t t . NEW ELEMENTARY ALGEBRA: EMBRACING THE FIRST PRINCIPLES OF THE SCIENCE. PKOFE880K OF HIGHER MATH KM ATIO8, COLUMBIA COLLEOB. NEW YORK : BARNES & BURR, PUBLISHERS, 51 & 53 JOHN ST. CHICAGO: GEORGE SHERWOOD. 118 LAKE ST. CINCINNATI: RICKEY AND CARROLL. ST. LOUIS : KEITH AND WOODS. 1804. A. S. BARNES AND BUHIi'S PUBLICATIONS. Da vies 1 Course of Mathematics. MATHEMATICAL WORKS, IS A SERIES OF THREE PAKT8 : ARITHMETICAL, ACADEMICAL, AND COLLEGIATE DAVIES' LOGIC AND UTILITY OF MATHEMATICS. Tins series, combining all that is most valuable in the various methods of Enropeas Instruction, improved and matured by the suggestions of more than thirty years' experience, now forms the only complete consecutive course of Mathematics. Ita methods, harmonizing as the works of one mind, carry the student onward by the same analogies, and the same laws of association, and are calculated to impart a com- prehensive knowledge of the science, combining clearness in the several branches, and unity and proportion in the whole ; being the system so long in use at West Point, through which so many men, eminent for their scientific attainments, have passed, and having been adopted, as Text Books, by most of the Colleges in the United States. I. THE ARITHMETICAL COURSE FOR SCHOOLS. 1. PRIMARY ARITHMETIC AND TABLE-BOOK. 2. INTELLECTUAL ARITHMETIC. 3. SCHOOL ARITHMETIC. (Key separate.) 4. GRAMMAR OF ARITHMETIC. II. THE ACADEMIC COURSE. 1. THE UNIVERSITY ARITHMETIC. (Key separate.) 2. PRACTICAL MATHEMATICS FOR PRACTICAL MEN. 3. ELEMENTARY ALGEBRA. (Key separate.) 4. ELEMENTARY GEOMETRY AND TRIGONOMETRY. 5. ELEMENTS OF SURVEYING. III. THE COLLEGIATE COURSE. 1. DAVIES' BOURDON'S ALGEBRA. 2. DAVIES' UNIVERSITY ALGEBRA. 3. DAVIES' LEGENDRE'S GEOMETRY AND TRIGONOMETRY. 4. DAVLES' ANALYTICAL GEOMETRY. 5. DAVIES' DESCRIPTIVE GEOMETRY. 6. DAVIES' SHADES, SHADOWS, AND PERSPECTIVE. 7. DAVIES' DIFFERENTIAL AND INTEGRAL CALCULUS. 8. MATHEMATICAL DICTIONARY, BY DAVIES & PECK. ENTEBED, according to the Act of Congress, in the year one thousand eight hundred and fifty-nine, by CHAELES DAVIES, in the .Clerk's Oflice of the District Court of the United States, for the Southern District of New York. 1TILI.IAM DEXYSE, STEREOTYPES. a P K E F A C E. ALGEBRA naturally follows Arithmetic in a course of scien- tilic studies. The language of figures, and the elementary combinations of numbers, are acquired at an early age. When the pupil passes to a new system, conducted by letters and signs, the change seems abrupt; and he often experiences much difficulty before perceiving that Algebra is but Arithmetic written in a different language. It is the design of this work to supply a connecting link between Arithmetic and Algebra ; to indicate the unity of the methods, and to conduct the pupil from the arithmetical processes to the more abstract methods of analysis, by easy and simple gradations. The work is also introductory to the University Algebra, and to the Algebra of M. Bourdon, which is justly considered, both in this country and in Europe, as the best text-book on the subject, which has yet appeared. In the Introduction, or Mental Exercises, the language of figures and letters are both employed. Each Lesson is so arranged as to introduce a single principle, not known iii IV PREFACE. before, and the whole is so combined as to prepare the pupil, by a thorough system of mental training, for those processes of reasoning which are peculiar to the algebraic analysis. It is about twenty years since the first publication of the ELEMKNTAEY ALGEBRA. Within that time, great changes have taken place in the schools of the country. The sys- tems of mathematical instruction have been improved, new methods have been developed, and these require correspond- ing modifications in the text-books. Those modifications have now been made, and this work will be permanent in its present form. Many changes have been made in the present edition, at the suggestion of teachers who have used the work, and favored me with their opinions, both of its defects and merits. I take this opportunity of thanking them for the valuable aid they have rendered me. The criticisms of those engaged in the daily business of teaching are invalu- able to an author ; and I shall feel myself under special obligation to all who will be at the trouble to communicate, to me, at any time, such changes, either in methods or lan- guage, as their experience may point out. It is only through the cordial co-operation of teachers and authors by joint labors and mutual efforts that the text-books of the country can be brought to any reasonable degree of perfection. COLUMBIA COLLEGE, NEW YORK, March, 1859. CONTENTS. CHAPTER I. DEFINITIONS AND EXPLANATORY SIGHS. PAQM. Algebra Definitions Explanation of the Signs 83-41 Examples in writing Algebraic expressions 41 Interpretation of Algebraic language 42 CHAPTER II. FUNDAMENTAL OPERATIONS. Addition Rule Examples 48-50 Subtraction Rule Examples Remarks 60-56 Multiplication Monomials Polynomials 56-63 Division Monomials 63-68 Signification of the symbol 68-70 Division of Polynomials Examples 71-76 CHAPTER III. USEFUL FORMULAS. FACTORING, ETC. Formulaa (1), (2), (3), (4), (5), and (6) 76-79 Factoring 79-81 Greatest Common Divisor 81-84 Least Common Multiple 84-87 CHAPTER IV FRACTIONS. Transformation of Fractions .' 89 fo Reduce an Entire Quantity to a Fractional Form 90 V ^ VI CONTENTS. To Reduce a Fraction to its Lowest Terms To Reduce a Fraction to a Mixed Quantity To Reduce a Mixed Quantity to a Fraction 93 To Reduce Fractions to a Common Denominator 94 Addition of Fractions 96-98 Subtraction of Fractions 98-99 Multiplication of Fractions 99-102 Division of Fractions 102-105 CHAPTER V. EQUATIONS OF THE FIRST DEGRIB. Definition of an Equation Different Kinds 105-106 Transformation ot Equations First and Second 106-110 Solution of Equations Rule 110-1 14 Problems involving Equations of the First Degree 115-130 Equations involving Two Unknown Quantities 130-131 Elimination By Addition By Subtraction By Comparison. . 131-143 Problems involving Two Unknown Quantities 143-148 Equations involving Three or more Unknown Quantities 148-159 CHAPTER VI. FORMATION OF POWERS. Definition of Powers 160-161 Powers of Monomials 161-163 Powers of Fractions 163-165 Powers of Binomials 1 65-1 67 Of the Terms Exponents Coefficients 167-170 Binomial Formula Examples 170-172 CHAPTER VII. SQUARE ROOT. RADICALS OF THE SECOND DEGREE. Definition Perfect Squares Ride Examples 173-179 Square Root of Fractions 179-181 Square Root of Monomials 181-183 Imperfect Squares, or Radicals 1 83-1 87 Addition of Radicals '. 187-189 Subtraction of Radicals .. 189-190 CONTENTS. Vll Multiplication of Radicals ............................... 190-191 Division of Radicals .................................... 191-192 Square Root of Polynomials ............................. 193-197 CHAPTER VIII. EQUATIONS OF THE SECOND DEGREE. Equations of the Second Degree Definition Form ......... 198-200 Incomplete Equations ____ . .............................. 200-209 Complete Equations Rule ............................... 209-211 Four Forms ........................................... 21 1-227 Four Properties ........................................ 227-229 formation of Equations of the Second Degree .............. 229-231 Numerical Values of the Roots ........................... 231-236 Problems .................. ............................ 236-240 Equations involving more than One Unknown Quantity . ..... 241250 Problems ............................................. 250-254 CHAPTER IX. ARITHMETICAL AND GEOMETRICAL PROPORTION. Ways in which Two Quantities may be Compared ........... 255 Arithmetical Proportion and Progression .................. 256-257 Last Term ......................... . ................... 257-260 Sum of the Extremes Sum of Series ..................... 260-262 The Five Numbers To find any number of Means .......... 262-265 Geometrical Proportion ................................. 267 Various Kinds of Proportion ................... . ......... 268-278 Geometrical Progression ................................. 278-280 Last Term Sum of Series ............................... 280-285 Progression having an Infinite Number of Terms ............ 285-288 The Five Numbers To find One Mean .................... 288-289 CHAPTER X. LOGARITHMS. Theory of Logarithms ................................... 290-291 SUGGESTIONS TO TEACHERS. 1. THE Introduction is designed as a mental exercise. If thoroughly taught, it "will train and prepare the mind of the pupil for those higher processes of reasoning, which it is the peculiar province of the algebraic analysis to develop. 2. The statement of each question should be made, and every step in the solution gone through with, without the aid of a slate or black-board ; though perhaps, in the begin- ning, some aid may be necessary to those unaccustomed to such exercises. 3. Great care must be taken to have every principle on which the statement depends, carefully analyzed ; and equal care is necessary to have every step in the solution distinctly explained. 4. The reasoning process is the logical connection of dis- tinct apprehensions, and the deduction of the consequences which follow from such a connection. Hence, the basis of all reasoning must lie in distinct elementary ideas. 5. Therefore, to teach one thing at a time to teach that thing well to explain its connections with other things, and the consequences which follow from such connexions, would seem to embrace the whole art of instruction. via ELEMENTARY ALGEBRA. INTRODUCTION. MENTAL EXERCISES. LESSON I. 1. JOHN and Charles have the same number of apples ; both together have twelve : how many has each ? ANALYSIS. Let x denote the number which John has ; then, since they have an equal number, x will also denote the number which Charles has, and twice x, or 2aj, will denote the number which both have, which is 1 2. If twice x is equal to 12, x will be equal to 12 divided by 2, which is 6 ; therefore, each has 6 apples. WRITTEX. * Let x denote the number of apples which John has; then, 12 x + x = 2x = 12; hence, x = -- = 6. 2t NOTE. When x is written with the sign + before it, it is read plu-s x : and the line above, is read, x phis ar equals 12. 10 INTRODUCTION. NOTE. When x is written by itself, it is read one x and is the same as, la; ; x or he, means once x, or one x, 2x, " twice x, or two x, 3x, " three times x, or three x, 4x, " four times x, or four , g6b cost? Why ? 8 oranges ? 4. Charles bought 3 lemons and 4 ora.ige?, for which he paid 22 cents. He paid twice as much for an orange as for a lemon : what was the price of each ? ANALYSIS. Let x denote the price of a lemon ; then, 2x will denote the price of an orange ; 3x will denote the cost, of 3 lemons, and 8x the cost of 4 oranges ; hence, 3x plus 8z, or lla;, will denote the cost of the lemons and oranges, which is 22 cents. If lla: is equal to 22 cents, x is equal to 22 cents divided by 11, which is 2 cents: therefore, the price of 1 lemon is 2 cents, and that of 1 orange 4 cents. WRITTEN". Let x denote the price of 1 lemon ; then, 2x = 1 orange ; and, 3o; -f 8x == llx = 22 cts., the cost of lemons and oranges; 22 cts hence, x = - = 2 cts., the price of 1 lemon ; and, 2x2 = 4 cts., the price of 1 orange. VERIFICATION. 3x2= 6 cents, cost of lemons, 4 x 4 = 16 cents, cost of oranges. 22 cents, total cost. 5. James bought 8 apples and 3 oranges, for which he paid 20 cents. He paid as much for 1 orange as for 4 apples: what did he pay for one of each ? 6. A farmer bought 3 calves and 7 pigs, for which he paid 1 9 dollars. He paid four times as much for a calf as for a pig : what was the price of each ? 7. James bought an apple, a peach, and a pear, for which he paid 6 cents. He paid twice as much for the peach as for M ]: N T A L EXERCISES. 19 the apple, and three times as much for the pear as for the apple : what was the price of each ? 8. William bought an apple, a lemon, and an orange, for which he paid 24 cents. He paid twice as much for the lemon as for the apple, and 3 times as much for the orange as for the apple : what was the price of each ? 9. A farmer sold 4 calves and 5 cows, for which he received 120 dollars. He received as much for 1 cow as for 4 calves : what was the price of each ? 10. Lucy bought 3 pears and 5 oranges, for which she paid 26 cents, giving twice as much for each orange as for each pear: what was the price of each? 11. Ann bought 2 skeins of silk, 3 pieces of tape, and a penknife, for which she paid 80 cents. She paid the same for the silk as for the tape, and as much for the penknife as for both : what was the cost of each ? 12. James, John, and Charles are to divide 56 cents among them, so that John shall have twice as many as James, and Charles twice as many as John: what is the share of each ? 13. Put 54 apples into three baskets, so that the second shall contain twice as many as the first, and the third as many as the first and second : how many will there be in each. 14. Divide 60 into four such parts that the second shall be double the first, the third double the second, and the fourth double the third : what are the numbers ? LESSON VL 1. If 2x + x is equal to 3cc, what is Sx x equal to ? Written, 3x x 2x. 2. What is 4x x equal to ? Written, 4x x = 3x. 20 INTRODUCTION. 3. What is Sx minus Qx equal to ? Written, 8x 6x = 2x. 4. What is 12a; 9x equal to? Ans. 3% 5. What is 15x Ix equal to ? 6. What is 11 x ISx equal to? Ans. 4a\ 7. Two men, who are 30 miles apart, travel towards each other ; one at the rate of 2 miles an hour, and the other at the rate of 3 miles an hour : how long before they will meet? ANALYSIS. Let x denote the number of hours. Then, since the time, multiplied by the rate, will give the distance, 2x will denote the distance traveled by the first, and 3x the distance traveled by the second. But the sum of the distances is 30 miles ; hence, 2x + 3x = 5x = 30 miles ; and if 5x is equal to 30, a; is equal to 30 divided by 6, which is 6 : hence, they will meet in 6 hours. WRITTEN". Let x denote the time in hours ; then, 2* = the distance traveled by the 1st; and 3* = " " 2d. By the conditions, 2a + 3x = 5x = 30 miles, the distance apart ; 30 hence, x = = 6 hours. 5 VERIFICATION. 2x6 = 12 miles, distance traveled by the first. 3x6 18 miles, distance traveled by the second. 30 miles, whole distance. 8. Two persons are 10 miles apart, and are traveling ID the same direction ; the first at the rate of 3 miles an hour, and the second at the rate of 5 miles : how long, before the second will overtake the first ? M K N T A L ]; X K U C I 8 E 8 . 21 ANALYSIS. Let x denote the time, in hours. Then, Sx will denote the distance traveled by the first in x hours; and 5x the distance traveled by the second. But Avhen the second overtakes the first, he will have traveled 10 miles more than the first : hence, 5x 3x 2x = 10 ; if 2x is equal to 10, x is equal to 5 : hence, the second will overtake the first in 5 hours. W1UTTEN. Let x denote the time, in hours : then, 3x = the distance traveled by the 1st; and, 5x = " " 2d; and, 5x 3x = 2x = 10 hours; 10 or, x = - - = 5 hours. 2i VERIFICATION. 3x5 = 15 miles, distance traveled by 1st. 5 X 5 = 25 miles, " " 2d. 25 15 = 10 miles, distance apart. 9. A cistern, holding 100 hogsheads, is filled by two pipes ; one discharges 8 hogsheads a minute, and the other 12 : in what time will they fill the cistern ? 10. A cistern, holding 120 hogsheads, is filled by 3 pipes ; the first discharges 4 hogsheads in a minute, the second 7, and the third 1 : in what time will they fill the cistern ? 11. A cistern which holds 90 hogsheads, is filled by a pipe which discharges 10 hogsheads a minute; but there is a waste pipe which loses 4 hogsheads a minute : how long will it take to fill the cistern ? m 12. Two pieces of cloth contain each an equal number of yards ; the first cost 3 dollars a yard, and the second 5, and both pieces cost 96 dollars : how many yards in each? 13. Two pieces of cloth contain each an equal number of yards ; the first cost 7 dollars a yard, and the second 5 ; the first 22 IN T R O D C C T I O X . cost CO dollars more than the second : how many yards in each piece ? 14. John bought an equal number of oranges and lemons the oranges cost him 5 cents apiece, and the lemons 3 ; and he paid 56 cents for the whole: how many did he buy of each kind ? 15. Charles bought an equal number of oranges and lemons; the oranges cost him 5 cents apiece, and the lemons 3 ; he paid 14 cents more for the oranges than for the lemons : how many did he buy of each ? 16. Two men work the sa.me number of days, the one receives 1 dollar a day, and the other two : at the end of the time they receive 54 dollars : how long did they work ? LESSON VIE. 1. John and Charles together have 25 cents, and Charles has 5 more than John : how many has each ? ANALYSIS. Let x denote the number which John has ; then, x + 5 will denote the number which Charles has, and x -f- x + 5, or 2x + 5, will be equal to 25, the number they both have. Since 2x -f 5 equals 25, 2x will be equal to 25 minus 5, or 20, and x will be equal to 20 divided by 2, or 10: therefore, John has 10 cents, and Charles 15. WRITTEN-. Let x denote the number of John's cents ; then, x 4- 5 = " Charles' cents ; and, * x + x + 5 = 25, the number they both have ; or, 2x + 5 = 25 ; and, 2x = 25 5 = 20 ; hence, 20 x = =10, John's number; and, I 10 + 5 z= 15, Charles' number. MENTAL EXERCISES. 23 VERIFICATION. John 1 !. Charles 1 . 10 +15 = 25, the sum. Charles'. John's. 15 - 10 = '5, the difference. 2. James and John have 30 marbles, and John has 4 more tnan James : how many has each ? 3. William bought 60 oranges and lemons ; there were 20 more lemons than oranges : how many were there of each sort ? 4. A farmer has 20 more cows than calves; in all he has 36 : how many of each sort ? 5. Lucy has 28 pieces of money in her purse, composed of cents and dimes ; the cents exceed the dimes in number by 16 : how many are there of each sort ? 6. What number added to itself, and to 9, will make 29 ? 7. What number added to twice itself, and to 4, will make 25 ? 8. What number added to three times itself, and to 12, will make 60 ? 9. John has five times as many marbles as Charles, and what they both have, added to 14, makes 44 : how many has each ? 10. There are three numbers, of which the second is twice the first, and the third twice the second, and when 9 is added to the sum, the result is 30 : what are the numbers? 11. Divide 13 into two such parts that the second shall be two more than double the first : what are the parts ? 12. Divide 50 into three such parts that the second shall be tAvice the first, and the third exceed six times the first by 4 : what are the parts? 13. Charles has twice as many cents as James, and John 24 I N T li O D U C T I O N . has twice as many as Charles ; if 7 be added to what they all have, the sum will be 28 : how many has each ? 14. Divide 15 into three such parts that the second shall be 3 times the first, the third twice the second, and 5 over : what are the numbers ? 15. An orchard contains three kinds of trees, apples, pears, and cherries; there are 4 times as many pears as apples, twice as many cherries as pears, and if 14 be added, the number will be 40 ; how many are there of each ? LESSON VIII. 1. John after giving away 5 marbles, had 12 left: how many had he at first ? ANALYSIS. Let x denote the number ; then, x minus 5 will denote what he had left, which was equal to 12. Since x diminished by 5 is equal to 12, x will be equal to 12, increased by 5 ; that is, to 1 7 : therefore, he had 1 7 marbles. WRITTEN. Let x denote the number he had at first; then, x 5 = 12, what he had left; and x = 12 + 5 = 17, what he first had. VERIFICATION. 17 5 = 12, what were left. 2. Charles lost 6 marbles and has 9 left : how many had he at first ? 3. William gave 15 cents to John, and had 9 left : how many had he at first ? 4. Ann plucked 8 buds from her rose bush, and there -?rere 1 9 left : how many were there at first ? M E X T A L EXERCISES. 25 5. William took 27 cents from his purse, and there \verr 1 3 left : how many were there at first ? 6. The sum of two numbers is 14, and their difference is 2: what are the numbers? ANALYSIS. The difference of two numbers, added to the less, will give the greater. Let x denote the less number ; then, x + 2, will denote the greater, and x + x + 2, will denote their sum, Avhich is 14. Then, 2x + 2 equals 14; and 2x equals 14 minus 2, or 12: hence, x equals 12 divided by 2, or 6 : hence, the numbers are 6 and 8. VERIFICATION. 6 + 8 = 14, their sum ; and 8 6 = 2, their difference. 7. The sum of two numbers is 18, and their difference : , what are the numbers ? 8. James and John have 26 marbles, and James has 4 more than John : how many has each ? 9. Jane and Lucy have 16 books, and Lxicy has 8 more than Jane : how many has each ? 10. William bought an equal number of oranges and lemons ; Charles took 5 lemons, after which William had but 25 of both sorts : how many did he buy of each ? 11. Mary has an equal number of roses on each of two bushes ; if she takes 4 from one bush, there will remain 24 on both : how many on each at first ? 12. The sum of two numbers is 20, and their difference is 6 : what are the numbers ? ANALYSIS. If x denotes the greater number, x 6 will denote the less, and x + x 6 will be equal to 20; hence, 2x equals 20 + 6, or 26, and x equals 26 divided by 2, equals 13 ; hence the numbers are 13 and 7. 2 I N T It O U U O T I O N . Let x denote the greater ; then, x 6 = the less ; and x -j- x 6 = 20, their sum ; hence, 2x = 20 + 6 = 26 ; or, 26 x = 13 ; and 13 6 = 7. VERIFICATION. 13 + 7 = 20; and, 13 7 = 6. 13. The sum of the ages of a father and son is 60 yeais, and their difference is just half that number : what are theij ages? 14. The sum of two numbers is 23, and the larger lacks 1 of being 7 times the smaller : what are the numbers ? 15. The sum of two numbers is 50 ; the larger is equal to 10 times the less, minus 5 : what are the numbers ? 16. John has a certain number of oranges, and Charles has four times as many, less seven ; together they have 53 : how many has each ? 17. An orchard contains a certain number of apple trees, and three times as many cherry trees, less 6 ; the whole num- ber is 30 : how many of each sort? LESSON IX. 1. If x denotes any number, and 1 be added to it, what will denote the sum ? Ans. x -f 1 . 2. If 2 be added to x, what will denote the sum ? If 3 be added, what ? If 4 be added ? < , is read, greater than, or less than. When placed between two quantities, it indicates that they are unequal, the greater one being placed at the opening of the sign. Thus, the expression, a > b, indicates that is greater than b ; and the expres- sion, c < d, indicates that c is less than d. 12. The sign . . means, therefore, or consequently. 13. A COEFFICIENT is a number written before a quan- tity, to show how many times it is taken. Thus, a + a-\-a + a + a = 5a, in which 5 is the coefficient of a. A coefficient may be denoted either by a number, or a letter. Thus, 5x indicates that x is taken 5 times, and ax 10. What is the sign of equality? When placed between two quanti- ties, what does it indicate ? 11. How is the sign of inequality read ? Which quantity is placed on rite side of the opening ? 1 2. What does . . indicate ? 13. What is a coefficient? How many times is a taken in 5a. By -.nnt may a coefficient be denoted? If no coefficient is written, what coefficient is understood ? In 5ar, how many times is ax taken? How many times is x taken ? D K F I N 1 T I O X OF T K R M 8 . 37 indicates that x is taken a, times. If no coefficient is writ- ten^ the coefficient 1 is understood. Thus, a is the same as la. 14. AN EXPONENT is a number written at the right and above a quantity, to indicate how many times it is taken as a factor. Thus, a x a is written a 2 , a x a x a " a 3 , a X a x a x a " a*, &c., &c., in which 2, 3, and 4, are exponents. The expressions are read, a square, a cube or a third, a fourth ; and if we have a m , in which a enters ra times as a factor, it is read, a to the mth, or simply a, mth. The exponent 1 is generally omitted. Thus, a 1 is the same as a, each denoting that a enters but once as a factor. 15. A POWER is a product which arises from the multi- plication of equal factors. Thus, a X = a 2 is the square, or second power of a. a X a X a = a 3 is the cube, or third power of a. a X a x a x a = a 4 is the fourth power of a. a x a X . . . . = a m is the mih power of a. 16. A ROOT of a quantity is one of the equal factors. Flic radical sign, -y/ , when placed over a quantity, indi- cates that a root of that quantity is to be extracted. The root is indicated by a number written over the radical sign, 14. What is an exponent? In a 3 , how many times is a taken as a fac- tor? When no exponent is written, what is understood? 1 5. What is a power of a quantity ? What is the third power af 2 ? Of 4 Of 6 ? 16. What is the root of a quantity? What indicates a root? What indicates the kind of root? What is the index of the square root? Of the cube root ? Of the mih root ? 38 ELEMENTARY ALGEBRA.. called an index. When the index is 2, it is generally omit- ted. Thus, ^/a, or y/a, indicates the square root of a. $/a indicates the cube root of a. \/a indicates the fourth root of a. 'H/a indicates the with root of a. 17. An ALGEBRAIC EXPRESSION is a quantity written in algebraic language. Thus, j is the algebraic expression of three tunes ( the number denoted by a ; 2 1 is the algebraic expression of five times ( the square of a ; is the algebraic expression of seven times the the cube of a multiplied by the ( square of b ; ( is the algebraic expression of the differ- Sa 5b< ence between three times a and five ( times b ; is the algebraic expression of twice the . square of ff, diminished by three times the product of a by >, augmented by four times the square of b. 18. A TERM is an algebraic expression of a single quan- tity. Thus, 3a, 2a, 5a 2 2 , are terms. 19. The DEGREE of a term is the number of its literal factors. Thus, j is a term of the first degree, because it contains but ( one literal factor. 17. W T hat is an algebraic expression 18. What is a term? 19. What is the degree of a term? What determines the degree of a term? DEFINITION OF TEKMS. 39 _ 2 j is of the second degree, because it contains two lite- ( ral factors. is of the fourth degree, because it contains four literal factors. The degree of a term is deteraiined by the sum of the exponents of all its letters. 20. A MONOMIAL is a single term, unconnected with any other by the signs + or ; thus, 3a 2 , 3 J 3 a, are monomials. 21. A POLYNOMIAL is a collection of terms connected by the signs + or ; as, 3a - 5, or, 2a 3 3b + 4b z . 22. A BINOMIAL is a polynomial of two terms ; as, a + b, 3a 2 c 2 , Gab c 2 . 23. A TRINOMIAL is a polynomial of three terms ; as, abc a 3 4- c 3 , ab gh f. 24. HOMOGENEOUS TERMS are those which contain the same number of literal factors. Thus, the terms, abc, a 3 , 4- c 3 , are homogeneous ; as are the terms, a>, gh. 25. A POLYNOMIAL is HOMOGENEOUS, when all its terms aio homogeneous. Thus, the polynomial, abc a 3 + c 3 , is homogeneous ; but the polynomial, ab gh f is not ho- mogeneous. i6. SIMILAR TERMS are those which contain the same literal factors affected with the same exponents. Thus, lab -f- Bab 2ab, 20. What is a monomial ? 21. What is a polynomial? 9.2. What is a binomial ? 23. What is a trincmial? 24. What are homogeneous terms ? *>.5. When is a polynomial homogeneous? 26. What are similar terms? 40 ELEMENTARY ALGEBRA.. are similar terms ; and so also are, but the terms of the first polynomial and of the last, are not similar. 27. THE VrxcuLUM, - , the Bar \ , the Paren- thesis, ( ) , and the Brackets, [ ] , are each used to con- nect several quantities, which are to be operated upon in the same manner. Thus, each of the expressions, 6-f-cxcc, + b (a + & -f c) x and [a + # -f c] x x, indicates, that the sum of a, , and c, is to be multiplied by x. 2. THE RECIPROCAL of a quantity is 1, divided by that quantity; thus, 1 1 c a' o~+~ft' d' are the reciprocals of A d a . a + o . - c 29. THE NUMERICAL VALUE of an algebraic expression, is the result obtained by assigning a numerical value to each letter, and then performing the operations indicated. Thus, the numerical value of the expression, ab + be + t7, when, a = 1, b = 2, c = 3, and d = 4, is 1X2 + 2X34-4 12; by performing the indicated operations. 27. For what is the vincular used ? Point out the other ways ir. -whici this may be done ? 28. "What is the reciprocal of a quantity? 29. What is the numerical value (fan algebraical expression? ALGEBRAIC KXPKE8SION8. 41 EXAMPLES IX WHITING ALGEBRAIC EXPRESSIONS. 1. Write a added to b. Ans. a -f b. 2. Write b subtracted from a. Ans. a b. Write the following : 3. Six times the square of a, minus twice the square of b. 4. Six times a multiplied by , diminished by 5 times c cube multiplied by d. 5. Nine times a, multiplied by c plus d, diminished by 8 times b multiplied by d cube. 6. Five times a minus #, plus 6 times a cube into 3 cube. 7. Eight times a cube into d fourth, into c fourth, plus 9 times c cube into d fifth, minus 6 times a into , into c square. , 8. Fourteen times a plus 5, multiplied by a minus b, plus 5 times , into c plus c?. 9. Six times a, into c plus 3/-73 Ana 14 CK A DDITION. CHAPTER II. FUNDAMENTAL OPERATIONS. ADDITION. 30. ADDITION is the operation of finding the simplest equivalent expression for the aggregate of two or more algebraic quantities. Such expression is called their SUM. When the terms are similar and have like signs. + 31. 1. What is the sum of a, 2a, 3a, and 4a? Take the sum of the coefficients, and annex the literal parts. The first term, or, has a coefficient, _|_ 1, understood (Art. 13). 2. What is the sum of 2#, 3a#, 6a>, and ab. When no sign is writtten, the sign -f is under- stood (Art 5). Add the following : (3.) (4.) (5.) a Sab lac -f- a lab - 10a 2ab 3ab Gab ab I2ab (6.) I5ab Sac I2ac 3abc + 'Jabc 30. What is addition ? SI. What is the rule for addition when the terms are similar and hare like sicrns ? 44 ELEMENTARY ALGEBRA. (7.) (8.) (9.) (10.) Sabc Sad 2adf Qabd 2abc 2ad Gadf I5abd 5abc 5ad 8adf 24abd Hence, when the terms are similar and have like signs : RULE. Add the coefficients, and to their sum prefix the common sign ; to this, annex the common literal part. EXAMPLES. (11.) (12.) (13.) Qab -{- ax 8c 2 Sb* loJ 3 c* I2abc* Sab + Sax lac 2 Sb 2 12aft 3 c* loabc* I2ab + 4ax Sac 2 Qb 2 ab 3 c* abc 2 When the terms are similar and have unlike signs. 32. The signs, + and , stand in direct opposition to each other. If a merchant writes + before his gains and before his losses, at the end of the year the sum of the plus numbers wall denote the gains, and the sum of the minus numbers the losses. If the gains exceed the losses, the difference, which is called the algebraic sum, will be plus ; but if the losses exceed the gains, the algebraic sum will be minus. 1. A merchant in trade gained $1500 in the first quarter of the year, 4000 in the second quarter, but lost $3000 in the third quarter, and $800 in the fourth : what was the re- sult of the year's business ? 1st quarter, + 1500 3d quarter, 3000 2cl " 3000 4th " 800 + 4500 3800 _|_ 4500 3800 = + 700, or $700 gain. 82. What is the rule when the terms are similar and hare unlike sitms ? A I) I) ITI O N. 45 2. A merchant in trade gained $1000 in the first quarter, and $2000 the second quarter ; in the third quarter he lost $1500, and in the fourth quarter $1800 : what was the result of the year's business ? 1st quarter, -f 1000 3d quarter 1500 2d " + 2000 4th " - 1800 -f 3000 - 3300 + 3000 3300 = 300, or $300 loss. 3. A merchant in the first half-year gained a dollars and lost b dollars ; in the second half-year he lost a dollars and gained b dollars : what is the result of the year's business ? 1st half-year, + a b 2d " a -f b Result, Hence, the algebraic siim of a positive and negative quan- tity is their arithmetical difference, icith the sign of the greater prefixed. Add the following: Bab 4acb z Sab 8acb z (jab acb 2 2a z b 2 c 2 5ab 3acb z Hence, when the terms are similar and have unlike signs : I. Write the similar terms in the same column: B.'. Add the coefficients of the additive terms, and also the coefficients of the subtractive terms : HI. Take the difference of these sums, prefix the sign of the greater, and then annex the literal part. EXAMPLES. 1. What is the sum of 46 ELEMENTARY ALGEBRA. Having written the similar terms in the same column, we find the sum of the positive coeffi- cients to be 15, and the sum of the negative coefficients to be 16 : their difference is 1 ; hence, the sum is a z b 3 . 2. What is the sum of 5a 2 5 3a 2 S + 4a 2 5 Qa*b 2 J ? Ans. 3. What is the sum of 2a 3 ic 2 4a 3 bc 2 + 6a 3 c 2 8a 3 bc z + Ua 3 bc*? Ans. I7 4. What is the sum of 8a?b Qa z b + Ua z b? Ans. 5. What is the sum of labc 2 abc 2 labc 2 + Sabc 2 + Gabc z ? Ans. 13abc z . 6. What is the sum of Qcb 3 - 5cb 3 8ac 2 + 20c5 3 + 9ac 2 24c5 3 ? Ans. + ac , To add any Algebraic Quantities. 33. 1. What is the sum of 3a, 55, and 2c? Write the quantities, thus, 3a + 5b 2c; which denotes their sum, as there are no similar terms. 2. Let it be required to find the sum of the quantities, 2a 2 4ab 3a 2 Sab + b* 2ab 5b* 5a 2 5ab 4b 2 83. What is the rule for the addition of any algebraic quantities? A 1) i) I T I O .V . 47 From the preceding examples, we have, for the addition of algebraic quantities, the following RULE. I. Write the quantities to be added, placing similar terms in the same column, and giving to each its proper sign : IT. Add up each column separately and then annex the dissimilar terms with their proper signs. EXAMPLES. 1. Add together the polynomials, 3a 2 2b z 4ab, 5a z 5 2 -j- 2ab, and Sab 3c 2 2& 3 . The term 3a 2 being similar to C . 5a\ we write 8a 2 for the result 3 * of the reduction of these two <| terms, at the same time slightly ___ IT * ~_r". crossing them, as in the first term. I & a ~ + ab 5b z 3c 2 Passing then to the term 4ab, which is similar to + 2ab and + 3ab, the three reduce to + ab, which is placed after So 2 , and the terms crossed like the first term. Passing then to the terms involving b 2 , we find their sum to be 5b 2 , after which we write 3c 2 . The marks are drawn across the terms, that none of them may be overlooked and omitted. (2.) (3.) (4.) *labc + 9ax Sax -} Sb I2a 6c 3abc 3ax Sax 9b 3a Qc 4abc "f 6ax I3ax Qb Qa 15c NOTE. If a 5, 5 = 4, c = 2, x = 1, what are the numerical values of the several sums above found ? 48 ELEMENTARY ALGEBRA. (5.) (6.) (7.) 9a + / Qax Sao 3af + g + m 6a 4- <; 7oKC 9ac ?, 2 - IZaPm 4 95. ADDITION. 49 18. What is tlie sum of 4 3 ft 2 c IGa^x 9ax 3 d, f Ga 3 b 2 c - QaxPd + I7a l , + lGax 3 d a*x 9 3 ft 2 c? Ans. a?b~c + ax 3 d. 19. What is the sum of Iff + 3ft + 4 3 take Sa 3 b 2 . -4s. 3a 2 i 3 15. From 7a 2 5 4 take 6 4 Z 2 . Ans. 7a 2 ^ 4 16. From Sab 2 take 2 5 5 . ^4ws. 35 17. From a: 2 ?/ take y 2 a7. Ans. x^y y~x. 18. From Sx^y 3 take ccy. -4?w. 3x 2 y 3 xy. 19. From ci z y 3 x take a^7/2. -4?z5. 8a 2 y' J x xyz. 20. From 9a 2 # 2 take 3a 2 i 2 . Ans. 21. From 14a 2 y 2 take 20a 2 y 2 . J[n. 22. From -- 24a 4 5 5 take 16 4 5 . Ans. 23. From - ISx^y* take 14cc 2 y 4 . Ans. x z y*. 24. From 47a 3 # 2 y take 5a 3 x 2 v Ans. 42a 3 # 2 v. * V * 25. From 94 2 x 2 take 3a 2 ic 2 . Ans. gia z x z . 26. From a + cc 2 take y 3 . ^4^5. -f x 2 + y :i . 27. From a 3 + i 3 take a 3 5 3 . ^tns 2a 3 + 2b 3 . 28. From 16a 2 a; 3 y take 19a 2 x 3 y. Ans. + 3a 2 ic 3 y. 29. Froin a 2 x 2 take a 2 + r 2 . Ans. 2# 3 . SUBTRACTION. 53 GENERAL EXAMPLES. (1.) flj I 1 ') From 6c Sab + c 2 6ac 5ab + c 2 take Sac + Sab + 7c *.f~ Sac Sab 7c rfl . ' Rem. 3ac Sab + c 2 7c.^,gj>g ) 'Sac 8ab + c 2 7c. (2.) (3.) From Qax a -f 3# 2 6ya; 3x 2 + 5b take 9a x + b 2 yx 3 -f a Rem. Sax a + x + 2& 2 . 5ycc 3 + 2b 2 2bc. The last expression may be changed to the former, by changing the signs of the last three terms, inclosing them in a parenthesis, and prefixing the sign . Thus, 6* Sab + 25 2 2bc = Go? (Sab - 2b 2 + 25c). In like manner any polynomial may be transformed, as in- dicated below : 7a 3 8a 2 & 46 2 c + Qb 3 = 7a 3 (8a*b + We 65 3 ) = 7a 3 9a?b - (4& 2 c - Qb 3 ). So, 3 - W + c d = 8a 3 (76 2 - c + d) = So 3 - 75 2 - ( c + d). 95 3 a + 3a 2 - d = Qb 3 - (a 3a 2 -f d) = 96 s a - (- 3a 2 + d). NOTE. The sign of every quantity is changed when it is placed within a parenthesis, and also when it is brought out. 40. From the preceding principles, we have, a (+ b) = a b; and a (- b) = a + b. 39. How is the subtraction of a polynomial indicated ? How is thi* indicated operation performed ? How may the result be again put under the first form ? What is the general rule in regard to the parenthesis ? 40. What is the s'gn which immediately precedes a quantity called? What is the sign which precedes the parenthesis called ? What is the 66 ELEMENTARY ALGEBRA. The sign immediately preceding b is called the sign of the quantity ; the sign preceding the parenthesis is called the sign of operation ; and the sign resulting from the combin- ation of the signs,is called the essential sign. When the sign of operation is different from the sign of the quantity, the essential sign will be ; when the sign of operation is the same as the sign of the quantity, the essen- tial sign will be -f . MULTIPLICATIOlSr. 41. 1. If a man earns a dollars in 1 day, how much will he earn in 6 days? ANALYSIS. In 6 days he will earn six times as much as in 1 day. If he earns a dollars in 1 day, in 6 days he will earn 6 a dollars. 2. If one hat costs d dollars, what will 9 hats cost ? A.ns. 9d dollars. 3. If 1 yard of cloth costs c dollars, what will 10 yards cost? Ans. lOc dollars. 4. If 1 cravat costs b cents, what will 40 cost ? Ans. 40& cents. 5. If 1 pair of gloves coats b cents, what will a pairs cost? ANALYSIS. If 1 pair of gloves cost b cents, pairs will cost as many tunes b cents as there are units in a : that is, b taken a times, or ab ; which denotes the product of b by a, or of a by b. resulting sign called? "When the sign of opeiation is different from the sign of the quantity, what is the essential sign ? When the sign of ope- ration is the same as the sign of the quantity, what is thr. essential sign ? 41. What is Multiplication? What is the quantity to be multiplied called? What is that called by which it is multiplied? What ia the result called? MULTIPLICATION. 57 MULTIPLICATION is the operation of finding the product of two quantities. The quantity to be multiplied is called the Midtiplicand ; that by which it is multiplied is called the Multiplier ; and the result is called the Product. The Multiplier and Multi- plicand are called Factors of the Product. 6. If a man's income is 3a dollars a week, how much will he receive in 45 weeks ? 3a X 46 = I2ab. If we suppose a = 4 dollars, and b = 3 weeks, the pro- duct will be 144 dollars. NOTE. It is proved in Arithmetic (Davies' School, Art. 48. University, Art. 50), that the product is not altered by chang- ing the arrangement of the factors ; that is, I2ab = axbxl2 = bxaxl2 = axl2xb. MULTIPLICATION OF POSITIVE MONOMIALS. 42. Multiply 3a 2 b 2 by 2a 2 b. "We write, 3a?b 2 X 2 2 & = 3 X 2 X 0? X a? X b 2 X b 3 x 2 a a a a b b b; in which a is a factor 4 times, and b a factor 3 times ; hence (Art. 14), 3 2 6 2 X 2a?b = 3 X 2a*b 3 = 6a 4 i 3 , in which ice multiply the coefficients together, and add the exponents of the like letters. The product of any two positive monomials may be found in like manner; hence the RULE. I. Multiply the coefficients togetJier for a new coefficient: II. Write after this coefficient all the letters in both mono- id. V^hat is the rule for multiplying one monomial by another ? 3* 58 ELEMENTARY ALGEBRA. mials, giving to each letter an exponent equal to the sum of its exponents in the two factors. EXAMPLES. 1. 8a?bc z X labd 2 = 56a 3 b 2 c 2 d*. 2. 2la 3 b 2 cd X 8abc 3 168 4 5 3 c 4 V. ^4ws. 648a 4 #W 8 26. Multiply 70a 8 W 2 / by 12 MULTIPLICATION OF POLYNOMIALS. 43. 1. Multiply a b by c. It is required to take the difference a b between a and 5, c times ; or, to c take c, a 6 times. As we can not subtract b from c, we begin by taking , c times, which is ac ; but this product is too large by b taken c times, which is be ; 56 21 = 35 hence, the true product is ac be. If a, 5, and c, denote numbers, as a =. 8, c = 7, the operation may be written in figures. ac be 8 - 7 = 5 7 = 3, and Multiply a b by c d. It is required to take a b as many times as there are units in c - d. If we take a J, c times, we have ac bc\ but this product is too large by a b taken I2a 4 bcd. 8. Multiply a 3 + 3a 2 + Saw; 2 + a^ by a + x. Ans. a* 9. Multiply 2 + y 2 by a; + y. . x 3 -f ay 2 4- 2 y 4- y 3 . 10. Multiply cc 5 4- xy 6 -\- lax by ax 4- 4- 11. Multiply a 3 4- 3a 2 ft 4- Sab 2 + b 3 by a 4- b. Ans. a 4 4- 4a 3 5 4- 6a 2 5 2 '-f 4a5 3 + b 4 . 1 2. Multiply a 3 4- x*y 4- y 2 + y 3 by cc 4-. y. 4ns. cc 4 4- 2z 3 y 4- 2x- 2 y 2 4- 2iy 3 4- y 4 . 13. Multiply x 3 + 2x 2 + x + 3 by 3 4- 1. 4ns. 3a* 4- Ta 3 4- 5x 2 + lOx + 3. 62 ELEMENTARY ALGEBRA. GENERAL EXAMPLES. 1. Multiply ....... 2ax Soft by ........ 3x b. The product ....... Gax 2 Qabx becomes after ...... 2abx reducing ........ 6ax z llabx + 3a5 2 . 2. Multiply a* 2b 3 by a b. Ans. a s 2ab 3 a*b + 2b*. 3. Multiply x z 3x 7 by a; 2. Ans. x 3 5a; 2 85 + 14. 4. Multiply 3a 2 5ab + 2b* by a 2 7a&. ^15. 3a 4 26a 3 6 + 37a 2 * 2 - Uab 3 . 5. Multiply b 2 + b 4 + b 6 by 6 2 1. .4ns. * 8 - J 2 . 6. Multiply x* 2x*y + 4:X 2 y z 8xy 3 + 16y* by x + 2y. Ans. x 5 + 32y 5 . 7. Multiply 4a; 2 2y by 2y. -4ns. 8a 2 y 4y 2 . 8. Multiply 2x -f 4y by 2 4y. ^4/15. 4a 2 16y 2 . 9. Multiply x 3 + a z y + xy z + y 3 by x y. Ans. x* y*. 10. Multiply a 2 -f %y + y 2 by cc 2 ccy + y 2 . -4ns. jc 4 + # 2 y 2 4- y 4 . 11. Multiply 2a 2 Sax + 4a^ by 5a 2 6aa; 2a 2 . Ans. 10a 4 27 3 -f 34a 2 it- 2 18(7^ 8x*. 12. Multiply 3x 2 2xy + 5 by x" 2 + 2xy 3. Ans. So 4 + 4^y 4x* - x~if + IGay 15. 13. Multiply 3x 3 + 2ar ! y 2 + 3y 2 by 2X 3 3z 2 y 2 + oy 3 . ( Gx 6 - Sa^y 2 6^y 4 + HS ' I5x 3 y 3 - QaPy* + lOa^y 5 + 15y 5 . 14. Multiply 8ax Gab c by 2ax + ab + c. Ans. 16a 2 ce 2 4a 2 to 6a 2 J 2 + Gacx 7afto DIVISION. (S3 1. Mult.ply So 2 5b 2 + 3c 2 by 2 ^4n. 3a* 8aW + 3a 2 c 2 16. 3a 2 bbd + r/ 5a 2 + 45d - 8c/. Pro. red. 17. Multiply a m cc a 2 6 2 by a'cc". 18. Multiply m + b n by a m 5". Ans. a zm b 2 *. 19. Multiply a m + 5" by m + b n . Ans. a? m + 2a m b n + 5 2lt . DIVISION. 45. DIVISION is the operation of finding from two quan- tities a third, which being multiplied by the second, will produce the first. The first is called the Dividend, the second the Divisor, and the third, the Quotient. Division is the converse of Multiplication. In it, we have given the product and one factor, to find the other. The rules for Division are just the converse of those for Multi- plication. To divide one monomial by another. 46. Divide 72a 5 by 8a 3 . The division is indicated, thus : 720 5 ~8a?' The quotient must be such a monomial, as, being multiplied by the divisor, will give the dividend. Hence, the coefficient 45. What is division ? What is the first quantity called? The second? The third? What is given in division? What is required ? 46. What is the rule for the division of monomials? 64 ELEMENTART ALGEBRA. of the quotient must be 9, and the literal part a 2 ; fo; quantities multiplied by 8a 3 will give 72# s . Hence, 8 3 The coefficient 9 is obtained by dividing 72 by 3; and the literal part is found by giving to a a a 3 But since any quantity divided by itself is eqiial to 1, it fol- lows that, - = a = 1, , = a 2 - 2 = a = 1, &c. ; a a 2 or, finally, if we designate the exponent by m, we have, /"flft = a m ~ m =: a = 1 ; that is, a m The power of any quantity is equal to 1 : therefore, Any quantity may be retained in a term^ or introduced into a term, by giving it the exponent 0. EXAMPLES. 1. Divide 6 2 6 2 c* by 2a 2 2 . ftt&lfle* = 3c*. 2. Divide 8a 4 5 3 c 5 by 4a 4 5 3 c. ^tna. 2a6c 4 = 2c 4 . 3. Divide 32m 3 >i 2 a 2 y 2 by m z ri*xy. Ans. 8mnxy = Sxy. 49. When the exponents of the same letter in the dividend and divisor are equal, what takes place ? May the letter still be retained ? Wi Ji what exponent? What is the zero power of any quantity equal to? DIVISION. 69 4. Divide OGa'&V 1 by 24a 4 & 5 . Ans. 4aJc n = 4c*. 5. Introduce a, as a factor, into 6b 5 c i . Ans. 6ab 5 c*. 6. Introduce ab, as factors, into Oc 5 ^". Ans. 9abc 5 d n . 7. Introduce ale, as factors, into Sd\f m . A. 8aicf? 4 / m . 5O. "When the exponent of any letter is greater in the divisor than it is in the dividend, the exponent of that letter in the quotient may be written with a negative sign. Thus, = ', also, = a"- 6 = a" 3 , by the rule; a 5 a 3 a hence, a~ 3 = -5- Since, a~ 3 = =. we have, b x a~ 3 = =; a 3 a 3 ' that is, in the numerator, with a negative exponent, is equal to a in the denominator, with an equal positive ex- ponent; hence, Any quantity having a negative exponent, is equal to the reciprocal of the same quantity icith an equal positive ex- ponent. Hence, also, Any factor may be transferred from the denominator to the numerator of a fraction, or the reverse, by changing the sign of its exponent. EXAMPLES. 1. Divide 32a 2 5c by 16 5 i 2 . 16a 5 b 2 a 3 b 60. When the exponent of any letter in the divisor is greater than in the dividend, how may the exponent of that letter be written in the quo- tient ? What is a quantity with a negative exponent equal to ? How may a factor be transferred from the numerator to the denominator of a frn.-tion ? 70 ELEilENTARY ALGEBRA. 2 - 3. Reduce -rr-ns * Ans. -- , or 5 la; 4 ?/ 3 3 ' 3 4. In 5ay~ 3 x~ z , get rid of the negative exponents. 5a Ans. - K T .. 5. In _ 6 , get rid of the negative exponents. 4a ? i 8 Ans. - 3x 2 15a~ y c~ 4 d~ s 6. In - - - i , get rid of the negative exponents. - 3 - 5 ~ 2 ' 7. Reduce -^^r ^n*. - -^ -, or 8. Reduce 72a 5 5 2 -f- 8a 6 5 3 . -4wa. Oa- 1 ^- 1 . or 9. In _ 2 l , get rid of the negative exponents. _ 10. Reduce -- --j =- ^Ins. 3aJ 2 c 2 . ~ 6 ~ To divide a polynomial by a monomial. 51. To divide a polynomial by a monomial : Divide each term of the dividend, separately, by the divisor ; the, algebraic sum of the quotients will be the quo- tient sought. EXAMPLES. 1. Divide 3a 2 5 2 - a by a. Ans. 3ab 2 1. 61. How do you divide a polynomial by a monomial ? DIVISION. 71 2. Divide 5a 3 J 2 25a" 2 by 5aW. Ans. 1 5a. 3. Divide 35a 2 6 2 25ab by Sab. Ans. lab + 5. 4. Divide 105 I5ac by 5o. -4ra. 25 3e. 5. Divide Gab Sax + 4a 2 y by 2a. 6. Divide 15aa 2 + 6a 3 by 3x. Ans. 5ax 2x 2 . 7. Divide 2 lay 2 + 35a 2 6 2 y - 7c 2 y by - 7y. ^4?i5. 3xy 5a 2 5 2 -f- c 2 . 8. Divide 40a 8 5* + 8a 4 5 7 32a 4 **c* by 8a*b 4 . Ans. 5a* + 5 3 4c<. DIVISION OF POLYNOMIALS. 52. 1. Divide 2a + 6a 2 8 by 2 + 2a. Dividend. Divisor. 6a 2 2a 8 | 2a + 2 6a 3 + 6a 3a 4 Quotient. 8a 8 8a 8 Remainder. We first arrange the dividend and divisor with reference to a (Art. 44), placing the divisor on the left of the dividend. Divide the first term of the dividend by the first term of the divisor ; the result will be the first term of the quotient, which, for convenience, we place under the divisor. The product of the divisor by this term (6 2 -f- 6), being sub- tracted from the dividend, leaves a new dividend, which may be treated hi the same way as the original one, and so on to the end of the operation. 52. What is the rule for dividing one polynomial by another ? When is the division exact ? When is it not exact ? 72 ELEMENTARY ALGEBRA. Since all similar cases may be treated in the same way, we Lave, for the division of polynomials, the following K U L E . I. Arrange the dividend and divisor with reference to the same letter: II. Divide the first term of the dividend by the first term of the divisor, for the first term of the quotient. Multiply the divisor by this term of the quotient, and subtract the product from the dividend: m. Divide the first term of the remainder by the first term of the divisor, for the second term of the quotient. Multiply the divisor by this term, and subtract the product from the first remainder, and so on: TV. Continue the operation, until a remainder is found equal to 0, or one whose first term is not divisible by that of the divisor. NOTE. 1. When a remainder is found equal to 0, the division is exact. 2. "When a remainder is found whose first term is not divisible by the first term of the divisor, the exact division is impossible. In that case, write the last remainder after the quotient found, placing the divisor under it, in the form of a fraction. SECOND EXAMPLE. Let it be required to divide by 4ab 5a 2 -f 3J 3 . Wo first arrange the dividend and divisor with reference to a. DIVISION. Dividend. 73 Divisor. 10a*- t Sa 3 b 2a 2 + 8a6 Quotient. -f 25a z b z 20ab 3 l5b 4 25a?b 2 20ab 3 15b* + (3.) 4- ccy 2 2 + y_ xy 4- 4- 4 4- Here the division is not exact, and the quotient is frac- tional. (4.) 1 4- a 1 - a 1 - a 2a 3 4 , &c. + 2a 4- 2a 4- 4- 2a 3 4- 2a 3 In this example the operation does not terminate. It be continued to any extent. EXAMPLES. 1. Divide a 2 4- 2ax + x 2 by a + x. Ans. a 4- 2. Divide a 3 3a 2 y 4- 3ay 2 y 3 by a y, Ans. a 2 2ay 4- J/*. 74 ELEMENTARY ALGEBRA. 3. Divide 24a 2 I2a*cb 2 Qab by Gab. Ans. *- 4 a -f 2a : cb -}- 1. 4. Divide Gas 4 96 by 3x 6. Ans. 2x? -f 4* 2 + 8z + 16. 5. Divide a 5 5a 4 a + 10a 3 cc 2 -lOa 2 * 3 + Soa 4 ar" by a 2 6. Divide 48a^ 76aa;- 64a 2 a; + I05a 3 by 2x 3a. Ans. 24a; 2 2aa; 35 2 . 7. Divide y 6 3y 4 a; 2 + 3y 2 ic 4 a; 6 by y 3 3y 2 a + a? 3 . Ans. y 3 + 3y*x + 3?/x z -f a^. 8. Divide 64 4 5 6 - 25a 2 i 8 by 8a 2 5 3 + 5ai 4 . u4;?s. 8a 2 i 3 5a5 4 . 9. Divide 6a 3 + 28a 2 J + 22a5 2 +5J 3 by -4*. 2a 10. Divide Qax* + Gaa 2 ?/ 6 + 42a 2 a 2 by aa; + 11. Divide loa 4 + S1a z bd 29aV/ 205 2 ^ + 8c 2 / 2 by 3a 2 - 5bd + cf. Ans. - 5a 2 12. Divide a^ + aj 2 ^ 2 + y* by a; 2 ;ry + y 2 . ^L?25. a; 2 + io// + y 2 . 13. Divide ar 4 y* by a; y. Ans. x 3 4- a^y 4- %y~ + 2/ 3 - 14. Divide 3a 4 8a 2 ft 2 + 3a 2 c 2 + 55 4 35 2 c 2 by 2 - i 2 . ^^5. 3a 2 5b z + 3c 2 . 15. Divide 6a 6 - Sa^y 2 Ga^y 4 ^- 6x 3 y 2 4- 15a; 3 y 3 9.r 2 y 4 -f 10a; 2 y 5 + 15y 5 by 3a^ + 2a; 2 y 2 + 3y 2 . Ans. 2x 3 - 3a;V + 5y 3 . 16. Divide c 2 + 16a 2 a: 2 tabc Itfbx 6a 2 J 2 + Gaca: by 8aa; Qab c. Ans. 2ax + ab -}- c. 17. Divide Bx* + 4a; 3 y 4a; 2 4cc 2 y 2 + 16a;y 15 by 2a;y + a; 2 3. An-s. 3x z 2a*y -f 5. DIVISION. 75 18. Divide x 5 + 3y 5 by x + 1y. Ans. x* 2x 3 t/ + 4ce 2 y 2 8xy 3 + 16y*. 19. Divide 3a 4 26a 3 b 14a5 3 + 37a 2 5 2 by 2b 2 Sab 3a 2 . Ans. a 2 7. 20. Divide a 4 - 5 4 by a 3 + 2 # + 5 2 + b*. Ans. a b. 21. Divide 3 3a 2 y + y 3 by x -\- y. 3y 3 2 22. Divide 1 + 2a by 1 a a*. Ans. 1 + 3a + 4a 2 + fa 3 + , &c. 76 ELEMENTARY ALGEBRA. CHAPTER LIL tJSEFtn FORMULAS. FACTORING. GREATEST COMMON DIVISOR. LEAST COMMON MULTIPLE. USEFUL FORMULAS. 53. A FORMULA is an algebraic expression of a general rule, or principle. Formulas serve to shorten algebraic operations, and are also of muchjuse in the operation of factoring. When trans- lated into common language, they give rise to practical rules. The verification of the following formulas affords addi- tional exercises in Multiplication and Division. (1.) 54. To form the square of a + b, we have, (a + by = (a + b) (a + b) = ct + 2ab + b\ That is, The square of the sum of any two quantities is equal to the square of the first, plus twice the product of the first by the second, plus the square of the second. 1. Find the square of 2a -f 3b. "We have from the rule, (2a + 35) 2 = 4a 2 + I2ab + 9b z . 53. What is a formula? What are the uses of formulas ? 64. What is the square of the sum of two quantities equal to ? U 8 K F U L F t K M U L A 6 . 77 2. Find the square of 5ab + 3ac. Ans. 25 2 & 2 -f ZQtfbc -f- 9cc 2 c 2 . 3. Find the square of 5a 2 + 8a?b. Ans. 25a* + 80a 4 & -f 4. Find the square of 6ax + 9a 2 ce 2 . -f 108a 3 a; 3 + (2.) 55. To form the square of a difference, a b, we have, (a - J) 2 = (a - b) (a - b) = a 2 - 2ab + b\ That is, The square of the difference of any two quantities is equal to the square of the first, minus twice the product of the first by the second, plus the square of the second. 1. Find the square of 2a b. We have, (2a - ) 2 = 4 2 4 & + * 2 - 2. Find the square of 4ac be. Ans. 16a 2 c 2 3. Find the square of 7a 2 Z> 2 12aJ 3 . Ans. (3.) 56. Multiply a + b by a b. We have, (a + b) x (a - 5) = a 2 R Hence, TAe s?m q/" ^wo quantities, multiplied by their difference, is equal to the difference of their squares. 1. Multiply 2c + b by 2c b. Ans. 4c 2 6 2 . 2. Multiply 9ac + 35c by 9ac 35c. . 81a 2 c 2 - 55. What is the square of the difference of two quantities equal to ? 56. What is the sum of two quantities multiplied by the'.r difference equal to ? 78 ELEMENTARY ALGEBRA. 3. Multiply 8a 3 + 7aft 2 by 8a 3 7oft 2 . Ans. 64a 6 49a 2 ft (4.) 57. Multiply a 2 + ab + ft 2 by a b. We have, (a 2 + ab + ft 2 ) (a - ft) = a 3 - ft 3 . (5.) 5. Multiply a 2 -- aft + ft 2 by a -f ft. We have, (a 2 - aft + ft 2 ) (a + ft) = a 3 + ft 3 . (6.) 59. Multiply together, a -f- ft, a ft, and a 2 -f ft 2 . We have, (a + ft) (a - ft) (a 2 + ft 2 ) = a 4 - ft*. 60. Since every product is divisible by any of its factors, each formula establishes the principle set opposite its number. 1. The sum of the squares of any two quantities, plus twice their product, is divisible by their sum. 2. The sum of the squares of any two quantities, minus twice their product, is divisible by the difference of the quantities. 3. The difference of the squares of any two quantities is divisible by the sum of the quantities, and also by their difference. 4. The difference of the cubes of any two quantities is divisible by the difference of the quantities ; also, by the sum of their squares, plus their product. 5. The sum of the cubes of any two quantities is divisi- 60. By what is any product divisible ? By applying this principle, v,-hat follows from Formula (1) ? What from (2)? What from (S) ? What from (4) What from (5) ? Wh at from (6) ? FACTORING. 7 ble by the sum of the quantities ; oho, by the sum of their squares minus their product. 6. The difference between the fourth powers of any two quantities is divisible by the sum of the quantities, by their difference, by the sum of t/ieir squares, and by the dif- ference of their squares. FACTOEING. 61. Factoring is the operation of resolving a quantity into factors. The principles employed are the converse of those of Multiplication. The operations of factoring are performed by inspection. 1. What are the factors of the polynomial ac -f ab + ad. We see, by inspection, that a is a common factor of all the terms ; hence, it may be placed without a parenthesis, and the other parts within ; thus : ac + ab -f ad (c -f b + d). 2. Find the factors of the polynomial a z b z + a z d a z f. Am. a z (b z + . 10. Find the greatest common divisor of 25a 2 c 2 and 5acd 2 -f 3d z x^y z . Ans. 5ac NOTE. To find the greatest common divisor of three quantities. First find the greatest common divisor of two of them, and then the greatest common divisor between this result and the third. 1. What is the greatest common divisor of 4ic 2 y, IQabx 2 , and 24ae 2 ? Ans. 4ax. 2. Of 3ic 2 6a, 2a; 3 4# 2 , and a; 2 //- 2xy ? Ans. x 2 2x. 72. When is one quantity a multiple of another ? LEAST COMMON MULTIPLE. 85 LEAST COilMON MULTIPLE. 72. One quantity is a MULTIPLE of another, when it can be divided by that other without a remainder. Thus, 8a 2 5, is a multiple of 8, also of a 2 , and of b. * 73. A quantity is a Common Multiple of two or more quantities, when it can be divided by each, separately, with- out a remainder. Thus, 24a 3 x 3 , is a common multiple of 6ax and 74. The LEAST COMMON MULTIPLE of two or more quan- tities, is the simplest quantity that can be divided by each, without a remainder. Thus, 12a 2 J 2 tc 2 , is the least common multiple of 2a 2 cc, 45 2 , and 6 2 J 2 tc 2 . 75. Since the common multiple is a dividend of each of the quantities, and since the division is exact, the common multiple must contain every prime factor in all the quanti- ties ; and if the same factor enters more than once, it must enter an equal number of times into the common multiple. When the given quantities can be factored, by any of the methods already given, the least common multiple may bo found by the following EULE. I. Resolve each of the quantities into its prime factors . II. Take each factor as many times as it enters any ou6 of the quantities, and form the continued product of these factors ; it will be the least common multiple. 73. When is a quantity a common multiple of several others? 74. What is the least common multiple of two or more quantities? 75 What does the common multiple of two or more quantities contain, MS factors? How may the least common multiple be found? * The }>i ultiph ol a quantity, is siraplj \ dividend which will give an exact quotient 86 ELEMENTARY ALGEBRA. EXAMPLES. 1. Find the least common multiple pf 12a 3 & 2 c 2 and 8a 2 5 3 . I2a 3 b"*c 2 = 2.2.3.aaabbcc. 8a?b 3 =3 2.2.2.aabbb. Now, since 2 enters 3 times as a factor, it must enter 3 times in the common multiple : 3 must enter once ; a, 3 times ; , 3 times ; and c, twice ; hence, 2.2.2.3aaabbbcc = 24 3 5 3 c 2 , is the least common multiple. Find the least common multiples of the following : 2. 6a, 5 2 5, and 25ac 2 . Ans. loOa-bc 2 . 3. 3a 2 , 95c, and 27a 2 3 . Ans. 2 f la 2 bcx z . 4. 4a 2 ic 2 y 2 , 8a 3 tfy, 16a 4 y 3 , and 24a 5 y 4 a;. 5. ax bx, ay by, and cc 2 y 2 . ^Lws. (a b}x.x.yy = 6. a + b, a 2 5 2 , and a 2 + 2ab + J 2 . u4?w. (a + 6) 2 (a - 6). 7. 3a 3 6 2 , 9a 2 a; 2 , 18a*y 3 , 3a 2 y\ Ans. ISaWxty 5 . 8. 8a 2 (a *), 15a 5 (a - ) 2 , and 12a 3 (a 2 b z ). Ans. 120a 5 (a J) 2 (a + J). FRACTIONS. 37 CHAPTER IV. FRACTIONS. 76. IF the unit 1 be divided into any number of equal i>arts, each part is called a FRACTIONAL UNIT. Thus, - , -, - , - , are fractional units. 77. A FRACTION is a fractional unit, or a collection of fractional units. Thus, - , - , - , -= , are fractions. 2 4 7 o 78. Every fraction is composed of two parts, the De- nominator and Numerator. The Denominator shows into how many equal parts the unit 1 is divided ; and the Nu- merator how many of these .parts are taken. Thus, in the fraction - , the denominator J, shows that 1 is divided into b equal parts, and the numerator or > - = I ) * PI hence, pq \pl pq \q/ Multiplying the denominator of a fraction by any quan- tity, is equivalent to dividing the fraction by that quantity. 83. Since the operations of Multiplication and Division are the converse of each other, it follows, from the preced- ing principles, that, Dividing the numerator of a fraction by any quantity, is equivalent to dividing the fraction by that quantity / and, Dividing the denominator of a fraction by any quantity, is equivalent to multiplying the fraction by that quantity. 84. Since a quantity may be multiplied, and the result divided by the same quantity, without altering the value, it follows that, Hoth terms of a fraction may be multiplied by any quan- tity, or both divided by any quantity, without changing the value of the fraction. TRANSFORMATION OF FRACTIONS. 85. The transformation of a quantity, is the operation of changing its form, without altering its value. The term reduce has a technical signification, and means, to Trans- form. S3. What follows from the preceding principles ? 84. What operations may be performed without altering the value of a fraction? 85. What is the transformation of a quantity ? 90 E I. M E N T A R \ ALGEBRA. FIRST TRANSFORMATION. To reduce an entire quantity to a fractional form having a given denominator. 86. Let a be the quantity, and b the given denomi- nator. We have, evidently, a = -j- ; hence, the BULB. Multiply the quantity by the given denominator, and write the product over this given denominator. SECOND TRANSFORMATION. To reduce a fraction to its lowest terms. 87. A fraction is in its lowest terms, when the numerator and denominator contain no common factors. It has been shown, that both terms of a fraction may be divided by the same quantity, without altering its value. Hence, if they have any common factors, we may strike them out. EULE. Resolve each term of the fraction into its prime fac- tors / then strike out all that are common to both. The same result is attained by dividing both terms of the fraction by any quantity that will divide them, without a remainder ; or, by dividing them by their greatest common divisor. 86. How do you reduce an entire quantity to a fractional form having a given denominator '? 87. How do you reduce a fraction to its lowest terms ? TRANSFORMATION OF FRACTIONS. 91 EXAMPLES. 15 1 '^ *"i +1 . W 1 5. Reduce Ans. - n 2 1 7i + l ,,.,-. x 3 ax 2 ? 6. Reduce : - Ans. x 2 2ax + a 2 x a 9Qa 3 b 2 c 8 7. Reduce ,^ . 370 Ans. - = 8. 8. Reduce , ,-,, T---TTS- Ans. a 2 b 2 A a + b 9. Reduce . . T0 jffu. ') r 7 i IO -*.*vv _ a 2 2ao + o 2 a o 5 a 3 10a 2 & + 5ab 2 10. Reduce ~..^. 8a 3 8a 2 6 8a a 11. Reduce 12a 4 + 6a 3 c 2 ' 4a 2 + a 2 + 2aa + ce 2 12. Reduce ~-r n ^ Ans. 3(a 2 - x 2 ) ' 3(a - 92 ELEMENTARY A L G E U R A. . THIRD TRANSFORMATION. To reduce a fraction to a mixed quantity. 88. When any term of the numerator is divisible by any term of the denominator, the transformation can be effected by Division. RULE. Perform the indicated division, continuing the opei'ation as far as possible ; then write the remainder over the deno- minator, and annex the result to the quotient found. EXAMPLES. ax a 2 a? 1. Reduce Ans. a x x ax ar 5 2. Keduce Ans. a x. x ab 2a 2 2a 2 3. Reduce = Ans. a =- b o ($, yZ 4. Reduce An*, a + x. a x 5. Reduce ~ y Ans. x 2 + xy + y 2 . x- y 10s 2 - 5x + 3 3 6. Reduce Ans. 2x 1 + 5x 5x 7. Reduce - - '-T^~ . . 4s 2 - 8 + 9x y T? ^ IBaef Gbdcf 2ad 60 lie 2 o JL/GCIU.CG r - """ _ ^ - - - r~ ' " ^ 3ac?^ d a 3/ 3.2 _(_ g. ^ 2 9. Reduce - Ans. x 1 x + 2 88. How do you reduce a fractior to a mixed quantity " TRANSFORMATION OF FRACTIONS. 93 10. Reduce =- Ans. a b -\ . a + b a +b , _ , x 2 + 3x - 25 3 11. Reduce Ans. x -f- Y - 4 'a;-4 FOURTH TRANSFORMATION. To reduce a mixed quantity to a fractional form. 9. This transformation is the converse of the preced- ing, and may be effected by the following RULE. Multiply the entire part by the denominator of the frac- tion^ and add to the product the numerator / write the result over the denominator of the fraction. EXAMPLES. 1. Reduce 6| to the form of a fraction. 6 X 7 = 42 ; 42 + 1 = 43 ; hence, 6} = ~ Reduce the following to fractional forms : x z (a? a 2 ) 2 8 Qa 2 x ab 8 18a 2 x + 5J 8 + t%' 4 f W WUrw ~"|~* Ov/Cv V <^ ^1W5. FIFTH TRANSFORMATION. To reduce fractions having different denominators, to equi- valent fractions having the least common denominator. 9O. This transformation is effected by finding the least common multiple of the denominators. 13 5 1. Reduce -, -, and , to their least common denomi- 94 J. +* nators. The least common multiple of the denominators is 12, which is also the least common denominator of the required fractions. If each fraction be multiplied by 12, and the result divided by 12, the values of the fractions will not be changed. - X 12 = 4, 1st new numerator ; - X 12 = 9, 2d new numerator ; 4 X 12 = 5, 3rd new numerator ; hence, i *- 49 o , , and are the new equivalent fractions. 90. How do you reduce fractions having different denominators, to equl valent fractions having the least common denominator ? When the nu- merators have no common factor, bow do you reduce them ? TRANSFORMATION OF FRACTIONS. 95 RULE. I. Find the least common multiple of the denominators : n. Multiply each fraction by it, and cancel the denom- inator : in. Write each product over the common multiple, and the results will be the required fractions. GENERAL RULE. Multiply each numerator by all the denominators except its own, for the new numerators, and all the denominators together for a common denominator. EXAMPLES. ct c 1. Reduce ^ and = to their least common a 2 b 2 a + b denominator. The least common multiple of the denominators is (a + b) (a-b): l -j- X (a + b) (a - b) = a a z - c '= X (a + b) (a b) = c(a b ; hence, Ct ~f~ c(a b) ,, . , and / . v *w ' IM are the required (a + b) (a - b) (a + b) (a - b) fractions. Reduce the following to their least common denominators : Sx 4 12a 2 45x 40 48^ 2 - 7' 6' and 15- An8 ' Weo'-eo- 3b z 5c 3 12o 95 2 lOc 3 3. a, -, and -. Ans. _,_,_. , Sx 2b 9ccc 4ab 6acd 4. - , , and a. Ans. - , - , 2a' 3c Qac Qac Gac 96 E L E M K X T ART A L G E B U A . 3 2 2cc 9a Sax 12a 2 + 24a? a; a 2 a; 3 6. . - , TZ -- ^, and 1 -a;' (1 _a;)2 ^ (1 _a)3 as(l a;) 2 C 2 (1 as) x 3 Ans. - ~ y -77 r^, and 7- TV (1 a;) 3 ' (1 a;) 3 (1 x) 3 c c b , c > and * M ) ) *.*-* V* 5a c a + ac 2 + c 2 5 2 c 5 2 o + 5abc 5# 2 c -f* 5abc ' 5^ 2 c 4~ 5abc ' 5o& 2 c + 5abc ex dx z a; 3 8. , , and a x a + x a + x : cx(a-{- x) dx' 2 (a x) , x 3 (a x) i-iis. - , , and ADDITION OF FBACTIOKS. 91. Fractions can only be added when they have a com- mon unit, that is, when they have a common denominator. In that case, the sum of the numerators will indicate how many times that unit is taken in the entire collection. Hence, the KTJLE. L Jteduce the fractions to be added, to a common denom- inator : n. Add the numerators together for a new numerator^ and write the sum over the common denominator. EXAMPLES. 64 2 1. Add -, -, and -, together. 23 5 91. What is the rule for adding fractions? ADDITION OF FRACTIONS. 97 By reducing to a common denominator, we have, 6 x 3 x 5 i= 90, 1st numerator. 4 x 2 x 5 = 40, 2d numerator. 2x3x2 = 1 2, 3d numerator. 2 x 3 x 5 = 30, the denominator. Hence, the expression for the sum of the fractions becomes 90 40 12 142 1 30 30 + 30 Z: ~30~ ; which, being reduced to the simplest form, gives 4}J. CL G & 2. Find the sum of -, - , and - b u j Here, a X d x / = udf \ c X b x f = cbf > the new numerators. e X b x d = eld ) and b x d x f = bdf the common denominator. <*df , cbf , ebd adf -f- cbf + ebd Al IIencc ' w f + + = ^ ~ ' * sum ' Add the following : 3x 2 2ax 2abx 3. a -- r- , and b -\ -- Ans. a + b -\ , . o c oc . 85 X _ X . X 4. -, -, and -. Ans. * + - , x 2 . 4 19a; 14 5 . ___ and y . Ans. x2 , , 2x 3 , lOa; 17 6. x H -- - and 3x -\ -- - -- Ans: 4cc -\ -- - 3 4 12 5x 2 T x + a Sx 3 + ax -+- a* 7. 4cc, - , and -! . 2 7a; _ 2a + 1 49o; + 12 8. T , T , and g Ans. 2x + ^ -- YO* // 44:35 9. 4, --, and 2 H- - ylw*. 2 + 4* + 5 45 98 ELEMENTARY ALGEBRA. 10. 3x -f ~ and x - ^. Ans. 3x + 59 45 6b c 11. ac -- and 1 , 8a d wi/u. -f- Sac Ans. 1 -f ac Sad 3? - 5x + 4 " 4(1 + a)' 4(1 -a)' "2(f~^) " l> SUBTRACTION OF FEACTIOIfS. 92. Fractions can only be subtracted when they have the same unit; that is, a common denominator. In that case, the numerator of the minuend, minus that of the sub- trahend, will indicate the number of times that the common unit is to be taken in the difference. Hence, the RULE. I. Reduce the two fractions to a common denomi- inator : n. Then subtract the numerator of the subtrahend from that of the minuend for a new numerator, and write the remainder over the common denominator. EXAMPLES. 3 2 1. What is the difference between - and - 7 8 3 2 24 14 10 5 _ _ __ ,_ ._ ^__ ._ _ . 7 8 "" 56 56 " 56 ~" 28 92. What is the rule for subtracting fractions ? MULTIPLICATION OF FRACTIONS. 9t 3* Cl 2l7 4J* 2. Find the difference of the fractions r - and 20 . 3c j (cc a) x 3c = Sex Sac [ . Here, i . v .j: rt . . > the numerators, ( (2a 4aj) x 20 = 4ao 8&c ) and, 2b X 3c = Qbc the common denominator. 3cx Sac 4ab8bx 3cx3ac4:ab+8bx Hence, -- - --- -= - = -- Ans. Qbc Qbo 6bc , -D j ^ *# f * 3x A 39a; 3. Required the difference of - and Ans. 7 5 3o 4. Required the difference of 5y and Ans. - 8 '^. 8 5. Required the difference of and Ans. - t \s Do e . From ZV subtract ^ - ^n*. x y x + y __ _ 7. From - subtract - - Ans. - - -- ;- y 2 y 2 s 2 y 2 2 2 Find the differences of the following : 3* + a , 2a; -f 7 24 + 8 105* 35o 8. -r and - Ans. -- ; -- ob 8 406 x , x a , ex + bx ab 9. 3x + - and x --- Ans. 2x -\ -- = -- be be a x , a + a 4a5 10. a -i -. - r and . ! - : Ans. a ( ciu.v& f fc ^.m./tw# \Af ,. ,j a + x) a(a 05) a* or MULTIPLICATION OF FRACTIONS. f* y 93. Let v and -^, represent any two fractions. It has O W been shown (Art. 81), that any quantity may be multiplied 93. What 13 the rule for the multiplication of fractions ? -* lUU E L K M K X T A K Y A L G K B K A . by a fraction, by first multiplying by the numerator, and then dividing the result by the denominator. CL G CIC To multiply j- by -^, we first multiply by e, giving ; Ctr then, we divide this result by d, which is done by multiply- ac ing the denominator by d; this gives for the product, = ; OCl/ that is, a c as , T X -, = r^; hence, b d bd RULE. I. If there are mixed quantities, reduce them to a frac- tional form ; then, IL Multiply the numerators together for a new numera- tor, and the denominators for a new denominator. EXAMPLES. bx , c _,. to a 2 + to 1. Multiply a -\ by - First, a H = , a J d a a , 2 + to c a 2 c + bcx hence, x -, = -= Ans. a a ad Find the products of the following quantities : , 2x Sab , Sac 2. , , and j- Ans. 9ax. a c 2b to a ab -f to 3. b -\ and - Ans. - ax x 4. and Ans. ^ - r-=- be b + c b 2 c + bo 3 x + 1 , x 1 ax* ax + x z 1 5. x -\ , and 7 Ant. a a + b a 2 + ab ax , a 2 x z a 3 + a 2 x 6. a -i and Ans. a x x + ar x + x 2 MULTIPLICATION OF FRACTIONS. 101 2d a? J2 7. Multiply i by -^ 1 J a b 3 la a* b 2 2a(a i i 2 ) 2a(a + b) (a - b~) T X " a J 3 3 (a 6) 3 (a After indicating the operation, we factored both numera- tor and denominator, and then canceled the common factors, before performing the multiplication. This should be done, whenever there are common factors. 9. 0-2 4/2 9,( QI x y 7c 12 9 Is* 3. Let be divided by : Ans. - 5 60 4. Let be divided by 5x. , 7 3o T a: + 1 ,-..,., 2a; a; + 1 5. Let be divided by Ans. - 6 3 4x x x A 2 6. Let be divided by - Ans. - x I 2 x 1 c/j* 26R 7. Let be divided by ~ 3 30 2 x b . ., , , 3ca; x I 8. Let =- be divided by = Ans. Divide the following fractions: - **-** ' a2 - 4 Ans. 4X 3 3 * + 2 10 bv Ans. x + iv/ ' .9. nX^. I XS A / _ _ _ X 4x z (a + o) 2 11. 2a*( + *) by -= Ans. ' > a + o *x '-- \- 1)7/ 2 (x I) 2 i2L . ^1S. .. - i y 3 ^2 fla . 3^ a;") 4a(a 2 cc 2 ) ' 6c + bx bj 4(^+^) ' 4W *' 3i(c 2 - a 2 ) 104 ELEMENTARY A L G E B K A . 14. 15. a x by 1 + X X t y Ans. Ans. x - a* -as" I x x 2 by a + 2xy 1 i_ a; 2 * + 1 x -y 16. b 3a by Qa - -2b Ans. 2a - b 2ab 62 - 2ab 4a T7 rt -y 4 l-nr _L y Ana X* - y 2 x'y y x 18. m 2 + 1 -r -^ by m H hi. m* J m Ans. m -\ 1. m IP (y x \ I i aj + f-r ) by (l - xf I + xyJ J \ 1 (x + 2y x\ . /x + 2y a; \ 20. I 7^ + - by I - ^ -- ) V* + y y/ Vy * + y/ EQUATIONS OF THE FIRST DEGREE. 105 CHAPTER V. EQUATIONS OP THE FIRST DEGREE. 95. AN EQUATION is the expression of equality between two quantities. Thus, x = b + c, is an equation, expressing the fact that the quantity ic, is equal to the sum of the quantities b and c. 9O. Every equation is composed of two parts, connected by the sign of equality. These parts are called members : the part on the left of the sign of equality, is called the first member ; that on the right, the second member. Thus, in the equation, x -\- a = b c, x + a is the first member, and b c, the second member. 97. An equation of the first degree is one which involves only the first power of the unknown quantity ; thus, Qx + 3x 5 = 13; (1 ) and ax + bx + c =. d ; (2) are equations of the first degree. 95. What is an equation ? 96. Of how many parts is every equation composed? How are the parts connected ? What are the parts called ? What is the part on the left called? The part on the right ? 97. What is an equation of the first degree ? 5* 103 ELEMENTARY ALGEBRA. 98. A NUMERICAL EQUATION is one in which the ^effi- cients of the unknown quantity are denoted by numbers. 99. A LITERAL EQUATION is one in which the coefficients of the unknown quantity are denoted by letters. Equation ( 1 ) is a numerical equation ; Equation ( 2 ) is a literal equation. EQUATIONS OF THE FIRST DEGREE CONTAINING BUT ONE UNKNTOWN QUANTITY. 100. The TRANSFORMATION of an equation, is the opera- tion of changing its form without destroying the equality of its members. 101. An AXIOM is a self-evident proposition. 102. The transformation of equations depends upon tin following axioms: 1. If equal quantities be added to both members of an equation, the equality icill not be destroyed. 2. If equal quantities be subtracted from both members of an equation, the equality will not be destroyed. 3. If both members of an equation be multiplied by the same quantity, the equality will not be destroyed. 4. If both members of an equation be divided by the same quantity, the equality will not be destroyed. 5. Like powers of the two members of an equation are equal. 6. Like roots of the two members of an equation are equal. 98. What is a numerical equation ? 99. What is a literal equation ? 100. What is the transformation of an equation ? 101. What is an axiom ? 102. Name the axioms on which the transformation of an equation depends. CLEARING OF FRACTIONS. 107 103. T\vo principal transformations are employed in the solution of equations of the first degree: Clearing of frac- tions^ and Transposing, CLEARING OF FRACTIONS. 1. Take the equation, 2# 35 C 3~ ' ' 7 " f 6 = The least common multiple of the denominators is 12. If we multiply both members of the equation by 12, each term will reduce to an entire form, giving, Sx Qx + 2 = 132. Any equation may be reduced to entire terms in the same manner. 104. Hence for clearing of fractions, we have the fol- lowing RULE. I. Find the least common multiple of the denominators: II. Multiply both members of the equation by it, reduc- ing the fractional to entire terms. NOTE. 1. The reduction will be effected, if we divide the least common multiple by each of the denominators, and then multiply the corresponding numerator, dropping the denominator. 2. The transformation may be effected by multiplying each numerator into the product of all the denominators except its own, omitting denominators. 103. How many transformations are employed in the solution of equa- tions of the first degree ? What are they ? 104. Give the rule for clearing an equation of fractions? In what throe vays may the reduction be effected? 108 ELEMENTAKY ALGEBRA. 3. The transformation may also be effected, by multiplying both members of the equation by any multiple of the de- nominators. EXAMPLES. Clear the following equations of fractions : 1. f + f 4 = 3. Ans. ?x + 5x 140 = 105. 5 7 2. '; + - 8. Ans. 925 + Qx 2x = 432. O J / x x x x 3. - H ----- 1 -- = 20. 2^3 9 ^ 12 Ans. I8x + 12a; 4x + 3* = 720. C C SB 4. - + - - = 4. .4ws. 14a; + lOa; 3ox = 280. o i Z 5. + = 15. ,4w*. 15aj 12a; + lOaj = 900. x 4 x 2 5 6 " . 2a; + 8 cc -f- 2 = 10. x 3 7. - -- h 4 = - -4ns. 5z + 60 20a = 9 3x. x ,a5,_ 19 8 ' 4 ~ 6 + 8 + 9 - 12 ' -4ns. 18 12a; + 9x + 8 = 864. ft ft 9. - -j -\- f = g. Ans. ad be -f bdf = bdg o u axe, 2c 2 ai , 4bo-x 5a 3 <*% ^_^^^___^ I _______ _ iw. _ a + a a cc a 2 x 2 Sax b 35 c 16. -^ ---- 2 = 4-5. 56 + 95 . a; = - -- 16a x x 2 x 13 17 5 -~3- + 2 = T - ^' = 114 ELEMENTARY ALGEBRA. ?: X f c~ d~ f ' Ans. x = , oca aca + aoa abc NOTE. What is the numerical value of , when a = 1, = 2, c = 3, Therefore, x = - - :_ 43 ; 2i and, consequently, x 19 = 43 - 19 = 24. GENERAL SOLUTION OF THIS PROBLEM. The sum of two numbers is s, their difference ia d: what are the tAVO numbers ? 120 ELEMENTARY ALGEBRA. ~ Let x denote the less number. Then, a; + d will denote the greater, and 2x + d = s, their sum. Whence, s d s d ~2~ = 2~2*' and, consequently, s d , s d +*=j- 5 +*= 5 +V As these two results are not dependent on particular values attributed to s or d, it follows that : 1. The greater of two numbers is equal to half their sum, plus half their difference : 2. The less is equal to half their sum, minus half their difference. Thus, if the sum of two numbers is 32, and their differ- ence 16, 32 16 the greater is, + -- ==16 + 8 = 24 ; and 2t 2i 32 16 the less, =16 8= 8. 2i 2i VERIFICATION. 24 + 8 = 32; and 24 8 = 16. 8. A jv*rson engaged a workman for 48 days. For each day that he labored he received 24 cents, and for each day that he vras idle, he paid 12 cents for his board. At the end of the 48 days, the account was settled, when the laborer received 504 cents. Required, the number of working days, and the number of days he was idle. If the number of working days, and the number of idle days, were known, and the first multiplied by 24, and the r u o B L E M s . 121 second by 12, the difference of these products wouldbe. 504. Let us indicate these operations by means of algebraic signs. Let x denote the number of working days. Then, 48 x = the number of idle days, 24 x x = the amount earned, and, 12(48 x} = the amount paid for board. Then, 24a - 12(48 a;) = 504, what was received, which is the statement. Then, performing the operations indicated, 24* 576 + 12as = 504, or, 3Qx 504 + 570 = 1080, and, x = - - = 30, the number of working days ; 3G whence, 48 30 =18, the number of idle days. VERIFICATION. Thirty days' labor, at 24 cents ) n J ' 1 30 X 24 = 720 cents, a day, amounts to ) And 18 days' board, at 12 cents ) J } 18 X 12 = 216 cents, a day, amounts to ) The difference is the amount received . 504 cents. GENERAL SOLUTION. * This problem may be made general, by denoting the whole number of working and idle days, by n ; The amount received for each day's work, by a ; The amount paid for board, for each idle day, by b ; And what was due the laborer, or the balance of the account, by c. 6 122 K L E M K N T A R Y ALGEBRA. ^ft before, let the number of working days be denoted by x. The number of idle days will then be denoted by n x. Hence, what is earned will be expressed by ax, and- the sura to be deducted, on account of board, by b(n x). The statement of the problem, therefore, is, ax b(n x) = c. Performing indicated operations, ax bn -f bx c, or, (a 4- f>}x = c -f- bn ; whence, x = - - number of working days ; (Ju ~*|~ (s c + bn an-\-bncbn and, n x = n j- = - v , a+ b . + o or, n x = = number of idle davs. a + b Let us suppose n = 48, a 24, b = 12, and c 504 ; these numbers will give for x the same value as before found. 9. A person dying leaves half of his property to his wife, one-sixth to each of two daughters, one-twelfth to a servant, and the remaining $600 to the poor ; what was the amount of the property ? Let x denote the amount, in dollars, M Then, ^ what he left to his wife, 2t JM - = what he left to one daughter, * O/k* / and, = - what he left to both daughters, u 3 X also, = what he left to his servant, and, $600 = what he left to the poor. PROBLEMS. 123 Then, by the conditions, /j / /> - + - 4- + 600 x, the amount of the property, . t > 1 w which gives, x = $7200. 10. A and J3 play together at cards. A sits down with $84, and _Z> with $48. Each loses and wins in turn, w r hen it appears that A has five times as much as J5. How much did A win ? Let x denote the number of dollars A won. Then, A rose with 84 + x dollars, and J? rose with 48 x dollars. But, by the conditions, we have, 84 + x = 5(48 aj), hence, 84 + x 240 5x; and, Qx = 156, consequently, x = 26 ; or A won $26. VERIFICATION. 84 -f 26 = 110 ; 48 26 = 22; 110 = 5(22) = 110. 11. A can do a piece of work alone in 10 days, JB in 15 days ; in what time can they do it if they work together ? Denote the time by &, and the work to be done, ly 1. Then, in 1 day, A can do of the work, and JB can do of the work ; and in 13 C x days, A can do of the work, and C B can do ~ of the work. 13 124 K L E M E N T A 14 Y ALGEBRA. Hence, by the conditions, C *C - (- = 1, which gives, 13x + Wx 130; 10 13 130 hence, 23x =130, x = = 5f days. 23 12. A fox, pursued by a hound, has a start of 60 of his own leaps. Three leaps of the hound are equivalent to 7 of the fox ; but while the hound makes 6 leaps, the fox makes 9 : how many leaps must the hound make to overtake the fox? There is some difficulty in this problem, arising from the different units which enter into it. Since 3 leaps of the hound are equal to 7 leaps of the fox, 7 1 leap of the hound is equal to - fox leaps. 3 Since, while the hound makes 6 leaps, the fox makes 9, 9 3 while the hound makes 1 leap, the fox will make - , or - leaps. Let x denote the number of leaps which the hound makes before he overtakes the fox ; and let 1 fox leap denote the unit of distance. 7 Since 1 leap of the hound is equal to - of a fox leap, x 7 leaps will be equal to -x fox leaps ; and this will denote the distance passed over by the hound, in fox leaps. ' 3 Since, while the hound makes 1 leap, the fox makes ^ 3 leaps, while the hound makes x leaps, the fox makes -x leaps ; m and this added to 60, his distance ahead, Avill give g -x + CO, for the whole distance passed over by the fox. I PROBLEMS. 125 Hence, from the conditions, 7 3 -a; = -x + 60 ; whence, Ux = Qx + 360; x = 72. The hound, therefore, makes 72 leaps before overtaking le fo leaps. Q the fox; in the same time, the fox makes 72 x - = 108 m VERIFICATION. 108 + 60 =: 168, whole number of fox leaps, 72 X I - 168. o 13. A father leaves his property, amounting to $2520, to four sons, A, _Z?, (7, and D. C is to have $360, J3 as much as C and D together, and A twice as much as .#, less $1000 : how much do A, B-, and D receive ? Am. A, $760; .#, $880; Z>, $520. 14. An estate of $7500 is to be divided among a widow, two sons, and three daughters, so that each son shall receive twice as much as each daughter, and the widow herself $500 more than all the children : what was her share, and what the share of each child ? {Widow's share, $4000. Each son's, 1000. Each daughter's, 500. 15. A company of 180 persons consists of men, women, and children. The men are 8 more in number than the women, and the children 20 more than the men and women together : how many of each sort in the company ? Ans. 44 men, 36 women, 100 children. 126 E L E M E N T A II Y A L E BRA. 16. A father divides $2000 among five sons, so that each elder should receive $40 more than his next younger bro- ther : what is the share of the youngest? Ans. $320. 17. A purse of $2850 is to be divided among three per- sons, A, J3, and C. A's share is to be to jB's as 6 to 1 1 , and C is to have $300 more than A and B together : what is each one's share? A>s, $450 ; B's, 8825 ; (7's, $1575. 18. T\vo pedestrians start from the same point and travel in the same direction ; the first steps twice as far as the second, but the second makes 5 steps while the first makes but one. At the end of a certain time they are 300 feet apart. Now, allowing each of the longer paces to be 3 feet, how far will each have traveled ? Ans. 1st, 200 feet ; 2d, 500. 19. Two carpenters, 24 journeymen, and 8 apprentices received at the end of a certain time $144. The carpenters received $1 per day, each journeyman, half a dollar, and each apprentice, 25 cents : how many days were they em- ployed? Ans. 9 days. 20. A capitalist receives a yearly income of $2940 ; four- fifths of his money bears an interest of 4 per cent., and the remainder of 5 per cent. : how much has he at interest ? Ans. $70000. 21. A cistern containing 60 gallons of water has three unequal cocks for discharging it ; the largest will empty it in one>hour, the second in two hours, and the third, in three: in what time will the cistern be ehipticd if they all run to- gether ? Ans. 32 T 8 T min. 22. In a certain orchard, one-half are apple trees, one- fourth peach trees, one-sixth plum trees; there are also, 120 cherry trees, and 80 pear trees : how many trees in the orchard? Ans. 2400. 23. A farmer being asked ho\\ many sheep ho had, P ii ) B L K M 8 . 127 answered, that he had them in five fields ; in the 1st he had |, in the 2d, , in the 3d, |, and in the 4th, T V, and in the 5th, 450 : how many had he ? Ans. 1200. 24. My horse and saddle together are worth $132, and the horse is worth ten times as much as the saddle : what is the value of the horse ? Ans. $120. 25. The rent of an estate is this year 8 per cent, greater than it was last. This year it is $1890: what was it last year? Ans. $1750. 26. What number is that, from which if 5 be subtracted, | of the remainder will be 40 ? Ans. 65. 27. A post is | in the mud, in the water, and 10 feet above the water : Avhat is the whole length of the post ? Ans. 24 feet. 28. After paying | and i of my money, I had 66 guineas left in my purse : how many guineas were in it at first ? Ans. 120. 29. A person was desirous of giving 3 pence apiece to some beggars, but found he had not money enough in his pocket by 8 pence; he therefore gave them each 2 pence and had 3 pence remaining : required the number of beg- gars. Ans. 11. 30. A person, in play, lost % of his money, and then won 3 shillings ; after which he lost of what he then had ; and this done, found that he had but 12 shillings remaining: what had he at first ? Ans. 20s. 31. Two persons, A and J5, lay out equal sums of money in trade; A gains $126, and B loses $87, and A's money is then double of B's : what did each lay out? Ans. $300. 32. A person goes to a tavern with a certain sum of money in his pocket, where he spends 2 shillings : he then borrows as much money as he had left, and going to another tavern, he there spends 2 shillings also; then borrowing 128 ELK MEN TAUT ALGEBRA. again as much money as was left, he went to a third tavern, where likewise he spent 2 shillings, and borrowed as much as he had left : and again spending 2 shillings at a fourth tavern, he then had nothing remaining. What had he at first ? Ans. 3s. 9 d. 33. A tailor cut 19 yards from each of three equal pieces of cloth, and 17 yards from another of the same length, and found that the four remnants were together equal to 142 yards. How many yards in each piece ? Ans 54. 34. A fortress is garrisoned by 2600 men, consisting of infantry, artillery, and cavalry. Now, there are nine times as many infantry, and three times as many artillery soldiers as there are cavalry. How many are there of each corps ? Ans. 200 cavalry; COO artillery ; 1800 infantry. 35. All the journeyings of an individual amounted to 2970 miles. Of these he traveled 3| times as many by water as on horseback, and 2| times as many on foot as by water. How many miles did he travel in each way ? Ans. 240 miles; 840 m. ; 1890 m. 36. A sum of money was divided between two persons, A and J5. A's share was to J?'s in the proportion of 5 to 3, and exceeded five-ninths of the entire sum by 50. "What was the share of each? Ans. A's share, 450; JFs, 270. 37. Divide a number into three such parts that the second shall be n times the first, and the third m times as great as the first. a na ma 2d T-T-; rrrr, '> 3d > OC ^*U* .5 wi 1 + m + n 1 + m + n 1 + m + n 38. A father directs that $1170 shall be divided among his three sons, in proportion to their ages. The oldest ia t\vice as old as the youngest, and the second is one-third older than the youngest. How much was each to receive? A/tv. $'270, youngest ; *:!UO, second ; $540, < Ulest. PROBLEMS. 129 39. Three regiments are to furnish 594 men, and each to furnish in proportion to its strength. Now, the strength of the first is to the second as 3 to 5 ; and that of the second to the third as 8 to 7. How many must each furnish ? Ans. 1st, 144 men ; 2d, 240 ; 3d, 210. 40. Five heirs, A, J?, (7, D, and E, are to divide an inher- itance of $5600. -B is to receive twice as much as A, and 8200 more ; C three times as much as A, less $400 ; D the half of what -Z> and C receive together, and 150 more ; and E the fourth part of what the four others get, plus $475. How much did each receive ? A>s, $500; .#'s, 1200; C"a, 1100; D's, 1300; E's, 1500. 41. A person has four casks, the second of which being filled from the first, leaves the first four-sevenths full. The third being filled from the second, leaves it one-fourth full, and when the third is emptied into the fourth, it is found to fill only nine-sixteenths of it. But the first Avill fill the third and fourth, and leave 15 quarts remaining. How many gallons does each hold ? Ans. 1st, 35 gal. ; 2d, 15 gal. ; 3d, 11} gal. ; 4th, 20 gal. 42. A courier having started from a place, is pursued by a second after the lapse of 10 days. The first travels 4 miles a day, the other 9. How many days before the second will overtake the first ? Ans. 8. 43. A courier goes 3H miles every five hours, and is fol- lowed by another after he had been gone eight hours. The second travels 221 miles every three hoxirs. How many hours before he will overtake the first ? Ans. 42. 44. Two places are eighty miles apart, and a person leaves one of them and travels towards the other at the rate of Si- miles per hour. Eight hours after, a person departs from 6* 130 ELEMENTARY ALGEBRA. the second place, and travels at the rate of 5| miles per hour. How long before they will be together? A.ns. 6 hours. EQUATIONS CONTAINING TWO UNKNOWN QUANTITIES. 11O. If we have a single equation, as, 2x + 3y = 21, containing two unknown quantities, % and y, we may find the value of one of them in terms of the other, as, 21 - 3 x = (1.) Now, if the value of y is unknown, that of x will also be unknown. Hence, from a single equation, containing two unknown quantities, the value of x cannot be determined. If we have a second equation, as, 5x + 4y = 35, we may, as before, find the value of x in terms of y, giving, 35 4y "-- (2 Now, if the values of x and y are the same .in Equations ( 1 ) and ( 2 ), the second members may be placed equal to each other, giving, 21 3y 35 4y t = ^-2 , or 105 - 15y = 70 - Sy ; . O from which we find, y = 5. 110. In one equation containing two unknown quantities, can you find the value of either ? I you hare a second equation involving the same two unknown quantities, can YOU find their values? What are such equa- tions called ? ELIMINATION. 131 Subtituting this value for y in Equations (1) or (2), we 6iid x = 3. Such equations are called Simultaneous equations. Hence, 111. SIMULTANEOUS EQUATIONS are those in which the Talues of the unknown quantity are the same in both. ELIMINATION. 112. ELIMINATION is the operation of combining two equations, containing tAvo unknown quantities, and deducing therefrom a single equation, containing but one. There are three principal methods of elimination : 1st. By addition or subtraction. 2d. By substitution. 3d. By comparison. We shall consider these methods separately. Elimination by Addition or Subtraction. 1. Take the two equations, 3x 2y = 7, 8x -f- 2y = 48. If we add these two equations, member to member, we obtain, llaj = 55; which gives, by dividing by 11, x = 5; and substituting this value in either of the given equations, we find, y = 4. 111. What are simultaneous equations? 112. What is elimination? How many methods of elimination are there ? What are they ? 132 ELEMENTARY ALGEBRA. 2. Again, take the equations, Sx + 2y = 48, 3z + 2y = 23. If we subtract the 2d equation from the 1st, we obtain, 5x =. 25; which gives, by dividing by 5, x 5; and by substituting this value, we find, V = 4. 3. Given the sum ot two numbers equal to s, and theif difference equal to <7, to find the numbers. Let x = the greater, and y the less number. Then, by the conditions, x + y = g. and, x y = d. By adding (Ait. 102, Ax. 1), 2x = s + d. By subtracting (Art, 102, Ax. 2), . . . 2y = * d. Each of these equations contains but one unknown quantity. From the first, we obtain, x = S - o ' and from the second, y These are the same values as were found in Prob. 7, page 120. 4. A person engaged a workman for 48 days. For each day that he labored he was to receive 24 cents, and for each day that he was idle he was to pay 1 2 cents for his board. At the end of the 48 days the account was settled, when the laborer received 504 cents. Required the number of work- ing days, and the number of days he was idle. ELIMINATION. 133 Let a; = the number of working days, y the number of idle days. Then. 24a = what he earned, and, 12y = what he paid for his board. Then, by the conditions of the question, we have, x + y = 48, and, 24 12y = 504. This is the statement of the problem. It has already been sho~wn (Art. 102, Ax. 3), that the two members of an equation may be multiplied by the same num- ber, without destroying the equality. Let, then, the first equation be multiplied by 24, the coefficient of x in the second ; we shall then have, 24ly = 22 ) 6. { Ans. x = 5, y = 6. ( 5x + 2y = 37 ) j 2x + Gy = 42 ) 7. -j Ans. x = 44, y = 54. ( 8x Qy = 3 ) j 8x 9y = 1 ) 8. i Ans. x 4, y i. ( 6cc 3 = 4z ) . J a; = 6, y = 9. 11. X 7 V 4ns. < x = 14, y = 16. 113. What is the rule fa? elimination by addition or subtraction? ELIMINATION. 135 12. Says A to .Z?, you give me $40 of jour money, and I shall then have five times as much as you will have left. Xow they both had $120 : how much had each? Ans. Each had $60. 13. A father says to his son, " twenty years ago, my age was four times yours ; now it is just double : " what were their ages ? / A i Father's, 60 years. Ans. S ,., ( Son's, 30 years. 14. A father divided his property between his two sons. At the end of the first year the elder had spent one-quarter of his, and the younger had made $1000, and their property was then equal. After this the elder spent $500, and the younger made $2000, when it appeared that the younger had just double the elder: what had each from the father? . j Elder, $4000. ' ( Younger, $2000. 15. If John give Charles 15 apples, they will have the same number; but if Charles give 15 to John, John will have 15 times as many, wanting 10, as Charles will have left. How many has each ? j j J^ n > 50. S - 1 Charles, 20. 16. Two clerks, A and _Z?, have salaries which are together equal to $900. A spends T ^ per year of what he receives, and _B adds as much to his as A spends. At the end of the year they have equal sums: what was the salary of each? A>s = $500. Elimination by Substitution. 114. Let us again take the equations, Sa; + ty = 43, ( 1.) Ux + 9y = 69. (2.) 1 1 4. Give the rule for eliminaticm by substitution. When is this method t>sed to the greatest advantage? 136 ELEMENTAKY ALGEBRA. Find the value of x in the first equation, which gives, Substitute this value of x in the second equation, and w have, 11 x -L_J + gy = 69; or,, 473 77y -f 45y = 345 ; or, 32y = 128. Here, ar has been eliminated by substitution. In a similar manner, we can eliminate any unknown quan- tity ; hence, the RULE. I. Find from either equation the value of the unknown quantity to be eliminated: II. Substitute this value for that quantity in the other equation. NOTE. This method of elimination is used to great advan- tage when the coefficient of either of the unknown quantities is 1. EXAMPLES. Find, by the last method, the values of x and y in the following equations : 1. 3cc y = 1, and 3y 2x 4. Ans. x = 1, y = 2. 2. 5y 4x = 22, and 3y + 4x = 38. Ans. x = 8, y =. 2. 3. x + 8y = 18, and y Sx = 29. Ans. x = 10, y = 1. ELIMINATION. 137 2 4. 5x - y = 13, and Sx -f -y = 29. y Ans. x = 3, y = 5. 103 = 69, and lOy " '= 49. 5 7 6. x + * - = 10, and + = 2. x = 8, y = 10. 7. | - | + 5 = 2, a; + | = I7f Ans. x = 15, y = 14. 8 - 1 + 5 + 3 = 8*. and 7~* = I' . = 3|, y = 4. y n _ i . . K 8 4 H 5) 12 16 Ans. x = 12, y = 16. 10. | - y - 1 = - 9, and 5a: - ^ = 29. ^Iws. a; = 6, y = 7. 11. Two misers, -4 and jB, sit down to count over their money. They both have $20000, and B has three times as much as A : how much has each ? , A A \ ^3., ' (^, $15000. 12. A person has two purses. If he puts 87 into the first, the whole is worth three times as much as the second purse : but if he puts $7 into the second, the whole is worth five times as much as the first: what is the value of each purse? Ans. 1st, $2 ; 2d, $3. 138 K L E M K N T A Ii Y A L G K B K A . 13. Two numbers have the following properties : if the first be multiplied by 6, the product will be equal to the second multiplied by 5 ; and 1 subtracted from the first leaves the same remainder as 2 subtracted from the second : what are the numbers ? Ans. 5 and 6. 14. Find two numbers with the following properties : the first increased by 2 is 3 times as great as the second ; and the second increased by 4 gives a number equal to half the first : what are the numbers ? Ans. 24 and 8. 15. A father says to his son, "twelve years ago, I was twice as old as you are now: four times your age at that time, plus twelve years, will express my age twelve years hence : " what were their ages ? ( Father, 72 years. AnS ' \ Son, 30 Elimination by Comparison. 115. Take the same equations, 5x + 1y = 43 llx + Qy = 69. Finding the value of x from the first equation, we have, 43 iy iyt _ __ 5_ ~T~ and finding the value of x from the second, we obtain, 69 9y 115. Give the rule for climinatnn by comparison. ELIMINATION. 139 Let these two values of x be placed equal to each other, and we have, 43 7y __ 69 9y 5 11 Or, 473 77y = 345 - 45y; or, 32y = 128. Hence, y 4. 69 36 And, x = = 3. This method of elimination is called the method by com- parison, for which we have the following BULE. I. Find, from each equation, the value of the same unknown quantity to be eliminated: II. Place these values equal to each other. EXAMPLES. Find, by the last rule, the values of x and y, from the following equations, 1. 3a5 4- ^ -f 6 = 42, and y ^- = 14- Ans. 35 = 11, y = 15. 2. | - | + 5 = 6, and | -f 4 = 2- + 6. Ans. x 28, y = 20. > c 22 3 ' 15 - 4 + T = '' and 3y ~ * = 6 ' ,4ns. 35 = 9, y = 5. 4. y 3 = -a + 5, and ~= = y 3. A ^ ^lw*. a; = 2, y = 9, 14:0 ELEMENTARY ALGEBRA. y-x x y_ Ans. x = 16, y = 7. 2y = as - -|, and s + y = H. . a; = 10, y = 6. 7. - --S = a> - 2f, BB - L = 0. . a; = 1, y * 8. 2y + 3 = y + 43, y - ^= = y - ^- o o Ans. x 10, y = 13. 9. 4y x ~ y = x + 18, and 27 y X + y + t. 2 . a; = 9, y =. 7. f-2 =|. . a; = 10, y = 20. 116. Having explained the principal methods of elimina- tion, we shall add a few examples which may be solved by any one of them ; and often indeed, it may be advantageous to employ them all, even in the same example. GENERAL EXAMPLES. Find the values of x and y in the following simultaneous equations : 1. 2a; 4- 3y = 16, and 3x 2y = 11. Ans. x = 5, y = 2. ELIMINATION. HI to ty _ j 2 5 4 - 20' 3x 2y 61 T 5" = T2~6" Ans. x = -, y = - 3. + 7y = 99, and + 7s = 51. 7 7 4 *_i2- y * 2 "4 .4ns. a; = 7, y = 14. ?L+y + 1 _ 8 = 2 _y^ + 27 . 5. 4x 5 x V x 4- V 6 ., -^rV^- 8 * j- 7. 8. 9. 3y - as 2a; - y _ g e y + o*K ~~~ o 4 Sx 3 o 6 - y 4a; 4 y 5 ! -i*-iy + = 79 6 = f_ Ans. x = CO, y =. 40. a; = 6. y-5. E L E M E N T A li Y A L G K B K A . c + ab bd I x = Ans. IT. P li O B L K M 8 . 143 PROBLEMS. 1. What fraction is that, to the numerator of which if 1 be added, the value will be - , but if 1 be added to ita 1 denominator, the value will be - ? m Let the fraction be denoted by - y Then, by the conditions, x -f 1 1 x I ;> : " d ' y + i--r whence, Sx + 3 = y, and 4x = y +1. Therefore, by subtracting, x 3 = 1, and x = 4. Hence, 12 + 3 = y; .-. y = 15. 2. A market-woman bought a certain number of eggs at 2 for a penny, and as many others at 3 for a penny ; and having sold them all together, at the rate of 5 for 2c?, found that she had lost 4d: how many of both kinds did she buy ? Let 2x denote the whole number of eggs. Then, x the number of eggs of each sort. Then will, -x = the cost of the first sort, " and, -x = the cost of the second sort. 3 But, by the conditions of the question, hence, will denote the amounts for which the eggs O were sold. 144 ELEMENTARY ALGEBBA. But, by the conditions, 1 1 4 S"+s?:-T-- 41 therefore, 15 -f 10x 24ce = 120; .*. x = 120 ; the number of eggs of each sort. 3. A person possessed a capital of 30,000 dollars, for which he received a certain interest ; but he owed the sum of 20,000 dollars, for which he paid a certain annual interest. The interest that he received exceeded that which he paid by 800 dollars. Another person possessed 35,000 dollars, for which he received interest at the second of the above rates ; but he owed 24,000 dollars, for which he paid interest at the first of the above rates. The interest that he received, an- nually, exceeded that which he paid, by 310 dollars. Re- quired the two rates of interest. Let x denote the number of units in the first rate of interest, and y the unit in the second rate. Then each may be regarded as denoting the interest on $100 for 1 year. To obtain the interest of $30,000 at the first rate, denoted by , we form the proportion, 100 : 30,000 : : x : '- , or 300a;. And for the interest of $20,000, the rate being y, 100 : 20,000 : : y : - O r 200y. But, by the conditions, the difference between these two amounts is equal to 800 dollars. We have, then, for the first equation of the problem, SOOaj 200y = 800. PROBLEMS. 145 By expressing, algebraically, the second condition of the problem, we obtain a second equation, 350y 240o; = 310. Both members of the first equation being divisible by 100, and those of the second by 10, we have, $x 2y = 8, 35y 24 = 31. To eliminate jc, multiply the first equation by 8, and then add the result to the second ; there results, 19y = 95, whence, y = 5. Substituting for y, in the first equation, this value, and that equation becomes, 3x 10 = 8, whence, x = 6. Therefore, the first rate is 6 per cent, and the second 5. VERIFICATION. $30,000, at 6 per cent, gives 30,000 X .06 = $1800. $20,000, 5 " " 20,000 X .05 $1000. And we have, 1800 1000 = 800. The second condition can be verified in the same manner. 4. "What two numbers are those, whose difference is 7, and sum 33 ? Ans. 13 and 20. 5. Divide the number 75 into two such parts, that three times the greater may exceed seven times the less by 15. Ans. 54 and 21. 6. In a mixture of wine and cider, | of the whole plus 25 gallons was wine, and part minus 5 gallons was cider : how many gallons were there of each ? Ans. 85 of wine, and 35 of cider. 7 146 ELEMENTARY ALGEBRA. 7. A bill of 120 was paid in guineas and moidores, and the number of pieces used, of both sorts, was just 100. If the guinea be estimated at 21s, and the moidore at 2Vs, how many pieces were there of each sort ? Ans. 50. 8. Two travelers set out at the same time from London and York, whose distance apart is 150 miles. One of them travels 8 miles a day, and the other 7 : in what time will they meet? Ans. In 10 days. 9. At a certain election, 375 persons voted for two candi- dates, and the candidate chosen had a majority of 91 : how many voted for oaeli ? Ans. 233 for one, and 142 for the other. 10. A person has two horses, and a saddle worth 50. Now, if the saddle be put on the back of the first horse, it makes their joint value double that of the second horse ; but if it be put on the back of the second, it makes their joint value triple that of the first : what is the value of each horse ? Ans. One 30, and the other 40. 11. The hour and minute hands of a clock are exactly to- gether at 12 o'clock : when will they be again together? Ans. Ih. 5/ T ni' 12. A man and his wife usually drank out a cask of beer in 12 days ; but when the man was from home, it lasted the woman 30 days : how many days would the man alone be in drinking it ? Ans. 20 days. 13. If 32 pounds of sea-water contain 1 pound of salt, how much fresh water must be added to these 32 pounds, in order that the quantity of salt contained in 32 pounds of the new mixture shall be reduced to 2 ounces, or | of a pound ? Ans. 224 Ibs. 14. A person who possessed 100,000 dollars, placed the greater part of it out at 5 per cent interest, and the other PUOBLKMS. 14:7 at 4 per cent. The interest which he received for the whole, amounted to 4640 dollars. Required the two parts. Ans. $64,000 and $36,000. 15. At the close of an election, the successful candidate had a majority of 1500 votes. Had a fourth of the votes of the unsuccessful candidate been also given to him, he would have received three times as many as his competitor, want- ing three thousand five hundred : how many votes did each receive? i 1st, 6500. 2d, 5000. - \ 16. A gentleman bought a gold and a silver watch, and a chain worth $25. When he put the chain on the gold watch it and the chain became worth, three and a half times more than the silver watch ; but when he put the chain on the silver watch, they became worth one-half the gold watch and 15 dollars over : what was the value of each watch ? j Gold watch, $80. (Silver $30. 1 7. There is a certain number expressed by two figures, which figures are called digits. The sum of the digits is 11, and if 13 be added to the first digit the sum will be three times the second: what is the number? Ans. 56. 18. From a company of ladies and gentlemen 15 ladies retire; there are then left two gentlemen to each lady. After which 45 gentlemen depart, when there are left 5 ladies to each gentleman : how many were there of each at first ? ( 50 gentlemen. Ans. 1 . f ,. } 40 ladies. 19. A person wishes to dispose of his horse by lottery. If he sells the tickets at $2 each, he will lose $30 on his horse ; but if he sells them at $3 each, he will receive $30 14:8 KLKMKNTAIIY ALGEBRA. more than his horse cost him. "What is the value of the horse, and number of tickets? . j Horse, $150. ' (No. of tickets, CO. 20. A person purchases a lot of wheat at $1, and a lot of rye at 75 cents per bushel ; the whole costing him 1 17.50. He then sells of his wheat and 1 of his rye at the same rate, and realizes $27.50. How much did he buy of each ? 80 bush, of wheat. Ans. -, , , 50 bush, of rye. 21. There are 52 pieces of money in each of two bags. A. takes from one, and IB from the other. A takes twice as much as J3 left, and J? takes 7 times as much as A left. How much did each take? ( A. 48 pieces. Ans. < ' ( B, 28 pieces. 22. Two persons, A and B, purchase a house together, Worth $1200. Says A to _Z?, give me two-thirds of your money and I can purchase it alone ; but, says J5 to A, if you will give me three-fourths of your money I shall be able to purchase it alone. How much had each ? Ans. A, $800 ; J5, $600. 23. A grocer finds that if he mixes sherry and brandy in the proportion of 2 to 1, the mixture will be worth 78s. per dozen ; but if he mixes them in the proportion of 7 to 2, he can get 79s. a dozen. "What is the price of each liquor per dozen? Ans. Sherry, 81s. ; brandy, 72s. Equations containing three or more unknown quantities. Let us now consider equations involving three or more unknown quantities. Take the group of simultaneous equations, 117. Give the rule for solving any group of simultaneous equations? EXAMPLES. 149 5x - Qy + 42 = 15, . . (1.) 7a + 4y 3z = 19, . . (2.) 2x + y -f- 62 = 46. ... (3.) To eliminate z by means of the first two equations, multi- ply the first by 3, and the second by 4 ; then, since the coefficients of z have contrary signs, add the two results together. This gives a new equation : 43z 2y = 121 . . . . . (4.) Multiplying the second equation by 2 (a factor of the coefficient of z in the third equation), and adding the result to the third equation, we have, 16a + 9y = 84 ' (5.) The question is then reduced to finding the values of x and y, which will satisfy the new Equations (4) and (5). Now, if the first be multiplied by 9, the second by 2, and the results added together, we find, 419a; = 1257; whence, x = 3. "We might, by means of Equations ( 4 ) and ( 5 ) deter- mine y in the same way that we have determined x ; but the value of y may be determined more simply, by substi- tuting the value of x in Equation ( 5 ) ; thus, 04 40 48 + Qy = 84. .-. y = - = 4. In the same manner, the first of the three given equations becomes, by substituting the values of x and y, 15 24 -f 4s = 15. .'. z = = 6. 4 In the same way, any group of simultaneous equations may be solved. Hence, the 150 ELEMENTARY ALGEBRA. EULE. 1. Combine one equation of the group icith each of the others, by eliminating one unknown quantity; there will result a neio group containing one equation less than the original group : II. Combine one equation of this new group with each of the others, by eliminating a second unknown quantity ; there will result a new group containing two equations less than the original group : HI. Continue the operation until a single equation is found, containing but one unknotcn quantity : IV. find the value of this unknown, quantity by the preceding rules ; substitute this in one of the group of two equations, and find tlie value of a second unknown quantity ; substitute these in either of the group of three, finding a third unknown quantity ; and so on, till the values of all are found. NOTES. 1. lu order that the value of the unknown quan- tities may be determined, there must be just as many inde- pendent equations of condition as there are unknown quan- tities. If there are fewer equations than unknown quantities, the resulting equation will contain at least two unknown quantities, and hence, their values cannot be found (Art. 110). If there are more equations than unknown quantities, the conditions maybe contradictory, and the equations impossible. 2. It often happens that each of the proposed equations does not contain all the unknown quantities. In this case, with a little address, the elimination is very quickly per- formed. Take the four equations involving four unknown quanti- ties : 2x 3y + 2z = 13. (1.) 4y + 2z = 14. (3.) 4u 2x = 30. (2.) 5y + 3u 32. (4.) EXAMPLES. 151 By inspecting these equations, we see that the elimination of s in the two Equations, ( 1 ) and ( 3 ), will give an equa- tion involving x and y\ and if we eliminate u in Equa- tions ( 2 ) and ( 4 ), we shall obtain a second equation, in- volving x and y. These last two unknown quantities may therefore be easily determined. In the first place, the elimination of z from ( 1 ) and ( 3 ) gives, ty - 2x = 1 ; That of u from ( 2 ) and ( 4 ) gives, 20y + Gx = 38. Multiplying the first of these equations by 3, and adding, 41y = 41; Whence, y = 1. Substituting this value in 1y 2x = 1, we find, x = 3. Substituting for x its value in Equation ( 2 ), it becomes 4u 6 = 30. Whence, u = 9. And substituting for y its value in Equation (8), there results, z 5. EXAMPLES. X + y + 2 = 29 x + 2y + 3z = 62 1. Given < > to find , y, and z. + ^ + f = :o Ans. x = 8, y = 9, z 12. 152 ELEMENT AKY ALGEBRA. f 2x + 4y 33 = 22 ~\ 2. Given < 4x 2y + 5z 18 I to find cr, y, and a. [ 6z + 7y - z = 63 J Ans. x = 3, y = 7, z = 4. 3. Given ^B + -V 4- -8 = L f to find a?, y, and z. 2 = 12 . a; = 12, y = 20, z = 30. 4. Given < x + y z = 18$ > to find a, y, and z. x - y + z = 13|J ^ln. x = 16, y V, s = 5^ {3a; + 5y = 161 ^ 7* + 23 = 209 V to find a, y, and 2. 2y + z - 89 J . x = 17, y = 22, z = 45. 6. Given x = 1 1 I = a x y - 1 + 1 = 6 X Z to find a;, y, and z. a + b c' a + c ' z = b+c NOTE. In this example we should not proceed to clear the equation of fractions; but subtract immediately the second equation from the first, and then add the third : we thus find the value of y. PROBLEMS. 153 PKOBLEMS. 1. Divide the number 90 into four such parts, that the first increased by 2, the second diminished by 2, the third multiplied by 2, and the fourth divided by 2, shall be equal each to each. This problem may be easily solved by introducing a new unknown quantity. Let a;, y, z, and u, denote the required parts, and desig- nate by m the several equal quantities which arise from the conditions. We shall then have, u x + 2 = m, y 2 = m, 2 = m, - = m. 2 From which we find, x = m 2, y = m + 2, z = , u = 2m. And, by adding the equations, x -{- y -\- z + u = m -f m H -f 2m = 4m. And since, by the conditions of the problem, the first member is equal to 90, we have, 4|m = 90, or fw 90; hence, m = 20. Having the value of m, we easily find the other values; viz.: a; 18, y = 22, z = 10, u = 40. 2. There are three ingots, composed of different metals mixed together. A pound of the first contains 7 ounces of silver, 3 ounces of copper, and 6 of pewter. A pound of the second contains 12 ounces of silver, 3 ounces of copper, and 1 of pewter. A pound of the third contains 4 ounces of silver, 7 ounces of copper, and 5 of pewter. It is required 7* EL K M K x T A 11 y A L G K B K A . to find how much it will take of each of the three ingots to form a fourth, which shall contain in a pound, 8 ounces of silver, 3J of copper, and 4 of pewter. Let a, y, and 2, denote the number of ounces which it is necessary to take from the three ingots respectively, in order to form a pound of the required ingot. Since there are 7 ounces of silver in a pound, or 16 ounces, of the first ingot, it follows that one ounce of it contains T ^- of an ounce of silver, and, consequently, in a number of ounces denoted 1x by a*, there is -- ounces of silver. In the same manner, 16 12v 4 z we find that, - , and , denote the number of ounces 16 16 of silver taken from the second and third ; but, from the enunciation, one pound of the fourth ingot contains 8 ounces of silver. We have, then, for the first equation, "tx 12 y 42 16 4 ' T6~ " 16 = or, clearing fractions, Va; + 12y + 42 = 128. As respects the copper, we should find, 3aj -f 3y + 72 = 60 ; and with reference to the pewter, 6 + y + 02 = 68. As the coefficients of y in these three equations are the most simple, it is convenient to eliminate this unknown quantity first. Multiplying the second equation by 4, and subtracting the first from it, member from member, we have, 5x + 242 r= 112. PROBLEMS. 15ft Multiplying the third equation by 3, and subtracting the second from the resulting equation, we have, I5x + Sz = 144. Multiplying this last equation by 3, and subtracting the preceding one, we obtain, 40a; = 320; whence, x = 8. Substitute this value for x in the equation, I5x + Sz = 144; it becomes, 120 -f 8z = 144, whence, 2 = 3. Lastly, the two values, x = 8, z = 3, being substituted in the equation, 6aj + y + 5z = 68, give, 48 + y + 15 = 68, whence, y = 5. Therefore, in order to form a pound of the fourth ingot, we must take 8 ounces of the first, 5 ounces of the second, and 3 of the third. VERIFICATION. If there be 1 ounces of silver in 16 ounces of the first ingot, in eight ounces of it there should be a number of ounces of silver expressed by 7x8 16 In like manner, 12 x 5 ,4x3 onri _ ~T6 ' ' 16 ' will express the quantity of silver contained in 5 ounces of the second ingot, and 3 ounces of the third. 156 E L K M E N T A K Y ALGEBRA. Now, we have, 7X8 12 X 5 4x3 128 16 16 16 16 therefore, a pound of the fourth ingot contains 8 oiui ;es of silver, as required by the enunciation. The same conditions may be verified with respect to the copper and pewter. 3. A>s age is double .Z?'s, and IPs is triple of <7's, and the sum of all their ages is 140 : what is the age of each? Ans. A's - 84; B's = 42; and C's = 14. 4. A person bought a chaise, horse, and harness, for 60 ; the horse came to twice the price of the harness, and the chaise to twice the cost of the horse and harness : what did he give for each? ( 13 6s. 8d. for the horse. Ans. i 6 135. 4d. for the harness. ( 40 for the chaise. 5. Divide the number 36 into three such parts that 1 of the first, i of the second, and 1 of the third, may be all equal to each other. Ans. 8, 12, and 16. 6. If A and B together can do a piece of work in 8 days, A and C together in 9 days, and B and C in ten days, how many days would it take each to perform the same work alone? Ans. A, 14|f ; B, I7ff; C, 23/ T . 7. Three persons. A, B, and (7, begin to play together, having among them all $600. At the end of the first game A has won one-half of _B's money, which, added to his own, makes double the amount B had at first. In the second game, A loses and B wins just as much as C had at the be- ginning, when A leaves off with exactly what he had at first : how much had each at the beginning ? Ans. A, $300 ; B, 200 ; C $100. 8. Three persons, A, B, and C, together possess $3640. PROBLEMS. 157 If B gives A $400 of his money, then A will have $320 more than B\ but if B takes 140 of C 's money, then B and C will have equal sums : how much has each ? Ans. A, $800 ; B, $1280; <7, $1560. 9. Three persons have a bill to pay, which neither alone is able to discharge. A says to B, " Give me the, 4th of your money, and then I can pay the bill." B says to (7, " Give me the 8th of yours, and I can pay it." But C says to A) " You must give me the half of yours before I can pay it, as I have but $8 " : what was the amount of their bill, and how much money had A and B ? . ( Amount of the bill, $13. S ' (A had $10, and .B $12. 10. A person possessed a certain capital, which he placed out at a certain interest. Another person, who possessed 10000 dollars more than the first, and who put out his capital 1 per cent, more advantageously, had an annual income greater by 800 dollars. A third person, who possessed 15000 dollars more than the first, putting out his capital 2 per cent, more advantageously, had an annual income greater by 1500 dollars. Required, the capitals of the three per- sons, and the rates of interest. . j Sums at interest, $30000, $40000, $45000. ' ( Rates of interest, 4 5 6 pr. ct. 11. A widow receives an estate of $15000 from her de- ceased husband, with directions to divide it among two sons and three daughters, so that each son may receive twice as much as each daughter, and she herself to receive $1000 more than all the children together : what was her share, and what the share of each child ? ( The widow's share, $8000 Ans. < Each son's, $2000 ' Each daughter's, $1000 158 ELEMENTARY ALGKBRA. 12. A certain sum of money is to be divided between three persons, A, B, and C. A is to receive $3000 les? than half of it, B $1000 less than one-third part, and C to receive $800 more than the fourth part of the whole : what is the sum to be divided, and what does each receive ? {Sum, $38400. A receives $16200. B " $11800. G " $10400. 13. A person has three horses, and a saddle which is worth $220. If the saddle be put on the back of the first horse, it will make his value equal to that of the second and third ; if it be put on the back of the second, it will make his value double that of the first and third ; if it be put on the back of the third, it will make his value triple that of the first and second : what is the value of each horse ? Ans. 1st, $20 ; 2d, $100 ; 3d, $140. 14. The crew of a ship consisted of her complement of sailors, and a number of soldiers. There w r ere 22 sailors to every three guns, and 10 over ; also, the whole number of hands was five times the number of soldiers and guns to- gether. But after an engagement, in which the slain were one-fourth of the survivors, there wanted 5 men to make 13 men to every two guns: required, the number of guns, soldiers and sailors. Ans. 90 guns, 55 soldiers, and 670 sailors. 15. Three persons have $96, which they wish to divide equally between them. In order to do this, A, who has the most, gives to B and C as much as they have already ; then B divides with A and C in the same manner, that is, by giving to each as much as he had after A had divided with them C then makes a division with A and J9, when it is PROBLK MS. 159 found that they all have equal sums: how ui.ich had each at first? Ans. 1st, $52; 2d, $28; 3d, $16. 16. Divide the number a into three such parts, that the first shall be to the second as m to n, and the second to the third as p to q. amp anp anq mp+np-rnq mp+np+nq mp+np+nq 17. Three masons, A, J3, and C", are to build a wall. A and B together can do it in 12 days ; B and C in 20 days ; and A and C in 15 days : in what time can each do it alone, and in what time can they all do it if they work together ? Ans. A, in 20 days ; J5, in 30 ; and C, in 60 ; all, in 10. 160 ELEMENTARY ALGEBRA. CHAPTER VI. FORMATION OF POWERS. 118. A POWER of a quantity is the product obtained by taking that quantity any number of times as a factor. If the quantity be taken once as a factor, we have the first power ; if taken twice, we have the second power ; if three tunes, the third power; if n tunes, the n th power, n being any whole number whatever. A power is indicated by means of the exponential sign thus, a = a 1 denotes first power of a.* axa = a? " square, or 2d power of a. a X a x a = a 3 " cube, or third power of a. axaxaxa = a* " fourth power of a. axaxaxaxa = a 5 " fifth power of a. axaxaxa = a m " m'* power of a. In every power there are three things to be considered : 1st. The quantity which enters as a factor, and which is called the first power. 2d. The small figure which is placed at the right, and a little above the letter, is called the exponent of the Since a = 1 (Art. 49), a 9 X a = 1 X a = a 1 ; so that the two factors of a 1 , are 1 and a. 118. What is a power of a quantity? What is the power when the quantity is taken once as a factor ? When taken twice ? Three times ? n times ? How is a power indicated ? In every power, how many things are considered ? Name them. POWERS OF MONOMIALS. 161 power, and shows how many times the letter enters as a factor. 3d. The power itself, which is the final product, or result of the multiplications. POWERS OF MONOMIALS. 1 19. Let it be required to raise the monomial 2a 3 > 2 to the fourth power. We have, (2a 3 S 2 ) 4 = 2a 3 5 2 x 2a 3 b z X 2a?I>* X 2a 3 & 2 , which merely expresses that the fourth power is equal to the product which arises from taking the quantity four times as a factor. By the rules for multiplication, this pro- duct is from which we see, 1st. That the coefficient 2 must be raised to the 4th power ; and, 2d. That the exponent of each letter must be multiplied by 4, the exponent of the power. As the same reasoning applies to every example, we have, for the raising of monomials to any power, the following RULE. I. JRaise the coefficient to the required power : IT. Multiply the exponent of each letter by the exponent of the power. EXAMPLES. 1. What is the square of 3o 2 y 3 ? Ans. Oa 4 ^ 6 . 119. What is the rule for raising a monomial to any power? When the monomial is positive, what will be the sign of its powers ? When negative, what powers will be plus? what minus? 162 ELEMENTARY ALGEBRA. 2. What is the cube of 6a 5 y-x ? Ans. 216a i yx 3 . 3. What is the fourth power of 2aV** ? 16a 12 y 12 20 . 4. What is the square of a?b s y 3 ? ^4.ns. a 4 5 10 y 6 . 5. What is the seventh power of a?bcd 3 ? Ans. 6. What is the sixth power of a 2 3 c 2 . 4. Find the fifth power of a b. Ans. a 5 5a 4 b + Wa 3 b 2 10a 2 ft 3 + 5ab* - b 5 . 5. Find the sixth power of a + b. Gab 5 + b s . 8. Find the sixth power of a b. a 6 6a 5 5 + 15a 4 5 2 20a 3 J 3 + 15aW - 6ab 5 127. When the terms of the binomial have coefficients, we may still write out any power of it by means of the Binomial Formula. X Let it be required to find the cube of 2c + 3d. (a + b) 3 a 3 + Sa z b + 3ab z + b 3 . POWERS OF BINOMIALS. 171 Here, 2c takes the place of a in the formula, and 3d the place of b. Hence, we have, . (I.) and by performing the indicated operations, we have, (2c + 3d) 3 = 8c 3 + 36c 9 -J If we examine the second member of Equation ( 1 ), we see that each term is made up of three factors: 1st, the numerical factor ; 2d, some power of 2c ; and 3d, some power of 3d. The powers of 2c are arranged in descend- ing order towards the right, the last term involving the power of 2c or 1 ; the powers of 3d are arranged in ascend- ing order from the first term, where the power enters, to the last term. The operation of raising a binomial involving coefficients, is most readily effected by writing the three factors of each term in a vertical column, and then performing the multipli- cations as indicated below. Find, by this method, the cube of 2c + 3d. OPERATION. 1+3 +3 +1 Coefficients. 8c 3 -f 4c 2 + 2c +1 Powers of 2c 1 + 3d + Qd 2 + 2ld 3 Powers of 3d (2c + d) 3 = 8c 3 + 3Qc z d + 54cc? 2 + 27d 3 The preceding operation hardly requires explanation. In the first line, write the numerical coefficients corresponding to the particular power ; in the second line, write the de- scending powers of the leading term to the power ; in the third line, write the ascending powers of the following term from the power upwards. It will be easiest to commence 172 K I. K M E N T A U Y ALGEBRA. the second line on the right hand. The multiplication should be performed from above, downwards. 8. Find the 4th power of 3a 2 < 2bd. (a + b)< = a 4 + 4 3 5 + 6 2 & 2 + 4a5 3 + b*. 1+4 +0 +4 '+ 1 81a 8 c 4 + 27a 6 c 3 + 9a*c 2 + 3a 2 c + 1 1 2bd + 4b*d 2 - 8b 3 d 3 + 16b*d . BlaV 9. What is the cube of 3x - Gy ? Ans. 27cc 3 162cc 2 y 10. What is the fourth power of a 35? Ans. a* I2a 3 b + 54a 2 i 2 108i 3 + Sib 4 . 11. What is the fifth power of c Ans. c 5 lOcV? + 40c 3 t? 2 8Qc"d 3 + BOcd* 32f7. 12. What is the cube of 5a 3d? Ans. I25a 3 225a?d + I35ad z - 2 * This Ingenious method of writing the development of a binomial Is due to fr-.ofewor WILLIAM G. PECK, of Columbia College. EXTRACTION OF ROOTS. 173 CHAPTER VH. SQUARE ROOT. RADICALS OF THE SECOND DEGREE. 12 . THE SQUARE ROOT of a number is one of its two equal factors. Thus, 6 X 6 = 36; therefore, 6 is the square root of 36. The symbol for the square root, is y' , or the fractional exponent \ ; thus, r * ya, or a , indicates the square root of a, or that one of the two equal factors of a is to be found. The operation of finding sucli factor is called, Extracting the Square Hoot. 129. Any number which can be resolved into two equal integral factors, is called a perfect square. The following Table, verified by actual multiplication, in- dicates all the perfect squares between 1 and 100. TABLE. 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, squares. 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, roots. 128. What is the square root of a number? Wha is the operation of finding the equal factor called ? 129. What is a perfect square ? How many perfect squares are there between 1 and 100, including both numbers ? What arc they? 174 ELEMENTARY ALGEBRA. "We may employ this table for finding the square root of any perfect square between 1 and 100. Look for the number in the first line / if it is found there, its square root will be found immediately under it. If the given number is less than 100, and not a perfect square, it will fall between two numbers of the ujyper line, and its square root will be found between the ttco numbers directly below ; the lesser of the two will be the entire part of the root, and will be the true root to within less than 1. Thus, if the given number is 55, it is found between the perfect squares 49 and 64, and its root is 7 and a decimal fraction. NOTE. There are ten perfect squares between 1 and 100, if -we include both numbers ; and eight, if we exclude both. If a number is greater than 100, its square root will be greater than 10, that is, it will contain tens and units. Let JV denote such a number, x the tens of its square root, and y the units ; then will, XT = (x + yY = a 2 + 2a:y + t/ 2 = 2 + (2x + y)y. That is, the number is equal to the square of the tens in its roots, plus twice the product of the tens by the units, plus the square of the units. EXAMPLE. 1. Extract the square root of 6084. Since this number is composed of more than two places of figures, its root will contain more GO 84 than one. But since it is less than 10000, which is the square of 100, the root will contain but two figures j that is, units and tens. SQUARE ROOT OF NUMBERS. 175 Now, the square of the tens must be found in the two left-hand figures, which we will separate from the other two by putting a point over the place of units, and a second over the place of hundreds. These parts, of two figures each, are called periods. The part 60 is comprised between the two squares 49 and 64, of which the roots are 7 and 8 ; hence, 7 expresses the number of tens sought / and the required root is composed of 7 tens and a certain number of units. The figure 7 being found, we write it on the right of the given 60 84 78 number, from which we separate 49 it by a vertical line : then we 7 X 2 = 14 8|118~4~ subtract its square, 49, from 60, 118 4 which leaves a remainder of 11, to which we biing down the two next figures, 84. The result of this operation, 1184, con- tains twice the product of the tens by the units, plus the square of the units. But since tens multiplied by units cannot give a product of a less unit than tens, it follows that the last figure, 4, can form no part of the double product of the tens by the units ; this double product is therefore found in the part 118, which we separate from the units' place, 4. Now if we double the tens, which gives 14, and then divide 118 by 14, the quotient 8 will express the units, or a number greater than the units. This quotient can never be too small, since the part 118 will be at least equal to twice the product of the tens by the units ; but it may be too large, for the 118, besides the double product of the tens by the units, may likewise contain tens arising from the square of the units. To ascertain if the quotient 8 expresses the right number of units, we write the 8 on the right of the 14, which gives 148, and then we multiply 148 by 8. This multiplication being effected, gives for a product, 1184, a 176 ELEMENTARY ALGEBRA. number equal to the result of the first operation. Hav- ing subtracted the product, we find the remainder equal to 0; hence, 78 is the root required. In this operation, we form, 1st, the square of the tens ; 2nd, the double product ot the tens by the units ; and 3d, the square of the units. Indeed, in the operations, we have merely subtracted from the given number GO 84 : 1st, the square of 7 tens, or of 70 ; 2d, t\vice the product of 70 by 8 ; and, 3d, the square of 8 ; that is, the three parts which enter into the composition of the square, 70 -f 8, or 78 ; and since the result of the sub- traction is 0, it follows that 78 is the square root of 6084. ISO. The operations in the last example have been per- formed on but two periods, but it is plain that the same methods of reasoning are equally applicable to larger num- bers, for by changing the order of. the units, AVC do not change the relation in which they stand to each other. Thus, in the number 60 84 95, the two periods 60 84, havo the same relation to each other as in the number 60 84 ; and hence the methods used in the last example are equally applicable to larger numbers. 131. Hence, for the ex-traction of the square root of numbers, we have the following E u L E . I. Point off the given number into periods of two figures each, beginning at tlie right hand: . H. Note the greatest perfect square in the first period on tJie left, andpkice its root on the right, after the manner of 131. Give the rule for the extraction of the square root of numbers? What is the first step ? What the second ? What the third ? What the fourth? What the filth? SQUARE KOOT OF NUMBERS. 177 a quotient in division y then subtract the square of this root from the first period, and bring down the second period for a remainder: III. Double the root already found, and place the result on the left for a divisor. Seek hoic many times the divisor is contained in the remainder, exclusive of the right-hand figure, and place the figure in the root and also at the right of the divisor : IV. Multiply the divisor, thus augmented, by the last figure of the root, and subtract the product from the re- mainder, and bring down the next period for a new remain- der. J3ut if any of the products should be greater than the remainder, diminish the last figure of the root by one : V. Double the ichole root already found, for a new di- visor, and continue the operation as before, until all the periods are brought down. 132. NOTE. 1. If, after all the periods are brought down, there is no remainder, the given number is a perfect square. 2. The number of places of figures iu the root will always be equal to the number of periods into which the given number is divided. 3. If the given number has not an exact root, there will be a remainder after all the periods are brought down, in which case ciphers may be annexed, forming new periods, for each of which there will be one decimal place in the root. 132. What takes place v.-hen the given number id a perfect square ! How many places of figures will there be in the root? If the given num- ber is not a perfect square, what may be done after all the periods are brought dowii ? 178 ELEMENTARY ALGEBKA EXAMPLES. 1. What is the square root of 36729 ? In this example there are two periods of decimals, and, hence, two places of decimals in the root. 3 67 29 191.64 + 1 29 261 381 3826 629 381 24800 22956 38324 31104 Rem. 2. To find the square root of 7225. Ans. 85. 3. To find the square root of 17689. Ans. 133. 4. To find the square root of 994009. Ans. 997. 5. To find the square root of 85673536. Ans. 9256. 6. To find the square root of 67798756. Ans. 8234. 7. To find the square root of 978121. Ans. 989. 8. To find the square root of 956484. Ans. 978. 9. What is the square root of 36372961 ? - Ans. 6031. 10. What is the square root of 22071204 ? Ans. 4698. 11. What is the square root of 106929? Ans. 327. 12. What of 12088868379025? Ans. 3476905. 13. What of 2268741 ? Ans. 1506.23 +. 14. What of 7596796? Ans. 2756.22 +, 15. What is the square root of 96 ? Ans. 9.79795 +. 16. What is the square root of 153? Ans. 12.36931 -f . 17. What is the square root of 101 ? Ans. 10.04987 -K SQUARE ROOT OF FRACTIONS. 179 18. What of 285970396644 ? Ans. 534762. 19. What of 41605800625 ? Ans. 203975. 20. What of 48303584206084? Ans. 6950078. EKTRACTION OF THE SQUARE ROOT OF FRACTIONS. 133. Since the square or second power of a fraction is obtained by squaring the numerator and denominator sepa- rately, it follows that The square root of a fraction will be equal to the square root of the numerator divided by the square root of tJie denominator. For example, the square root of 7-5- is equal to r: for, a a __ a^ V ^ T TO * b b b z 1. What is the square root of -? Ans. - 2. What is the square root of 16' Ans. - - 4 8. What is the square root of 81 8 Ans. - 4. What is the square root of 256 361 16 Ans. -. 5. 6. What is the What is the square root square root of of 64* 4096 Ans. I. 64 A division until the number of decimal places is double the number of places required in the root. TJien extract the root of the decimal by the last rule. EXAMPLES. 1. Extract the square of to within .001. This num- ber, reduced to decimals, is 0.785714 to within 0.000001 ; but the root of 0.785714 to the nearest unit, is .886; hence, 0.886 is the root of to within .001. 14 / 1 3 2. Find the \/ 2 ~ to within 0.0001. Ans. 1.6931 +. 3. What is the square root of ? Ans. 0.24253 +. 7 4. What is the square root of -? Ans. 0.93541 +. 8 g 5. What is the square root of - ? Ans. 1.29099 +. 3 EXTRACTION OF THE SQUARE ROOT OF MONOMIALS. 136. In order to discover the process for extracting the square root of a monomial, we must see how its square ia formed. By the rule for the multiplication of monomials (Art. 42), we have, x 5a 2 3 c = 25a 4 i 6 c 2 ; 135. What is a second method of finding the approximate root? 136. Give the rule for extracting the square root of monomials? 182 ELEMENTARY ALGEBRA. that is, in order to square a monomial, it is necessary to- square its coefficient and double the exponent of each oftht letters. Hence, to find the square root of a monomial, we have the following RULE. I. Extract the square root of the coefficient for a new coefficient : H. Divide tJie exponent of each letter by 2, and then annex all the letters with their new exponents. Since like signs in two factors give a plus sign in the pro- duct, the square of a, as well as that of -f a, will be + a 2 ; hence, the square root of a 2 is either + 01 a. Also, the square root of 25 2 5 4 , is either + 5aJ 2 , or 5ab z . Whence we conclude, that if a monomial is positive, its square root may be affected either with the sign + or ; thus, 1/9O 4 = 3a 2 ; for, + 3a 2 or 3a 2 , squared, gives -'- 9a 4 . The double sign , with which the root is affected, is read plus and minus. EXAMPLES. 1. What is the square root of 64a 6 & 4 ? and, -y/64a 6 6 4 = 8a 3 5 2 ; for 8a 3 5 2 x 8a 3 5 2 = Hence, -y/64a 6 6 4 = Sa 3 b*. 2. Find the square root of 625a 2 5 8 c 6 . 25a 4 c 3 . 3. Find the square root of 576a 4 5 6 c 8 . 24a 2 3 c 4 . 4. Find the square root of 196a; 6 y 2 s 4 . 14x 3 yz 2 . 5. Find the square root of 441a 8 6 c 10 J 16 . 21a 4 5 3 c 5 J 8 . 6. Find the square root of 784a 12 & 14 c 16 d' 2 . 2Sa e b'c e d. 7. Find the square root of 8la*b*c*. 9o 4 2 c 3 . IMPERFECT 8 Q \I A K E S . 1 83 NOTES. 137. 1. From the preceding rtuo it follows, that when a monomial is a perfect square, its numerical coefficient is a perfect square, arid all its exponents even numbers. Tims, 25 4 & 2 is a perfect square. 2. If the proposed monomial were negative, it would be impossible to extract its square root, since it has just been shown (Art. 136) that the square of every quantity, whether positive or negative, is essentially positive. Therefore, are algebraic symbols which indicate operations that cannot be performed. They are called imaginary quantities, or rather, imaginary expressions, and are frequently met with in the resolution of equations of the second degree. IMPERFECT SQUARES. 13. When the coefficient is not a perfect square, or when the exponent of any letter is uneven, the monomial is an imperfect square : thus, QSab* is an imperfect square. Its root is then indifcated by means of the 'adical sign ; thus. Such quantities are called, radical quantities, or radicals of the second degree : hence, A RADICAL QUANTITY, is the indicated root of an imperfect power. 137. When is a monomial a perfect square ? What monomials are these whose square roots cannot be extracted ? What are such expres- sions called? 138. When is a monomial an imperfect square ? What are such 'Jtios called ? What is A radical quantity ? 184 ELEMENTARY ALGEBRA. TRANSFORMATION OF RADICALS. 139. Let a and b denote any two numbers, and J5 the product of their square roots: then, V^ X V* = P CD Squaring both members, we have, a X b = p* .... (2.) Then, extracting the square root of both members of (2), i/ao = p (3.) And since the second members are the same in Equations ( 1 ) and ( 3 ), the first members are equal : that is, The square root of the product oftico quantities is equal to the product of their square roots. 140. Let a and b denote any two numbers, and q the quotient of their square roots ; then, ^ = q (1.) Squaring both members, we have, a . . = = q 2 (2.) o then extracting the square root of both members of (2), ..... and since the second members are the same in Equations ( 1 ) and (3 ), the first members are equal ; that is, 139. To what is the square root of the product of two quantities equal? 140. To what is the square root of the quotient of two quau tiling ual? equal? TRANSFORMATION CF RADICALS. 185 The, square root of the quotient of two quantities is equal to the quotient of their square roots. These principles enable us to transform radical expres- sions, or to reduce them to simpler forms ; thus, the expres- sion, 98&* = 495* X 2a ; hence, yOSab* = -v/49i 4 x 2a; and by the principle of (Art. 139), x 2a = In like manner, X 5bd = V/864a 2 5 c n = -v/144a 2 6 4 c 10 x Qbc 12ab*c 5 The COEFFICIENT of a radical is the quantity without the sign ; thus, in the expressions, the quantities 75 2 , 3abc, 12a5 2 c 5 , are coefficients of the radicals. 141. Hence, to simplify a radical of the second degree, we have the following RULE. I. Divide the expression under the radical sign into two factors, one of which shall be a perfect square : II. ^Extract the square root of the perfect square, and then multiply this root by the indicated square root of the, remaining factor. 141. Give the rule for simplifying radicals of the second degree. How do you determine whether a given number has a factor which is a perfect square ? 186 ELEMENTAKY A L G K B E A . -To determine if a given number has any factor which is a perfect square, we examine and see if it is divi- sible by either of the perfect squares, 4, 9, 1C, 25, 36, 49, 64, 81, &c. ; if it is not, we conclude that it does not contain a factor vhich is a perfect square. EXAMPLES. Reduce the following radicals to their simplest form : Ans. 5a-\/3abc. 2. Vl2Sb 5 a 6 d z . Ans. 3* / QO/797j8/ A ,vj o \t O C(, U C -/A/co. 4. i/256a 2 & 4 c 8 . Ans. 5. -v/1024a 9 6 7 c*. Ans. 32a l b 3 c 2 yabc. 6. 7. \/675a 7 5 5 c 2 , and 5c^/b are similar, and so also are 2, and 144. Radicals are added like other algebraic quantities ; hence, the following 143. \V h;;t are similar radicals of the second degree ? 144. Give the rule for the addition of radicals of the second degree ? 188 ELEMENTARY ALGEBKA BULE. L If the radicals are similar, add their coefficients, and to the sum annex the common radical : II. If the radicals are not similar, connect them together icith their proper signs. Thus, 3ay^ + Sc-y/J = (3a 4- In like manner, li/2a + 3-/2a = (7 + 3)-v/2 NOTES. 1. Two radicals, which do not appear to be sim- ilar at first sight, may become so by transformation (Art. 141.) For example, 2-V/45 + 3-v/5 = 6-v/S + 3-/5 = 9^5- 2. When the radicals are not similar, the addition or sub- traction can only be indicated. Thus, La order to add 3 -y/J to 5-v/a, we write, 5-v/a + 3-/J. Add together the following : 2. -/SOo^ 2 and yY2a 4 Z> 2 . 3. - - and 5 7^5 \/ V 15 4. -y/125 and v^OOo 2 . ^/w. (5 + 10a)-v/5 / V oo 10 294 11. SUBTRACTION OF RADICALS. 189 '9Sa z x and Ans. 1a^/2x + 6 i/x 2 a 2 . 12. -v243 and 10-/363. Ans. 13. - v /320o 2 i 2 and -\/245a 8 b 6 . Ans. (Sab + 7a*5 3 ) i/5. 14. y^oa 6 * 7 and -v/300a 6 6 5 . Ans. (5a 3 5 3 SUBTRACTION OF RADICALS. 145. Radicals are subtracted like other algebraic quan- tities ; hence, the following 'RULE. I. If the radicals are similar, subtract the coefficient of the subtrahend from that of the minuend, and to tJie differ- ence annex the common radical : II. If the radicals are not similar, indicate the operation by the minus sign. EXAMPLES. 1. What is the difference between 3a-\/b and a Here, 3a\/b a^fb = 1a-\/b. Ans. 14 K. Give the rule for the subtraction of radicals. 190 ELEMENTARY ALGEBRA. 2. From Oav/276 2 subtract 6ay275 2 . First, Qa-^27b 2 27a-y/3, and Ga-v/276 2 = 1 and, 27aJy / 3 l8ab\/3 9a6-v/3. ^4.ns. Find the differences between the following : 3. -1/75 and 4. and . (2ab 27 and j_ 45 3 6 3 and and 6. 7. 8. **\ and 10. -/3200 2 and -SOa 2 . Ans. . 4a-/5". 12. 13. -v/ll2a 8 6 6 and -v/28a 8 6 6 . MULTIPLICATION OF RADICALS. 146. Radicals are multiplied like other algebraic quan- tities ; hence, we have the following EULE. L Multiply the coefficients together for a new coefficient: 146. Give the rule for the multiplication of radicals. DIVISION OF KADICAL8-. 391 II. Multiply together the quantities under the radical signs: HI. Then reduce the result to its simplest form. 1. Multiply 3a\/bc by 2\/ab. 3a^/bc X 2<\/ab = 3a X 2 x -\/bc x ^/ab. which, by Art. 139, = 6a\/b 2 ac = dbi/ac. Multiply the following : 2. 3\/5ab and 4y^0a. Ans. 120v^- 3. 2a-[/bc aad Sa^bc. Ans. 6a 2 5c. 4. Za^/aT+W and 3a<^a?~+&. A. - 5. Zabi/a + 6 and 6. 3-V/2 and 2-v/8. ^ws. 24. 7. fy^^ and T S O 8. 2x + -\/b and 2 ^/b. Ans. 4cc 2 b. 9. 10. 3-/27a 3 by DIVISION OF EADICALS. 147. Radical quantities are divided like other algebraic quantities ; hence, we have the following RULE. I. Divide the coefficient of the dividend by the coefficient of the divisor, for a new coefficient : 147. Give the rule for the division of radicals. 192 ELEMENTARY ALGEBRA. II. Divide the quantities under the radicals, in the same manner : HI. Then reduce the result to its simplest form. 1. Divide EXAMPLES. by < ' ' = 2, new coefficient. Art. 140, hence, the quotient is 2 X - = - 2. Divide by 25-y/c. Ans. V/-- 3. Divide 12ac-y/6Jc by 4. Divide 6a\X%6* by 5. Divide 4a 2 -/506 5 by 6. Divide 26a 3 5-/81a 2 & 2 by 13a\/Qab. A. 1. Divide 84a 3 5 4 v^7ac by 42a5-v/3ff- ^L 8. Divide -v/i^ ^7 V^- 9. Divide 6a 2 J 2 -/20a 3 by 12-v/5. 10. Divide 6a-v/10i 2 by 3-v/sl 11. Divide 485* -/15 by 2J 2 y^ 12. Divide Sa 2 ^ 3 -/^ 3 by 2a-v/28d 13. Divide 96a 4 c 3 /98i 5 by 4a5 v / 3". a 3 i 2 . 360R . 2ai*c 3 J. . 14a 3 ic 2 . SQUARE BOOT OF POLYNOMIALS. 193 14. Divide 27a 5 6 6 -/2" 3 by ^Ja. Ans. 15. Divide 18a B b 6 ^/8a* by 6ab 2 , squaring 4 2 , and taking the sum, we find the polynomial 40a 2 2 24ab 3 + 16& 4 , which, subtracted from the first remainder, gives 0. There- fore, 5a 2 Sab + 4 2 is the required root. 2. Find the square root of a*+ 4a 3 +6a 2 a; 2 +4cfa^+ cc 4 * -4ns. 2 + 2a + a 2 . 3. Find the square root of a 4 4a 3 a+6a 2 cc 2 4a# 3 + JB*. Ans. a 2 2ax + 2 . 4. Find the square root of 4cc 6 + 12x 5 + So 4 - 2x 3 + Va: 2 2aj + 1. Ans. 2x? + 3a; 2 x + 1. 5. Find the square root of 3a 2 2ab + 4i 2 . 6. What is the square root of a 4 4ax 3 + 4a 2 a 2 4a 2 + 8aa; + 4 ? -4;i5. a 2 2aa 2. 7. What is the square root of 9 2 12aj + 6xy + y z 4y + 4 ? -4ws. 3a; + y 2. 8. What is the square root of y 4 2y 2 a; 2 + 2x 2 2y 2 1 + a* ? ^tn5. y 2 & 1. 9. What is the square root of 9a 4 4 30a s 3 + 25a 2 5 2 ? Ans. 3a 2 b 2 Sab. 10. Find the square root of 48a5 2 c 3 f 365 2 c* - 36a 2 5c 3 + 9a*c 2 . Ans. 5a?b - 3 2 c 4abc -f 196 ELEMENTARY ALGEBKA. I5O. Wo will conclude this subject with the following remarks : 1st. A binomial can never be a perfect square, since we know that the square of the most simple polynomial, viz., a binomial, contains three distinct parts, which cannot ex- perience any reduction amongst themselves. Thus, the expression a 2 -f b z , is not a perfect square ; it wants the term 2a, in order that it should be the square of a b. 2d. In order that a trinomial, when arranged, may be a perfect square, its two extreme -terms must be squares, and the middle term must be the double product of the square roots of the two others. Therefore, to obtain the square root of a trinomial when it is a perfect square : Extract the roots of the two extreme terms, and give these roots the same or contrary signs, according as the middle term is positive or negative. To verify it, see if the double product of the two roots is the same as the middle term of the trinomial. Thus, 9 6 48a 4 2 + 64a 2 J 4 , is a perfect square, since, y^ 6 = 3c/ 3 , and -/64 2 i 4 Sab 2 ; and also, 2 x 3a 3 X Sab 2 = 4Sa*b 2 = the middle term. But, 4a 2 4- 14a# + 95 2 is not a perfect square : for, although 4a 2 and + 9b z are the squares of 2a and 3#, yet 2 x 2a X 35 is not equal to I4ab. 3d. In the series of operations required by the general rule, when the first term of one of the remainders is not exactly divisible by twice the first term of the root, we may 150. Can a binomial ever be a perfect power? Why not? When is a trinomial a perfect square ? When, in extracting the square root, we find that the first term of the remainder is not divisible by twice the root, is the polynomial a perfect power or not? SQUARE ROOT OK PCLYNOMIAI, S. 197 conclude that the proposed polynomial is not a perfect square. This is an evident consequence of the course of reasoning by which we have arrived at the general rule for extracting the square root. 4th. When the polynomial is not a perfect square, it may sometimes be simplified (See Art. 139). Take, for example, the expression, <^d*b + 4a 2 5 2 -f- 4ab 3 . The quantity under the radical is not a perfect square ; but it can be put under the form ab(a? + 4a5 + 4& 2 .) Now, the factor within the parenthesis is evidently the square of a -f 25, whence, we may conclude that, + 4a 2 5 2 + 4a6 3 = (a + 25) 2. Reduce -y/2a 2 6 4a5 2 + 25 3 to its simplest form. Ans. (a b) 1#S ELEMENTARY ALGEBRA CHAPTER VILL EQUATIONS OF THE SECOND DEGREE. EQUATIONS CONTAINING ONE UNKNOWN QUANTITY. 151. AN EQUATION of the second degree containing bu^ one unknown quantity, is one in which the greatest exponent: is equal to 2. Thus, a 2 = a, ax 2 + bx = c, are equations of the second degree. 152. Let us see to what form every equation of the second degree may be reduced. Take any equation of the second degree, as, (1 +)>-?*- 10 = 5 -? + - . Clearing of fractions, and performing indicated operations, we have, 4 + 8x + 4 2 3x 40 = 20 + 2x\ Transposing the unknown terms to the first member, the known terms to the second, and arranging with reference to the powers of #, we have, 4a 2 2a; 2 + 8x 3x + x = 20 + 40 4 ; 151. What is an equation of the second degree ? Give an example. 152. To what form may every equation of the second degree be reduced? EQUATIONS OF THE SECOND DEGREE. and, by reducing, 2x 2 + 6x = 56 ; dividing by the coefficient of # 2 , we have, x 2 + 3x = 28. If we denote the coefficient of x by 2p, and the second member by q, we have, x z + 2px = q. This is called the reduced equation. 153. When the reduced equation is of this form, it con- tains three terms, and is called a complete equation. The terms are, FIRST TERM. The second power of the unknown quan- tity, with a plus sign. SECOND TERM. The first power of the unknown quantity, Math a coefficient. THIRD TERM. A known term, in the second member. Every equation of the second degree may be reduced to this form, by the following BULK. I. Clear the equation of fractions, and perform all the indicated operations : n. Transpose all the unknown terms to the first member, and all the known terms to the second member : 153. How many terms are there in a complete equation ? What is the first term ? "What is the second term ? What is the third term ? How many operations are there in reducing an equation of the second degree to the required form ? What is the first ? What the second ? What the third ? What the fourth ? 200 ELEMENTAKY ALGEBKA. TTT. Reduce all the terms containing the square of the unknown quantity to a single term, one factor of which is the square of the unknown quantity ; reduce, also, all thd terms containing the first power of the unknown quantity, to a single term : IV. Divide both members of the resulting equation by the coefficient of the square of tJie unknown quantity. 154. A ROOT of an equation is such a value of the un- known quantity as, being substituted for it, will satisfy the equation ; that is, make the two members equal. The SOLUTION of an equation is the operation of finding its roots. INCOMPLETE EQUATIONS. 155. It may happen, that 2/>, the coefficient of the first power of x, in the equation x + 2px = q, is equal to 0. In this case, the first power of x will disappear, and the equation will take the form, *=! (1.) This is called an incomplete equation ; hence, AN INCOMPLETE EQUATION, when reduced, contains but two terms; the square of the unknown quantity, and a known term. 156. Extracting the square root of both members of Equation ( 1 ), we have, x = i/q. 154. What is the root of an equation ? What is the solution of an equation ? 155. What form will the reduced equation take when the coefficient ol a: is ? What is the equation then called ? How many terms are there iu an incomplete equation ? What are they ? 156. What is the rule for the solution of an incomplete equation T How many roots are there in every incomplete equation ? How do tb roots compare with each other ? EQUATIONS OF THE SECOND DEGREE. 201 Hence, for the solution of incomplete equations : RULE. I. Reduce the equation to the form x 2 = q: II. Then extract the square root of both members. NOTE. There will be two roots, numerically equal, but having contrary signs. Denoting the first by a;', and the second by ", we have, x' +i/q, and x" = * + 8x = 64 ? If we first divide by the coefficient 2, we obtain 2 -f- 4x = 32. and q in the first form? What in the second? What in the third? WViat in the fourth ? 212 ELEMENTARY ALGEBKA. Then, completing the square, x 2 + 4x + 4 = 32 + 4 = 36. Extracting the root, x + 2 = -v/36 = + 6, and 6. Hence, x' = 2 + 6 = +4; and, x" 2 6 = 8. Hence, in this form, the smaller root, numerically, is positive, and the larger negative. VERIFICATION. If we take the positive value, viz. : x f = + 4, the equation, a; 2 + 4& = 32, gives 4 2 + 4 x 4 = 32 ; and if we take the negative value of , viz. :,"=. 8, the equation, a 2 + 4x = 32, gives ( 8) 2 + 4( 8) = 64 - 32 = 32 ; from which we see that either of the values of x, viz.: x' = + 4, or x" 8, will satisfy the equation. 2. What are the values of x in the equation, 3 2 + 12aj 19 = a 2 12 + 89? By transposing the terms, we have, 3a 2 + y? + 12cc + I2x = 89 + 19 ; and by reducing, 4z 2 + 24o; = 108; and dividing by the coefficient of jc 2 , a; 2 + Qx = 27. C O M P L K T K EQUATIONS. 213 Now, by completing the square, x 2 + 6x + 9 = 36 ; extracting the square root, x + 3 = -/36 = + 6, and - 6; hence, x r =+6 3 = -f 3; and, x" 6 3 = 9. VERIFICATION. If we take the plus root, the equation, cc 2 -f 6z = 27, gives (3) 2 + 6(3) = 27; and for the negative root, x z + Qx = 27, gives (- 9) 2 + 6( 9) = 81 54 = 27. 3. What are the values of a; in the equation, x z - lOaj + 15 = ^ - 34a; + 155 ? O By cleaiing of fractions, we have, 5x z 50x +75 = x z I70a + 775; by transposing and reducing, we obtain, 4x z + 12Cte = 700 ; then, dividing by the coefficient of 2 , we have, x z + 30x = 175; and bj completuig the square, y? + 30a -f- 225 = 400 ; 214 ELEMENTARY ALOEBKA. and by extracting the square root, x + 15 = -y / 40() = + 20, and 20. Hence, x f = + 5, and x" = 35. VERIFICATION. For the plus value of #, the equation, x 2 + 30x = 175, gives, (5) 2 -f 30 X 5 = 25 + 150 = 175. And for the negative value of C, we have, (_ 35)2 _j_ 30 (_ 35) _ 12 25 1050 175. 4. What are the values of x in the equation, s*-3+! = '-! *+T?' Clearing of fractions, we have, 10a 2 - 6x + 9 = 96 Sx I2x z + 273; transposing and reducing, 22a 2 + 2x = 360 ; dividing both members by 22, 2 360 C 2 H X = r 22 22 (1 \ 2 I to both members, and the equation becomes, 2 / 1 \ 2 360 / 1 \ 2 />* _l _,__/> J, | __ _ I ^22* h \22/ ' 22 h l22/' whence, by extracting the square root, - J_ /360 / 1 y h 22 Z : V 22 " h \22/' COMPLETE EQUATIONS. 215 therefore, / 1 \ 2 \22/' and, " = __ It remains to perform the numerical operations. In the first place, 860/1 \ 2 22 " h \22/' must be reduced to a single number, having (22) 2 for its denominator. Now, 360 /J_\ 2 _ "22" \22/ : 360 X 22 + 1 7921 22 \22/ (22) 2 " (22) 2 ' extracting the square root of 7921, we find it to be 89; therefore, 89 22 \22 Consequently, the plus value of x is, -1 L o 2 + 22 ~ 22 ~~ ' and the negative value is, 1 _ 89 45 22 ~ 22 ~ ~ TI that is, one of the two values of x which will satisfy the proposed equation is a positive whole number, and the other a negative fraction. NOTE. Let the pupil be exercised in writing the roots, in Jie last five, and in the following examples, without com- pleting the square. 216 ELEMENTARY ALQEBKA. 6. What are the values of x in the equation, SaJ 2 + 2x 9 = 76 ? x' = 5. 6. What are the values of a; in the equation, K/v. /vt2 2x 2 + 8a; + 7 = - ^ + 197 ? 4 8 ' = 8. 7. What are the values of x in the equation, =: 9. ' ( x" 64}. 8. What are the values of a; in the equation, x 2 5x x T 4~ " 2 ~ ' *' ' = 2. 9. What are the values of a; in the equation, 9? . x x z x 13 EXAMPLES OF THE SECOND FORM. 1. What are the values of x in the equation, a- 2 Sx -f- 10 = 19? COMPLETE EQUATIONS. 217 By transposing, 2 - Sx = 19 10 = 9; then, by completing the square, a; 2 - 8x + 16 = 9 + 16 = 25 ; and by extracting the root, x 4 = i/25 = + 5, or 5. Hence, x' = 4 + 5 9, and x" = 4 5 =. 1. That is, in this form, the larger root, numerically, u positive, and the lesser negative. VERIFICATION. If we take the positive value of x, the equation, a 2 8x = 9, gives (9) 2 8x9 = 81 72 = 9; and if we take the negative value, the equation, jc 2 - 8a; = 9, gives (- I) 2 8 ( 1) = 1 + 8 = 9; from which we see that both roots alike satisfy the equa- tion. 2. What are the values of x in the equation, By cleaiing of fractions, we have, 6* 2 + 4x 180 = 3x z + 12* 177 , and by transposing and reducing, 3x 2 8x = 3 ; and dividing by the coefficient of a; 2 , we obtain, *-!* = ' 218 ELEMENTARY ALGEBRA. Then, by completing the square, we have, ^LV+21 h^--- Vit **J ^ .1 1 39 99 and by extracting the square root, 4 /25 5 Hence, = 3 + 3 = 3) "3 3 ~ S VEKIFICATION. For the positive root of a;, the equation, Q gives 3 2 -- x3 = 9 8 = 1; 3 and for the negative root, the equation, x z - ~x = 1, 1\ 2 8 1 1,8 3J - 3 X ~ 3 = 9 + 9 = l 3. What are the values of x in the equation, Clearing of fractions, and dividing by the coefficient of x 2 , we have, *- = n- COMPLETE EQUATIONS. 219 Completing the square, we have, 2 - - + - I " 1?. 3 9 ~ 9 ~~ 36 ' then, by extracting the square root, we have, 1 /49 7 7 3~~ V 36 ~ 6' 6 J hence, ,_17 9 _ \ 7 5 O O O O O O VERIFICATION. If we take the positive root of JB, the equation, x ^jc =. 1 4 , 2 gives (H) 2 X 1 = 2J 1 = Ij.; O and for the negative root, the equation, 5 2 2 5 25 10 45 4. What are the values of a; in the equation, 4a 2 2a 2 + 2craj = I8at> By transposing, changing the signs, and dividing tfj- 2, the equation becomes, a 2 ax = 2a 2 9a& + 95 2 ; 220 ELEMENTARY ALGEBRA. whence, completing the square, - ** + 7 = -T ~ extracting the square root, /y . VJ Now, the square root of Qab + 95'^, is evidently i 35. Therefore, s' = 2a 3b. f j'-- a + 3b. What will be the numerical values of cc, if we suppose a = 6, and 5 = 1? 5. What are the values of a: in the equation, 4 x z 45 5 within (x' = 7.12 Ho withi AnS '\x= -5.73f 0.01. 6. What are the values of x in the equation, 8x z Ux +10 = 22 CI JL'T* - O [^ t7*O ^^ ^^ f*^ """"" 1 "0" 5 45 4 Ans. 10. What are the values of x in the equation, x - y? 3 Qx + 1 ? j x' = 1. AnS ' \ x" = - 4! 11. What are the values of * in the equation, x* + 4a; - 90 = 93 ? EXAMPLES OF THE FOURTH FOKM. 1. What are the values of x in the equation, x 2 8a; = 7 ? 224: ELEMENTARY ALGEBRA. By completing the square,' we have, x 2 - Sx + 16 = 7 + 16 = 0; then, by extracting the square root, x 4 = i/9 = + 3, and 3 ; hence, x' = + 7, and x" = + ! That is, in this form, both the roots are positive. VERIFICATION. If we take the greater root, the equation, a; 2 8a; = 7, gives, 7 2 8x7 = 49 56 : z 7 ; and for the lesser, the equation, x* - Sx - - 7, gives, I 2 -8x1 = 1 8= -7; hence, both of the roots wih 1 satisfy the equation. 2. What are the values of x in the equation, 40 lia; 2 + 3x - 10 = Ha 2 - IBx + ? 2 By clearing of fractions, we have, So; 2 + Qx 20 Sx 2 - 36z + 40 ; then, by collecting the similar terms, _ 6cc 2 + 42a; = 60 ; then, by dividing by the coefficient of cc 2 , which is 0, we have, a: 2 - Vaj = - 10. By completing the square, we have, x 7 ' 7z + 12.25 =- 2.25, COMPLETE EQUATIONS. 225 and by extracting the square root of both members, x 3.5 = y^25 = + 1.5, and 1.5; hence, x' = 3.5 + 1.5 = 5, and x" = 3.5 - 1.5 = 2. VERIFICATION. If we take the greater root, the equation. x z fce = - 10, gives, 5 2 - 7 X 5 = 25 35 - 10; and if we take the lesser root, the equation, x 2 7x = 10, gives, 2 2 7 X 2 = 4 '14 = - 10. 3. What are the values of x in the equation, - Bx + 2a 2 + 1 = 17fe - 2z 2 - 3 ? By transposing and collecting the terms, we have, 4 2 20|a; = 4 ; then dividing by the coefficient of * 2 , we have, C 2 - 5}X = - I. By completing the square, we obtain, , 169 169 144 -i4,+ _ == -i + - 5 - = ; and by extracting the root, hence, 12 12 1 * = 2f -f = 5, and, x" = 2J - -- = -. VERIFICATION. If we take the greater root, the equation, a; 2 5$x = 1, gives, 5 2 5 x 5 = 25 26 = 226 ELEMENTARY ALGEBRA. and if we take the lesser root, the equation, * -5lx=- 1, gives, Q) 2 - 6 jxJ=^-! = - 4. What are the values of x in the equation, fa' = 3. 5. What are the values of a in the equation, 4a 2 x + 1 = 5a 2 + 8a? 6. What are the values of a in the equation, 7. What are the values of a in the equation, a 2 10 T yc = 1 ? (a' = AllS - \ v n _ I a = 8. What are ;he values of a in the equation, = 10. 2a 2 27a ^ + 100 = - + 12a 26 ? 5 5 f = ' Ans. i ( x" = 6. 9. What are the values of a in the equation, 22a -f 15 = - -f 28a 30 ? 8 3 x' = 9. PROPERTIES OF EQUATIONS. 227 10. What are the values of x in the equation, 2 2 -30a;-|-3 = - 2 -f- 3-frx ^- ? f 11 fc J. A PROPERTIES OF EQUATIONS OF THE SECOlfD DEGREE. FIRST PROPERTY. 161. We have seen (Art. 153), that every complete equation of the second degree may be reduced to the form, 2 + 2px = q ..... ( 1.) Completing the square, we have, x 2 + 2px + p 2 q -f p z ; transposing q -f p 2 to the first member, tc 2 + 2px + p* (q + j(? 2 ) = 0. . ( 2.) Now, since as 2 + 2px + ^> 2 is the square of x + ^?, and q + j9 2 the square of ^/g~+ p z , we may regard the first member as the difference between two squares. Factoring, (Art. 56), we have, (x+p + V 3 + P z ) (x+p V<1 + P*) = 0. . ( 3.) This equation can be satisfied only in two ways : 1st. By attributing such a value to x as shall render the first factor equal to ; or, 161. To what form may every equation of the second degree be re- duced? What form will this equation take after completing the square and transposing to the first member ? After factoring ? In how many wajc may Equation ( 3 ) be satisfied ? What are they ? How many roots has every equation of the s^ond degree? 228 ELEMENTARY ALGEBRA. 2d. By attributing such a value to x as shall raider ih second factor equal to 0. Placing the second factor equal to 0, we have, x+p -\/q+p z = 0; and x' = p+ \/Q +P 2 - ( 4 -) Placing the first factor equal to 0, we have, " = p Since every supposition that will satisfy Equation ( 3 ), also satisfy Equation ( 1 ), from which it was derived, it fol- lows, that x' and x" are roots of Equation ( 1 ) ; also, that Every equation of the second degree has two roots, and only two. NOTE. The two roots denoted by x' and ", are the same as found in Art. 158. SECOND PROPERTY. 162. We have seen (Art. 161), that every equation ol the second degree may be placed under the form, (x + p + Vg + P Z ) ( x + P V 234 ELEMENTARY ALGEBRA. EXAMPLES. 1. What are the values of jc in the equation, x 6 + Qx 3 = 112? Completing the square, x 6 + Qx 3 + 9 = 112 + 9 = 121 ; then, extracting the square root of both members, x 3 + 3 = -/121 = 11 J hence, x' = *f 3 + 11, and x" = y 3 11 ; hence, x' = ^/8 = 2, and x" = */ 14 = -- 2. What are the values of a; in the equation, x* 8x z = 9 ? Completing the square, we have, x* 8x 2 + 16 = 9 + 16 = 25. Extracting the square root of both members, x* 4 = -y/25 = 5 ; hence, x' = -y/4 + 5, and cc" = -y/4 5 ; hence, = + 3 and 3 ; and a" = + -/ 1 and -/ 3. What afe the values of x in the equation, SB 6 + 20Z 3 = 69 ? Completing the square, a; 6 + 20a: 3 + 100 = 69 + 100 = 169. Extracting the square root of both members, a 3 + 10 = yl69 = 13 ; hence, x' = \/- 10 + 13, and x" = 3 ^/ 10 - 13. x' = 3/3, and a" = ^/- 2o. TRINOMIAL, EQUATIONS. 235 4. Wha: are the values of x in the equation, * 2x z = 3 ? Ans. x' = y's, and a" = y' i. 5. What are the values of x in the equation, x* + 8a; 3 9 ? . x' = 1, and SB" = \~ $ 6. Given x + -/Gee + 4 = 12, to find cc. Transposing cc to the second member, and then squaring, Qx + 4 = cc 2 24a; + 144 ; .-. a; 2 33a; = 140; and, x' = 28, and x" = 5. 7. 4a; -f 4 + 2 = 7. ^ln. SB' = 4J-, x" 8. a; + y5x +10 8. Ans. x' = 18, ce" = 3. NUMERICAL VALUES OF THE ROOTS. 166. We have seen (Art. 160), that by attributing all possible signs to 2p and p z , the quantity under the radical sign will become negative ; hence, its square root cannot be extracted (Art. 137). Under this supposition, the values of x are imaginary. How are these results to be interpreted ? If a given number be divided into two parts, their pro- duct will be the greatest possible, when the parts are equal. Denote the number by 2p, and the difference of the parts by d', then, p + - == the greater part, (Page 120.) 2 and, and, p- P 2 - 2 d z 4 the less part, P, their product. 171. If we make q = 0, to what does the first form reduce? What, then, are its roots ? Under the same supposition, to what does the second form reduce ? What are, then, its roots? 1*72. If q > p a , in the third and fourth forms, what takes place? If a number be divided into two parts, when will the product be the greatest possible ? 238 ELEMENTARY ALGEBRA It is plain, that the product P will increase, as d dimin- ishes, and that it Avill be the greatest possible when d = ; for then there will be no negative quantity to be subtracted from p 2 , in the first member of the equation. But when d = 0, the parts are equal ; hence, the product of the two parts is the greatest when they are equal. In the equations, x 2 + 2px = <7, x 2 2px = q, 2p is the sum of the roots, and q their product; and hence, by the principle just established, the product q, can never be greater than p 7 . This condition fixes a limit to the value of q. If, then, we make q > ^> 2 , we pass thie limit, and express, by the equation, a condition which cannot be fulfilled ; and this incompatibility of the conditions is made apparent by the values of x becoming imaginary. Hence, we conclude that, When the values of the unknown quantity are imaginary, the conditions of the proposition are incompatible with each other. EXAMPLES. 1. Find two numbers, whose sum shall be 12 and pro duct 46. Let x and y be the numbers. . By the 1st condition, x + y = 12 ; and by the 2d, xy = 46. The first eqiiation gives, x = 12 y. Substituting this value for x in the second, we have, 12y - y 2 = 46 ; and changing the signs of the terms, we have, _ 12 = - 46 NUMERICAL VALUE OF THE KOOTS. 239 Then, by completing the square, 2/ 3 12y + 36 = 46 + 36 = 10; which gives, y' = 6 + ^/ 10, and, y" = 6 / 10 ; both of which values are imaginary, as indeed they should be, since the conditions are incompatible. 2. The sum of two numbers is 8, and their product 20: what are the numbers ? Denote the numbers by x and y. By the first condition, x + y = 8; and by the second, xy = 20. The first equation gives, x = 8 y. Substituting this value of x in the second, we have, 8y - y 2 = 20 ; changing the signs, and completing the square, we have, 2/ 2 - 8y + 16 = - 4 ; and by extracting the root, y' = 4 + >/ 4, and y" = 4 y' 4. These values of y may be put under the forms (Art. 142), y = 4 -f- 2-v/ ^j and y = 4 2^ 1. 3 What are the values of x in the equation, x* 2x = 10? ( 85' rr 1+3 15, and RCl ~ 2 x' = oU x" = 2 - 12. NOTES. 1. The value, x' = 15, satisfies the c nunciation in its arithmetical sense. For, if 15 yards cost 240 cents, 11 242 ELEMENTARY ALGEBKA. 240 -4-15 = 16 cents, the price of 1 yard ; and 240 -f- 12 = 20 cents, the price of 1 yard under the second supposition. 2. The second value of x is an answer to the following Problem : A certain person purchased a number of yards of cloth for 240 cents. If he had paid the same for three yards more, it would have cost him 4 cents less per yard : how many yards did he buy ? This would give the equation of condition, 240 240 -i^'iTi = 4; or ' x z 3x = 180; the same equation as found before ; hence, A single equation will often state two or more arith- metical problems. This arises from the fact that the language of Algebra is more comprehensive than that of Arithmetic. 3. A man having bought a horse, sold it for $24. At the sale he lost as much per cent, on the price of the horse, as the horse cost him dollars : what did he pay for the horse ? Let x denote the number of dollars that he paid for the horse. Then, x 24 will denote the loss he sustained. But C as he lost x per cent, by the sale, he must have lost upon each dollar, and upon x dollars he lost a sum denoted x z b\ : we have, then, the equation, * 100 - = x 24; whence, a 3 lOOa z= 2400, 100 PROBLEMS. il4:i) and, x = 50 -v/2500 2400 = 50 10. Therefore, x' = 60, and " = 40. Both of these roots will satisfy the problem. For, if the man gave $60 for the horse, and sold him for $24, he lost $36. From the enunciation, he should have lost, 60 per cent, of $60 ; that is, 60 60 X 60 100 f6 -loo" >6 ' therefore, $60 satisfies the enunciation. Had he paid $40 for the horse, he would have lost by the sale, $16. From the enunciation, he should have lost 40 per cent, of $40 ; that is, 40 40 X 40 of 40 = = 16 : 100 100 therefore, $40 satisfies the enunciation. 4. The sum of two numbers is 11, and the sum of their squares is 61 : what are the numbers? Ans, 5 and 6. 5. The difference of two numbers is 3, and the sum of their squares is 89 : what are the numbers ? Ans. 5 and 8. 6. A grazier bought as many sheep as cost him 60, and after reserving fifteen out of the number, he sold the re- mainder for 54, and gained 25. a head on those he sold : how many did he buy ? Ans. 75. 7. A merchant bought cloth, for which he paid 33 15s., which he sold again at 2 8s. per piece, and gamed by the bargain as much as one piece cost him : how many pieces did he buy? Ans. 15. 8. The difference of two numbers is 9, and their sum, multiplied by the greater, is equal to 266 : what are the numbers? Ans. 14 and 5. 24-i ELEMENTARY A L G E B K A . 9. To find a number, such that if you subtract it from 10, and multiply the remainder by the number itself, the pro- duct will be 21. Ans. 7 or 3. 10. A person traveled 105 miles. If he had traveled 2 miles an hour slower, he would have been 6 hours longer in completing the same distance : how many miles did he travel per hour ? Ans. 7 miles. 11. A person purchased a number of sheep, for which he paid $224. Had he paid for each twice as much, plus 2 dol- lars, the number bought would have been denoted by twice what was paid for each : how many sheep were purchased ? Ans. 32. 12. The difference of two numbers is 7, and their sum multiplied by the greater, is equal to 130: what are the numbers? Ans. 10 and 3. 13. Divide 100 into two such parts, that the sum of their squares shall be 5392. Ans. 64 and 38. 14. Two square courts are paved with stones a foot square ; the larger court is 12 feet larger than the smaller one, and the number of stones in both pavements is 2120 : how long is the smaller pavement? Ans. 26 feet. 15. Two hundred and forty dollars are equally distributed among a certain number of persons. The same sum is again distributed amongst a number greater by 4. In the latter case each receives 10 dollars less than in the former: how many persons were there in each case. Ans. 8 and 12. 16. Two partners, A and _B, gained 360 dollars. A 1 a money was in trade 12 months, and he received, for prin- cipal and profit, 520 dollars. JPs money was 600 dollars, nnd was in trade 16 months: how much capital had A ? Ans. 400 dollars. MOKE THAN UNK UNKNOWN QUANTITY. 2-i; r > EQUATIONS INVOLVING ilOKE THAN ONE UNKNOWN QUANTITY. 173. Two simultaneous equations, each of the second degree, and containing two unknown quantities, will, when combined, generally give rise to an equation of the fourth degree. Hence, only particular cases of such equations can be solved by the methods already given. FIRST. Two simultaneous equations, involving two unknown quantities, can always be solved when one is of the first and the other of the second degree. EXAMPLES. 1. Given , , , ^ r to find x and y. ( x z + y = 100 ) By transposing y in the first equation, we have, x = 14 - y; and by squaring both members, x 2 = 190 28y + y 2 . Substituting this value for x 2 in the second equation, we have, 196 28y + y 2 + y 2 = 100 ; from Avhich we have, By completing the square, y 2 - 14y + 49 = 1 ; 173. When may two simultaneous equations of the second degree be solved ? 246 ELEMENTARY ALGEBRA. and by extracting the square root, y 7 -v/1 = + 1> and 1 ; hence, y' 7 + 1 = 8, and y" 1 . 1 = 6. If we -take the greater value, we find x = 6 ; and if we fake the lesser, we find x = 8. ( v' s " fi j 1 '. U *o U* AnS '\y>= 6, y"^ 8 VERIFICATION. For the greater value, y = 8, the equation, a; + y = 14, gives 6 + 8 14 ; and, a 2 + y 2 = 100, gives 36 + 64 = 100. For the value y 6, the equation, x + y = 14, gives 8 + 6 = 14 ; and, a 8 + y 2 = 100, gives 64 + 36 = 100. Hence, both sets of values satisfy the given equation. 2. Given j [ to find x and y. ( x z y 2 = 45 ) Transposing y in the first equation, we have, x = 3 + y; then, squaring both members, a; 2 = 9 + 6y + y 2 . Substituting this value for a; 2 , in the second equation, we have, 9 + 6y + y 2 y 2 = 45 ; whence, we have, 6y = 36, and y = 6. SIMULTANEOUS EQUATIONS. 247 Substituting this value of y, in the first equation, we have, x 6 = 3, and, consequently, x' = 3 + 6 = 9. VERIFICATION. x y = 3, gives 9 6 = 8 ; and, x 2 y 2 = 45, gives 81 36 = 45. Solve the following simultaneous equations : + y = 12 r ''= 7. g ' = 5. ix y = 3) . ( a;' = 9, x"= 6. 4. J 2,0 , , H J- -A7W. -I , ( a; 2 + y 2 =z 117 f ( y' = 6, y"= 9. r /> 4. / Q \ //>' *j " ; _4^-ry y A i x o, o. 5. 4 , , , L -4^5. 4 . .. ( a;- 2xy -\- y 2 = 1 $ ) y = 4, y == 4. raj-y = 5 i ( a$ 2 + 2a?y + y 2 = 225 [ yf = 10, x"= - 5. y' = 5, y"= - 10. SECOND. 174. Two simultaneous equations of the second degree, which are homogeneous with respect to the unknown quan- tity r , can always be solved. EXAMPLES. r . j a? + Sxy = 22 (1.) ' (a?+ 3a:y + 2y 2 = 40 (2.) to find x and y. 174. When may two simultaneous equations of the second degree be solved ? 2i8 ELEMENTARY ALGEBRA. Assume x = ty, t being any auxiliary unknown quantity. Substituting this value of x in Equations ( 1 ) and ( 2 ), we have, 40 y+ 3ty*+w = 40, .-. 7/2 = 2 ; (*0 22 40 hence, t 2 + 3t t* + 3t + 2 ' hence, 22 2 + 66J + 44 = 40f 2 + 120; 22 reducing, <2 _j_ 3^ _ . y 2 11 whence, ' = -, and t"= Substituting either of these values in Equations ( 3 ) or ( 4 ), we find, y' +3, and y" 3 Substituting the plus value of y, in Equation ( 1 ), we have, y? + 9x = 22 ; from which we find, x' = + 2, and x" = 11. If we take the negative value, y" = 3, we have, from Equation ( 1 ), x z 9 = 22 ; from which we find, x' = + 11, and x" = 2. VERIFICATION. For the values y' = +3, and x' = +2, the giveii equation, * 2 4- 3ay = 22, SIMULTANEOUS EQUATIvNS. 249 gives, 2 2 +3X2X3 = 4+18 = 22; and for the second value, x" 11, the same equation, x 2 + Zxy = 22, gives, ( 11) 2 + 3 X - 11 X 3 = 121 - 99 = 22. If, now, we take the second value of y, that is, y" = 3, and the corresponding values of ic, viz., x' = +11, and x" = 2; for x f = +11, the given equation, x 2 + Sxy = 22, gives, II 2 + 3 x 11 x 3 = 121 99 = 22; and for x" = 2, the same equation, x z + 3xy = 22, gives, ( 2) 2 + 3 X - 2 X 3 ^4 + 18 = 22. The verifications could be made in the same way by em- ploying Equation ( 2 ). NOTE. In equations similar to the above, we generally find-but a single pair of values, corresponding to the values hi this equation, of y' = +3, and x' +2. The complete solution would give four pairs of values. ' X 2 - y2 = _ 9 j j_. -| * y = 5. a; = 6. -<9L7^. -{ ( y = 7. = 470 I .o a I (23 4 2. .J _ t 4n. J /* - xy = 5 f ( y = 5. 3. V? 7 ^o . i -4w. rz: HO 4. ^ T " 7 ' 5 xy ( 5xy 3y 2 32 . 1 i -L. 2 o , 5- - 4 * 1 '*' 11* 250 ELEMENTARY ALGEBRA. THIRD. PARTICULAR CASES. 175. Many other equations of the second degree may be so transformed, as to be brought under the rules of solution already given. The seven following formulas will aid in such transformation. (1.) When the sum and difference are known: x + y = s x y d. Then, page 132, Example 3, s + d 1,1, s d 1 1 , x = _- = -s + -d, and y = -^- = -s - -d. (2.) When the sum and product are known: x -{- y = s (l.) xy = p (2.) 2 + 2#y + y 2 = s 2 , by squaring ( 1 ) ; 4xy = 4p, by mult. ( 2 ) by 4. K 2 2xy + y z = s 2 4/>, by subtraction. x y = yV 4p, by ext. root. But, x + y s\ hence and, I7i5. What is the first formula of this article? What the second? Third? Fourth? Fifth? Sixth? Seventh? SIMULTANEOUS EQUATIONS. 251 (3.) When the difference and product are known: - y = d ....... . ( 1.) xy = p . . % f . . . (2.) x 2 2xy + y 2 = of 2 , by squaring ( 1 ) 4xy = 4p, mult. ( 2 ) by 4. 2 -f 2;cy + y 7 = ..(2.) .-. 2ay = 2^> . . (3.) Adding ( 1 ) and ( 3 ), x 2 + 2y + y 2 = s + 2p ; hence, SB + y= yT+~2^ (4.) Subtracthig (3) from ( 1 ), x z 2xy + y 2 = 5 2p ; hence, x y = / 2p (5.) Combining (4) and (5), x = i-y/s + 2p + %-Js 2p, (5.) When the sum and sum of the squares are known : x + y = s ..... ( 1.) cc 2 + y 2 = s' ..... (2.) x 2 + 2xy + y 2 = s 2 by squaring ( I ) 2cey = s z s' s 2 - s' *y = 7T- = p. (3-) ELEMENTABY ALGEBKA. By putting xy = p, and combining Equations ( 1 ) and (3 ), by Formula (2), we find the values of x and y. (6.) "When the sum and sum of the cubes are known : x + y = 8 .... (1.) 3.3 _|_ y3 i 52 . . . . (2.) JB 3 + 3x 2 y + 3xy z + y 3 = 512 by cubing ( 1 ). 3a: 2 y + 3xy 2 =360 by subtraction. 3xy(x + y) = 360 by factoring. 3ay(8) = 360 from Equa. ( 1 ). 24xy = 360 hence, xy 15 .... (8.) Combining ( 1 ) and ( 3 ), we find a; = 5 and y = 3 |0 When we have an equation of the form, ( + yY + (x + y) = q. Let us assume x 4- y = 2. Then the given equation becomes, + 2 = ?; and s = - - \ q + - EXAMPLES. 2 t ~i \ \ ir I 1. Given { cc + y + z = 7 ( 2 )> to find a, y, and as, ' + 2,2+22 = 21 (3)) SIMULTANEOUS EQUATIONS. 253 Transposing y in Equation ( 2 ), we have, x + z = 7-y; ... (4.) then, squaring the members, we have, K 2 -f 2cez + z 2 = 49 14y + y 2 . If now we substitute for 2a:z, its value taken from Equa- tion ( 1 ), we have, a;2 + 2 y 2 + z 2 = 49 14y -f y 2 ; and cancelling y 2 , in each member, there results, 2 -f y 2 + z 2 = 49 14 y. But, from Equation ( 3 ), we see that each member of the last equation is equal to 21 ; hence, 49 14y = 21, and, 14y = 49 21 28 , 28 hence, y = = 2. Substituting this value of y in Equation (1 ), gives, xz = 4 ; and substituting it in Equation (4 ), gives, x + z = 5, or a; = 5 z. Substituting this value of , in the previous equation, we obtain 5z - z z =4, or z z fiz = 4 ; and by completing the square, we have, 2 2 - 5z + 6.25 = 2.5. and, 2 2.5 = \/2^ = + 1.5, or 1.5; hence, z = 2.5 -f 1.5 = 4, and 2 = -f 2.5 1.6 = 1 254 ELEMENTARi' ALGEBRA. _ \ 2. Given x + yxy + y = 19 ) ~ , , , , , f to find x and y. and x 2 + ay + y 2 = 133 ) Dividing the second equation by the first, AVC have, x i/xy + y = 7 but, a; + Va-y + y = 19 hence, by addition, 2 + 2y = 26 or, a; + y = 13 and substituting in 1st Equa., -v/ay + 13 = 19 or, by transposing, V^y ^ and by squaring, ay = 36. Equation 2d, is a 2 + xy + y 2 = 133 and from the last, we have, 3a;y =108 Subtracting, x 2 2xy + y 2 = 25 hence, x y = db 5 but, x + y = 13 hence, x = 9 and 4 ; and y = 4 and 9. PROBLEMS. 1. Find two numbers, such that their sum shall be 15 and the sum of their squares 113. Let x and y denote the numbers ; then, x 4- y = 15, (1.) and a 2 + y 2 = 113. (2.) From Equation ( 1 ), \ve have, a; 2 = 225 30y -f y 2 Substituting this value in Equation ( 2 ), 225 -80y + y 2 + y 2 = 113; PROBLEMS. 256 hence, 2y 2 30y = 112; 2/2 - I5y = - 56, . hence, ' y' =. 8, and y" 7. The first value of y being substituted in Equation ( 1 ), gives x' = 7 ; and the second, x" = 8. Hence, the num- bers are 7 and 8. 2. To find two numbers, such that their product added to their sum shall be 1 7, and their sum taken from the sum of their squares shall leave 22. Let x and y denote the numbers; then, from the con- ditions, (x + y) + xy = 17. ... (1.) x* + y> - (x + y) = 22. . . . (2.) Multiplying Equation ( 1 ) by 2, we have, <2xy + 2(x + y) = 34. ... (3.) Adding ( 2 ) and ( 3 ), we have, a 2 + Ixy + y 2 + (a? + y) = 56 ; hence, (x + y} z + (x + y) - 66. . . (4.) Regarding (x -\- y) as a single unknown quantity (page 248), Substituting this value in Equation ( 1 ), we have, 7 + xy = 17, and y = 5. Hence, the numbers are 2 and 5. 3. What two numbers are those whose sum is 8, and sum of their squares 34 ? Ans. 5 and 3. 256 ELEMENTARY ALGEBIIA. 4. It is required to find two such numbers, that the first shall be to the second as the second is to 16, and the sum oi whose squares shall be 225 ? Ans. 9 and 12. 5. "What two numbers are those which are to each othei as 3 to 5, and whose squares added together make 1666 ? Ans. 21 and 35. 6. There are two numbers whose difference is 7, and half their product plus 30 is equal to the square of the less number: what are the numbers? Ans. 12 and 19. 7. What two numbers are those whose sum is 5, and the sum of their cubes 35 ? Ans. 2 and 3. * 8. What two numbers are those whose sum is to the greater as 11 to 7, and the difference of whose squares is 132? Ans. 14 and 8. 9. Divide the number 100 into two such parts, that the product may be to the sum of their squares as 6 to 13. Ans. 40 and 60. 10. Two persons, A and _B, departed from different places at the same time, and traveled towards each other. On meeting, it appeared that A had traveled 18 miles more than ; and that A could have gone I$'s journey in 15 days, but JB would have been 28 days in performing A's journey : how far did each travel ? $ A, 72 miles. Ans. j ' . ( J3, 54 miles. 11. There are two numbers whose difference is 15, and half their product is equal to the cube of the lesser number : what are those numbers ? Ans. 3 and 18. 12. What two numbers are those whose sum, multiplied by the greater, is equal to 77 ; and whose difference, multi- plied by the less, is equal to 12 ? Am. 4 and 7, or f \/2 and y V^- PEOBLKMS. 257 13. Divide 100 into two such parts, that the sum of their square roots may be 14. Ans. 64 and 36. 14. It is required to divide the number 24 into two such parts, that their product may be equal to 35 times their dif- ference. Ans. 10 and 14. 15. The sum of two numbers is 8, and the sum of their cubes is 152 : what are the numbers ? Ans. 3 and 5. 16. Two merchants each sold the same kind of stuff; the second sold 3 yards more of it than the first, and together they receive 35 dollars. The first said to the second, " I would have received 24 dollars for your stuff;" the other replied, "And I should have received 12^ dollars for yours :" how many yards did each of them sell ? 1st merchant x' = 15, x" = 5. 17. A widow possessed 13,000 dollars, which she divided into two parts, and placed them at interest in such a manner that the incomes from them were equal. If she had put out the first portion at the same rate as the second, she would have drawn for this part 360 dollars interest ; and if she had placed the second outsat the same rate as the first, she would have drawn for it 490 dollars interest : what Avere the two rates of interest ? Ans. 7 and 6 per cent. 18. Find three numbers, such that the difference between the third and second shall exceed the difference between the second and first by 6 ; that the sum of the numbers shall be 83, and the sum of their squares 467. Ans. 5, 9, and 19. 19. "What number is that which, being divided by the product of its two digits, the quotient will be 3 ; and if 1 8 be added to i$ the resulting number will be expressed by the digits inverted? Ans. 24. 258 ELEMENTARY A L G E Ji K A . 20. What two numbers are those which are to each other as m to n, and the sum of whose squares is b ? myb n-\fb Ans. 21. What t\\ r o numbers are those which are to each other as m to ft, and the difference of whose squares is b ? Ans. . - , 22. Required to find three numbers, such that the product of the first and second shall be equal to 2 ; the product of the first and third equal to 4, and the sum of the squares of the second and third equal to 20. Ans. 1, 2, and 4. 23. It is required to find three numbers, whose sum shall be 38, the sum of their squares 634, and the difference between the sQcond and first greater by 7 than the difference between the third and second. Ans. 3, 15, and 20. 24. Required to find three numbers, such that the product of the first and second shall be equal to a ; the product of the first and third equal to b ; and the sum of the squares of the second and third equal to c. Ans. x =. y = 25. What two numbers are those, whose sum, multiplied by the greater, gives 144; and whose difference, multiplied by the less, gives 14 ? Ans. 9 and 7. PROPORTIONS AND PROGRESSIONS. 259 CHAPTER IX. OF PROPORTIONS AND PROGRESSIONS. 176. Two quantities of the same kind may be compared, the one with the other, in two ways : 1st. By considering how much one is greater or less than the other, which is shown by their difference ; and, 2d. By considering hoic many times one is greater or less than the other, which is shown by their quotient. Thus, in comparing the numbers 3 and 12 together, with respect to their difference, we find that 12 exceeds 3, by 9; and in comparing them together with respect to their quo- tient, we find that 12 contains 3, four times, or that 12 is 4 tunes as great as 3. The first of these methods of comparison is called Arith- metical Proportion, and the second, Geometrical Propor- tion. Hence, Arithmetical Proportion considers the relation of quantities with respect to their difference, and Geometrical Proportion the relation of quantities with respect to their quotient. 176. In how many ways may two quantities be compared the one with the other? What does the first method consider? What the second? What is the first of these methods called? What is the second called? How then do you define the two proportions? 260 ELEMENTARY ALGEBRA. OP AlimniETICAL I'KOrOKTION AXD PIIOGKESSION. If we have four numbers, 2, 4, 8, and 10, of which the difference betM'een the first and second is equal to the difference between the third and fourth, these numbers are said to be in arithmetical proportion. The first term 2 is called an antecedent, and the second term 4, with which it is compared, a consequent. The number 8 is also called an antecedent, and the number 10, with which it is com- pared, a consequent. When the difference between the first and second is equal to the difference between the third and fourth, the four mimbers are said to be in proportion. Thus, the numbers, 2, 4, 8, 10, are in arithmetical proportion. 178. When the difference between the first antecedent and consequent is the same as between any two consecutive terms of the proportion, the proportion is called an arith- metical progression. Hence, a progression by differences, or an arithmetical %>rogression, is a series in which the suc- cessive terms are continually increased or decreased by a constant number, which is called the common difference of the progression. Thus, in the two series, 1, 4, 7, 10, 13, 16, 19, 22, 25, ... 60, 56, 52, 48, 44, 40, 36, 32, 28, ... 177. When are four numbers in arithmetical proportion ? What is the first called? What is the second called? What is the third called? What is the fourth called ? 178. What is an arithmetical progression ? What is the number called by which the terms are increased or diminished ? What is an increasing progression? What is a decreasing progression? Which term is only an antecedent ? Which only a consequent ? A K I T II M K T I C A L P It O G K K S S I O N. 26 1 the first is called an increasing progression, of which the common difference is 3, and the second, a decreasing pro- gression, of which the common difference is 4. In general, let a, #, c, d, e, f, ... denote the terms of a progression by differences ; it has been agreed to write them thus : a . b . c . d . e . f . g . h . i . k . . . This series is read, a is to b, as b is to c, as c is to d, as d is to e, &c. This is a series of continued equi-differences, in which each term is at the same time an antecedent and a consequent, with the exception of the first term, which is only an antecedent, and the last, which is only a consequent. 179. Let d denote the common difference of the . pro- gresion, a.b.c.e.f.g.h. &c., which we will consider increasing. From the definition of the progression, it evidently fol lows that, b = a + d, c = b + d a -f 2d, e = c + d = a + 3d; and, in general, any term of the series is equal to the first term, plus as many times the common difference as there are preceding terms. Thus, let I be any term, and n the number which marks the place of it ; the expression for this general term is, I = a + (n l)d. Hence, for finding the last term, we have the following 179. Give the rule for finding the last term of a series when the pro- gression is increasing. 262 ELEMENTARY ALGEBRA. KULE. I. Multiply the common difference by the number of terms less one: IT. To the product acid the first term y the sum will be the last term. EXAMPLES. The formula, I = a + (n !)<, serves to find any term whatever, without determining all those which precede it. 1. If we make n = 1, we have, I = a ; that is, the series will have but one term. 2. If we make n = 2, we have, I = a + d ; that is, the series will have two terms, and the second term is equal to the first, plus the common difference. 3. If a = 3, and d 2, what is the 3d term? Ans. 7, 4. If a = 5, and d = 4, what is the 6th term? Ans. 25. 5. If a 7, and d = 5, what is the 9th term ? Ans. 47. 6. If a = 8, and d = 5, what is the 10th term ? Ans. 53. 7. If a = 20, and d 4, what is the 12th term? Ans. 64. 8. If a = 40, and d = 20, what is the 50th term ? Ans. 1020. If a = 45, and d = 30, what is the 40th term? Ans. 1215. ARITHMETICAL PROGRESSION. 263 10. If a = 30, and d = 20, what is the 60th term? Ans. 1210. 11. If a 50, and d = 10, what is the 100th term? Ans. 1040. 12. To find the 50th term of the progression, 1 . 4 . 7 . 10 . 13 . 16 . 19 . . . , and consequently give rise to the following general problem, viz. : Any three 182. How many numbers are considered in arithmetical proportion? What are they ? In every arithmetical progression, \vhat is the common difference equal to ? ARITHMETICAL PROGRESSION. 2(17 of these Jive quantities briny given, to determine the other tico. We already know the value of S in terms of a, n, and I. From the formula, I = a + (n I)c7, we find, a = I (n \}d. That is : The first term of an increasing arithmetical pro- gression is equal to the last term, minus the product of the common difference by the number of terms less one. From the same formula, we also find, I a d = n 1 That is : In any arithmetical progression, the common dif- ference is equal to the last term, minus the first term, divided by the number of terms less one. The last term is 10, the first term 4, and the number of terms 5 : what is the common difference ? The formula, n 1 16-4 gives, d = - - 3. 4 2. The last term is 22, the first term 4, and the number of terms 10 : what is the common difference? Ans. 2. 183. The last principle affords a solution to the follow- ing question : To find a number m of arithmetical means between two given numbers a and b. 183. How do you find any number of arithmetical means between two given numbers ? 268 E 1 K M K X T A K Y A L G K B R A . To resolve this question, it is first necessary to find the common difference. Now, we may regard -a as the first term of an arithmetical progression, b as the last term, and the required means as intermediate terms. The number of terms of this progression will be expressed by m + 2. Now, by substituting in the above formula, b for ?, and m + 2 for n, it becomes, , b a b m + 2 1 ' ~ m + 1 ' that is : The common difference of the required progression is obtained by dividing the difference betioeen the given numbers, a and b, by the required number of means 2>hts one. Having obtained the common difference, d, form the second term of the progression, or the first arithmetical mean, by adding d to the first term a. The second mean is obtained by augmenting the first mean by d, &c. 1. Find three arithmetical means between the extremes 2 and 18. The formula, d = - . m + 1 18-2 gives, = = 4 ; hence, the progression is, 2 . 6 . 10 . 14 . 18. 2. Find twelve arithmetical means between 12 and 77. The formula, d = m 77 - 12 gives, d = -- - = 5 ; hence, the progression is, 12 . 17 . 22 . 27 . . 77. A R I T H M K T I C A L 1' K O G R K S S I O N . 209 184. REMARK. If the same number of arithmetical means are inserted between all the terms, taken two and two, these terms, and the arithmetical means united, will form one and the same progression. For, let a . b . c . e . f . . . be the proposed progression, and m the number of means to be inserted between a and >, b and c, c and e . . . . &c. From what has just been said, the common difference of each partial progression Avill be expressed by b a c b e c m + 1 ' m + 1 ' m + 1 ' expressions which are equal to each other, since , 5, c . . . are in progression ; therefore, the common difference is the same in each of the partial progressions ; and, since the last term of the first forms the first term of the second, &c., we may conclude, that all of these partial progressions form a single progression. EXAMPLES. 1. Find the sum of the first fifty terms of the progression 2 . 9 . 16 . 23 ... For the 50th term, we have, I = 2 + 49 X 7 = 345. 50 Hence, 8 = (2 + 345) x -- = 347 X 25 = 8675. 2. Find the 100th term of the series 2 . 9 . 16 . 23 .... Ans. 695. 3. Find the sum of 100 terms of the series 1.3.5.7. 9 . Ans. 10000. 270 K L ]-: M K X T A K Y ALGEBRA. 4. The greatest term is 70, the common difference 3, nnd the number of terms 21 : what is the least term and the sum of the series ? Ans. Least term, 10 ; sum of series, 840. 5. The first term is 4, the common difference 8, and the number of terms 8 : what is the last term, and the sum of the series? Ans. j Last term, GO. ( Sum = 25G. 6. The first term is 2, the last term 20, and the number of terms 10 : what is the common difference ? Ans. 2. 7. Insert four means between the two numbers 4 and 1 9 : what is the series? Ans. 4 . 7 . 10 . 13 . 16 . 19. 8. The first term of a decreasing arithmetical progression is 10, .the common difference one-third, and the number of terms 21 : required the sum of the series. Ans. 140. 9. In a progression by differences, having given the com- mon difference 6, the last term 185, and the sum of the terms 2945 : find the first term, and the number of terms. Ans. First term = 5 ; number of terms, 31. 10. Find nine arithmetical means between each antecedent and consequent of the. progression 2. 5. 8. 11. 14... Ans. Common diff., or d = 0.3. 11. Find the number of men contained in a triangular battalion, the first rank containing one man, the second 2, the third 3, and so on to the n th , which contains n. In other words, find the expression for the sum of the natural num- bers 1, 2, 3 . . ., from 1 to n inclusively. Ans. S = *-L>. 2 12. Find the sum of the n first terms of the progression of uneven numbers, 1.3.5.7.9,... Ans. 8 = ;i 2 . GEOMETRICAL PROFOKTION. 271 13. One hundred stones being placed on the ground in a straight line, at the distance of 2 yards apart, how far will a person travel who shall bring' them one by one to a basket, placed at a distance of 2 yards from the first stone ? Ans. 11 miles, 840 yards. GEOMETRICAL PROPORTION AND PROGRESSION. 185. jRatio is the quotient arising from dividing one quantity by another quantity of the same kind, regarded as a standard. Thus, if the numbers 3 and 6 have the same unit, the ratio of 3 to 6 will be expressed by And in general, if A and It represent quantities of the same kind, the ratio of A to J5 will be expressed by B A' 16. The character cc indicates that one quantity is proportional to another. Thus, A cc .B, is read, A proportional to JB. If there be four numbers, 2, 4, 8, 16, having such values that the second divided by the first is equal to the fourth divided by the third, the numbers are 185. What is ratio ? What is the ratio of 3 to G ? Of 4 to 12 ? 186. What is proportion? How do you express that four numbers are in proportion ? What are the numbers callnd? What are the first ud fourth terms cilled ? What the second and third ? 272 ELEMENTAKY ALGEBRA said to form a proportion. And in general, if there be four quantities, A, J?, <7, and .Z), having such values that, B D A '"'' xL That is : If there are two sets of proportions having an an- tecedent and consequent in the one, equal to an antecedent and consequent of the other, the remaining terms will be proportional. If we have the two proportions, 2 : 6 : : 8 : 24, and 2 : 6 : : 10 : 30, we shall also have, 8 : 24 : : 10 : 30. 200. If we have, 7? J\ A '. B : : C : D, and consequently, -j = -^, we have, by dividing 1 by each member of the equation, A C -= = _ , and consequently, B : A : : D : C. 199. If you have two sets of proportions having an antecedent and con- sequent in each, equal ; what will follow ? 200. If four quantities are in proportion, will they be in proportion when taken inversely ? 278 ELEMENTARY ALGEBRA. That is : Four proportional quantities will b$ in proportion^ when taken inversely. To give an example iu numbers, take the proportion, 7 : 14 : : 8 : 16; then, the inverse proportion will be, 14 . 7 : : 16 : 8, hi which the ratio is one-half. 2O1. The proportion, A : B : : C : D, gives, A x D = B x C. To each member of the last equation add H x D. "We shall then have, (A + J?) x D = (C + D) x JB; and by separating the factors, we obtain, A + B : B : : C + D : D. If, instead of adding, we subtract JB x D from both members, we have, (A - B) x D = (C- D) x J?; which gives, A - B : JB : : C - D : Z>. That is: If four quantities are proportional, they will be in proportion by composition or division. Thus, if we have the proportion, 9 : 27 : : 16 : 48, 2 1 '!. If four quantities are in proportion, will they be in proportion by composition ? Will they be in proportion by division ? What is th difl'ereiice between composition and division ? GEOMETRICAL PROPORTION. 279 tre shall have, by composition, 9 + 27 : 27 : : 16 + 48 : 48; that .is, 36 : 27 : : 64 : 48, in which the ratio is three-fourths. The same proportion gives us, by division, 27 9 : 27 :: 48 16 : 48; that is, 18 : 27 : : 32 : 48, in which the ratio is one and one-half. 2O2. If we have, and multiply the numerator and denominator of the first member by any number m, we obtain, j = -= , and mA : mJ3 : : C : D. mA C That is : Equal multiples of two quantities have the same ratio as the quantities themselves. For example, if we have the proportion, 5 : 10 : : 12 : 24, and multiply the first antecedent and consequent by 6, we have, 30 : 60 : : 12 : 24, in which the ratio is still 2. 2O3. The proportions, A : B : : C : D, and A : B : : E : F, 202. Have equal multiples of two quantities the same ratio as the quantities ? 203 Suppose the antecedent and consequent be augmented or dimin- ished by quantities having the same ratio ? 280 ELEMENTARY ALGEBRA. give, A x D = JB x <7, and AxF=JSxE\ adding and subtracting these equations, we obtain, A(D+F) = E(CE], or A : B : : G E : D F. That is : If C and D, the antecedent and consequent, be augmented or diminished by quantities E and F, which have the same ratio as C to D, the resulting quantities will also have the same ratio. Let us take, as an example, the proportion, 9 : 18 : : 20 : 40, in which the ratio is 2. If we augment the antecedent and consequent by the numbers 15 and 30, which have the same ratio, we shall have, 9 -f 15 : 18 + 30 : : 20 : 40; that is, 24 : 48 : : 20 : 40, in which the ratio is still 2. If we diminish the second antecedent and consequent by these numbers respectively, we have, 9 : 18 :: 20 15 : 40 30; that is, 9 : 18 : : 5 : 10, in which the ratio is till 2. 2O4. If we have several proportions, A : B : : C : D, which gives A x D B x (7, A : B : : E : F, which gives A x F = B x E, A : B : : G : H, which gives A x H = B x G, &c., &c., 204. In any number of proportions having the same ratio, how viD any one antecedent be to its consequent? GEOMETRICAL PROPORTION. 281 we shall have, by addition, A(D + F+ II) = B(C + E + G); and by separating the factors, A : B : : C + E + G : D + F + H. That is: In any number of proportions having the same ratio, any antecedent will be to its consequent as the sum of the antecedents to the sum of the consequents. Let us take, for example, 2 : 4 : : 6 : 12, and 1 : 2 : : 3 : 6, &c. Then 2:4::6 + 3:12 + 6; that is, 2 : 4 : : 9 : 18, in which the ratio is still 2. 2O5. If we have four proportional quantities, A : B : : C : D, we have, -. -^ ; ^3L and raising both members to any power whose exponent is n, or extracting any root whose index is n, we have, j? D -^ =-. --, and consequently, A n : B n : : C n : D n . That is : If four quantities are proportional, their like powers or roots will be proportional. If we have, for example, 2:4 : : 3 : 6, we shall have, 2 2 : 4 2 : : 3 2 : 6 2 ; 205. In four proportional quantities, how arc like powers or roots? ELEMENTARY A L G .E B H A . that is, 4 : 16 : : 9 : 36, in which the terms are proportional, the ratio being 4. 2O6. Let there be two sets of proportions, 7? T) A : J5 : : C : D, which gives r = -~ ; jfl F II E : F : : G : JET, which gives -=, = -^ .& Or Multiply them together, member by member, we have, E x F _ D x H A x E z: G x~' A x E : B x F:: C x G : D x H. Tliat is : In two sets of proportional quantities, the product* of the corresponding terms are proportional. Thus, if we hav^ the two proportions, 8 : 16 : : 10 : 20, and, 3 : 4 : : 6 : 8, we shall have, 24 : 64 : : 60 : 160. GEOMETRICAL PROGRESSION. 2O7 1 . We have thus far only considered the case in which the ratio of the first term to the second is the same as that of the third to the fourth. 20C. In two sets of proportions, how arc the products of the correspond- ing terms ? 207. What is a geometrical progression? What is the ratio of the progression ? If any term of a progression be multiplied by the ratio, what will the product be ? If any term bo divided by the ratio, w hat G E U M K T K I C A L I' K O G B K 6 S I O N. 283 If we 1m e the farther condition, that the ratio of the second term to the third shall also be the same as that of the first to the second, or of the third to the fourth, we shall have a series of numbers, each one of which, divided by the preceding one, will give the same ratio. Hence, if any term be multiplied by this quotient, the product will be the succeeding term. A series of numbers so formed, is called a geometrical progression. Hence, A Geometrical Progression, or progression, by quotients, is a series of terms, each of which is equal to the preceding term multiplied by a constant number, which number is called the ratio of the progression. Thus, 1 : 3 : 9 : 27 : 81 : 243, &c., is a geometrical progression, in which the ratio is 3. It is written by merely placing two dots between the terms. Also, 64 : 32 : 16 : 8 : 4 : 2 : 1, is u geometrical progression in which the ratio is one-half. In the first progression each term is contained three times in the one that follows, and hence the ratio is 3. In the second, each term is contained one-half times in the one which follows, and hence the ratio is one-half. The first is called an increasing progression, and the second a decreasing progression. Let a, b, c, d, e, f, ... be numbers, in a progression by quotients ; they are written thus : a : b : c : d : e : f : g . . . and it is enunciated in the same manner as a progression by differences. It is necessary, however, to make the distinc- will the quotient be? How is a p-ogression by quotients written? Which of the terms is only nu antecedent? Which only a consequent? Hovt laav each of the others be considered? 284 ELEMENTARY ALGEBRA.. tion, that one is a series formed by equal differences, and the other a series formed by equal quotients or ratios. It should be remarked that each term is at the same time an antecedent and a consequent, except the first, which is only an antecedent, and the last, which is only a consequent. 2O8. Let r denote the ratio of the progression, a : b : c : d . . . r being > 1 when the progression is increasing, and r< 1 when it is decreasing. Then, since, b _ c d e . a b c d ~ we have, b = ar, c = br ar 2 , d = cr = ar 3 , e = dr = ar*, f = er = ar 5 . . . that is, the second term is equal to ar, the third to ar 2 , the fourth to ar 3 , the fifth to ar 4 , &c. ; and in general, the nth term, that is, one which has n I terms before it, is ex- pressed by ar n ~ 1 . Let I be this term we then have the formula, I = ar n ~\ by means of which we can obtain any term without being obliged to find all the terms which precede it. Hence, to find the last term of a progression, we have the following E u L E . I. Raise the ratio to a power whose exponent is one less than the number of terms. H. Multiply the power thus found by the first term : the product will be the required term. 208. By what letter do we denote the ratio of a progression? In an increasing progression is r greater or less than 1 ? In a decreasing pro- GEOMETRICAL PROGRESSION. 285 EXAMPLES. 1. Find the 5th term of the progression, 2 : 4 : 8 : 16 . . . in which the first term is 2, and the common ratio 2. 5th term = 2 X 2* = 2 X 16 = 32. Ans. 2. Find the 8th tetm of the progression, 2 : 6 : 18 : 54 ... 8th term = 2 X 3 7 = 2 X 2187 = 4374. Ans. 3. Find the 6th term of the progression, 2 : 8 : 32 : 128 ... 6th term = 2 X 4 5 = 2 X 1024 = 2048. Ant 4. Find the 7th term of the progression, 3 : 9 : 27 : 81 . . . 7th term = 3 x 3 6 = 3 x 729 = 2187. Ans. 5. Find the 6th term of the progression, 4 : 12 : 36 : 108 ... 6th term = 4 X 3 s 4 x 243 = 972. Ans. 6. A person agreed to pay his servant 1 cent for the first day, two for the second, and four for the third, doubling every day for ten days : how much did he receive on the tenth day? Ans. $5.12. grcssiou is r greater or less than 1 ? If a is the first term and r the ratio, what is the second term equal to ? What the third ? What the fourth ? What is the la?t term equal to ? Give the rule for finding the last terra. 286 ELK MJ:> ;TA i: Y ALGEBRA. 7. What is the 8th term of the progression, 9 : 36 : 144 : 576 . . . 8th term = 9 X 4 7 = 9 x 16384 - 147456. Ans. 8. Find the 12th term of the progression, 64 : 16 : 4 : 1 : 7 . 4 /l\ u 4 3 1 1 12th term 64-1 = = - = - Ans. \4/ 4 11 4 8 60086 2O9. We will now proceed to determine the sum of n terms of a progression, a : b : c : d : e : f : . . . : i : Jc : /; I denoting the nih term. We have the equations (Art. 208), b ar, c = br, d = cr, e dr, . . . k = ir, I = AT, and by adding them all together, member to member, we deduce, Sum of 1st members. Sum of M members. b+c+d+e+ . . . +&+l=(a + b + c in which we see that the first member contains all the terms but , and the polynomial, within the parenthesis in the second member, contains all the terms but I. Hence, if we call the sum of the terms S, we have, S - a = (S - l)r = Sr - Ir, .-. Sr - S = Ir - a; Ir a whence, o = -- r 1 209. Give the rule for finding the sum of the series. What is the first sttp? What the second? What the third ? G E O M K T li I C A L PROGRESSION. 287 Therefore, to obtain the sum of all the terms, or sum of the series of a geometrical progression, we have the KULE. I. Multiply the last term by the ratio : II. Subtract the first term from the product : III. Divide the remainder by the ratio diminished by 1 and the quotient will be the sum of the series. 1. Find the sum of eight terms of the progression, 2 : 6 : 18 : 54 : 1C2 . . . 2 x 3 7 = 4374. = lr_-a = 13IM-. _ 656fl _ ' r 1 2 2. Find the sum of the progression, 2 : 4 : 8 : 16 : 32. 8= lr ~ Cl - 64 ~ 2 -- 62 ' r - 1 1 3. Find the sum of ten terms of the progression, 2 : 6 : 18 : 54 : 16^ . . . 2 X 3 9 = 39366. Ans. 59048. 4. What debt may be discharged in a year, or twelve months, by paying $1 the first month, $2 the second month, $4 the third month, and so on, 'each succeeding payment being double the last ; and what will be the last payment ? i Debt, . $4095. Ans. *\ -. ( Last payment, $2048. 5. A daughter was married on New-Year's day. Her father gave her Is., w r ith an agreement to double it on the first of the next month, and at the beginning of each succeed- ing month to double what she had previously received. How much did she receive ? Ans. 204 15*. 288 ELEMENTARY ALGEBRA. 6. A man bought ten bushels of wheat, on the condition that he should pay 1 cent for the first bushel, 3 for the second, 9 for the third, and so on to the last : what did he pay for the last bushel, and for the ten bushels ? j Last bushel, $196 83. HS ' ( Total cost, $295 24. 7. A man plants 4 bushels of barley, which, at the first harvest, produced 32 bushels ; these he also plants, which, in like manner, produce 8 fold ; he again plants all his crop, and again gets 8 fold, and so on for 16 years : what is his last crop, and what the sum of the series ? . s j Last, 140737488355328 bush. m ' ( Sum, 160842843834660. 21O. When the progression is decreasing, we have, r < 1, and ?< a ; the above formula, _ lr-g - for the sum, is then written under the form, a Ir o ^ - 1 1 r in order that the two terms of the fraction may be positive. 1. Find the sum of the terms of the progression, 32 : 16 : 8 : 4 : 2 32 2 X a Ir 2 31 S = - - = -- - -- zr 62. 1 r 1 1 i 210. What is the formula for the sum of the series of a decreasing ' progression ? GEOMETRICAL PROGRESSION. 289 2. Find the sum of the first twelve terms of the pro- gression, 64 : 16 : 4 : 1 : 4 84/7 ), or l \4/ 65536 X- 256-- S a "~ lr _ 65536 4 _ 65536 _ 65535 = 1 - r ~ '~3~~ ~~ *" 196008 4 211. REMARK. We perceive that the principal difficulty consists in obtaining the numerical value of the last term, a tedious operation, even when the number of terms is not very great. 3. Find the sum of six terms of the progression, 512 : 128 : 32 ... AM. 682f 4. Find the sum of seven terms of the progression, 2187 : 729 : 243 ... Ans. 3279. 5. Find the sum of six tenns of the progression, 972 : 324 : 108 ... Ans. 1456. 6. Find the sum of eight terms of the progression, 147456 : 36864 : 9216 . . . Ans. 196605. OF PROGRESSIONS HAVING AN INFINITE NUMBER OF TERMS. 212. Let there be the decreasing progression, a : b : o : d : e : f : . . . 212. When the progression is decreasing, and the number of termg In- finite, what is the expression for the value of the sum of the series ? 18 290 ELEMENTARY A L O E B K A . containing an indefinite number of terms. In the formula, a - Ir S = T^T' substitute for I its value, ar n ~ l , (Art. 208), and we have, a ar n S = 1 r which expresses the sum of n terms of the progression. This may be put under the form, a ar* & == i~ 1 r I r Now, since the progression is decreasing, r is a proper fraction ; and r* is also a fraction, which diminishes as n increases. Therefore, the greater the number of terms we take, the more will - - X r n diminish, and consequently, -L ^* / the more will the entire sum of all the terms approximate to an equality with the first part of S, that is, to - Finally, when n is taken greater than any given number, or n infinity, then - - x r* will be less than any L ~ T given number, or will become equal to ; and the expres- t will then represent the true value of the sum 1 "~ 7* of all the terms of the series. Whence we may conclude, that the expression for the sum of the terms of a decreasing progression, in which the number of terms is infinite, is, that is, equal to the first term, divided by 1 minus the ratio. GEOMETRICAL P KOU KE8SI ON . L".l This is, properly speaking, the limit to which the partial Bums approach, as we take a greater number of terms in the progression. The difference between these sums and - - , may be made as small as we please, but will only become nothing when the number of terms is infinite. EXAMPLES. 1. Find the sum of "We have, for the expression of the sum of the terms, * = r^ = i=l = * *& ~3 The error committed by taking this expression for the value of the sum of the n first terms, is expressed by X r* = -(-V- 2\3/ 1 r First take n = 5 ; it becomes, 3/l\ 5 _ 1 JL_ 2W := 2 . 3* " 162* When n = 6, we find, 3/l\ 6 _ _1_ 1 1 2\3/ : = 162 X 3 == 486' q Hence, we see, that the error committed by taking - for m the sum of a certain number of terms, is less in proportion as this number is greater. 29*3 K L K M K X 'C A K Y A I. Ci E FJ K A . 2. Again, take the progression, - - l - - '& : 2 : 4 : 8 : 16 : 32 : We have, S = --^ = - : = 2. I r \ Ans. _ ~~ 2 3. What is the sum of the progression, 111 1 1, , - , - , - , &c., to infinity. 10' 100' 1000' 10000' K - a l - i 1 Av< 8 - r=~r - ~r V Ans - 10 a 13. In the several questions of geometrical progres- sion, there are five numbers to be considered : 1st. The first term, . . a. 2d. The ratio, . . . . r. 3d. The number of terms, n. 4th. The last term, . . /. 5th. The sum of the terms, S. 214. We shall terminate this subject by solving this problem : To find a mean proportional between any two numbers, as m and n. Denote the required mean by x. We shall then have (Art. 197), a; 2 = m x n ; and hence, x = \/m x n. 213. How many numbers are considered in a gcometrlcnJ progression? What are they? 214. How do you find a mean proportional between two numbers ? GEOMETRICAL PROGRESSION. 293 That is : Multiply the two numbers together, and extract the square root of the product. 1. What is the geometrical mean between the numbers 2 and 8 ? Mean = 18 x 2 = /lQ 4. Ans. 2. What is the mean between 4 and 16 ? Ans. 8. 3. What is the mean between 3 and 27 ? Ans. 9. 4. What is the mean between 2 and 72 ? -4ws. 12. 5. What is the mean between 4 and 64 ? ^1*. 16. therefore, $40 satisfies the enunciation. 294: ELEMENTARY ALGEBRA. CHAPTER X. OF LOGARITHMS. 215. THE nature and properties of the logarithms in common use, will be readily understood by considering attentively the different powers of the number 10. They are, 10 = 1 10 1 1= 10 10 2 = 100 10 3 = 1000 10 4 = 10000 10 s = 10000* &c., &c. It is plain that the exponents 0, 1, 2, 3, 4, 5, . What is a table of logarithms? Explain the manner of finding the logarithms of numbers between 1 and 100? 13* 298 ELEMENTARY ALGEBRA. TABLE. N. Log. N. Log. N. Log. N. Log. ,1 2 3 4 5 0.000000 0.301030 0.477121 0.602060 0.698970 26 27 28 29 30 1.414973 1.431364 1.447158 1.462398 1.477121 51 52 53 54 55 1.707570 1.716003 1.724276 1.732394 1.740363 7(5 77 78 79 80 1.880814 1.886491 1.892095 1.897627 1.903090 6 7 8 1 10 0.778151 0.845098 0.903090 0.954243 1.000000 31 32 33 34 35 1.491362 1.505150 1.518514 1.531479 1.544068 56 57 58 59 60 1.748188 1.755875 1.763428 1.770852 1.778151 81 82 83 84 85 1.908485 1.913814 1.919078 1.924279 1.929419 11 12 13 14 15 1.041393 1.079181 1.113943 1.146128 1.176091 36 37 38 39 40 1.556303 1.568202 1.579784 1.591065 1.602060 61 62 63 64 65 1.785330 1.792392 1.799341 1.806180 1.812913 86 87 88 89 90 1.934498 1.939519 1.944483 1.949390 1 . 954243 16 17 18 19 20 1.204120 1.230449 1.255273 1.278754 1.301030 41 42 43 44 45 1.621784 1.623249 1.633468 1.643453 1.653213 66 67 68 C9 70 1.819544 1.826075 1 . 832509 1.838849 1 . 845098 91 92 93 94 95 1.959041 1.963788 1.968483 1.973128 1.977724 21 22 23 24 25 1.322219 1 . 342423 1.361728 1.380211 1.3979-10 46 47 48 49 ' 50 1.662758 1.672098 1.681241 1.690196 1.698970 71 72 73 74 75 1.851258 1 . 857333 1.863323 1.869232 1.875061 96 97 98 99 100 1.982271 1.986772 1.991226 1.995635 2.000000 EXAMPLES. 1 . Let it be required to multiply 8 by 9, by means of logarithms. We have seen, Art. 216, that the sum of the logarithms is equal to the logarithm of the product. There- fore, find the logarithm of 8 from the table, which is 0.903090, ami then the logarithm of 9, which is 0.954243 ; and their sum, which is 1.857333, will be the logarithm of the product. In searching along in the table, we find that 72 stands oppo- site this logarithm ; hence, 72 is the product of 8 by 9. OF LOGABITI1MS. 299 2. What is the product of V by 12? Logarithm of 7 is, Logarithm of 12 is, . . . Logarithm of their product, and the corresponding number is 84. 3. What is the product of 9 by 11 ? Logarithm of 9 is, Logarithm of 11 is, Logarithm of their product, and the corresponding number is 99. 0.845098 1.079181 1.924279 0.954243 1.041393 1.995636 4. Let it be required to divide 84 by 3. We have seen in Art. 218, that the subtraction of Logarithms corresponds to the division of their numbers. Hence, if we find the logarithm of 84, and then subtract from it the logarithm of 3, the remainder will be the logarithm of the quotient. The logarithm of 84 is, The logarithm of 3 is, Their difference is, ... and the corresponding number is 28. 5. What is the product of 6 by 7? Logarithm of 6 is, Logarithm of 7 is, Their sum is, 1.623249 and the corresponding number of the table, 42. 1.924279 0.477121 1.447158 0.778151 0.845098 RECOMMENDATIONS OF DAVIES' MATHEMATICS. DAVIKS' COUBKE OF MATHEMATICS are the prominent Texl-Jiooks in ntott of the Golltgts of the United Sltite.i, and also^in the various Scho< is in. I Acitdttinics throughout the Union. YORK, PA., Auy.'2\ !>. Davits' Seriet of Mathematict 1 deem the very best I ever saw. From a nunihei if authors I selected it, after a careful perusal, as a course of stndy to be pursued by Hie Teachers attending the sessions of the Turk Co. Normal School believing It also n b.- well adapted to the wants of the schools throughout our country. Already r.vo indred schools are supplied with DAVI&' valuable Sit-ins of Arithmetic* ; and I Mlv believe that in very short time the Teachers of our country en ma*e will i* i .rs.'ed in imparting instruction throngh the medium of this new and easy mett-v- 1 .-itijtlysls of numbers. A. E. BLAIR, Principal of York Co. Normal Scbm JACKSON UNION SCHOOL. MICIIIGAX. Xei>t. 2.\ K,-. Mrc*i:8 A. S. BARSRS & Co. : I take pleasure in adding my testimony in favor ol I'.irifx' y/n-ifg of Mathfmativ*, ax published by yon. We" have used these work> in this school for more than four years ; a d i v)l satisfied are we of their nperioHtl over any other Series, that we 'neither contemplate making, iior de-ire to make, HHJ change 'in that direction. Yours truly, E. L. IJIl'LKY. NKW UKITAIK. June I'M. 1V.S. MI-.SHKS. A. S. P,AI:XI:S & Co . : I h:ive eNamincd I>-if!i.- , '" with some cure. They appear well adapted for the different o]> t'ur which they aro designed. The language is clear and precise : e.cch principle i* thoroughly analyzed, and the whole so arranged as to facilitate the work of instruc- tion. Having observed the satisfaction and success with which the different ho,.k> have been used by eminent tenchers, it srivc,* me pleasure to commend them to others. DAVID N. C.\Ml\f'riii(.-ij>fConn. State Normal Scho'. I have long regarded Davits' Series of Mathematical Tert-Books as far superioi to any now before the public. We find them in every way adapted to the wants n'. the Normal School, and we use no other. A unity of sys'tem and method runs through- out the series, and constitutes one of its preat excellences. Especially in the Arith- metics the author has earnestly endeavored to supply the wants of our Common and Union Schools: and his success is complete, and undeniable. 1 know of no Arith- metics which exhibit so clearly the philosophy of numbers, and at the same time lead the pupil surely on to readiness and practice. A. 8. WELCH. From PBOF. G. W. PLVMPTON, late of the Sbite Normal School, X. Y. Out of a great number of Arithmetics that I have examined during the past year, 1 find none that will compare with Datie/f Intellectual and Dar.iej Analytical and Practical Arithmetics, in clearness of demonstration or philosophical arrangement. I shall with pleasure recommend the use of these two excellent works to those who go from our institution to teach. From C. MAY, JK., School Commissioner, Keene, 2V. H. I bare carefully examined Da-tie*' Seriei of Arithmetics, and Higher Mat'if- tnaticit, and am prepared to say that I consider them far superior to any with which I am acquainted. ffiim Joim L CAMPBBLL, Professor uf Jfitt?ifmatic*, Natural PhUovophy, an/I Astronomy, in Wabash College, Iwlidna. W ABASH COLI.KOK. Jntif 22. !-.> MSRS. A. S. BAKHES & Co. : GF.NTLKMKN : Every text-book on Science pvo|" r'y < IINJSIS of two jiarts the philosophical and the illuxtratitt. A proper cointiitiiiior. . -i pstract reasoning and practical illustration is the chief excellence in Prof. l)vi- Malhcmatical Works. I prefer his Arithmetics, Algebras. Geometry, and Trisronoii: try. to all others now in use. and cordially recommend them to all who d<-Miv i> . vhancemcnt of sound learning, Yours, very truly, JOHN L. CAMPBELL PK.IKKSSORS MAH.SN, BAHTLETT, and CtirROii. of the United States Military Aca.l<-n. West 1'oint. HfvtDnvfa? L'niei-ryiti/ Arithmetic: " In the distinctness wi^h whieh the various definitions are given, the clear nd strictly mathematical demonstration of the rules, the convenient form and well-chosen matter of the tables, as well as in the complete and much-desired application of all to the business of the country, th*> Uv>rfity Arithmetic of Prof. Davits is st perior to *!!} other work of the kind wi;h h'.-h ' c rp sc'jnainti-.i " RECOMMENDATIONS OF PARKER & WATSON'S READERS from PROF. FREDERICK S. JKWKLI., of the _\> &>r commendation. From n brief examination of them. I urn led to believe tin.' -nm IIox. TiiKonoRK FRKUXGIIUYSKN, President of Uutgeri? College, 2f J. .\ cursory examination leads me to the conclusion that the system contained ID >.li>.e volumes deserves the patronage of our schools, and I have no doubt that il w!t W-come extensively used in the education of children and youth. fnan N. A. HAMILTON, President of Teacher*' Union, Whitewater, n'i*. Tin- National Readers and Speller I have examined, and carefully compared witb others, and must pronounce them decidedly superior, in respect to literary merit, style. :in-xcel lence of the National Series of Readers, by PAKKF.K and WATSON. The gradation of the books of the series is very fine : we have reading in its elements and in its highest style. The fine taste displayed in the selections ami in the collocation of the piece? it'serves much praise. A distinguishing feature of the series is the variety of thf subject-matter and of the style. The practical teacher knows the value of this charm: terisik for the development of the voice. The authors seein to have kept constantly in view tlie fact that a reading-book is designed for children, and therefore they have surceeded in forming a very interesting and improving collection of reading-matter, highly adapted to the wants and purposes of the school-room. In short, I look upon the National Scries of ReaUers as a great success. from \. P. HARRINGTON, Principal of Union School, MaraUion, N. }'. These Renders, in my opinion, aie the best I have ever examined. The rhetorical exercises, in particular, are superior Ui any thing of the kind I have ever seen. I have iiad tn-tter success with my reading classes since I commenced training them i.n those ilnii 1 ever met with before. The marked vowels in the reading exori-ises coiury 10 '.lie reaiicr's iniiid ut once the astonishing fct that he has been accustomed to im-pro- :.nnve more than one-third of the words of the English language. f"rrm OriAr.i.KS S. HALSF.V. Prinrijml C<>Ufyi< it V-ij ii'u! typiwriipby. and the general neat and inviting appearance of it* several !H>. I.* I predict tor it a cordial Ve'.coine and a general introduction by many of 0111 nv>st *t,tei(.] isinj ton hers. RECOMMENDATIONS CLARK'S ENGLISH GRAMMAR. We cannot better set forth the merits of this work than by quoting a part of a com- munication from Prot F. 8. JEWKLL, of tlie New York State Normal School, in wblct school this Grammar is now used as the text book on this subject : ; CT.ARK'S SYSTKM or GRAMMAR is worthy of the marked attention of the friends 01 jilucntion. Its points of excellence are of the most decided character, and will nc*. *>'>n be surpassed. Among them are 1st. '-The justness of its around principle of classification. There is no simple, phi!- isophieal. and practical classification of the elements of language, other than thnt bui't ;, iheir use or i;fltce. Our tendencias hitherto to follow tho analogies of the classical .!!gu.-is:-s. and classify extensively according to forms, have been mischievous and ab- minl. It is time we corrected them. 2d. ' Its thorough and yt simple and transparent analysis of the elements of the language according to its ground principle. Without such an analysis, no broad and comprehensive view of the structure and power of the language can be attained. The al.setx-e of this analysis has hitherto precipitated the study of Grammar upon a surface of dry details and bare authorities, and useless technicalities. 3d." " Its happy method of illustrating the relations of elements by diagrams. These, however uncouth they may appear to the novice, sre really simple and philosophical. Of their utility there can be no question. It is supported by the usage of other sci- ences, and has bc-n demonstrated by experience in this. 4th "The tenden'-y of the system, when rightly taught and faithfully carried out, to cultivate habits of nice discrimination and close reasoning, together with skill in illustrating truth. In this it is not excelled by any, unless it be the mathematical sci- ences, and even there it hn> this advantage. th:it it deals with elements more within tire present gra>p of the intellect On this point I speak advisedly. 5th. -'The system is thoroughly progressive and practical, and as such. American in [f character. "It does not adhere to old usages, merely because thev are veneraliy musty; anil yet it does ma discard things merely because they are old. or are in un- Importxni minutie not prudishly perfect. It does nut overlook details and technicali- ties, nor does it allow them to interfere with plain philosophy or practical utility. ' Let any clear-headed, independent-minded teacher master the system, and then give it a fa'ir trial, and there.wiil be no doubt as to his testimony." A Testimonial from tlie Pr in ci juris of the Public School* of Rochester, N. Y. We reg.-ird CLARK'S GRAMMAR as the clearest in its analysis, the most natural and logical itt its arrangement, the most concise and accurate in its definitions, the mos-. M .-lemittic in design, and the best adapted to the use of schools of any Grammar tth which we are acquainted. C C. MKSKRVE, WM. C. FEGLES. M D. ROWLEY, OHN ATWATKK, C. R. BUBRICK, EDWARD WEBSTER, J. R. VOSBURG, 8. W. STARKWEATHER, E. K. ARMSTRONG PHILIP CURTISS. LAWEF.XCR INSTITUTE, Brooklyn, Jan 15, 1859. MESSRS. A. S. BARXFS & Co: Having used Clark's New Grammar since Its publica- tion, i do most unhesitatingly recommend it as a work of cupvrior merit By the ut* of ii-i other work, ami I have used several, have I been enabled to advance uiy pupil* v. njiidly and thoroughly. The author has, by an "Etymological Chart and a system of Diagrams, made Gram :iar the study that it ought to be, interesting as well as nseful. MARGARET S. LAWRENCE, Principal. WELCH'S ENGLISH SENTENCE, frtttK P*or. J. R BOISE, A. M., Profetxor of the Latin and Greek Langvaget etna Literature in the University of Michigan. Tliis work belongs to a new era in the grammatical study of our own language. \V r n/.Mrk without that constant turning over of leave*, or refrr- nce to a separate book, as Is necessary with most othor Geographies. Yours. &c. Messrs. A. S. BAKNKS A Co., New York. V. W. DAVIS. From CHARLES BARNRS, late Preaiilfnt State Ttavktr*' Amouiutiim, and Stijifhi- tendent qfth, Public Softools at Neic Albany, Indiana. MKSSHS. A. S. BAKNES in every way worthy of a place in every Public School in the State. Maine Tettcker. This is a most capital work : just the thing for children. Our boy commenced ,he "tudy of it the day it came to hand. It is arranged in the catechetical form, and is fiiK-ly illustrated with maps, with special reference to the matter discussed in the text, It begins with the first discoveries of America. nd comes down to the laying of the Atlantic Telegrnph Cable. Many spirited engravings are given to illustrate the work. It also contains brief Biographies of all prominent men who have identified them- selves with the history of this country. It is the best work of the kind we have Been. Cluster County Times. WILLARD'S HISTORIES. from RET. HOWARD MALCOLM, D. D., President .. vrnu-nt in (size that contains so full and elegant an exposition of mythology ax thi one before us. It will be found at once a most interesting and a most useful book to any one who wiskos an acquaintance with the splendid myths and fables with whiob the great masters of ancient learning amused their leisure and cheated tbtir faith - tficMgan Journal offduca'ien. , UCSB LIBRARY LIBRARY FACILITY A 000607197 1 m I IMMHH I 1