-5, a3 + 2/=6. (10.) 2,' + 2=a, z + x=h, x-\-y=c. ( 129 ) CHAPTER XX. !), or (3) LI. 10. =21. 2=21. PROBLEMS. 158. In the following problems the various unknowns are expressed m terms of separate and distinct symbols Td th^ ot equations. If two symbols x and y be employed thA tit and 1 *'' -"f ^^". "^^* '"^^ two'indepentn tiua! tions, and three mdependent equations will be required to determine three unknowns, x, y, and z, ^ *"* Examples^ (1.) A fraction becomes equal to 1 when 1 is added fn fli« rni^rir Lot* '''^" '^^-^^ *^^-^- Let - be the fraction. y Then by the conditions of the question, a+l_ y X 1, the solution of which is x=5, y=.Q, Hence the fraction is |. ^^«Z^^? ^^""^ numbers such that the sum of the first one-fifth the secondhand one-tenth the third, shall be equal K 130 PROBLEMS, li 'I H : to 4; the sum of one-half the first, the second, and one- tenth the third equal to 7 ; and the sum of one-half the first, one-fifth the second, and the third equal to I'i. Let aj= the first, y=. the second, and 2= the third number. Then by the conditions of the question, 2 5 Multiply these severally by 10, 10a;+ %j-\- 2= 40 ... . (1), 5x + 10y-f 2~ 70 . . . . (2), hx\ 2?/-|-10;^=120 .... (3), 2x(2)-(l), 182/-h2=100 .... (4), (3)-~(2), -87/ + 92= 50 ... . (5>, 9x(4)-(5), 170y/=850. .•.?/=5. Therefore from (4) 2=100-18?/=10 ; and from (1) a; =2. The required numbers are thus 2, 5, and 10. Exercise LI. (1.) One of the digits of a number is greater by 5 than the other. When the digits are inverted, the number becomes I of the original number. Find the digits. (2.) In a division the majority was 162, which was i\ of the whole number of votes ; how many voted on each side ? (3.) The sum of two digits is 9. Six times one of the numbers they form is equal to 5 times the other number, i'ind the digits. PROBLEMS. 131 md one- tho first, lumber. than tlio becomes as -i\ of side? of the Qumber. (4.) If the numei^ator and denominator of a fraction be each increased by 3, the fraction becomes 2 ; if each be in- creased by 11, it becomes f . Find the fraction. (5.) A number consists of two digits whose sum is 12, and Btich that, if the digits be reversed in order, the number pro- duced will be less by 36. Find the number. (6.) Three towns A, B, and C are at the angles of a triangle. From A to through B, the distance is 82 miles ; from B to A through C, is 97 miles; and from C to B through A, is 89 miles. Find the direct distances between the townSk (7.) The diameter of a five-franc piece is 37 milUmetres^ and of a two-franc piece is 27 milUmetres. Thirty pieces laid in contact in a straight line measure one metre exactly. How many of each kind are there ? (8.) At a contested election there are two members to bo returned and three candidates. A, B, C. A obtains 2112 votes, B 1974, C 1866. Now 170 voted for B and C, 1500 for C and A, 316 for A and B. How many plumped for A, B, C, respectively ? (9.) A boat goes up stream 30 miles and down stream 44 miles in 10 hour >. Again, it goes up stream 40 miles and down stream 55 .irui>k stream and boat. in 13 hours. Find the rates of the (10.) At a contested election there are two members to be returned, and three candidates, A, B, C. A obtains 1056 votes, B 987, and C 933. Now 85 voted for B and C, 744 for B only, 98 for C only. How many voted for C and A, how many for A and B, how many for A only ? (11.) Seventeen gold coins, all of equal value, and as many silver coins, all of equal value, are placed in a row at random. A is to have one-half of the row, B the other half. A's share is found to include seven gold coins, and the value of it is £6. The value of B's share is £6 15s. Find the value of each gold and silver coin. k2 it V »32 PROBLEMS. (12. road and )m A to D passes through cessively. The distance between A and B is six miles greater than that between C and D, the distance between A and C is ^g of a mile short of being half as great again as that between B atd D, and the point half-way from A to D is between B and C half a mile from B. Determine the dis- tances between A and B, B and C, C and D. (13.) Fifteen octavos and twelve duodecimo volumes are arranged on a table, occupying the whole of it. After six of the octavos and four of the duodecimos are removed, only | of the table is occupied. How many duodecimos only, or octavos only, might be arranged similarly on the table ? (14.) Three thalers are worth \d. more than 11 francs. Five francs are worth \d. more than 2 fijrins. One thaler is worth %l. more than a franc and a florin together. Find the value of each coin in English money. (15.) Six Prussian poimds weigh \ oz. more than 5 Austrian pounds. Twenty-five Austrian pounds weigh \ oz. more than 14 kilogrammes. One kilogramme weighs 1 oz, less than the sum of the weights of a Prussian and an Austrian pound. Find the number of ounces in each foreign measure of weight. (16.) A person walks from A to B, a distance of 9i^ miles, in 2 hours and 52 minutes, and returns in 2 hours and 44 minutes, his rates of walking up hill, down hill, and on the level being 3, 3*, and 31- miles an hour, respectively. Find the length of level ground between A and B. ( 133 ) CHAPTER XXI * EXPONENTIAL NOTATION. u 159. Although the notation adopted in the preceding pages is sufficient for the purposes of the operations herein treated of, yet it is found expedient, before proceeding farther, to employ another notation to express roots, powers of roots, and their reciprocals. This notation, which consists in em- ploying fractional exponents instead of radical signs and integral exponents, and negative exponents instead of reci- procal forms, possesses the great advantage of reducing to a few uniform laws the operations of Multiplication, Division, Involution and Evolution, with respect to powers, roots, and powers of roots, of a quantity, and their reciprocals. 160. The exponential notation consists in writing and a" instead of y^a'", 1 a-P Thus, according to this notation. a^=^a, a^=^a\ a^=ya», a 1 a ■3 1 -8 a = 1 a ■^ 1 1 a •f_ _1 1 * This Chapter may be omitted by those who do not intend to read more advanced works on Algebra, % « ! 1 1 :l jr,:i 134 EXPONENTIAL NOTATION, -3 2 2a =^,' 161. When the exponential notation is employed, the quantity is said to be raised to the -power indicated by tho exponent. Thus a^ is read a to the 'power ^ ; a it if i» a X X it n a a f; a -3; a -|. The term poiver in this extended Algebraical sense thus includes the terms power, root, root of a power, reciprocal of a power, reciprocal of a root, reciprocal of a root of a power, as used in tho ordinary or Arithmetical sense. U i:: EXEBOISE LII. Express in the Arithmetical notation— ,_ ^ .1. A 2. 3. (1.) a^, a«, «», a2. (2.) x-^, x-\ x-'^'^. (3.) m 3, n '^, p *. (4.) 2a*, 3a:- 2, Gm'^. Express in the Exponential notation — (5.) ^x, ^m, ^n. X a'^ a'' a* (7.) ^'] ^^% V^, ^^'^ ^x ^x j;/^' EXPONENTIAL NOTATION. 135 (9.) sd, the I by tho ise thus )coX of a oweif as 2 3 10 m rr p^ (10.) -!,. 4. ' ^x ^x ^x^' 162. The utility of employing the exponential notation will be exhibited in the statement of the three following rules, which are usually called Index Laws. I. The product of any two powers of the same quantity is a power whose exponent is tlie Algebraic sum of the exponents of the factors. Since the product of a quantity and its reciprocals 1, this rule cannot be applied when the exponents of the two factors are numerically equal and of opposite signs, unless tlie zero power of a quantity be considered=l. Examples. (1.) a'^ ' a3 = a2 3=^6. ^8 8. 3 + 6. XS (2.) a* • a*^=a'^ ^=a^^, (3.) a^ .a-i=a3-i=a2. (4.) a-2 . a-3=a-2-3=ra-6. -^ ^ —A 1 — .' i (5.) a- a •^=a '■'=a'\ 5. -3 5_8 JL (6.) a^ • a * = a« * = ai2. (7.) a- a-i=ai-i=a°=l. (8.) a^ ' a"2=ra" = l. This law may also be expressed briefly as follows :— where m and n are any quantities whatsoever, positive or ne^Of Uve^ integral or fractiotial, including zero ifsP=l. 136 EXPONENTIAL NOTATION. I' ■'%''■ m' % m 163. Proofs of the preceding rule will be exhibited in the following particular cases. (1.) If m and n be positive integers, a"* • a.^-:^aaa . . . . (w factors) xaaa . . . . (w. factors), = aaa .... (w+n factors), _ „♦»+» (2.) aa . a8=Va ^a^^~Q^ ^a2 = ^a'^=o« = a2^3. Here v^ = >v/^> because each of them when multiplied by itself six times produces a*. So also ^a = y~€?-\ and generally, if m, %, j9, are integers, because each of these quantities when multiplied by itself «p times produces a'"^. (3.) If m, n^p^qhe positive integers, then = ^a»»«+p», by Ex. 1, m p (4.)a».a-»~=^=a->=a'^-». a* a Exercise LIII. Find the products of (1.) 2a;, Sec"; x^, 405*"; dac", a;^™, (2.) a;2, 2a;*; 3a; s, 2x2; 6a;*, 5a;^. (3.) as", aj*; 2a;2, Sec"; a;2^ a;}f. EXPONENTIAL NOTATION. >37 (4.) 2a», a- a ; a-\ Sa^; 5a, ^a'\ (5.) Q^, a" 3; 2a*, a"«; a, a'^. (6.) aa, a"3; a«, a~^; a^", a"^. (7.) a», a-3; a», a-; 2a, 3a-»; wa^, Ma-«. 164. n. When one power of a quantity is divided hy another y the quotient is a power whose exponent is the Algebraic difference letween the exponents of the dividend and divisor. Since the quotient is=l when the dividend and divisor are the same, this rule cannot be applied in the case of equal powers unless the zero power of a quantity he considered=.l. a (1.) -3=a«-s=a8. 2 (3.) -3= a =a • a (5.) ''=|=a-^t=aA. Examples. (2.) «^=at-t=a*. ai 4-i A ai (4.) — r— a =a^ a a" (6.) -,=a»-fi=aO=l. This rule may also be expressed briefly as follows :— a a" TO -Qfn^n where m and n are any quantities ivhatsoever, positive or nega- tive, integral or fractional^ including zero ifaP=l. 165. The proof of this rule rests on that of the preceding. For since a"*-". «"=«"', it follows that a"* il If ¥' 1*1 I3S EXPONENTIAL NOTATION. Also, since t^ =«"•*, the second rule may be considered to be included under the first. Thus a" W y^— M __ /»W1"»II ssa™. a "=«' .■T 'iSf' Divide Exercise LIV. (1.) a^"* by a*"; a"" by a*. 1. 3 . 2 .3 (2.) or by a^ ; a by o^ (3.) a;^ by a; ; ^ by a*. (4.) 01? by a;-^ ; x^ by a;'^. (6.) a;-^ by a;-2 ; a;-^ by as'". (6.) a;^ by a;'^; a;^ bya;-^ (7.) a;"" by aj-^"; a; 2 by a?"". 166. m. The power of a power of a quantity is a power whose exponent is the product of the numbers expressing those powers. In other words, where m and n are any quantities whatsoever^ positive or nega- tive j integral or fractional, including zero ifaP=X. Examples. (1.) (a^y=a^; {a'y=--a'^. (2.) (a^)2=a«; (««> = «*. (3.) (a"^)a = a«; (a^y = a^. (4.) (a-l)2=a-2. (^-2)3^^-6^ (5.) (a-i)-2=a2; (a-3)-4=ai2. (60 ia^r^ = a~^; (a^y^z=a'K EXPONENTIAL NOTATION. 139 167. Proofs of the preceding rule will be exhibited in the following particular cases : — (1.) («")=^=a2 . a2 . a2=««=a3'<2. (2.) (a3)2=a'3 • a3=a^"'3=a« (3.) If n be a positive integer, (»"»)•»= a*" • a" — /ym+w-f =«•»". . to n factors, to n terms (4.) If ^ and i? be positive integers, (a")«=a«, because each of these quantities when raised to the ^th power produces a"^. EXEEOISE LV. Express the following as powers of a : — (1.) (aO^(«^y;W^ (a-i)2; (a-2)3. (^-3)4^ (a2)-3; (a-2)-3; (a-3)-*. (2.) (8.) (4.) (5.) (6.) (a3)2; (0^4)3 J (a"4)3. -tx-1 (c.-^-^ ; (a-t)-^ ; (a-t)-!. 168. The Index Laws (I., IL, III.) are thus seen to be true on the assumption that -1' 1 for all values of m and w, positive or negative, integral or fractional, including zero. Instead, however, of treating the subject of the Index Laws and notation as in the preceding Articles, we may proceed as follows : — If m and n be positive integers, it may bo proved, as is done in Arts. 163, 165, 167, that i I n if h" ■ 140 EXPONENTIAL NOTATION. L II. III. a m ^n—^m+n^ ^ sa"'-", w being greater than n. a (a"')'»=a»'*. These laws, which are thus proved to hold in the particular case where m and n are positive integers, are then assumed to be true when m and w are any quantities whatsoever, posi- tive or negative, integral or fractional, including zero ; and from this extension of these laws we deduce that a" must=v a"*, a" '^=0^, and a°=l. Thus, by I., But a/ a. /s/a =a. /. a^=^a. By m., (afy=:a\ But (*/'^y=a\ .-. a^=Va\ ByL, a* . a^=a\ But a' X l=a'. /. a«=l. ByL, a3 . a-3=a»=l. But a^ • rt-3 — -^ a*" 169. From the remarks of this chapter it will be thus seen that, although there is no absolute necessity for using such 3. 4, —5. symbols as a*, a" , a 3, still their introduction gives rise to a uniformity in certain Algebraical processes ; and the number of rules which otherwise would be required to meet the different cases that arise in those operations thus becomes largely reduced. ( 141 ) ANSWERS. 1. (1.) 5r 10+15; 10+12+f ; 2+i+f ; a+^+|i+^. (2.)i 5; 2i-li; 2-5-1-6. (3.) 11 + 35 + 6-17; 81+76-69-4a (4) 15-7+8 + 9. (50 28-16 + 10-4. U. (1.) A's -60, +20; B's +60, -30; C's +30, -20. (2.) A's +20, +20, -30,-40; B's -20, +30, +30, C's -20, -30, +40, +40. (3.) A -10, +4; B -7, +10; -4, +7. (4.) -2°, +5°, -3°. (6.) +1°, -P, +P, -P, +P. (6.) -2°, +2°, -2^ +2°. (7.) +25°, -7°. m. -40; (1.) 30. (2.) 20. C3.) 22. (4.) 22. (5.) 24, 120, 0. (6.) 14. (7.) 7. (8.) 16. (9.) ^' El 14^ ANSWERS. (iO.) h (11.) M. (13.) 2a^3a^2(i2 ^3^3 +4^4^ IV. (1.) 2, 3a, I, &c, 4a*-^6, k. (2.) -], +8, -f, +26, -5i»2. (3.) +1, -1, -8a, +|c(£. (4.) -1, +3a^ -faz. (6.) 1, -1, -% -3, +i -f. (8.) -a2, +|a^ +2a2a;, ^a^x. (9.) -Sa^^a;, +2a*£c; aos^, ^ax\ (5.)^>|,4, +f, (7.) 2£c, oj. 'f* if (1.) +25. (4.) -21^. (7.) +^. (10.) +1L (13.) -H. (1.) V6a. (4.) -13a. (7.) -3a. (10.) -4x2. (13.) -bV*'- (1.) 2ct-3&. (4.) 3a + 2a;— 5!/. (6.) d'-^li'^h (8.) -«;+22/-2 + l. V. (2.) +43^. (8.) -28. (5.) +5. (6.) -6. (8.) -aV (9.) -0-7. (11.) +2. (12.) -12. (H.) +1/5. (15.) +0*342. VI. (2.) +15a. (3.) -5a. (5.) +2x=». (6.) +3a2. (8.) +10c. (9.) -5c. (11.) -*-7a&. (12.) +> (14.) -fa. (15.) -f|a. VII. I.) -a; + 82/. (3.) -f:!;z;-3?/-2'. (5.) 4-a) +2^. (7.) a-2ft + 3c-(?. (9.) 4a-26. (10.) a2-45c. (12.) -2a-&-c. (14.) 4a + 07 + 5. (16.) 2(1 + 26+ 2c. (18.) 4a;l X ANSWERS. M3 (21.) |+i%2/+i^^. (11.) 4a-9&. (13.) 2.x-.6y-2!. (15.) 5a— 4x. (17.) lOx + Sy-.-, (19.) 4a' +46^ (20.) a + &. (22.) 2a+|5-|c. (1.) 4. (4.) -3. (7.) 10-6. (10.) -0-96G. (13.) lOa^ (16.) -W. (19.) 2«. (22.) — aa;— 5&2/+4:C2. (24.) -7ccH2az;2-5x. (26.) ia+i&--ic. (2.) 2. (5.) 6. (8.) -3-39* (11.) -a. (14.) 2:;. (17.) \\x\ (20.) 3a-cc+4. (3.) -9. (6.) ~3. (9.) -21G. (12.) -7a;. (15.) 2a-|. (180 3a + &. (21.) -5a& + 26=^, (23.) -6a + &+«-4. (25.) 1+14^. IX. (1.) 2*2+-l; 3a; + -42^+-5; 2a+-.36++4c. (2.) 2a- -5a. (4.) 2a+-8/>-+7. (6.) 5a + (6-4). (8.) a-4 + (26-r). (3.) ~6-+5x. (5.) 5++a: 3a. (7.) -a +(-6 + 5). (9.) a;2 + (2i/ + 5)-;^. (10.) a-l + (36 + 5)--3c-. (11.) ic + (2x''-l) + (-3x^-8). (12.) 4a2-(62-c). (13.) «2 + 4-(-26 + 3). (14.) 2a-5-(a2-2a-+ 3). (15.) a + 6 + c + (a-6-c)— (-a + 26-3c). ■I 144 ANSIVERS, J*' (1.) 2a+36-(;. (3.) x'^^x-l. (5.) 8a— ft— c. (7.) 8a+26-3c (9.) a;+ll+4y. (10.) h. XI. (2.) o&— 6c+c. (4.) 5a;»-3a;2+7xB-8. (6.) 8a-6+c. (8.) 2a-6 + 6+c. (11.) 2aiH3a;. <■ 1 1 (1.) (2.) |i (3.) 11 (4.) (1.) (a-l)(2a2-3); (-2+a)(-3-a2); (a5-5)(-2x+7): (2.) -2a2(62«i); (a2-l)(_3a); 5a;(~a;H3). (3.) K^'-l); K2^-3); -1(0)^-5). (4.) -5<a;-l)(a; + 2); (a;«-4)( + 5iB)(2a+3). (50 +8<-52/)(a:y-l); -7a(a6-3)(+86). xn. — 6a&; — 5ac; +6a*6; — SOajy. — 14a&c''; — 20a'*6c; — 16a3yz; +48a. -fa^?^; -fa&; +iajy; -fa^J. — 6a5^«/^; +3aa"''«/''; — 4a'fe^c*. w a^b* (5.) -g ; 15 ; ~" "20"' ''""12^ xm. (1.) -8aa; + 66aj-2cx; 1203=2/ -8a;y+43^; -2a6c?-|-3c(f. (2.) 3x»-6»2-15i»; -2a'''+3a*-7a^ -4aa;H4aV-8a»a;. (3.) 4a;V + 2^y'^^-6ic^2^ -28a362+ 4^253 _ 4^2 j2^ (4.) 2fi26 + a?>2-fci&c; -a3 + fa2-2a; iaa;^- ia*^;* +|a'a;. (5.) -12a + 10a6-15; ^-^ax^^\^ax-\-^%\ (6.) 5x'*-5a; + 20; -2a+2a6-6; -a'+2a'»aj-a. (7.) ax -a;*; ac -&c + c*; -Saft + Sa^i^-lSaSj, (8.) 4a3-6a2+8a; +2x'-§a;H3a;». ANSWERS. »4S XIV. (1.) 2a;2+6a;-12; -^x^^x+h; 2-!c-.3a;2. (2.) 2a;«-a;2_4^^2; 2-2x+3a:2-.3a)«; 3+3x2-a;3-.a;« (3.) 6a3-7a2+14a-8; l+a^; i_^3^ (4.) am-an ^bm-hn; am +hm^cm + 2an+ 2bn^2cn' (5.) a;y-x*; a;3+a,3y-a;2^2_2a;y-2a;2/H2y3. (6.) 2aHa3-22a2+23a-4; a2-4&H126c-9c2. (7.) 3a|-4^&+8ac-46H8&c-3c2; 3a.*-4a;3y+6xV+4a;y» (8.) a)*-fa.Hi; a^^a-i; 2a='-i (9.) 2^3-|x2+|ia.-i; 9a.«-|a;2^.|^_^^ (11.) ia^-^H|a-l. (12.) a;«+3a:y-2a:+y3-2y+l. (13.) a»-3a6c + 63+c3. XV. (1.) +30a«J»; -J^a;7. ^i3^y,^',^ (2.) -24a;V+30a;V; -2a'>6+3a*62-a85. (3.) 8x3-26x«-17a.+6. (4.) a.*.a:3+^_i, (6.) a;8-2aV+a8. XVI. (1.) (2a-6)-r-.3a; (4a2..,3a + l)-f.(3a-4). (2.) 2a-f— 3&; -a^-j- + 2x'; 3a;-r-2a. (3.) 4a;2-j-(2c-5); -aaj2_j>(a;-a). xvn. (1) -4; -6; +f ; -V. (2.) ^±. -?^' . "^ 2a;' 26 + 2«y a 146 ANSWERS. (3-) -|; 6«a5. W 56' (5.) -2a; 6a6V. 14a2- (4) 2a'; a»; 4a^. (6.) -ia;?/'; +fay. xvm. (1.) -2a+36-4; aa;--3+2a2. (2.) -4a;+3-a; 4a;2_ar + 3. (3.) -aH4a-5; a;2-3a;i/+4/. (4) 2a-36+c. (5.) -4ac2+3&c'-l. (6.) (a— 6)0?, (2a— c+l)a;. (7.) (4— a)a;s', (3a3— y)a;y. XIX. (1.) a!-4; 3x+l. (2.) «+!; 2a;-3. (3.) 3a5+2; 3a:2-2(»+6. (4.) a;-l; oj^-jb+I. (5.) x^-Zx->^\, (6.) a;Ha;Ha;+l; (c^-ac'+cc^-aj + l. (7.) a;-3/; x^-xy-\-y\ (8.) aHaJ+ft^. (9.) 5a; + 6y-3. (10.) aj^-aaj+a''. (11.) a;2-2i»2/+2/'. (12.) faa;-2a;2. XX. (1.) ar2-3a5+ V , - f . (2.) x-a, '2a\ (3.) !r2_aa;+aa, -2a3. (4.) a-2-a;-f4, -ac-4. (5.) 2a;2+3, -5a;2-3a;-3. XXI. (1.) 3a3— amiles. (2.) 50 +x dollars. (3.) 26 miles. (4.) x^—y^ square feet (5.) - hours. a (6.) .+|+3^ <7-) -+^+4- /Q \ lOoj (8.) ^ . ANSWERS, 1 147 (9.) |. (!«•) 1^00- 100* (11.) ^+f -5. (12.) 3& acres. (13.) ^^ miles. V (14.) 1^ hours. (15.) 6aa;+ a dollars. XXII. L. (1.) 2. (2.) 10. (3.) 7. •y)xy. (4.) -5. (5.) 2. (6.) 5. (7.) "h (8.) -5. (9.) f . 1 (10.) 2. (11.) 3. (12.) 2. j (13.) 7. (14.) 10. (15.) 2f 1 (16.) i. (17.) 6. (18.) 61 1 (19.) 7. (20.) 13. (21.) «. f (22.) 3a. (23.) a +6. (24.) %. 1 I (25.) 6+c. (26.) 1. (27.) a2+a& + &^ 1 1 (2«.) '^*-^'. •1 I a— c i; 1 1 XXIII. s 1 (1.) 12. (2.) 12. (3.) 24. 1 (6.)6A. ! 1 (4.) 30. (5.) 23i. ^^^1 (7.) 6. (8.) 3. (9.) 3. " i I (10.) 9. (11.) 4. (12.) 5. ' I (13.) 3i. (14.) -2. (15.) 33. I (16.) 2i. (17.) 2. (18.) 2. 1 (19.) 120. (20.) 13. 1 XXIV. { 1 (1.) 15 and 10. (2.) 60 and 75. (3.) 20 and 17. 1 (4.) 14 lbs. (5.) 23, 17. (6.) 181 and 145. l2 148 ANSWERS, (7.) 5. (8.) 14 years. (9.) In 9 years. (10.) 18. (11.) 66 years. (12.) 400. (13.) 700. (14.) 30 for translation, 5 for mathematics. 4 for Latin prose. (15.) 21 shillings. (16.) A's £800, B's £100. (17.) 400 inches. (18.) 18, 11 and 8. (19.) 35. (20.) 200 quarters. (21.) 12 lbs. (22.) 150 lbs. (23.) 240 sovereigns, (24.) 13. 480 shillings, 720 pence. (25.) £3000 at 5 per cent., (26.) £450 at 4i per cent. £10,000 at 4 per cent. £350 at 5i per cent. (27.) 1800 infantry, (28.) 17 years. 600 artillery, 200 cavalry. (29.) 26. (30.) 12. (31.) 56 workmen; 150 shillings. (32.) 1330. (33.) 4290 feet. (34.) 30,000 men. (35.) 200 miles from Edinburgh. (36.) In 56 hours. XXV. (1.) a;2-2i» + l, a;2 + 2aa; + a^ a;2-10a;+25, a;H6a; + 9. (2.) 4a;H4a5 + l, 9a;2-.6a; + l, 4a;H12a; + 9, 9a;2-12a!H-4. (3.) £c*--2aa;2 + a2^ 4xV + 4a;2/ + l, ^x^ -VHax^ ^-^a^, a'x'' -8a6x2 + 1662. (4.) a;2 + 2/H2^—2iC2/ + 2x2-2^2, ix^ + dy^ + z^ + 12x1/ -4:xz -62/z, x^+4ty^+2bz^'-ixy^lOxz-\-20yz, 4a;- + 16y + l-16a7/ +4a3— 8y. (5.) 4a* + 4a3 + ISa^ + 6a + 9, 9a* - 24a» + 22a^- 8a + 1, a*-.4a3-4aH16a + 16. (6.) 2401, 9604, 990025. XXVI. (1.) ar^-l; a2-9;f4-a;2. (2.) 4ar»-l; 25a2-4; lex^-a^ ANSWERS. 149 years. ) 12 lbs. ) 13. )er cent. »er cent. .) 12. .) 1330. ) men. hours. 9. H-4. \xy—4:xz 1— 16ay [• (8.) a*-a^; a«-l; a»-a;*. (4.) 9a*-462 ; 16a«-4a;< ; 49a8-25a«. (5.) 2496 ; 9975 ; 489975. xxvn. (1.) mP-{-n^; p^—g^. (2.) m^+l; 1-2'. (3.) a;»+27; a^-QL (4.) 8as + l; 64a;»-a3. (5.) 8a»+276^; 27x3-1252^^ (6.) x^-l; x^^-a^ xxvni. (1.) 03^—03 + 1; aj*— 03^+03'^— aj+1. (2.) x^+x+1; x^+x^+x'^+x+1, (3.) x-1; a;3-a!2+a;-l. (4.) x + 1; x^+x^+x + l. (5.) 2a-3&. (6.) 3a;3+2a. (7.) W-x\ (8.) 4a2— 6a5 + 9&2; a*-2a2+4. (9.) 9a2 + 3ab + &2 ; 4,^4 ^ 5^2 j ^ 95a, XXIX. (1.) (-a)3, (2x)3, (ari/'^)^, {2a*hhy. (2.) (2a-l)^ («-& + in (x3-l)2. (3.) { (xy ] \ { (-2ay } \ { {^axf \ \ { {^aVf ] (4.) { (a-hf ] ^ { (x'-iy ] ^ { {x-'-3x^2f ] \ (5.) (a^^)^ U-2x)M^ {{a'^hfW [(x-afW {{x'-ax^Vf]K (6.) {(-a)M^ {{x'^fW { (403.1)2 }», {(a^-a»)M A3 I 2 XXX. (1.) jbS 8< a;», 8ic», 81a;«. (2.) aV, a^r^ oVy'. \ 150 ANSWERS. ^. (3.) aWc\ a«6V, Sa'fe'c". (4.) x^Y^z^y a'&'*c". (5.) «H2a; + l, 4a;2-12a; + 9, a!*-10a;2+25, a;«-4aV+4a*. (6.) a;*+4a;H10a;2 + 12a; + 9, a;*-6a!'+17a;2-24a; + 16, 4a;«~4a;»+a;*+20cc3-10xH25. XXXI. (1.) V^, Ja^\ V^^^. Va:2-3a;+4. (2.) ^":::^«, ^"3^3, ^^^ir^, ^(a3-3a+4). (3.) V^^x, V^^^-l, Vl/'^^l . (4.) \/^2, V4^3^^1, A//y2(?^^^^iH3. ^ (5.) 4^"^, ^'^, y W, !p^ . xxxn. (1.) 2a'&, 5a;2/', 9a; V. (2.) 4aj + 5. (3.) 6a;-3. (4.) l+3a^. (5.) x^h (6.) a;-|. (7.) 2a;-J^. (8.) 2a;-32/. (9.) a;2+2a;+l. (10.) jcHaj + l. (11.) aj2-2a;i/+2/''. (12.) 2a;»-aj2-3a;+2. (13.) 1+|-|'; remainder l"-!^, (1.) ^l\ (5.) 7aV2/. (9.) 4a6. XXXIII. (2.) 5a&. (3.) ^xy. (6.) aftwv. (7.) 4a26. (10.) baW. (11.) 26^ (4.) 3aaj. (8.) 6xV. (12.) 4ww;. (1.) 05 + 3. (4.) x+3. XXXIV. X- (2.) (5.) 2ir-5. (3.) a;-3. (6.) x-3. ANSWERS. iS« (7.) (B2+10a;+25. (10.) 05-2. (13.) 2a3+5. (16.) aj-3y. (19.) a;+3y. (1.) a!r(2a;+8). (4.) a;-l. (1.) a(aj— a). (4.) aj+l. (8.) a;2-5a;+6,. (11.) a;-l (14) a;-2. (17.) a; +2/. (20.) £c2-/. XXXV. (2.) 2(i»+3). (5.) 3x-2. XXXVI. (2.) 2a(aj-8). (5.) a; +2. XXXVII. (2.) I^^xhj. (5.) 240a2&2c2«fcVM;2 (9.) cc2-9. (12.) a;2_3. (15.) a;2-2a; + l. (18.) a;2+y. (3.) x(a;-2). (3.) a(a;-l). (6.) cc-a (3.) a^W. (1.) 6a&scy. (4) 24a26V. (6.) (a;2-.7 + 12)(a; + 2) = (a;2-a;-6)(a;-4). (7.) (2a;2-5a;-3) (2iK + 1) = (4*2 +4a; + 1) (a;-3). (8.) (3a;2-lla;+6)(2x-l) = (2x'-^-7x+3)(3x-2). (9.) (x^ - 4aa;2 + ^o?x - 2a») (x" f 2aa; + 2a2) = {x' - 2a2a; .4a3)(aj2-2aa; + a2)- (10.) 24(a? + l)(a;-l)2. (11.) (!r-l)n«^+l)'. (12.) xhfix'-y^), (13.) a3&(«'-&')- (1^-) 12(a;=^-l)(a'Ha;+l). (15.) ip^i-q'')(p'-q'Kp'-pq+f)' (16.) (i>*-l)(i>*+pHl). (17.) (b^c)(c^a)(a~b), (18.) 24a2&2(a2-62). XXXVIII. a <^-) fc (2.) - cy (4.) 2a& a :::&■ (^•) ^f^ 362-6a« (3.) (6.) [JB 3_ 4^2' x::! 152 ANSWERS, bvw I (7.) 5+2* (10.) J,^. (13.) (16.) 3+aj 3-a;' 3(4a;~ l) 2(3a;!*+l) (8.) (11.) (14.) (9.) _^±y,. (12.) (15.) (17 ^ ^'(^+2y) ris^ g! + 3 05 — 5* a;-3 cc+y' a>. 6 2a XXXIX. (3.)?^^, 3?., 4 (5.) ::?- , y a;?/^ xyz xyz -2a 3 aajy axy axy ,«x a(a; + l ) 2a ^^^ a; + 3 2( g; + l) ^''•' a;2-l ' x^-r ^''^ (a; + l)(a; + 3) ' (x + 1) (x + 3) ' (9.) a (10.) (11.) X JC — 1 * 03 — 1* 4(a;'^-l) 3a;(g;-l) gi'-l g;'^-a; + l 3a; g-l a—b * a—b' nS.^ 2(«^~2) 3(a;+2) ^ ^ (a;-l)(a;-2)(x+2) ' (a;-l)(£c-2)(a;+2) * (13.) a;-l (a;-l)(a;-2)(cc + 2)* a—c (fc«c)(c-a)(a-fe) ' (b-rc)ic-d)(a-b) ' 6— a (6— c) (c— a) (a— &) * .,.^ bx(x—b) ax(a—x) > ^^ abx(a-b)lx-d)(x~b) ' a&a;(a-&)(a;-a)(c»-6) ' (q— &)(ag~a)(ag— &) qhy(a-^h) (r —a)(.r~ h) ANSWERS, «S3 (1.) (4.) (6.) (8.) (11.) (14.) (17.) (19.) (21.) '+6» ah 3a;~-2 a;2~3a;+6 2x^ ' 2a'' +26' a»-5a * a+h a V-1) 2a; + 2y (2.) XL. aa;+2 2aa • (3.) 4a2a; + 6a+6 12aa (5.) (7.) 8x«+12ag;- 1 66-17a 60 (9.) (12.) (15.) 4:db ~2 a;(4a;2-l) (10.) (13.) 1- a 2a;» a;* + a;2 + l" a5-2y (16.) 2x + 6 a;*-l' x^-)rxy-^y^ X -9 (aj-l)(a;-2)(aJ+2) (18.) (20.) 2a;(2a;Hl) £»' «-.l a^ («—•&)(«— c) (a+6+c)aj (a; — a) (a; — 6) (as — c) (22.) 0, (23.) 0. a.) (4.) (6.) (8.) (10.) (12.) 3+5cc a+aic— 05 2xHg;~l ^^ • ^-«2+2a;-3 ^^ g— •a'-f a& a— 6 4a;g+9a;-4 (c + S • (2.) XLI. cc— a'^ a' (5.) (7.) (3.) 3a— aa;+a3 as 2x2 + 1 2a— OS a OS 2 • ^n\ l + 3a? a+36 a + 6 ' x^—x y (11.) (13.) «S4 ANSWERS. *■ S;i: (14.) S+|. a 0.7.) 2aj-l+?. a; 3a;+5 (19.) 14 a;2-l' (21.) 4-j-^^. (23.) 8- ^"^ aaj^-oj+l' (1.) ^. (4.) ^y (7.) (9.) (a+6)2 • 1 (11.) 2/»+2aj+ajy. (13.) ^. (17.) oj+y' (4.) «:::^. (15.) a-±. a; (16.) ^-1. (18.) ^-1+2.. (20.) 5--i-^. (22.) .-3+1=7, (24.) a;2- XLH. (2.) 1. a+b x^+x—5 x^+l ' (5.) 6 • (6.) x^yz" ^±abVb^ a^-ab + b^' (10.) ,4^Y2- (a;'*— 05 + 1/ (12.) aj-a. (14.) -^ . (16.) 1. 2 (a~6) (7.)- a+a? (2.) (5.) (8.) (18.) XLUI. 4ax^y a^+ab + b^ a^^ab + b^' 2 (a^br r (3.) (6.) (9.) x+r a 2ap ' ANSH^EKS. m (10.) (12.) ax (11.) ^^rf. (1.) 10. (4.) 6. (7.) -^ (10.) 20. (13.)f|. (16.) f . (19.) a+&. a+6 ' (22.) (13.) M. XLIV. (2.) 8. (5.) 8. (8.) -1. (11.) V. (14.) -i (17.) i^. (20.) a. XLV. (2.) 40. (14.) 1^. (3.) 12. (6.) 2. (9.) -f. (12.) 15. (15.) 3. a+6 * (18.) 2. (^^•) 4- (1.) 40. (2.) 40. (3.) 355. (4.) 4 bowled, 3 run out. (5.) ^■^. (6.) £5. (7.) £9150 of 3 per cents., £5820 of 3i per cents. (8.) A £1250, B £1500. (9.) 100. (10.) 7 miles an hour. (11.) 45 and 30 miles an hour. (12.) C 42 days, B 84 days, A 168 days. (13.) 240, 180, 144 days. (14.) 5^^ minutes past 7. (15.) 26^ minutes past 2. (16.) 49^^ minutes past 3. (17.) 32^^ minutes past 3. (19.) 30 gallons. (21.) 70 grains. (23.) 3 miles an hour. (25.) 10 yards in 200. (27.) Imile; half-an-hour. (29.) ^6 of a mile behind. (18.) 10 gallons. (20.) 112 oz. (22.) 21 oz. (24.) 4i miles. (26.) 1430 yards. (28.) 5f . 156 ANSWERS. mk a.) ^ -6. (4.) 4, -4. (7.) 0, 3. (10.) 0, f. (13.) 3, 5. (16.) i, f. (19.) ^, -1 (1.) 2, 4. (4) 2, h. (J.) h -f. (10.) I, f . (13.) 2, «^. (16.) -1, -I-. (19.) 1, 2. (21.) 6, f . XLVL (2.) 3, -3. (5.) 2, -2. (8.) 0, -12. (11.) 0, -^\. (14.) -5,7. (17.) h -h (20.) 0, a, XLVn. (2.) 5, -1. (5.) h -2. (8.) i |. (11.) 2, i. (14.) 14, -10. (17.) 4, 4. (20.) 3, -|. (22.) a, 6. (3.) 9, -9. (6.) V26, -V26. (9.) 0, 19. (12.) 0, -|. (15.) -1,-3. (18.) -f, -h (3.) 3, -7. (6.) h h (9.) 6, -4. (12.) V, -10. (15.) 4, -1. (18.) 1, V. XLVIII, (1.) 60 ft. by 30 ft. (2.) 13 and 7. (3.) 2 or 10. (4.) 12 and 3. (5.) 4 or 6. (6.) 5, 4, 3 ; or -1, 0, 1. (7.) 30 ft. (8.) £20. (9.) £90 or £10. (10.) A's rate 4 miles ; B's 2^ nail^s an hour. (11.) |. (12.) 10. (13.) 87. (14.) 2 ft. (15.) 11. (16.) 23. (17.) 12. (18.) 56. (19.) 1296. ANSWERS. 157 (1.) 2, 3. (4) 6, 12. (7.) 4, -1. (10.) 2, 3. (13.) 13, 3. (16.) a, f . XLIX. (2.) 3, 5. (5.) 30, 12. (8.) 3, 5. (11.) 9, 4. (11) 4, 1. aib--aG (3.) 6, 4. (6.) 4, 3. (9.) 2, 3. (12.) 60, 40. nx!^\ a+& a—h ^^^'^ "2"' "2"* (17.) ah— he b—a * a—h L. (1.) 2, 1, 3. (2.) 1, 3, 2. (4.) 3, 4, -3. (5.) 3, 2, 1. (7.) 4, 3, 2. (8.) 19, 7, 4. nn^ ^+c— a c+a— & a+6--c y.^^-) — o — * — o — » o • (3.) 5, 4, 3. (6.) 4, 5, 6. (9.) 2, 4, 7. LI. (1.) 72. (2.) 378 and 216. (3.) 5 and 4. (4.) V (5) 84. (6.) A to B 37 miles, B to C 45 miles, C to A 52 miles. (7.) 19 five-franc pieces, 11 two-franc pieces. (8.) 296 for A, 1488 for B, 198 for C. (9.) The stream 3 miles an hour; the boat 8 miles an hour. (10.) 148 for A, 750 for C and A, 158 for A and B. (11.) The gold coins are half-sovereigns, the silver coins are crowns. (12.) A to B Hi miles, B to C 7 miles, C to D 5| miles. (13.) 24 octavos or 32 duodecimos. (14.) A thaler = 2s. lid. ; a franc = W. ; a florin = Is. lUd. 158 ANSWERS. ll ,«.r m (15.) A Prussian pound = 16^ oz. ; an Austrian pound = 191 oz. ; a kilogramme = 35i oz. (16.) 3i miles. (1.: '/a, «/a, ^a\ ^a^ (3.) Ml. (2.) L L L. ^m V»i° tIP' (4.) 2^ a - 3 cc- ./wi^ 1. 1. 1 (5.) x^, m^, n (7.) a. 3L i a;2 ass gjt (6.) (8.) x~\ a~\ a'\ a -8 -1 -i. X -3 (9.) 2m-i, 3»-2, lOp (1.) 6a;"+*; 4a;'»+2; 4a;*». a; -1 _a (10.) 2«"2, 5a;"3, 7a;~» Lin. (2.) 2.x* ; 6x6 . 30a; :l 2n+l 3n fin (3.) jc 2 ; 6a;=^ ; a;« (5.) a6; 2a"'^; o^. (7.) 1; 1; 6; wiw. (4) 2a; Ba'^; 30a -1 (6.) a^; a^; a\ (1.) a"; a2«. (4.) a? ; (B^ Sn (7.) as"; a;2. 12. «8 a': a (1.) a (3.) a . . (5.) a; a^; a*. LIV. (2.) a»; ai (6.) a?; aj^. LV. (!2.) a-= (3.) x^; X. (6.) (c; (B*. a -6. „-12 -«»• a«: ai2. a=« ; a"; a 9. „io (4.) (6.) a3 ; as ; a^. id =