Jllllllllllll l llllllllllllf UC-NRLF hi i i? 1111111 I^^HIBiiiAilil^ ^ B 4 2S0 38S Ijl 11 I 111111 ill MffifWi lii* iliii! 1 ft liHl HLpuIiUi ' ii-i !i i'l 'i Www ill 11 nil 111! II MM 1 1 1 WImP ffi IN MEMORIAM FLOR1AN CAJORI Ia&w ELEMENTARY ALGEBRA BY FREDERICK H. SOMERVILLE, B.S. THE WILLIAM PENN CHARTER SCHOOL, PHILADELPHIA NEW YORK .^CINCINNATI-:. CHICAGO AMERICAN BOOK COMPANY Copyright, 1908, by FREDERICK H. SOMERVILLE. Entered at Stationers' Hall, London. el. alo. W. P. I \JMf PREFACE The plan of the early pages of this text is to offer a gradual introduction to the subject without plunging the young student too deeply into new difficulties. The treatment of negative numbers as a natural extension of the familiar arithmetical numbers, and a generous amount of detail in the explanations of the early chapters, serve to clarify the beginnings of a sub- ject so often troublesome. New processes are accomplished by the use of the simplest of symbols, and the undivided atten- tion of the student is centered upon the new elements of the operation at hand. Definitions are introduced only as rapidly as new processes call for them, and the young mind is not confused and discouraged at the outset by the attempt to learn the meaning of a bewildering mass of strange terms. In arrangement the book does not differ widely from the general scheme of the standard texts, but in some details there are changes that have been found to be of genuine value in the classroom. For example, in Factoring, the simple and the difficult types are separated ; an elementary course gives a treatment free from complex forms, while a supplementary section provides for the preparation of college requirements. The Lowest Common Multiple and its immediate application to Addition of Fractions form a single chapter, and the plan suggested by this order has proved to be natural, practical, and sound. The classification of Simple Simultaneous Equations and of the Theory of Exponents is both new and teachable, and the logical arrangement of Affected Quadratic Equations provides a chapter that, while omitting no essential, gives a brief and clear general treatment of this important subject. Throughout the early chapters exercises for oral drill are frequent, and their introduction confines the simplest types 3 rz*yf\£yc\*Jtz 4 PREFACE of the problems to the oral discussion of the classroom. Such discussions are of great value in a live class, and if supple- mented by a practice of reading by members of the class each step of every new illustrative solution, a most gratifying progress results. The written exercises consist of new prob- lems carefully graded, and the frequent reviews are constructed on the lines of recent entrance questions of the leading col- leges and universities. The treatment of Graphs is full and complete, but is free from those elements of the advanced discussions that so frequently confuse the young student. A comprehensive introduction of the common Physical Formulas familiarizes the student with a practical branch of applied algebra; and this is accomplished without assuming that the teacher is an expert in the laboratory practice of that science. In those subjects where several methods of procedure are possible, the text offers, as a rule, but one. To select arbi- trarily and to suggest one method as the best of many is a matter of personal choice and opinion, and the only claim for the single methods chosen in the following chapters is that they are uniformly practical. The text is planned on the theory that one practical method thoroughly mastered is sufficient for the needs of the young student, and that the elementary classroom has neither time nor need for those comparative discussions that interest only the mature mind. The author gratefully acknowledges his indebtedness to those friends whose suggestions and encouragement have been of material aid in the preparation of this text. FREDERICK H. SOMERVILLE. The William Penn Charter School, Philadelphia . CONTENTS CHAPTER PAGE I. Introduction, Symbols, Negative Numbers ... 9 Symbols of Operation 11 Symbols of Quality 13 Negative Numbers 14 Algebraic Expressions 20 II. Addition, Parentheses 22 Addition of Monomials 22 Addition of Polynomials 24 Parentheses 26 III. Subtraction, Review 32 Subtraction of Polynomials . . .... 34 General Review 35 IV. Multiplication . . . . 38 Multiplication of a Monomial by a Monomial ... 42 Multiplication of a Polynomial by a Monomial . . 43 Multiplication of a Polynomial by a Polynomial . . 45 Multiplication of Miscellaneous Types .... 47 V. Division, Review 52 Division of a Monomial by a Monomial .... 54 Division of a Polynomial by a Monomial .... 55 Division of a Polynomial by a Polynomial . . . ,56 General Review^ 62 VI. The Linear Equation, The Problem . . .64 General Solution of the Linear Equation .... 68 The Solution of Problems 70 VII. Substitution 84 The Use of Formulas 87 VIII. Special Cases in Multiplication and Division ... 89 Multiplication 89 Division 95 5 CONTENTS IX. Factoring Expressions having the Same Monomial Factor Trinomial Expressions Binomial Expressions Expressions of Four or More Terms factored by Grouping Supplementary Factoring X. Highest Common Factor Highest Common Factor of Monomials . Highest Common Factor of Polynomials by Factoring XI. Fractions, Transformations . The Signs of a Fraction . Transformations of Fractions XII. Fractions (Continued), Lowest Common Multiple, Lowest Common Denominator, Addition of Fractions Lowest Common Multiple Lowest Common Multiple of Monomials . Lowest Common Multiple of Polynomials by Factoring Lowest Common Denominator . . Addition and Subtraction of Fractions . XIII. Fractions (Continued), Multiplication, Division, The Com plex Form Multiplication of Fractions ..... Division of Fractions The Complex Form XIV. Fractional and Literal Linear Equations, Problems Special Forms . . . . * . Problems leading to Fractional Linear Equations XV. Applications of General Symbols, Review The General Statement of a Problem Use of Physical Formulas Transformation of Formulas . General Review XVI. Simultaneous Linear Equations, Problems Elimination by Substitution CONTENTS CHAPTER PAGE Elimination by Comparison 176 Elimination by Addition or Subtraction .... 176 Systems involving Three or More Unknown Quantities . 178 Fractional Forms 181 Literal Forms . 185 Problems producing Simultaneous Linear Equations . 187 The Discussion of a Problem 193 XVII. Graphical Representation of Linear Equations . . .197 Graph of a Point 197 Graph of a Linear Equation in Two Unknown Numbers . 200 Graphs of Simultaneous Linear Equations . . . 202 XVIII. Involution and Evolution 206 Involution 205 Evolution 209 Square Root of Polynomials 213 Square Root of Arithmetical Numbers . . . .217 XIX. Theory of Exponents 221 The Zero Exponent 221 The Negative Exponent . . . . . . 222 The Fractional Form of the Exponent .... 223 Applications 227 XX. Radicals, Imaginary Numbers, Review .... 236 Transformation of Radicals 237 Operations with Radicals 242 Equations involving Radical Expressions . • . . . 252 Imaginary and Complex Numbers 254 Operations with Imaginary Numbers .... 255 General Review . . 261 XXI. Quadratic Equations 266 Pure Quadratic Equations 266 Affected Quadratic Equations 268 Discussion of Affected Quadratic Equations . . .277 Graphs of Affected Quadratic Equations .... 282 XXII. The Quadratic Form,. Higher Equations, Irrational Equa- tions 286 The Quadratic Form ' . .286 CONTENTS CHAPTER PAGE Higher Equations solved by Quadratic Methods . . 286 Irrational Equations 292 Physical Formulas involving Quadratic Equations . 293 XXIII. Simultaneous Quadratic Equations, Problems . 294 Solution by Substitution 294 Solution by Comparison and Factoring . . . 295 Solution of Symmetrical Types . . . . 296 Solution of Miscellaneous Types 298 Graphs of Quadratic Equations in Two Variables . 302 Problems producing Quadratic Equations . . . 310 XXIV. Ratio, Proportion, Variation 315 Katio 315 Proportion . 317 Variation 330 XXV. The Progressions 338 Arithmetical Progression 338 Geometrical Progression 348 XXVI. The Binomial Theorem — Positive Integral Exponent . 358 Proof of the Binomial Formula 359 Applications 361 XXVII. Logarithms 364 The Parts of a Logarithm 365 The Use of the Four-place Table 367 The Properties of Logarithms 374 The Cologarithm 376 Use of Logarithms in Computations .... 377 Miscellaneous Applications of Logarithms . . . 379 XXVIII. Supplementary Topics, Review . . . . .385 The Remainder Theorem 385 The Factor Theorem 385 The Theory of Divisors of Binomials . . . . 387 H. C. F. of Expressions not readily Factorable . . 389 L. C. M. of Expressions not readily Factorable . . 392 Cube Root of Polynomials 394 Cube Root of Arithmetical Numbers .... 396 General Review 398 INDEX . 405 ELEMENTARY ALGEBRA CHAPTER I INTRODUCTION. SYMBOLS. NEGATIVE NUMBERS 1. The definite number symbols of arithmetic, 1, 2, 3, 4, 5, etc., are symbols that express in each case a number with one definite value. Thus, 3 units, 5 units, 7 units, etc., represent particular groups, the symbols, 3, 5, and 7 having each a particular name and value in the number system that we have learned to use. 2. The general number symbols of algebra are symbols that may represent not one alone but many values, and for these general symbols the letters of the alphabet are in common use. THE ADVANTAGE OF THE GENERAL NUMBER SYMBOL 3. Many of the familiar principles of arithmetic may be stated much more briefly, and usually with greater clearness, if general or literal number symbols are employed. To illustrate : (a) The area of a rectangle is equal to the product of its height, or altitude, by its length, or base. Or, arith- metically, Area = altitude x base. ** Using only the first letters of each word, we may write, Area = a x 6, b and this latter expression, while equally clear in meaning, serves as a general expression for obtaining the area of any rectangle with any values of a and b. 9 10 INTRODUCTION. SYMBOLS. NEGATIVE NUMBERS (6) The familiar problem of simple interest gives arithmetically, Interest = principal x rate x time. . A much simpler form of expressing the same principle is obtained here as above, using only the first letters of each element for the general expression. Thus : I = p x r x t. (c) Two principles relating to the operation of division in arithmetic may be clearly and briefly expressed by the use of literal symbols. Thus : Dividend -f- divisor = quotient. That is, — = q. d divisor x quotient = Dividend. Or, d x q = D. 4. In a more extended manner the literal symbol permits a breadth and power of expression not hitherto possible with the number symbols of arithmetic. Many problems involving unknown quantities are readily stated and solved by means of general symbols, and clearness of expression in such problems is invariably gained by their use. To illustrate a common use of literal symbols, consider the following problem : Two brothers, John and William, possess together 200 books, and John has 20 books more than William. Write expressions that clearly state these conditions. (Compare carefully the two methods of expression.) The Arithmetical Expression (a) The number of John's books + the number of William's books =200 (6) The number of John's books — the number of William's books = 20 The Algebraic Expression Let us assume that x = the number of books that John has, and that y = the number of books that William has. Then from the conditions given in the problem : (a) x + y = 200. (6) x-y = 20. THE SYMBOLS OF OPERATION 11 By a simple process the values of x and y are readily determined. From this parallel between arithmetical and algebraic forms of ex- pression the brevity and the advantage of the literal or general symbol, for number is clearly manifest. The later processes of algebra will con- stantly furnish the means wherewith we may broaden our power of expression, and the meaning of algebra will be interpreted as merely an extension of our processes with number. THE SYMBOLS OF OPERATION 5. The principal signs for operations in algebra are identical with those of the corresponding operations in arithmetic. 6. Addition is indicated by the " pins " sign, + . Thus, a + b is the indicated sum of the quantity a and the quantity b. The expression is read " a plus 6." 7. Subtraction is indicated by the " minus " sign, — . Thus, a — b is the indicated difference between the quantity b and the quantity a. The expression is read " a minus 6." 8. Multiplication is usually indicated by an absence of sign between the quantities to be multiplied. Thus, ab is the indicated product of the quantities a and b. abx is the indicated product of the quantities a, 6, and x. Sometimes a dot is used to indicate a multiplication. Thus, a • b is the product of a and b. The ordinary symbol, " x ," is occasionally used in algebraic expression. An indicated product may be read by the use of the word " times " or by reading the literal symbols only. Thus, ab may be read "a times 6," or simply " a&." 9. Division is indicated by the sign " -j-," or by writing in the fractional form. 12 INTRODUCTION. SYMBOLS. NEGATIVE NUMBERS Thus, a -*- b is the indicated quotient of the quantity a divided by the quantity b. - is the fractional form for the same indicated quotient. b Both forms are read u a divided by 6." 10. Indicated operations are of constant occurrence in alge- braic processes, for the literal symbols do not permit the combining of two or more into a single symbol as in the case of numerals. Thus: Arithmetically, 5 + 3 + 7 may be written " 15," for the symbol 15 is the symbol for the group made up of the three groups, 5, 3, and 7. Algebraically, a + b + c cannot be rewritten unless particular values are assigned to the symbols a, &, and c. The sum is an indicated result. Algebraic expression, therefore, confines us to a constant use of indicated operations, and we must clearly understand the meaning of : I. An Indicated Addition, a +- b. II. An Indicated Subtraction, a — b. III. An Indicated Multiplication, ab. IV. An Indicated Division, -• b 11. Equality of quantities or expressions in algebra is indi- cated by the sign of equality, =, read "equals," or "is equal to." Thus, a + 6 = c + d is an indicated equality between two quantities, a + b and c + d. 12. If two or more numbers are multiplied together, each of them, or the product of two or more of them, is a factor of the product ; and any factor of a product may be considered the coefficient of the product of the other factors. Thus : In 5 a, 5 is the coefficient of a. In acmx, a is the coefficient of cmx, In ax, a is the coefficient of x, or ac may be the coefficient of or x is the coefficient of a. mx, etc. THE SYMBOLS OF QUALITY 13 Coefficients are the direct results of additions, for 5 a is merely an abbreviation of a + a + a + a + a. 4 xy is an abbreviation of xy + xy + xy + xy. If the coefficient of a quantity is " unity " or " 1," it is not usually written or read. Thus, a is the same as 1 a. xy is the same as 1 xy. 13. The parenthesis is used to indicate that two or more quantities are to be treated as a single quantity. The ordinary form, ( ), is most common. For clearness in the discussion of elementary principles, the parenthesis will frequently be made use of to inclose single quantities. 14. An axiom is a statement of a truth so simple as to be accepted without proof. Two of the axioms necessary in early discussions are : Axiom 1. If equals are added to equals, the sums are equal. Axiom 2. If equals are subtracted from equals, the re- mainders are equal. THE SYMBOLS OF QUALITY 15. In scientific and in many everyday discussions greater clearness and convenience have resulted from a definite method of indicating opposition of quality. For example : Temperature above and below the zero point, Latitude north and south of the equator, Assets, or possessions, and liabilities, or debts, in business, etc., represent cases in which direct opposites of quality or kind are under discussion; hence, a need exists for a form of ex- pression that shall indicate kind as well as amount, quality as well as quantity. To supply this need the plus and minus signs are in general use, their direct opposition making them useful as signs of 14 INTRODUCTION. SYMBOLS. NEGATIVE NUMBERS quality as well as of operation; and we will now consider their use as THE + AND - SIGNS OP QUALITY 16. A few selected examples of common occurrence clearly illustrate the application of quality signs to opposites in kind. In the laboratory : % Temperature above 0° is considered as +. Temperature below 0° is considered as — . In navigation : Latitude north of the equator is considered as +. Latitude south of the equator is considered as — . In business administration : Assets, or possessions, are considered as + . Liabilities, or debts, are considered as — . The following illustration emphasizes the advantage of in- dicating opposites in kind by the use of the -J- and — signs of quality. A thermometer registers 10° above at 8 a.m., 15° above at 11 a.m., 5° below at 4 p.m., and ' •■ ' , M 1am -4- 1 o 10° below at 10 p.m. At the right we have tabu- lated the conditions in a concise form made possible .,_ , _ i ^ u .1. £ ,. 4 • 10 p.m. — 10° only through the use of quality signs. By applying this idea of opposition to arithmetical numbers we may establish NEGATIVE NUMBERS 17. It is first necessary to show that there exists a need for a definite method of indicating opposition of quality in number. Consider the subtractions : (1) 5-4= (2) 5-5= (3) 5-6 = There are three definite cases included. In (1) a subtrahend less than the minuend. In (2) a subtrahend equal to the minuend. In (3) a subtrahend greater than the minuend. NEGATIVE NUMBERS 15 The first two cases are familiar in arithmetical processes, but the third raises a new question, and we ask, 5 — 6 = what number ? 18. On any convenient straight line denote the middle point by " 0," and mark off equal points of division both to the right and to the left of 0. The two directions from are dearly defined cases of opposition, and this / opposition may be indicated by marking the suc- cessive division points at the right of with numerals having plus signs, -5 -4 -3 -2 -1 O +1 +2 +3 +4 +5 1 and, similarly, the division points left of with numerals hav- ing minus signs. The result is a series of positive and negative numbers established from a given point which we may call " zero." With this extended number system we may at once obtain a clear and logical answer for the question raised above. The minuend remaining the same in each case, the result for each subtraction is established by merely counting off the subtrahend from the minuend, the direction of counting being toward zero. Therefore, For (1) The subtraction of a positive number from a greater positive number gives a positive result. Or, 5-4 = 1. For (2) The subtraction of a positive number from an equal positive number gives a zero result. Or, 5-5 = 0. For (3) The subtraction of a positive number from a less posi- tive number gives a negative result. Or, 5 - 6 = - 1. 16 INTRODUCTION. SYMBOLS. NEGATIVE NUMBERS We may conclude, therefore, that : 19. The need of a negative number system is that subtraction may be always possible. On the principle of opposition the idea of negative number is as firmly established as that of positive number, and the definition of algebra as an extension of number is even further warranted. 20. The extension of arithmetical number to include nega- tive as well as positive number establishes algebraic number. 21. It is often necessary to refer to the magnitude of a num- ber regardless of its quality, the number of units in the group being the only consideration. Thus, -f 5 and — 5 have the same magnitude, or absolute value, for each stands for the same group idea in the numeral classification. They differ in their qualities, however, being exact opposites. In like manner, -f a and — a, + xy and — xy, are of the same absolute value in each respective case, no matter what values are represented by a or by xy. POSITIVE AND NEGATIVE NUMBERS COMBINED The simplest combination of algebraic numbers is addition, and since two groups of opposite kinds result from the positive and negative qualities, we must consider an elementary dis- cussion of algebraic addition under three heads. I. Positive Units + Positive Units. If a rise in temperature is 16° -f 15° and a further rise of 10° occurs, 4- 10° we have a final reading of . . . . . + 25° II. Negative Units + Negative Units If a fall in temperature is 12° — 12° and a further fall of 16° occurs, — 16° we have a final reading of — 28 c POSITIVE AND NEGATIVE NUMBERS COMBINED 17 III. Positive Units + Negative Units. If a rise in temperature is 20° and -f 20° an immediate fall of 15° occurs, — 15° we have finally + 5° If a fall in temperature is 30° and _ 30° an immediate rise of 24° occurs, + 24° we have finally — 6° The student should verify these illustrations by making a sketch of a thermometer and applying each case given above. In general, addition of algebraic numbers results as follows : (1) If a and b are positive numbers : (+«) + (+&) = +« + &. i . . . Numerical Illustration : V * 5 , p " f " 3 J (+5) + (+3) = + 5 + 3=+8. (2) If a and b are negative numbers : (-«) + (- 6) = -a- b , ^ , , , 1 , , O Numerical Illustration : V^ 3 Q 5 J (_5) + (-3) = -5-3= -8. (3) If a is positive and b negative : (+«) + (-&)= +a-b. +5. . . Numerical Illustration : ^-+2 (+5) + (-3)=+5-3=+2. (4) If a is negative and b positive : (-a) + (+6) = -a + 6. -5 Q Numerical Illustration: r +3 ' — 2— '" (_5) + (+3)=-5 + 3 = -2. From the four cases we may state the general principles for combining by addition any given groups of positive quantities, negative quantities, or positive and negative quantities : SOM. EL. ALG. — 2 18 INTRODUCTION. SYMBOLS. NEGATIVE NUMBERS 22. The sum of two groups of plus, or positive, units is a posi- tive quantity. 23. TJie sum of two groups of minus, or negative, units is a negative quantity. 24. The sum of two groups of units of opposite quality is posi- tive if the number of units in the positive group is the greater, but negative if the number of units in the negative group is the greater. * JTrom Art. 24 we derive the following important principle : 25. The sum of two units of the same absolute value but oppo- site in sign is 0. In general: (+«) + (-«) = + a- a = 0. Numerical Illustration : (+5) + (-5) = +5-5 = 0. Exercise 1 1. If distances to the right are to be considered as posi- tive in a discussion, what shall we consider distances to the left? 2. If the year 20 a.d. is considered as the year " -f 20," how shall we express the year 20 b.c. ? 3. Draw a sketch of a thermometer, and indicate upon it the temperature points + 35°, — 18°, + 12°, and — 12°. 4. On your sketch determine the number of degrees passed through if the temperature rises from 8° to 35° ; * from to 15°; from - 15° to 10°. 5. On your sketch determine the number of degrees passed through in a fall of temperature from 35° to 10° ; from 20° to -10°. * Note that the quality of a unit is considered plus when no sign of quality is given. POSITIVE AND NEGATIVE NUMBERS COMBINED 19 6. Show by your sketch that a rise of 25° followed by a fall of 15° results in an actual change of 10° from the starting point. 7. In example 6 is the conditi6n true for the single case when you begin at 0°, or is the result the same no matter at what point you begin ? Illustrate. 8. Show that a fall of 18° from the zero point succeeded by a rise of 30° results in a final reading of + 12°. 9. Show that a rise of 10° from the zero point succeeded by a further rise of 18°, and later by a fall of 40°, results in a final reading of 12° below zero. Express " 12° below zero " in a simpler form. 10. From your sketch determine the final reading when, after a rise of 40° from 0, there occurs a fall of 15°, a succeed- ing rise of 7°, and another fall of 35°. Express these changes with proper signs. 11. C is a point on the line AC B AB. A traveler starts at C, goes +10 ^ 10 miles toward B, turns back 7 < ~+9 miles toward A, and returns 9 miles toward B. Determine his final distance from G, and also his position at either the right or the left of C. (Assume distances to the right of C as +.) Would the information given be sufficient to determine the result without using the sketch ? 12. Determine the result of a journey 8 miles from C toward B, returning 6 miles toward A. Give the total distance trav- eled and the final position. 13. Determine the result of a journey 9 miles from C toward B, 16 miles back toward A, and then 8 miles toward B. Make a drawing similar to that illustrating example 11, and prove your answer by reference to the drawing. 20 INTRODUCTION. SYMBOLS. NEGATIVE NUMBERS 14. Determine the result of a journey 17 miles toward B, 10 miles back toward A, 3 miles more toward A, back 19 miles toward B, the starting point being at G. Make a drawing illustrating the entire journey. 15. With the same distances and directions as in problem 14, determine the result if the starting point had been at A, giv- ing for the answer the final distance of the traveler from G as well as from A. (Assume that from G to A is 30 miles.) ALGEBRAIC EXPRESSIONS 26. An algebraic expression is an algebraic symbol, or group of algebraic symbols, representing some quantity. An ex- pression is numerical when made up wholly of numerical symbols, and is literal when made up wholly or in part of literal symbols. 20 + 10 — 13 x 2 is a numerical expression. ab + mn — xy and 5 a — 6 ex — 12 are literal expressions. 27. The parts of an expression connected by the + or - signs are the terms. In the expression ab + ( C - d) _ mn + « + » _ g .4 - « » -P b -y 2(d + k) the terms are ab, + (c — d), — mn, + a ~ x , and b - y 2(d + k) A parenthesis, or a sign of the same significance, may inclose a group as one term. A fraction as an indicated quotient is also a single term. The sign between two terms is the sign of the term following. The sign of the first term of an expression is not usually written if it is +. In the term ab of the given expression both the sign, +, and the coefficient, 1, are understood. ALGEBRAIC EXPRESSIONS 21 28. Terms not differing excepting as to their numerical coefficients are like or similar terms. 5 ax and — Sax are similar terms. 3 ab and 14 mn are dissimilar terms. 29. The common expressions of algebra are frequently- named in accordance with the number of terms composing them. The following names are generally used : A Monomial. An algebraic expression of one term. 4 a, 5 win, — Sxy, and 17 xyz are monomials. A Binomial. An algebraic expression of two terms. a + b, 3 m — x, 10 — 7 ay, — 4 mnx + 11 z are binomials. A Trinomial. An algebraic expression of three terms. Sx — 7 m + 8, 4 ab — 11 ac — 10 mny are trinomials. A Polynomial. Any expression having two or more terms. While the binomial and the trinomial both come under this head they occur so frequently that common practice gives each a distinct name. Expressions having four or more terms are ordinarily named polynomials. Oral Drill Kead the following algebraic expressions : 1. a + 3x — 4:mn-\-cdn — 3xy. 2. 2 mx — 3 acd -f- (a -f- x) — (m -f- w). 3. (a — x) — (c — d) + (m — y + z). 4. 5ayz — 2(2 m — n) -f- a(a — x) -f 3 a(2 a — 3y). 5. — mnx + 3 a(c — 2 d + 1) — (a — m + n)sc -f- 12(4 — sc). 6. (a - X )( c + y)-(x + 2)0/ - 3) - ■(«+ l)(a> + 2)(* + 3). & 2/ 3(c-d) "c-2/ CHAPTER II ADDITION. PARENTHESES 30. Addition in algebra, as in arithmetic, is the process of combining two or more expressions into an equivalent ex- pression or sum. The given expressions to be added are the addends. THE NUMBER PRINCIPLES OF ADDITION 31. The Law of Order. Algebraic numbers may be added in any order. In general : ' a + 6 = b + a. Numerical Illustration : 5 + 3 = 3 + 5. 32. The Law of Grouping. The sum of three or more alge-, braic numbers is the same in whatever manner the numbers are grouped. In general: a+b+c=a+(b+c) =(a+6) + c=(a+c) + 6. Numerical Illustration: 2+3 +4 =2 +(3 +4) = (2 +3) +4= (2 +4) +3. A rigid proof of these laws is not necessary at this point ; but may be reserved for later work in elementary algebra. The law of order is frequently called the commutative law, and the law of grouping is called the associative law. ADDITION OF MONOMIALS The principles underlying the addition of the simplest forms of algebraic expressions have already been developed, and they are readily applied in the more difficult forms of later work. 22 ADDITION OF MONOMIALS 23 (1) The sum of like quantities having the same sign, all -f or all - . By Articles 22 and 23 : + 7 + 12a -7 -12a + 3 + 5a -3 — 5a + 10 + 17 a -10 -17 a In general : 33. The coefficient of the sum of similar terms having like sigiis is the sum of the coefficients of the given terms with the common sign. (2) The sum of like quantities having different signs, some + and some — . By Article 24 : +7 + 12 a - 7 - 12 a -3 - 5q + 3 + 5q + 4 + la -4 - la In general : 34. The coefficient of the sum of similar terms having unlike signs is the arithmetical difference between the sum of the +- co- efficients and the sum of the — coefficients, with the sign of the greater. Oral Drill Add orally : 1. 5a Sa 2. 6a 9a 3. 4a; 7x 4. 9mn Smn 5. Sbcd 11 bed 6. 7 amx 19 amx 7. 17 cmy 19 cmy 8. -4a6 -3a6 9. — 8 en —5 en 10. — Sxz —xz 11. — 15cz -11 cz 12. —Sexy — 14: cxy 13. -12 abx —19 abx 14. — 5 dny — dny 15. 10 ac —6ac 16. — 19 mx 9mx 17. -Scy 15 cy 18. 12 ax — Sax 19. — 15 cz 15 cz 20. 27 axy — 16 axy 21. — Icdmn 45 cdmn 24 ADDITION. PARENTHESES If the sum of three or more terms is required, we apply the law of grouping (Art. 32), and separately add the + and the — terms. Thus, 5-8 + 11 -16 + 3 = 5+ 11 +3-8-16 = 19-24 = — 5. Result. Let the student apply this principle in the following : Oral Drill Add orally : 1. 2-7 + 6-9. 7. — 8x + 9x-6x + 5x. 2. -8 + 3 — 12 + 7. 8. -llac + 14ac-ac-2ac. 3. —9 + 8 — 15 — 11. 9. 3 mnx — 8 mnx — 9 mnx + 13 mnx. 4. 7-14 — 3 + 10. 10. -1 cy-§cy + 3ey-llcy. 5. 13-18 + 7-21. 11. 3a-5a + 8a-lla-3a + a. 6. 3 a — 4a + 6 a — 3a. 12. — 4cxy-\-3xy—7xy+xy—xy-{-8xy. 13. 5 am — 8 am — 24 am + 13 am — am + 6 am — 11 am. 14. 6 m — 7 — 4 m + 11 — 5 m — 17 — m + 5 ra + 13. 15. — 4 ex + 8 ex — 3 ex + 2 ex — 3 ex + c# — 15 ex — 14 'ex. ADDITION OF POLYNOMIALS The principles established for the addition of monomials apply directly to the addition of polynomials. Illustrations : 1. Find the sum of 5a + 76 — 2c, 2a — 36 + 8 c, and -3a + 26-10c. In the customary form : 5a + 76— 2c 2a-36+ 8c -3a + 26-10c 4a + 6&- 4c Result. For the sum : the coefficient ofa= 5 + 2— 3 = the coefficient of&= 7-3+2 = the coefficient ofc=-2+8-10 = 4, 6, -4. ADDITION OF POLYNOMIALS 25 It frequently happens that not all of the terms considered are found in each of the given expressions, in which case we arrange the work so that space will be given to such terms as an examination shows need for. 2. Add 4a + 36 + 3m, 2b + 3 c — d, 2a + 3d + 2m — x, and 5b — 5m — 3 x. 4a-f 36 + 3m + 26 + 3c- d 2a +3d+2m- x + 5 6 — 5 m — 3% 6a + 10 6 + 3c + 2(Z -4 a; Result. The coefficient of the wi-term in the sum being 0, that term disappears. In general, to add polynomials : 35. Write the given expressions so that similar terms shall stand in the same columns. Add separately, in each column, the positive coefficients and the negative coefficients, and to the arithmetical difference of their sums prefix the proper sign. Collecting terms is another expression for adding two or more given expressions. 36. Checking results. The accuracy of a result may be checked by substituting convenient numerical values for each of the given letters. The substitution is made both in the given expressions and in the sum obtained, and the work may be considered accurate if both results agree. Illustration : To check the sum of 5a + 76 + 2c and 2a — 36 — 5c, let a = 1, 6 = 2, and c = 1. Then 5a + 76 + 2c = 5 + 14 + 2=21 and 2a-36-5c = 2- 6-5 = -9 Whence, 7 a + 4 6 - -3c = 7+ 8- -3 = 12. • Exercise 2 Find the sum of : 1. 2. 3. 4. 3a + 76 _7a_f9 4 n — 9 cz — acm-f- 3xyz 5a-36 12^-1 — n -f- 9 cz 2 acm—YI xyz 26 ADDITION. PARENTHESES 5. 6. 7. 5ab- - 7 ac — ac? - 2cx-16 m — nx —z 11 ab + 2 ad ay + 20 c# 3nx — 5my + z 8. 4a+3 6, 5a — 2 6, — 7a+4 6, and 4 a —11 6. 9. 3a — 2c — jb, 4a — c + 7 x, and — a — 5c+9sc. 10. 4 x + 3 2/ — 11, — 5 a — 2 y — 8, and x + 19. 11. 4 m — 2 n — sc, — 3 n — 4 a?, 2 m + 3 n, and 7 % — 9 x. 12. 3a — 6 — 4c + 7, a — 4 6 — c+6, and —2a— 6 + 6c— 14. 13. — 3a+76— 2c — 7, c— 3a— 6+8, and 7—4 6 — a + 3 c. 14. a — 3y + m-2x + 7, 36-2?/ + 2a-4, 7x — 2a — y, and66 — 5a-2a-12. Collect similar terms in : 15. 5a6 + 66c — 3 am — 8a; + 4a6 — ac-f 3cd — am + 3ab—2cd—x. 16. 4m — 9 + d — 2/ + 3 — 2d — 6 — 5m — 3y-\-2m -y-d+16. 17. 4a6c— 8 6cd — 5cd*a; — 18 + 13 6cd* — 18a6c— 5cdx -7 + 12a6c. 18. — 9 dx — mn — 3 ay + 4 6w — 19 — 14 ay + 5 bn — 15 mn + 8 da; — 10 bn + 14 mn + 16 ay. PARENTHESES 37. The parenthesis is used in algebra to indicate that two or more quantities are to be treated as a single quantity. Thus, a + (b + c) is the indicated sum of a and the quantity, b + c. a — (6 + c) is the indicated difference between a and the quantity, b + c. 38. A sign + or — before a parenthesis indicates an opera- tion to be performed. a + (6 4- c) is an indicated addition, a — (6 + c) is an indicated subtraction. PARENTHESES 27 The Parenthesis preceded by the Plus Sign Consider the expression 20 + (10 + 5). By first adding 10 and 5, 20 + (10 + 5) = 20 + 15 = 35. Adding separately, 20 + (10 + 5) = 20 + 10 + 5= 35. The result is clearly the same whether the parenthesis is removed before or after adding. Again, consider the expression 20 + (10 — 5). By first subtracting 5 from 10, 20 + (10 - 5) = 20 + 5 = 25. Or, subtracting separately, 20 -f (10 - 5) = 20 + 10 - 5 = 25. And, again, the same result from each process. In general symbols : For the + parenthesis : a + (b + c) = a + b + c. a + (& — c) = a + b — c. In general : 39. If a + ( ) is removed, the signs of its terms are not changed. The Parenthesis preceded by the Minus Sign Consider the expression 20 — (10 + 5). By first adding 10 and 5, 20 - (10 + 5) = 20 - 15 = 5. Or, subtracting separately, 20 — (10 + 5) = 20 — 10 — 5 = 5. The result is the same from each process. Again, consider the expression 20 — (10 — 5). In this expression we are not to take all of 10 from 20, only the differ- ence between 10 and 5, i.e. (10 — 5), being actually subtracted. There- fore if all of 10 is subtracted, we must add 5 to the result. By first subtracting 5 from 10, 20 - (10 - 5) = 20 - 5 = 15. Or, separately, 20 - (10 - 5) = 20 - 10 + 5 = 15. And, again, the same result from each process. In general symbols : For the — parenthesis : a — (b + c) = a — b — c. a — (b — c) = a — b + c. In general : 40. Ifa—() is removed, the sign of every term in it must be changed. 28 ADDITION. PARENTHESES Two important principles must be kept constantly in mind : (1) The sign before a parenthesis indicates an operation of either addition or subtraction. (2) The sign of a parenthesis disappears when the parenthesis is removed. For example : 15 - (7 - 4 + 2) = 15 - 7 + 4 - 2. The — sign of 7 in the result is not the original — sign of the paren- thesis. The original sign of 7 was + , and this sign was changed to — by the law of Art. 40. The sign of a parenthesis shows the operation, and disappears when we perform it. Oral Drill Give orally the results on the following : 1. (+5) + (+2). io. (_7a)-(+5a). 2. (+5)-(+2). 11. _(_5a)-(-7a). 3. (+5) + (-2). 12. -(+6m) + (-6m). 4. (+5) -(-2). 13. (-9 a?) -(9 a;). 5- (-7) + (+6). 14. (-9x)-(-9x). 6. (+5)-(-9). 15. -(-19ran)-(-20ran). 7. (-10) -(-19). 16. -(Jxyz)-(-Sxyz). 8. (+16) + (-17). 17. '(-16 6c)-(-116c). 9. (-14) -(-13). 18. -(±2xy)-(-15xy). 19. (_5)-(-2)-(-3) + (-4). 20. a-(4a) + (-4a) + (-a). 21. -5x + (-2x)-(-x)-(3x) + (-x). 22. _(-l)-(l)_(_l)_(l) + (-l)-(-l). 23. 2-(+3)-(-4) + (-6)-(-6). 24. -(-2a:)-(+3a>)-(2as)-(-3a!) + (-2a>). 25. — (— mn) + (— 3 raw) — (— 4 ran) + (— raw). 26. 2 a?-(- 3 «)-(+2o;) + (- 3 *)-( + ») + (-«). PAKENTHESES 29 Parentheses within Parentheses It is frequently necessary to inclose in parentheses parts of expressions already inclosed in other parentheses, and to avoid confusion different forms of the parenthesis are used. These forms are: (a) the bracket [ ], (6) the brace { \, (c) the vin- culum — . Each has the same significance as the paren- thesis. That is, (a + &) = [« + 6] = {a + b] = a + b. In removing more than one of these signs of aggregation from an expression : 41. Remove the parentheses one at a time, beginning either with the innermost or with the outermost. If a parenthesis pre- ceded by a minus sign is removed, the signs of its terms are changed. Collect the terms of the result. Illustration : Simplify 4m- [3 m-(n+2)-4]-Sm+3n—(7i-l)j+n. 4 m - [3 m - (n + 2) - 4] - {m + 3 n - (n - 1)} + n = 4m- [3 m - n-2 - 4] - {m + 3 n — n + 1} + n = 4m- 3m + n + 2 +4 -{m + Sn - ra + 1} +w = 4m— 3 m + n + 2 +4 — m — 3 n + n — 1 + n = 6. Result. Exercise 3 Simplify : 1. 4a+(3a-8). 7. 9 a + [3 a; -(a -2)]. 2. 5x-(3x + 7). 8. 9z-[3a+-(a;+-2)]. 3. 2c+(c-8)-6. 9. ±a-\a-(-a + 7)\. 4. 3m — (2 m — ?i+-3). 10. 2ac — (ac — 7 + 3ac). 5. 4z-(8-3z)-ll. 11. 5n+[9-{3n-n + 2\]. 6. — 5m— (— m— p-6). 12. — 6 — (— xy+[ — 2 xy— (xy+3)~\). 13. -3a-[36-f2a-(6&-a)n. 14. m — [m + \m— (m — m — 1)}]. 30 ADDITION. PARENTHESES 15. -(2m + l)-(-[-m + m-3]). 16. 7x-[2+(-3x-)x-(-6x-x-7)-x\)]. 17. _(_ a _i)_2a-(a + l)-3a-5-8. 18. l_(_l) + )_l-lH-a-(-l)i, 19. 2-(-l)+j-l-l + a-(-l)-(-a)J. 20. -H-(_l)_{l + l-(a + l)+(-l)t. 21. 3m- [2m— \m — n — 3m— 9— 2 m— 7+3 nj — (— m— 5)]. 22. 2a-(36 + c)-2a + (c-36) -Jc-(2a-36)-[a-(6-c)] 23. l_(l+a)-[-j-[(-l-a-l-3a)-a]-lS-a]. 24. ax— (- aa;)— | — (—ax)\ — [ — (—\ — ( — ax)\)]. 25. (m- 1)— j— (— J — (— m + 1) — nj — m)— raj. Inclosing Terms in Parentheses This operation is the opposite of the removal of parentheses, hence we may invert the principles of Arts. 39 and 40 and obtain : 42. Any number of terms may be inclosed in a + ( ) without changing the signs of the terms inclosed. 43. Any number of terms may be inclosed in a — ( ) provided the sign of each term inclosed be changed. Illustrations : (Art. 42. ) 1. a + b + c-d + e = a + b + (c-d+e). (Art. 42.) 2. a + 6-c-d + e = a + & + (-c-d + e). (Art. 43.) 3. a + 6-c + d-e = a + &-(c-d + e). (Art. 43.) 4. a-b + c-d + e = a-b-(-c + d-e). PARENTHESES 31 Exercise 4 Inclose the last two terms of each of the following in a parenthesis preceded by the plus sign : 1. a + 3 6-|-9 c. 4 12 mn + 4 mx — 5 my -f- 16. 2. 4mH-3n-f-2j9 — 8. 5. 5c — 5d — x — mny. 3. 4ae + 6ad + 9ae-ll. 6. -9c + lld~6 + a». Inclose the last three terms in a parenthesis preceded by the minus sign : 7. 4a-56 + 3c-9. 11. 3x + ±y + 5z + 2. 8. 2a — 66 — c + x. 12. 4 ac — 5 6c + 3 bd + 10 ad. 9. 5a; — 8y + 3« — 11. 13. 12 — w + ra — 2p + x. 10. — 7 — 5ra + 3w — 2 p. 14. m + n—p + x — y + z. 15. — 2 — x -f a» — xy — yz. Without changing the order of the terms, write each of the following expressions first in binomials and then in trinomials. Before each parenthesis use as its sign the sign of the term which is to be' first in that parenthesis : 16. a-f-6 — c-fd — x — y. 17. a — c + d — m — n + z. 18. c — ra — x + y — z — 1. 19. a — d — e — m — n + x. 20. m — n—p — x — y — z. 21. ab — ac + cd — mn — np + mp. 22. 2a + 3c-f 4a , + 5a — 3y-2z. 23. 7a6 — 3ac + 4 6d' — 5 6c + 3 ad — 4ca\ 24. 5ay — 3az + 8z — 4:xy — 3xz + 2yz. 25. ama; — cny — bz — cnx — amy — dz. CHAPTER III SUBTRACTION. REVIEW 44. Subtraction is the inverse of addition. In subtraction we are given the algebraic sum and one of two numbers, and the other of the two numbers is required. The given sum is the minuend, the given number the subtrahend, and the required number the difference or remainder. 45. We may base the process of subtraction on the principle of Art. 40, for, The quantity to be subtracted may be considered as inclosed in a parenthesis preceded by a minus sign. Consider the example, From 10a + 36 + 7c take 6 a + b - c. By definition, the first expression is the niinuend, the second the subtrahend. Inclosing both expressions in parentheses, and replacing the word " take " by the sign of operation for subtraction, we have (10 a + 3 b + 7 c) - (6 a + 6 - c). Removing parentheses, 10a + 36 + 7c — 6a — b + c. Collecting terms, 4 a + 2 b + 8 c. Result. We have, therefore, changed the signs of the terms of the sub- trahend and added the resulting expression to the minuend. In practice the usual form would be Minuend : 10 a + 3 b + 7 c. Subtrahend : 6 a + b — c. Difference : 4a + 2& + 8c. Result. The change of signs should be made mentally. Under no cir- cumstances should the given signs of the subtrahend be actually altered. 32 SUBTRACTION 33 From the foregoing we make the general statement for sub- traction of one algebraic expression from another. 46. Place similar terms in vertical columns. Consider the sign of each term of the subtrahend to be changed, and proceed as in addition. Two important principles result from the process of sub- traction. 47. Subtracting a positive quantity is the same in effect as adding a negative quantity. In general : By Art. 40: (+ a) -(+&) = «-&. By Art. 39 : («j-a) + (- 6)=a-&. Numerical Illustrations : (+ 5) - (+ 3) = 5 - 3 = 2. (+ 5) + (- 3) = 5 - 3 = 2. 48. Subtracting a negative quantity is the same in effect as adding a positive quantity. In general : By Art. 40: (+«)-(- &) = « + b. By Art.39: (+ a) + (+ b)=a + b. Numerical Illustrations : (+ 6) - (- 3) = 5 + 3 = 8. (+ 5) + (+ 3) = 5 + 3 = 8. Oral DriU From 1. 13 2. 12 a 3. -9a 4. -3a 5. 13 ab 6. 15 a Take _4 5 a -2a — 6 a -1 ab -3x From 7. 9m 8. -7c 9. 1 xy 10. — Sac 11. — 19 mn 12. — 7 xz Take 13 m 8c - -5xy — Sac — 11 mn 23 xz From 13. — lac 14. + 7ac 15. 16. 17. abc 18. — 4 acx Take -f-7ac — 7ac 9x -9a — 9abc acx SOM. EL. ALG. 34 SUBTRACTION. REVIEW SUBTRACTION OF POLYNOMIALS 49. By application of the principle of Art. 46, we subtract one polynomial from another. Results in subtraction may be tested as shown in Art. 36. Exercise 5 1. From 5 a + 7 Take 3a-4 2. 8m-19 3 m- 11 3. - 7 c + 14 -9c + 14 4. -17ac-8 - 9ac + 9 5. From 3a + 46*-7 Take 3 6-9 6. 6x-$y x + 3y- z 7. -3m 2 a — 5 m + 7 8. a + b — c 9. From 7 a + 3 m — 9 take 2 a — 5 m 4- 8. 10. From 16a-|-6a;— 3y take a — 7 x — 15 y. 11. From 4a + 9a"-18 take -3a + 9d'-15. 12. From 5x—7y-\-3z take 3y — 1 z. 13. From — 7a + 6c-r-3cZ + 5 take 2 a — 5 c + 3. 14. From 11 a + 3 m — a; take 2 a + 7 # — m. 15. From 26 + 5c — 3n take — 2 a — 6 + 4 c — 3 n. 16. Subtract 15 aa? — 3 ay — 19 from 17 ax — hay — 11. 17. Subtract 10 — 3x — 1 y — z from z + ll — 2x-3y. 18. Subtract 5c + 6d — 4m + n from 3c — m. 19. Subtract 5a — 11 6 — 2c from 3a — 8 c + 2 ra. 20. From the sum of 3m — n + 2p and 3m-4n-5|) take the sum of m -f- 3 n —p and 3 m — 7 n + 6 p. 21. From the sum of 5x— 3y + 2 z -f 3 and 3 x + 7 subtract the sum of 3 a; — y 4- 2 and 7 — 2x — y-\-llz. 22. Subtract the sum of 4 a — 11 c + d and 3 b — c + 10 from the sum of36 + d — 8c and 4 a — 4 c + 9. 23. Take the sum of 3 m— n+2y, 2n— 3 m— 4^ and m+p— z from the sum of m — 3 3, 3 n +p, and p-f-m — 2y + n — z. GENERAL REVIEW 35 50. Addition and Subtraction with Dissimilar Coefficients. When the coefficients of similar terms are themselves dis- similar, the processes of addition and subtraction result in expressions with compound coefficients, i.e. coefficients having two or more terms. The principle is easily understood from the following illustrations : Addition Subtraction ax cm am cdz bx_ — acm cm — 9 amdz (a + b)x (1 — a) cm (a — c)m (c + 9 am) dz Exercise 6 Find the sum of : 1. am + dm. 4. — 17 mxy + 2pxy. 7. bcdx + 12 cdx. 2. cdx + 5 nx. 5. cdn — 7 dn. 8. 5 cdx — 12 bdx. 3. 3 bed — 2 mcd. 6. acx — 19 bx. 9. — mxy — 11 cxy. 10. 6a; + ay and ma; + ny. 13. 5 ab — 2 cd and 3 6 + 7 a\ 11. a# + nz and 6aj+pz. 14. 6 az -f- 5 by and raz — 3 y. 12. acm + 6n and dm + cfri. 15. 11 mxy + 5 ccz and ascy — acz. 16. 2 ftm + 3 a; + ?/ and 5m +cx — m?/. 17. 3 a6 + 7 ac + 11, m& — lie — 7, and 5 6 — nc + 3. 18. (3 + c)x + (2 c - 9)a> + (6 - 3 c)x + (a + c + 1)*. (Examples 1-12 inclusive will serve as an exercise for subtraction ; the first expression in each being the given minuend, the second expression, the subtrahend.) GENERAL REVIEW Exercise 7 1. Find the sum of 3 a + 2 6 — 6 m, 2 b — 5 c + x, — 4 a— 76 -f c — m, and a — 3 c + x. 2. Simplify 1-J1- [1- (1-1 + m)]-mj -m. 3. Add 3a + 26-c, 2a-36 + 3c, 4a + 36-7c, and -8a-46 + 5c. 36 SUBTRACTION. REVIEW 4. Subtract 4 x -f- 3 y — 2 from 5 as, and add x — 3 y + 2 to the result. 5. Subtract 2# — 3y from 4aj+"2y, subtract this result from 7 x — Sy, and add a; — ?/ to the final remainder. 6. What expression must be added to 4a — Sm-\-x to pro- duce la — 5 ra + 3 a; ? 7. What is the value of (-3) - (+ 2) - (-3) + (- 5) -(-2)? 8. Collect — 3a — & + 4 c, 2c — 3 a* -f #, a — 4a— d, 5 c + 2 a + 3 as, and 4 d + 6 - 7 c. 9. Simplify x — 1 — J# — 1 — [a — 1 — a; — 1 — ay } . 10. Subtract 4 a? — 3 y + 11 from unity, and add 5 a; — 3 y-h 12 to the result. 11. Combine the m-terms, the n-terms, and the avterms in the following, inclosing the resulting coefficients in parentheses : am -{■ 3 bn -\- x + m + n — ax + 2 n — ex. 12. A given minuend is 7 x + 12 y — 7, and the corresponding difference, 4 x — y + 2. Find the subtrahend. 13. To what expression must you add 2a — 3c + m to pro- duce 5 a -f 7 c — 9m? 14. Subtract 2a—7x+3 from the sum of 3 x + 2 a, 4 a — 10 #, 5 a — 7, and — 4 a; — 11 a. 15. From (a + c) y + (m -f- n) z take (a — c)i/ — (m — n)z. 16. Subtract a; + 17 y — 2 from 12 x -f- 3 z and add the result to3x+(y + llz)-4y. 17. Add the sum of 4 a; + 7 c and 2 a? + 3 c to the remainder that results when x -f 4 c is subtracted from 5 c — 11 x. 18. From (a + 4) x + (a -f 3) y subtract (a + 1) x -f (a + 2) ?/. 19. Add 3 a? + 2 m — 1 and 2x — 3m-\-7, and subtract the sum from 6 a; — m + 6. GENERAL REVIEW 37 20. Simplify a — 1 — a + 1 + [a — 1 — a — a — 1 — (a — 1) — 21. From 7ra + 3 - 2 z s - 3 x. MULTIPLICATION OF POLYNOMIAL BY POLYNOMIAL 45 67. The degree of a product is equal to the sum of the degrees of its factors. 68. A polynomial is called homogeneous when its terms are all of the same degree. Thus, sc* — 4 x s y + 6 x 2 y 2 — 4 xy z + y 4 is a homogeneous polynomial. MULTIPLICATION OF A POLYNOMIAL BY A POLYNOMIAL A further application of the law of distribution for multipli- cation (Art. 55) establishes the principle for multiplying a polynomial by a polynomial. By Art. 55 : (a + b) (as + y) = a (x + y) + b (x + y) = ax + ay + bx + by. The polynomial multiplicand, (x +- y), is multiplied by each separate term of the polynomial, (a + 6), and the resulting products are added. The process will be clearly understood from the following comparison : Numerical Illustration : Algebraic Illustration : a + 5 Multiplicand. a + 7 Multiplier. a 2 + ha + 7 q + 35 a 2 + 12 a + 35 Product. Explanation : a (a + 5) = a 2 + 5 a 7 (a + 5) = 7 a + 35 a 2 + (5 a + 7 a) + 35 = a 2 + 12 a + 35 We have, therefore, the following general process for multi- plying a polynomial by a polynomial : 69. Arrange the terms of each polynomial according to the ascending order or the descending order of the same letter. Multiply all the terms of the midtiplicand by each term of the multiplier. Add the partial products thus formed. 12 10+2 13 10+ 3 36 100 + 20 12 30 + 6 156 = 100 + 50 + 6 Explanation : (10 + 2) = 100 + 20 3 (10 + 2) = 30 + 6 00 + (20 + 30) + 6 = 156 46 MULTIPLICATION Illustrations : 1. Multiply 2 a +- 7 by 3a- 8. 2(1 + 7 T, , 3 _ 8 Explanation : 6 a 2 + 21 a 3 a (2 a + 7) = 6 a 2 + 21 a - 16 a - 56 - 8 (2 a + 7) = - 16 a - 56 6 a 2 + s a _56 Result. 6a 2 + (21a-16a) -56 = 6a 2 + 5a-56. 2. Multiply a 3 -2a 2 + 3a-2 by a 2 + 3a-2. a 3_ 2a 2 + 3 a _2 Explanation : a 2 +3q-2 a 2 (a 3 -2 a 2 +3 a-2)=a 6 -2 a 4 +3 a 8 -2a 2 a 5_2a 4 +3a 8 - 2 a 2 3 a (a 3 -2 a 2 +3 a-2)=3 a 4 -6 a 8 +9a 2 -6a +3a 4 -6a 8 + 9 a 2 - 6 a -2 (a 3 -2 a 2 +3 a-2) = -2 a 3 +4a 2 -6a+4. —2 a + 4a — 6 a+4 Adding the partial products, we have a 6 + a 4 -5a 8 +lla 2 -12a+4 a 5 + a* - 5a 8 + 11 a 2 - 12a + 4. Expressions given with their terms not arranged should both be arranged in the same order before multiplication. 3. Multiply l — 7x* + a? + 5x by -4z- 1 + 23J 2 . Arrange both multiplicand and multiplier in the descending order. x 8 — 7se 2 + 5se-}-l (Let the student complete this multiplication, 2x — 4a: — 1 writing out a complete explanation in the same form as those accompanying examples 1 and 2.) 70. Checking. A convenient check for work in multipli- cation can usually be made by the substitution of a small num- ber as shown in addition. Thus, in Ex. 2, if a = 2 : Multiplicand a 3 -2a 2 +3a-2= 8-8 + 6-2=4 Multiplier a 2 + 3a -2= 4+6-2=_8 Product a 5 + a 4 - 5 a 3 + 11 a 2 - 12 a + 4 = 32 + 16 - 40 + 44 -24 +4=32 It is well to remember that this check will not always serve as a test for both coefficients and exponents. If the value, 1, had been used above, only the coefficients would have been tested, for any power of 1 is 1. MULTIPLICATION OF MISCELLANEOUS TYPES 47 Exercise 9 Multiply : 1. 4 a + 7 by 3 a + 5. 7. 7 ac- 3 by 5 ac 4-1. 2. 3a + 4 by 7 14. 2 -n-f l)(n 2 + n + l)(n 4 -» 2 + l). 15. (x + 6)(x-7)(x-3). 16. (2x-5)(3x + l)(2x + 5)(3x-l). 17. (cd-3)(cc* + 7)(2cd-l)(3cd + 2). 18. (a 2 -l)(a 2 -5)(a 2 + l)(a 2 + 5). 19. (9x 2 -3x + l)(±x 2 + 2x + T)(3x + l)(2x-l). 20. (a + &)(a + m). 24. (a + c)(& + d). 21. (3a + x)(2a + y). 25. (3a + 2&)(2c-5d). 22. (am — x)(am — y). 26. (m 3 + 2)(m 2 — m). 23. (3cd-ra)(2cd + n). 27. (m + w+l)(m-y). Perform the indicated operations and simplify : 28. 5x 2 -(x + l)(2x-3)-3x(x-l). 29. ( a -2) 2 +(a + 3) 2 -2(a 2 + a + 4)-l. 30. (2m-l)(m + 3)-(4m + l)(2m-5)-(l-3m)(l+2m). 31. (2a-3) 2 -3a(a-2)-(3-a) 2 . 32. cd(cd + 1) + cd(cd + l)(od + 2) - Result. 7 a?x s y 2 Hence, to divide a monomial by a monomial: 81. Divide the coefficient of the dividend by the coefficient of the divisor, annexing to the result the literal factors, each with an exponent equal to its exponent in the dividend minus its exponent in the divisor. Oral Drill Give orally the quotients of : 1. 2. 3. 4. 5. a 3 )a[_ a *)- a* -x 2 )^ -m ^-m 7 -a b)-a 2 b 3 6. 7. 8. 9. 4q )-12a 8 -Sx ^Wx'y 11 x y)-U&y* - 7 a 8 b ) - 28 a s b 2 10. 11. 12. 13. 14. - 36 my - 39 c 3 dx 42 afyV - 108 ah 1 - 84 a s bc 2 d 9 m 2 y' 2 -13c 2 x -7xfz* - IS a 2 z - 12 a 2 bcd DIVISION OF POLYNOMIAL BY MONOMIAL 55 x DIVISION OF A POLYNOMIAL BY A MONOMIAL Applying the Law of Distribution (Art. 75), and the prin- ciples governing the division of monomials (Art. 81), we obtain a process of division when the dividend is a polynomial and the divisor a monomial. By Art. 75: ax + & ± az = ^ + & + ^ = x + y+ z. a a a a In practice the usual form of expression is a ) ax + a V + az x + y + z Quotient. Illustration : 1. Divide — 6 m A n -f 10 m 3 n 2 — 14 m 2 n 3 + 4 ran 4 by — 2 mn. - 2 mn ) - 6 m 4 n + 10 w 3 w 2 - 14 m 2 n 3 + 4 wn 4 3 w 3 - 5 m% + 7 mn 2 -2n 3 Kesult. Each term of the quotient is obtained by applying the principle of Art. 81, for in the division of each term by the monomial divisor we have a division of a monomial by a monomial. Hence, to divide a polynomial by a monomial : 82. Divide separately each term of the polynomial by the monomial and connect the terms of the resulting polynomial with the proper signs. Exercise 12 Obtain the quotient of : 1. 2. 3. x )x-7xy 4 m )12m 2 -16m 8 - 2 q )8 a 3 b - 10 ac 4. 5. - 9 a c)18ac-27a 3 c 3 a V-4a 3 + 2a 2 6. 7. - 4 m) 8 m 4 - 1 2 m 3 + 8 m 2 - 4 m - 3 a 2 xz ) - 9 a 3 x V - 6 a W 8. 14 x 3 y — 21xy 3 by 7 xy. 9. 4c 3 d-2cd + 6cd 3 by -2cd 56 DIVISION. REVIEW 10. - 25 a?x 2 + 20 a 2 x 2 - 15 aV by - 5 a¥. 11. 5 a 3 6 - 15 a 2 b 2 + 30 ab 3 by 5 ao. 12. -36m 4 -48m 5 H-60m 6 -72m 7 + 84m 9 by -12m 3 . DIVISION OF A POLYNOMIAL BY A POLYNOMIAL By Art. 69 : (a + 6) (z + y) = ax + ay + 6x + 6y. Then, by definition, ax + ay + 6x + fry may be considered a gri'vera divi- dend, and x + y a given divisor. It remains to determine the correspond- ing quotient. Both dividend and divisor are first arranged in order, a step imperative in all algebraic divisions with polynomials. The x-term is the selected letter for the arrangement. In algebraic divisions the divisor is written at the right of the dividend for convenience in the later steps of the process. Dividend Divisor ax -f- x = a ax + ay -+ bx + by (% + y a(x-\-y)= ax-\-ay+ a + b Quotient, or Result. bx -=- x = b +bx + by b(x + y)= +bx+by The steps of the operation are as follows : (1) The expressions are arranged in the same order. (2) The first term of the dividend, ax, is divided by the first term of the divisor, x ; and the quotient, a, is written as the first term of the quotient. (3) The divisor, x + y, is multiplied by the first term of the quotient, a, and the product, ax + ay, is subtracted from the given dividend. (4) The remainder, bx + by, is a new dividend, and the process is repeated with this new dividend, the divisor always being the first term of the given divisor. The process applies the law of distribution, for ax + ay + bx + by _ ax + ay . bx + by x + y x + y x + y = a(x + y) b(x + y) x+y + x+y DIVISION OF POLYNOMIAL BY POLYNOxMIAL 57 A simple comparison with a numerical process is frequently an aid to beginners. Numerical Illustration: Algebraic Illustration : 12)156(13 100 4- 50 + 6(10 + 2 12 100 + 20 10 + 3 36 30 + 6 36 30 + 6 a 2 + 5 a + 6{a + 2 a 2 + 2 a a + 3 Quotient 3a + 6 3a + 6 Explanation : Explanation : 100 -r- 10 = 10, 1st term of quo. a 2 -r- a = a, 1st term of quo. 10(10 + 2) = 100 + 20. (Subtract.) a (a + 2) = a 2 + 2 a. (Subtract.) 100 + 50 + 6 - (100 + 20) = a 2 + oa + 6-(a 2 + 2c) = 30 + 6, the new dividend. 3 a + 6, the new dividend. 30 -=- 10 = 3, 2d term of quo. 3 a + a = 3, 2d term of quo. 3(10 + 2) = 30 + 6. (Subtract.) 3(a + 2) = 3 a + 6. (Subtract.) 30 + 6 - (30 + 6) = 0. 3 a + 6 - (3 a + 6) = 0. Hence, the quotient is 10 + 3. Hence, the quotient is a + 3. We have, from these illustrations and from the general principle, the following process for the division of a polyno- mial by a polynomial : 83. Arrange both dividend and divisor in the same order of some common letter. Divide the first term of the dividend by the first term of the divisor, and write the result as the first term of the quotient. Multiply the whole divisor by the first term of the quotient just obtained, and subtract the product from the dividend. Regard the remainder as a new dividend and proceed in the same manner as before. Illustration : » 1. Divide a 4 + a 3 -5 a 2 + 13 a- 6 by a 2 -2a + 3. Dividend a*+ a 3 - 5 a 2 + 13 a - -6 (a 2 -2a + 3 Divisor a 4 - 2 a 3 + 3«2 a 2 + 3a -2 Quotient + 3 a 3 - 8 a 2 + 13 a + 3a 3 -6a 2 + 9 a - 2 a 2 + 4a- -2«2 + 4 a _ -6 -6 58 . DIVISION. REVIEW Explanation : (1) a 4 -j- a 2 — a 2 , the first term of the quotient. (2) Multiplying (a 2 - 2 a + 3) by a 2 , we obtain a* - 2 a 8 + 3 a 2 , which product is subtracted from the given dividend. The remainder, which must have three terms, is 3 a z — 8 a 2 + 13 a, and this remainder is the new dividend. (3) 3 a 8 ■+■ a 2 = 3 a, the second term of the quotient. (4) Multiplying (a 2 - 2 a + 3) by 3 a, we obtain 3 a 8 - 6 a 2 + 9 a, which product is subtracted from the last dividend. The remainder, with the last term of the given dividend, is — 2a 2 + 4a — 6, the new dividend. (5) — 2 a 2 -*- a 2 = — 2, the third term of the quotient. (6) Multiplying (a 2 -2a + 3) by - 2, we obtain - 2 a 2 -f 4 a - 6, which, subtracted from the last dividend, gives a remainder of 0, com- pleting the division. 2. Divide 7x 2 -x A -x s -\-2x 5 + 2-9xhj3x-2x 2 -2 + x s . Arranging both dividend and divisor in descending powers of x: Dividend 2x 5 - a*- X* + 7s 2 - -9s + 2(a; 8 - - 2 a 2 + 3 a - 2 Divisor 2x 5 -4x i + 6x 8 - 4x 2 2 a 5 + 3a; — 1 Quotient + 3**- -7z 3 + llx 2 - -9x + 3z*- -6z 8 + 9x 2 - -6x ■ SC 3 + 2x 2 - -3x + 2 . cc 8 + 2z 2 - -3x + 2 In certain expressions the intermediate powers of the terms of the dividend appear during the process. In such cases order must be observed in subtractions. 3. Divide x 3 — y 3 by x — y. 4. Divide a 4 — 16 by a + 2. X * _ yS (X-y a* -16(a + 2 x*-x 2 y* x 2 + xy + y 2 a* + 2 a 3 a 8 - 2 a 2 + 4 a - -8 + x 2 y* -2 a 8 + x 2 y - xy 2 - 2 a 8 - 4 a 2 + xy 2 - y* + 4 a 2 + xy 2 - y* + 4 a 2 + 8 a - 8 a - 16 -8a -16 DIVISION OF POLYNOMIAL BY POLYNOMIAL 59 5. Divide a 3 -3 a&c+-6 3 + c 3 by a + 6 + c. a 3 - 3 abc + 6 3 + c 3 ( q + &+c ' a 3 + d 2 b + a 2 c (a 2 - ab - ac + 6 2 - be + c 2 - a 2 6 - a 2 c - 3 a5c + 6 3 + c 3 - +-6 by x + 2. 2. 2c 2 + 5c-12by c + 4. 3. 3m 2 +-llm-4by 3m-l. 4. 3z 2 + 14z+-15by* + 3. 5. 16 - 8 a +- a 2 by 4 - a. 6. 72 + m-m 2 by 8 + m. 7. 7a 2 6 2 + 123a6-54by7a6-3. 8. 20a 2 -47a6 + 216 2 by 4a-76. 9. 21 c 4 d 4 -f 36 c 2 dV + 15 a 4 by 3 c 2 cP + 3 jb 2 . 10. 10 a 6 6 4 + 23 a 3 6 2 cd? - 21 c 2 ^ by 10 a?b 2 - 7 cd 2 . 60 DIVISION. REVIEW 11. a 3 +3a 2 + 3a + lby a + 1. 12. c 3 ^-\-c 2 x 2 -3cx-6 by cx-2. 13. 15 — 8 mn + 6 m 2 n 2 — m 3 7i 3 by 5 — mn. 14. a 4 + 2 a 3 6 4- 2 a 2 6 2 + ab 3 - 6 6 4 by a? + ab- 2 b\ 15. 2 ra 4 n 4 — m 3 n 3 x — 3 m 2 n¥ + 5 mna? 5 — 2 # 4 by 2 m 2 7i 2 — 3 mnx + 2x*. 16. 8 aW - 8 aVd - 4aW + 11 aW - 10 acd 4 + 3 d 5 by 2a 2 c 2 -3aca , + a*. 17. 9 a 2 - a 4 - 16 a- a 3 + 6 a 5 + 3 by 5 a -2 a 2 - 1 + 3 a 3 . 18. 16a; 6 + l-4a 2 -4a; 4 by 2 a 2 - 1 + 4 ^-2 a. 19. 5m 2 -6-llm 4 + 6m 6 -5m 8 + 2ra 10 by 2 + 2m 4 -m 2 . 20.5 mV - 13 m 6 ?i + 6 m 3 n 4 + 3 m¥ + 6mn 6 + 6m 7 4- 8 ti 7 — 15 ra 4 7i 3 by 3 m 4 — mti 3 — 2 w 3 ti -2n 4 + m 2 n 2 . 21. m 2 — w 2 by m - w. 24. ra 4 — ti 4 by ra + ti. 22. m 2 -7i 2 by ra + 7i. 25. 8 a 3 -27 by 2 a-3. 23. m 3 — n 3 by m — n. 26. m 6 — 32 n 5 by m — 2 w. 27. 81 -w 4 by 3 +%. 28. 27 + a 6 by 3 + x 2 . 29. a^-27^by a;-32/ 2 . 30. 27 a,- 3 + 64 y 5 by 3a + 4?/ 2 . 31. 16ra 4 n 4 -81aV 6 by 2m7i + 3cey. 32. m 2 + 2 mn + ti 2 — x 2 by m + n + x. 33. c 2 — 4c + 4 — d 3 by c-2 + d 34. m 2 + 6m + 9— 25 a 4 by m + S-Bx 2 . 35. 16 m 2 n 2 + 40 m7iy + 25 2/ 2 - 81 by 4ra7i + 5y + 9. 36. a 2 + m 2 + a? 2 + 2 am + 2 asc + 2 m# by a + m + aj. 37. 9m 2 n 2 + 4 : x 2 z 2 + 25-12mnxz-30mn + 20xz by 3 m» — 2 o& — 5. 38. m 3 + 7i 8 +p 3 — 3m7ip by m + n+p. 39. a 6 6 6 -2a 3 6 3 + lby a 2 6 2 -2a6 + l. LITERAL EXPONENTS AND DETACHED COEFFICIENTS 61 85. Division with Literal Exponents. (See Art. 72.) Illustration : Divide 8 of 1 * 4 - 18 af n+3 - 13 x m+2 + 9 a^ 1 +- 2 x™ by 4 x m + x" 1 - 1 — 2 z m - 2 . 8 x m+i - 18 sc m + 3 — 13 x m+ ' 2 + 9 x w+1 + 2x m (4 z OT + x m-1 - 2 x m ~ 2 8 x m+4_|_ 2x n> + 3 - 4x m + 2 2 x 4 - 5 x s - x 2 Result. — 20x w + 3 — 9x m + 2 + 9x m+1 -20x" t + 3 - 5x m + 2 +l0x m + 1 — 4x m + 2 — x m+1 + 2x m — 4 a? n + 2 — x m+1 4- 2 x m The exponent of cc in the first term of the quotient is : (m + 4) — m = m + 4 — ra = 4. Exercise 14 Divide : 1. ^ + 3^-18 by x m -S. 2. ar ?n -4ar Jn -20a; n +-3by x n + 3. 3. Sx 4m -Sx Zm -10x 2m -x m + lbj 3x^-^30^ — 1. 4. x^-S by » n -2. 5. x^ — y*" by x m + 2/ n . 6. 16 x m+1 - 46 a; m+2 + 39 x m+3 - 9 af +4 by 8 a; M+1 - 3 a w+2 . 7. # n+4 +- a n+3 - 4 z n + 2 + 5 x n+l - 3 # n by x n+2 + 2 a n+1 - 3 af . 8. 6a^+ 6 -a;^+ 6 +-4a^ +6 -5a^ w+6 -a; m+6 — 15 a; 6 by 2 a^ m — x m + 3. 86. Division with Detached Coefficients. The same process described in Art. 73 is of advantage in division, as two simple cases will illustrate. 1. Divide x*- 3 ar 5 - 36 x 2 - 71 x - 21 by x 2 - 8 x- 3. Detaching the coefficients and dividing, we obtain : 1 _ 3 _ 36 _ 71 - 21 ( 1-8-3 1-8-3 1 + 5 + 7 Coefficients. + 5_33_71 + 5-40-15 + 7-56-21 z 2 + 5 Z + 7. Result. + 7-56-21 62 DIVISION. REVIEW If occurs as a coefficient either in the given expressions or in the process, the same provision is made as in multiplication. 2. Divide # 5 + 2 a^+^a 3 - 31 a+-9 ^+-15 by 3-2 x-x 2 . Arranging the expressions in ascending powers of x and detaching coefficients, we have : 15 -31 + 9 + 4+2 + 1 (3- -2-1 15 -10-5 5 - -7 + 0- 5-7* 1 Coefficients. -21 + 14 + 4 -21 + 14 + 7 + 0-3+2+1 -3+2+1 — X s . Result. For practice in division with detached coefficients use Ex. 13, 11 to 20, inc. GENERAL REVIEW Exercise 15 1. Show that (x-l) 2 -(x-2)=l + (x-l)(x-2). 2. Simplify(a+l) 2 -(a+-l)(a-l)-[a(2-a)-(2a-l)]. 3. Prove that (a + m)(a — m) +- (m -f- 1) (m — 1) + (l-a)(l+a) = 0. 4. Divide4a) 6 -2^-a?-2^-2a;-lby2^-£ 2 -a;-l. 5. Add the quotient of (a 3 — 1) -f- (x — 1) to that of (tf-2x + l) + (x-l). 6. What is the coefficient of ac in the simplified form of (ac+-3) 2 -3ac(ac-l)? 7. Simplify (m + l)(m + 3)(m + 5) -(m-l)(m-3)(ra-5). 8. Show that (x +- y +- z — 1) (x +- y — z +- 1) — 4 (xy +- z) + (a-2/+-z + l)(l-a; + 2/ + z)=:0. 9. Divide 2 x 2 — 3 xy—5 xz — 2 y 2 — 5yz—Sz 2 by 2x + y + z. 10. A certain product is 6 a 4 +- 4 a 3 # — 9 a 2 y 2 — 3 ay s +- 2 ?/ 4 , and the multiplier 2 a 2 -{-2 ay — y 2 . Find the multiplicand. 11. Simplify 3[ap — 2 joj — 3(2 a? — 3 aj + 7){]. GENERAL REVIEW 63 12. Show that (1 - 3 x + x 2 ) 2 + x (1 - x) (2 - x) (3 - x) - 1 = . 13. Simplify 2 x 2 -3-(3 x -}-3 x 2 )-x(x 2 -3)-(x + 1)(2 - x") and subtract the result from 5 — 2 x. 14. Simplify (a + 4) (a + 3) (a + 2) - (a + 3) (a + 2) (a + 1) _( a +2)(a+l)-(a + l). 15. Prove that (1 + c 2 ) (1 4- a) 2 - 2 (1 - be) (a - c) = (l + c) 2 (l+a 2 ). 16. Subtract a + 3 from the square of a + 2, and multiply the result by the quotient obtained when a 5 — 1 is divided by a A + a z +a 2 + a + l. 17. Simplify a — x— [3 a— (x— a)] + [(2 x — 3 a) — (a— 2 a)]. 18. Divide c?-3cd+cP + l by c 2 - cd -c + d 2 -a' + l. 19. Find the continued product of a 2 — ab -f 6 2 , a 2 + a& -f- 6 2 , anda 4 -a 2 6 2 +6 4 . 20. Simplify (a¥ — a¥ -f aa — 1) (ax + 1) - (aV + 1) (a» + 1) (ax - 1). 21. Multiply 8 a 3 — 27 by a + 2 and divide the product by 2a-3. 22. Divide c + 36c 5 -18c 2 -73c 3 + 12 by _5c + 4-6c 8 . 23. Simplify [(x 2 + 3x + 2)(x 2 -9)]-*- [(* + 3)(^ -«- 6)]. 24. Show that a^ + ^+l — 3a#— (1 — ajy)— ?/(# 2 - «) — 25. Divide 82 m 4 n 4 + 40 - 67 mV + 18 mV - 45 mV by 3 m 4 n 4 — 4 m 2 n 2 + 5. 26/ What must be the value of m in order that x 2 + 18 x + m may be exactly divisible by x -f- 4 ? 27. Show that 2(4 + o^ + a 2 -ax-2a-2x) = (2-a) 2 + (a-2) 2 + (x-a)\ 28. What must be the value of m + nm order that x* + 3 X s + 2 x 2 + Tnx + n may be exactly divisible by x 2 + 2 x + 1? 29. By how much does (aV -f- 3 ax + 2) 2 exceed 2(3 a 3 x 3 + 2a*s* + 6aa>)? CHAPTER VI THE LINEAR EQUATION. THE PROBLEM 87. An equation is a statement that two numbers or two expressions are equal. Thus, 3x + 5 = x + 7. 88. The expression at the left of the sign of equality is the left member (or first member), and the expression at the right, the riglft member (or second member) of the equation. 89. An equation is a conditional equation if its members are equal for particular values of the unknown quantity. Thus, 3x + 5 = x + 7isan equation only when the value of x is 1. 90. An equation is an identical equation when i*s members are equal for any and all values of the unknown quantity. Thus, x 2 — l = (x+l)(x— 1) is an equation for any value of x whatsoever. A conditional equation is usually referred to as an equation ; an iden- tical equation, as an identity. 91. To solve an equation is to obtain the value of the un- known number that will, .when substituted for that unknown number, make the members of the equation equal. 92. The value found to make the members of an equation equal, or to satisfy the equation, is a root of the equation. A root of an equation when substituted for the unknown quan- tity reduces the original equation of an identity. 64 AXIOMS 65 93. A linear or simple equation is an equation which, when reduced to its simplest form, has no. power of the unknown quantity higher than the first power. Thus : 5 x = 15 is a linear or simple equation in x. 7 y — 35 is a linear or simple equation in y. 3aj + 2 = 2aj + 7isa linear equation in x, but is not reduced in form. (x + 5) 2 =# 2 + 7a; + 6isa linear equation in x, for, when simplified, the resulting equation will have only the first power of x. While the final letters, x, y, and z, are most commonly used for representing unknown quantities in equations, any other letters may and will be used in later practice. 94. The solution of equations is based upon the truths known as » AXIOMS 1. If equals are added to equals, the sums are equal. 2. If equals are subtracted from equals, the remainders are equal. 3. If equals are multiplied by equals, the products are equal. 4. If equals are divided by equals, the quotients are equal. 5. If two quantities are equal to the same quantity, they are equal to each other. In general, these axioms may be illustrated as follows. Given the equation A = B. By Axiom 1 A = B By Axiom 2 A = B Add C, C= C Subtract C, C= C A+ C = B+ C A- C = B- C Or, briefly : 95. The same number may be added to, or subtracted from, both members of an equation. By Axiom 3 A = B By Axiom 4 A-B Multiply by C, C= G Divide by C, C= C AC=BC 4=B C C SOM. EL. ALG. — 5 66 THE LINEAR EQUATION. THE PROBLEM Or, briefly : 96. Both members of an equation may be multiplied, or divided, by the same number. By Axiom 5 If A = B and B *= D ; we have, A — D. THE TRANSPOSITION OF TERMS 97. Most equations are given in such a form that the known and the unknown terms occur together in both members. Transposition is the process of changing the form of an equa- tion so that the unknown terms shall all be in one member, usually the left, and the known terms all in the other. The process is based on Art. 95. Given the general equation, ax — c = bx-\- d By Axiom 1, adding c = c (1) (2) (3) ax =bx + c + d By Axiom 2, subtracting bx = bx ax — bx — c + d ■ Compare carefully (1) and (3). In (3) we find c in the right member with its sign changed from — to +. In (3) we find bx in the left member with its sign changed from + to — . In general: 98. Any term in an equation may be transposed from one member to the other member if its sign is changed. As a direct consequence of the use of the axioms, we have : (1) The same term with the same sign in both members of an equation may be discarded. Given the equation, Sx + a — n = 2x-\-a + m. Whence, Sx — n = 2 x + m. THE TRANSPOSITION OF TERMS 67 (2) The sign y of every term in an equation may be changed without destroying the equality. Given the equation, — 5x + m = b — a. Multiplying by — 1, 5x — m = a — b. The sign of a root in a solution depends upon the law of \ signs for division. Thus : 5 a: = 20 5x=-20 -5a = 20 -5 a; =-20 x = 4: $e=— 4 x= — 4 se = 4 In general : 99. When both members of an equation are reduced to simplest form, lilce signs in both members give a positive root, unlike signs a negative root. # If the coefficient of the unknown quantity in a simplified equation is not exactly contained in the known quantity, the root is a fraction ; and if, in a simplified equation, the member containing the known quantities reduces to zero, the root is zero. That is: If 3 a = 5 - 4 x = 7 And if 5x = -9s = x = f x= -£ x = x = Oral Drill Give orally the roots of the following : 1. 5x = S0. 6. 6z = -36. 11. 8aj = 7. 2. 7z = 42. 7. -7x = 21. 12. 7y = 13. 3. 4a = 28. 8. -8x = -56. 13. -5a=16. 4. 3y = -lS. 9. -3a = -39. 14. 7#=0. 5. -3y=l$. 10. 52=3. 15. -3a = 0. THE VERIFICATION OF LINEAR EQUATIONS 100. To substitute a *root in an equation is to replace the unknown literal factor in each term by the value of the root obtained. 68 THE LINEAR EQUATION. THE PROBLEM 101. To verify a root is to show that, by the substitution of this value, the given equation reduces to an identity. The verification of a root, as illustrated in the solutions following, should always be made in the original equation. THE GENERAL SOLUTION OF THE LINEAR EQUATION Illustrations : 1. Solve 5x-4:= 3a + 12. 5x-4 = 3x + 12. Transposing 3 x to the left member and — 4 to the right member, 5-l)J = - 10. 5 x - [3 - (x - 2 x ~ 1)] = - 10. Removing parentheses, 5 x — 3 + x — 2 x + 1 = - 10. Transposing, &x + x — 2x = + S — 1 — 10. Uniting, 4 x = — 8. Dividing both members by 4, x = — 2, the root. Verification : In the original equation, 5 x - [3 -- (x — 2 x - 1)] = - 10. Substitute - 2 for x, 5( -2) - [3 - ( - 2 - 2(-2)-l)] = - 10. Simplifying, * - 10 - 3 - 2 + 4 + 1 = - 10. - 10 = - 10. GENERAL SOLUTION OE THE LINEAR EQUATION 69 3. Solve (x + 3)(2x-5) = 2(x-2) 2 -2(x + l). (x + 3) (2 x - 5) = 2 (x - 2) 2 - 2 (as + 1). Multiplying, 2 x 2 + X - 15 = 2 (x 2 - 4 x + 4) - (2 x + 2). Removing parentheses, 2x 2 + x — 15 = 2 x 2 — 8x + 8 — 2x — 2. Discarding x 2 -terms and transposing, x + 8x + 2x = 15 + 8-2. Uniting, 11 x = 21. x = ff , the root. Verification : Substituting \ \ for x in the original equation, (H + 3)(£i-5)=2(fi-2) 2 -2(fi + l). (ff)(-i!)=2(-T 1 T) 2 -2(!f). -m = tIt - Hf 4.' Solve .3 25 - (x - 3) 2 = 3.25 - (x - 3.5) (a + 2.1). .3 x - (x - 3) 2 = 3.25 - (x - 3.5) (x + 2.1). Multiplying, .3 x - (x 2 - 6 x + 9) = 3.25 - (x 2 - 1.4 x - 7.35). Removing ( ), .3 x - x 2 + 6 x - 9 = 3.25 - x 2 + 1.4 x + 7.35. Transposing, .3 x + 6 x - 1.4 x = 9 + 3.25 + 7.35. Uniting, 4.9 x = 19.6. Dividing by 4.9, x =s 4, the root. Verification : Substituting 4 for x in the original equation, .3 (4) - (4 - 3) 2 = 3.25 - (4 - 3.5)(4 + 2.1). 1.2 - 1 = 3.25 - (.5) (6.1). .2 = 3.25 - 3.05. .2 = .2. From the foregoing we may state the general method for the solution of linear equations: 102. Perform all indicated multiplications and remove all parentheses. Transpose the terms containing the unknown quantity to one member y and all known terms to the other member of the equation. Collect the terms in each member. Divide both members by the coefficient of the unknown quantity. 70 THE LINEAR EQUATION. THE PROBLEM Exercise 16 Find and verify the roots of the following : 1. 4a; + 5 = 3aj + 9. 6. 7 -2x = 9 x — 92. 2. 6 x -j- 7 = 5 x— 11. 7. 5a — l = 3x + l. 3. 7x-2-5x-12 = 0. 8. 3# + 7-a=-3. 4. 8a-12 = 3a; + 8. 9. - 7 a- 5 = - 5 + 2x. 5. 4« + 5 = x- 28. 10. 12a>-7 — 14a; = 12a>. 11. 5a>- (2 a + 3) = 12. 12. (2a-l)-15 = 4-(5-7a). 13. 13-(2a> + ll) = 6-(« + l). 14. 3«-2-(l-3a>) = 0. 15. 12-2(a>-5) + [3aj-(2-aj)] = l. 16. ll-(2«-3)--5=-3»-(5-7a>). 17. (x+l)(x + 3) = (x-2)(x-B). 18. (2a-3)(x-7) = (a?-l)(a + 4) + a; 2 . 19. (aj-3) 2 + 2(a-4) 2 -3(a-5)(a; + 5) = 7. 20. 5(a-l) 2 -3(a-2) 2 ==(2a-l)(3 + a;)-6. 21. 4[3«-2(aj 8 + l)] = 7-4s(2a>-16). 22. - [2 (x - 3)(a> - 5) - (a + 7)(3 -»)] = - 3 (a 2 + 3). 23. (a,- + 2) 3 -(x-l) 3 -(3a; + l)(3a;-4)=0. 24. 2.7 x - (11 - 1.3 x) - 6.7 x = .62 + .4 a - 11. 25. 0.007 x - 2 (.0035 g + .07) = .017- (.14 - .85 x). THE SOLUTION OP PROBLEMS 103. A problem is a question to be solved. In general, a problem is a statement of conditions involving an unknown number or numbers. We seek the value of that unknown number, and by assuming a literal symbol for the unknown we are able to state the given conditions in terms of that unknown. LITERAL SYMBOLS FOR UNKNOWN QUANTITIES 71 104. The solution of a problem is (1) a translation of the language into the symbol-expression of algebra by means of an assumed value for the unknown ; (2) the translation of the given conditions into equations ; and (3) the finding of the root of the derived equation. The result of a solution should always be verified in the given conditions. LITERAL SYMBOLS FOR UNKNOWN QUANTITIES 105. A few simple statements will illustrate the ease with which the conditions of a problem are translated, or expressed, in algebraic symbols. It will be seen that in each case we first assume a value for an unknown quantity upon which the statements seem to depend, and then write expressions for the different statements. Illustrations : 1. If a denotes a certain number, write an expression for 10 more than x. Since we may assume x = the given number, By adding 10, x + 10 = the required number. And, fulfilling the given condition, we have written a number greater than the given number by 10 ; the words "more than" being translated by the symbol of addition, + . 2. In a certain exercise John solved twice as many examples as William. Write an expression for the total number of ex- amples both together solved. We assume that x = the number that William solved. Then 2 x = the number that John solved. By addition, 3 x = the number both together solved. Here we assumed a literal symbol for that particular number upon which the problem seemed to depend ; that is, the number of examples solved by William. A simple application of multiplication and addition completes the translation of the conditions. 72 THE LINEAR EQUATION. THE PROBLEM 3. A man is y years of age. (a) How old was he 10 years ago ? (b) How old will he be in z years ? (a) Since y = the number of years in his present age, By subtraction, y — 10 = his age (in years) 10 years ago. And "years ago," a decrease, is translated by the sign — . (b) Since y = the number of years in his present age, By addition, y + z — his age (in years) z years from now. And " in " referring to the future is an increase translated by the sign +. 4. m yards of cloth cost $ 3 per yard, and n yards of silk, $ 5 per yard. Write an expression for the total cost of both in dollars. At 1 3 per yard m yards of cloth cost (3 x m) dollars = 3m dollars. At $ 5 per yard n yards of silk cost (5 x n) dollars =5w dollars. Therefore, adding, (3 ra + 5 ii) dollars equals the total cost of both. 5. Write three consecutive numbers, the least of the three being x. The difference between any two consecutive numbers is 1. Therefore, when x = the first and least number, x + 1 = the second number, x + 2 = the third number. 6. Write three consecutive even numbers, the least being m. The difference between consecutive even numbers is 2. Therefore, as above, m, m + 2, and m + 4, are the required numbers. 7. Write three consecutive odd numbers, the greatest being a ; and write an expression for their sum. i The difference between consecutive odd numbers is 2. Therefore, a = the first and greatest number, a — 2 = the second number, a — 4 = the third number. By addition, 3 a — = the sum of the three numbers, LITERAL SYMBOLS FOR UNKNOWN QUANTITIES 73 8. A boy has x dimes and y nickels and spends 10 cents. Write an expression for the amount he has remaining. Problems involving value must be so stated that different denominations are all expressed in one and the same denomination. Hence, the value of x dimes = (10 • x) cents = 10 a; cents. Hence, the value of y nickels = (5 • y) cents = 5 y cents. Therefore, he had at first, (10 x -f 5 y) cents. Subtracting the 10 cents he spent, we have for an expression of his final amount, (10 x + 5 y) — 10 cents. Result. Oral Drill . 1. If x denotes a certain number, give an expression for 15 less than x. 2. If y denotes a certain number, give an expression for 12 less than the double of y. .3. If x denotes the number of square inches in a certain surface, how many square inches are there in m similar surfaces ? 4. John has x marbles, William y marbles, and Charles z marbles. What is the expression for the total number all together have ? 5. A boy had m marbles and lost 1 of them. How many had he left? 6. A boy earned x cents, found three times as many, and spent c cents. How many cents had he finally ? 7. A boy solved n examples and his sister solved 5 more than twice as many. How many examples did the sister solve ? 8. John caught m trout and his brother caught 3 less than three times as many. How many did both together catch ? 9. If William solves x examples, how many examples must John solve so that both together shall solve y examples ? 74 THE LINEAR EQUATION. THE PROBLEM 10. Three men together buy a field. B pays twice as much as A, C four times as much as B. If A pays d dollars, what is the cost of the field ? 11. A horse cost y dollars, a harness x dollars, and a wagon as much as the combined cost of a horse and two harnesses. What did all three together cost ? 12. If a line 10 inches in length is increased by n inches, what is the length of the new line ? 13. How much remains of a line m inches long if n inches are cut from one end andp inches are cut from the other end ? 14. A rectangle is x inches long and y inches wide, and a strip z inches wide is cut from one end. What is the area of the part cut off and also of the part remaining ? 15. The sum of two numbers is 10 and the smaller number is n. What is the larger number ? 16. The sum of three numbers is 50; one is x, and another 35. What is the expression for the third number ? 17. Name three consecutive numbers, the least of the three being n. 18. Name three consecutive even numbers, the least number being n. 19. Name three consecutive odd numbers, the least of the three being m. 20. Name five consecutive numbers, the middle one being x. 21. If x is an even number, what is the next odd number above x ? 22. If a; is an odd number, what is the next even number above x ? 23. Name the three consecutive odd numbers below n, n being an odd number. , 24. Name the three consecutive odd numbers below n, n being an even number. LITERAL SYMBOLS FOR UNKNOWN QUANTITIES 75 25. What is the sum of the three consecutive even numbers below x + 1, x + 1 being odd ? 26. If x is the middle one of three consecutive odd numbers, what is the expression for. their indicated product? 27. If a man is x years old, how old will he be in a years ? 28. If a man is x years old now, how many years will pass before he is a years old ? 29. If a man is y years old now, how old was he z years ago? 30. What is the sum of the ages of a father and son if the son is x years old arid the father m times as old as the son ? 31. A man is m times as old as his son and n times as old as his daughter. If the daughter is x years old, what is the sum of the ages of all three ? 32. In a period of c year's a man earns d dollars. If he spends n dollars each one of the c years, what is the expression for his final saving ? 33. How many cents are represented by x dimes ? 34. Express the condition that x dimes shall equal y nickels. 35. One dollar is lost from a purse that had contained x dollars and y quarters. Write the expression for the amount remaining in cents. 36. A man travels a miles an hour for h hours. What is the total distance he travels ? 37. A boy rides x miles in a train, then y miles by boat, and, finally, by automobile twice as far as he has already traveled. Write the expression for the total number of miles in his journey. 38. What is the expression for the statement that the square of the difference of two numbers, a and b, is 2 less than the sum of the squares of two other numbers, m and n ? 76 THE LINEAR EQUATION. THE PROBLEM 39. A room is x yards long and y yards wide. What is the expression for its area in square yards ? in square feet ? 40. If a man is now x + 1 years old, how many years ago was he 30 years old? How many years must pass before he will be 50 years old ? When was he x — 1 years old ? When will he be 2 a? years old ? PROBLEMS LEADING TO LINEAR EQUATIONS 106. From the oral work j ust covered the student will see that no general directions can be given for the statement^of problems. The following suggestions will be of assistance in the state- ment of all problems, and will serve as a general outline of the method by which we attack the different types. 107. In stating a problem : 1. Study the problem to find that number whose value is re- quired. 2. Represent this unknown number or quantity by any conven- ient literal symbol. 3. The problem will state certain existing conditions or relations. Express those conditions in terms of your literal symbol. 4. Some statement in the problem will furnish a verbal equation. Translate this verbal equation into algebraic expression by means of your stated conditions. The following illustrations will show the ease with which certain common words and phrases may be translated into the common operations of algebra. Illustrations : 1. The greater of two numbers is 3 more than the less, and four times the less number exceeds twice the greater number by 8. Find the two numbers. PROBLEMS LEADING TO LINEAR EQUATIONS 77 Let x = the smaller number. Then x + 3 = the larger number. Now the word "exceeds," in this case, may be translated by the sign " — ," and the word " by " may be translated by the sign " =." Hence, we select the conditional sentence : " four times the less exceeds twice the greater by 8," and translate :4 a; — 2(x + 3) =8. Solving the equation 4x — 2(x + 3) = 8, x = 7, the smaller number. Consequently, adding 3, x + 3 = 10, the larger number. The results verify in the original condition, hence in the equation. 2. Find the number which, multiplied by 4, exceeds 40 by as much as the number itself is less than 40. Let x = the number. Then 4 x = four times the number. Translating the conditional sentence : " number . . . multiplied by 4 exceeds 40 by as much as 40 exceeds the number," x X 4-40 = 40 - x. Solving the equation, 4 x — 40 = 40 — as, x = 16, the number. Verification : 4(16) - -40: = 40- -16. 64- -40 = 40- -16. 24: = 24. 3. A father is four times as old as his son, but in 24 years the father will be only twice as old as the son. What is the present age of each ? Let x = the number of years in the son's present age. Then 4 x = the number of years in the father's present age. Therefore, x + 24 = the son's age after 24 years. 4 x -f 24 = the father's age after 24 years. Now, "... in 24 years the father will be only twice as old as the son." Or, -24 + Ax = 2(« + 24). Solving the equation, 24 + 4 x = 2(x + 24). x = 12, the son's age now. 4 x = 48, the father's age now. Verification : In 24 years the father will be (48 + 24) = 72 years of age. In 24 years the son will be (12 + 24) = 36 years of age. 78 THE LINEAR EQUATION. THE PROBLEM The problems in the following exercise are at first classi- fied in four groups involving only the simplest of commonly occurring conditions. Exercise 17 (I) Problems involving One Number. 1. Four times a certain number is 36. Find the number. Let x = the required number. From the problem 4 x = four times that number. But the problem states that 36 = four times that number. Hence, from our assumed condition and from the given condition, we have two expressions, 4 x and 36, representing the same quantity. Therefore, 4 x = 36. x = 9, the required number. Verification : 4(9) = 36. 36 = 36. 2. What is that number which, when decreased by 5, gives a remainder of 19 ? 3. Three times a certain number is diminished by 7 and the remainder is 11. What is the number ? 4. William has three times as many books as John, and both together have 32 volumes. How many has each ? 5. If four times a certain number is added to five times the same number, the sum is 36. What is the number? 6. Four times a certain number is subtracted from eleven times the same number, and the remainder is 42. Find the number. 7. I double a certain number and subtract 7 from the result, and my remainder is 1 more than the original number. What was the number ? 8. If a certain number is increased by 5, the sum is 8 less than twice the original number. Find the number. PROBLEMS LEADING TO LINEAR EQUATIONS 79 9. Twelve times a certain number is decreased by 5, and the remainder is 15 more than seven times the original num- ber. What was the number ? 10. Find that number which, if doubled, exceeds 60 by as much as the number itself is less than 75. 11. What number is that which, if doubled and subtracted from 50, gives a remainder 5 less than three times the origi- nal number ? (II) Problems involving Two or More Numbers. 12. The sum of two numbers is 24, and the greater number is 3 more than twice the smaller number. Find the numbers. Let x = the smaller number. Then 24 — x = the larger number. Now "the greater number is 3 more than twice the smaller," Hence, 24 - x = 3 + 2 x. Solving, x = 7, the smaller number. Whence, 24 — 7 = 17, the larger number. Verification : 24 - 7 = 3 + 2(7). 17 = 3 + 14. 17 = 17. 13. One number exceeds another number by 5, and their sum is 49. Find the numbers. 14. One number is four times as large as a second number, and their sum is 21 more than twice the smaller number. Find the numbers. 15. The sum of three numbers is 108. The second number is twice the first number, and the third is equal to twice the sum of the first and second. Find the three numbers. 16. Find the three consecutive numbers whose sum is 54. 17. Find the three consecutive odd numbers whose sum is 39. 18. Find the five consecutive odd numbers whose sum shall be equal to nine times the smallest of the numbers. 80 THE LINEAR EQUATION. THE PROBLEM 19. Find four consecutive odd numbers such that twice the sum of the three smallest shall be 15 more than three times the greatest one. 20. Divide 17 into two parts such that the smaller part plus four times the larger part shall be 50. (Hint : Let x = the smaller part • 17 — x = the larger part.) 21. Divide 64 into two parts such that three times the smaller part added to twice the larger part shall be 158. 22. Divide 100 into two parts such that twice the larger part shall be 50 more than three times the smaller part. 23. Divide 75 into two parts such that three times the larger part decreased by 6 shall equal four times the smaller part increased by 9. (Ill) Problems involving the Element of Time. 24. A man is twice as old as his brother, but 5 years ago he was three times as old. Find the present age of each. Let x = the number of years in the brother's present age. Then 2 x = the number of years in the present age of the man. Now x — 5 = the brother's age 5 years ago. And 2x — 5 = the man's age 5 years ago. From the statement in the problem : 2x-5 = 3(x-5). Solving, x = 10, the brother's age now. 2 x = 20, the man's age now. 25. A boy is 5 years older than his sister, and in 4 years the sum of their ages will be 29 years. Find the present age of each. 26. A man is twice as old as his son, but 10 years hence the sum of their ages will be 83 years. What is the present age of each ? i 27. Five years ago the sum of the ages of A and B was 50 years, but at the present time B is four times as old as A. How old is each now ? PROBLEMS LEADING TO ^LINEAR EQUATIONS 81 28. In 7 years the sum of the ages of A and B will be 26 years less than three times A's present age. If A is now three times as old as B, find the age of each after 7 years. 29. In 7 years a 'boy will be twice as old as his brother, and at the present time the sum of their ages is 13 years. Find the present age of each. 30. A boy is three times as old as his sister, but in 4 years he will be only twice as old. What is the present age of each ? 31. A young man is 23 years of age and his brother is 11 years old. How many years ago was the older brother three times as old as the younger ? 32. A man 50 years old has a boy of 9 years. In how many years will the father be three times as old as the son ? 33. The sum of the present ages of a man and his son is 60 years, and in 2 years the man will be three times as old as the son. What will be the age of each when the sum of their ages is 100 years ? (IV) Problems involving the Element of Value. 34. A man pays a bill of $49 with five-dollar and two-dol- lar bills, using the same number of each *kind. How many bills of each kind are used ? Let x = the number of bills of each kind. Then 5 x = the value of the fives in dollars, and 2x = the value of the twos in dollars. Therefore, 7 x = 49. From which x = 7, the number of bills of each kind. 35. Divide $100 among A, B, and C, so that B shall receive three times as much as A, and C $20 more than A and B together. 36. A has $ 16 less than B, and C has as many dollars as A and B together. All three have $ 60. How many dollars has each? SOM. EL. ALG. — 6 82 THE LINEAR EQUATION. THE PROBLEM 37. A number of yards of cloth cost $ 3 per yard, and the same number of yards of silk, $ 7 per yard. The cost of both pieces was $ 100. How many yards were there in each piece ? 38. $41 was paid to 16 men for a day's work, a part of the men receiving $2 per day and the other part $3 per day. How many men worked at each rate ? 39. A boy has $42 in two-dollar bills and half dollars, and there are three times as many coins as bills. How many has he of each kind ? 40. $2.10 was paid for 8 dozen oranges, part costing 20 cents a dozen and part costing 30 cents a dozen. How many dozen were there in each lot ? 41. A merchant bought 50 postage-stamps, the lot being made up of the five-cent and the two-cent denominations. Twice the cost of the two-cent stamps was 48 cents more than three times the cost of the five-cent stamps. How many of each kind were bought, and what was the total amount paid for them ? (V) Miscellaneous Problems. 42. Find the two numbers whose sum is 70 and whose dif- ference is 6. 43. A and B together have $90, and A has $12 more than B. How many dollars has each? 44. One number exceeds another by 3, and the difference between their squares is 51. What are the numbers ? 45. The difference between the ages of a father and son is 36 years, and the father is three times as old as the son. Find the age of each. 46. Divide 70 into two parts such that ten times the smaller part shall equal eight times the larger part. 47. A man divided $1500 among four sons, each receiving $ 50 more than the next younger. How much did each receive ? PROBLEMS LEADING TO LINEAR EQUATIONS 83 48. Divide 31 into two parts such that 1 less than eight times the smaller part shall equal five times the greater part. 49. Ten times a certain number is as much above 77 as 43 is above five times the number. What is the number ? 50. The difference between the squares of two consecutive even numbers is 52. Find the numbers. 51. The sum of two numbers is 16 and the difference of their squares is 32. What are the numbers ? 52. One number is three times another, and the remainder when the smaller is subtracted from 19 is the same as the re- mainder when the larger is subtracted from 43. Find the two numbers. 53. A man has three hours at his disposal and walks out into the country at a rate of 4 miles an hour. How many miles can he walk so that, by returning on a trolley car at the rate of 12 miles an hour, he will return within just 3 hours ? 54. A walks over a certain road at a rate of 3 miles an hour. Two hours after he leaves, B starts after him at a rate of 4 miles an hour. How many miles will A have gone when B overtakes him ? 55. A and B are 60 miles apart and start at the same time to travel towards each other. A travels 4 miles an hour and B 5 miles an hour. In how many hours will they meet and how far will each have traveled ? 56. A man walks 5 miles on a journey, rides a certain dis- tance, and then takes an automobile for a distance four times as great as he has already traveled. In all he travels 75 miles. How far does he go in the automobile ? 57. How can you pay a bill of $5.95 with the same number of coins of each kind, using only dimes and quarters? 58. The sum of the ages of a father and son is 96 years ; but if the son's age is trebled, it will be 8 years greater than the father's age. How old is each ? CHAPTER VII SUBSTITUTION 108. Substitution is the process of replacing literal factors in algebraic terms by numerical or by other literal values. Illustrations : 1. If a = 5 and 6 = 7: (« + &) = (5 + 7) 12. Result. 2. If a = 4 and b= -2: (2a-a 2 6) = . [2-4 -42 (-2)] 8 -16 (-2) 8 + 32 40. Result. 3. If a; = 4 m, y — 3 m, and Z= — 5m: (jc + y - z) = (4 m + 3 m + 5 m) 12 m. Result. 4. With the same values for x, y, and % as in (3) : (aj + 2/2)(2a;-^-2«2) = [4m+(3m)2][2(4m)-3m-2(-5m)2] = (4m + 9w 2 )[8ra-3m-2(25m 2 )] = (4 m + 9 m 2 ') (5 m - 50 m 2 ) = (20 m 2 - 155 m 8 - 450 wi 4 ). Result. From these illustrations we state the general method for substitution : 109. Replace the literal factors of the terms of the given ex- pression by their respective given values. Perform all indicated operations and simplify the result. 84 SUBSTITUTION OF NUMERICAL VALUES 85 I. SUBSTITUTION OF NUMERICAL VALUES Exercise 18 Find the numerical values of the following when a = 4, 6 = 3, c = 4, and d = 1. 1. a+b + c. 7. ab — 3 be + 5 ad. 2. a-26 + 5c. 8. 2ab-3cd + 2bd. 3. 46-3c + 2a\ 9. 5ad , -36a , + 8cd'. 4. 7a— a" + 9c — 6. 10. ab—(bc—cd). 5. 10a + 6-(3c-d). 11. 8a6c-5 6ca' + 2aca\ 6. 46 — [3 c — (2a + a")]. 12. acd* — (abc — bed — acd). 13. acd-2 6cd-3(a6-c(f). 14. 2a6- (3cd + 2a — a6 + 6c — bd). 15. 6c-[-6d-(a6-cd") + (a6 — 6d*)]. 16. Simplify a(a+ c) —c(c — a) when a = 4 and c = 2. 17. Simplify (a + 6) 2 - (a + 6) (a - b) - (a - 6) 2 when a = 5 and 6 = 4. 18. Simplify 10 (a -f l) 2 - (3 a - 2) 2 when a = -2. 19. Simplify 4 (2 a; -f- y) 2 — S(x +y) (x — y) when x = — 2 and 2/ = 0. 20. Simplify 3(ra -2x) 2 - 2(ra + 2 »)(m + 3 x) -f 4 x when m = — 5 and a; = 2. 21. Simplify d — 2(c + d)(c - 2 d) - 3(e - d) when c =2 and a" =-2. 22. Simplify a(a + 6) - b(a - 6) 2 - (a + &) 3 when a = - 3 and 6 = — 2. 23. Simplify (3 a + m)(a- 5 m) +[9 a 2 - |ra-a(2m-9a) j] when a = and m = 10. 86 SUBSTITUTION 24. Simplify 7 a(a + 1) - (2 + a) (3 a - 5) - 3 a(a - 1) when a = 0. 25. Simplify (3 m- 2) 2 +(w + n-l)»-(2w- 1) (m + 1) when m = n= — 3. II. SUBSTITUTION OF LITERAL VALUES When literal values are substituted, the results will be in terms of those literal values. Exercise 19 Simplify : 1. (a — x) 2 -\-(a — 2x) 2 when a = 2x. 2. (a — 3 x — x 2 ) when x = 2a. 3. (a + x)(a — 2x) 2 — (a — x) 2 when a= — x. 4. (a? + n)(x + ft — 1) when x = m and n = — m. 5. (2 a — 3 c -\- x)(x — 2 a) — 2 acx when or = x and c = 0. 6. 5 ac — (a + c)(a — 4 c) -f- (a — c) 2 when c = 0. 7. 2(ra — a) — 2 (m + 3 x)(m — 2 x) — 3 mx when ra = a and x = — 2 a. 8. (2a>+l)*-3a?-4a>(a!-5) whena= -2a. 9. (3x-l) 2 -(3x+l)(2x-5) whenx = ab. 10. (3a 2 -2a + l)-(3a;-2) 2 when »= — mnl 11. (m + a?)(m — #) — (m — a;) 2 + 2 m when m= —x. 12. (a-26 + 8) 2 -x(a; + 2a)-(a-2&) 3 -l when a = 2& and x = 0. 13. (6 ar> - 11 o?y - 10 y 2 ) - 6(a^ — 2xy) + 11 2/ 2 when aj = m and y——m. 14. (a? — a)(aj + a — y) — (x — y) (x -\» a) a — (x — a) 2 when x= — m and a = 0. 15. c(c + c7 + j) — (c — d-f-l)(c+d— 1) — cd— 1 whenc=d=0. THE USE OF FORMULAS 87 HI. THE USE OF FORMULAS 110. A formula is an algebraic expression for a general principle. For example : If the altitude of a triangle is represented by a, the base by b, and the area by S, we have the general expression for its area in the formula o ab Given the values of a and b for a par- ticular triangle, we obtain its area by substituting those values in this general formula and simplifying. A few common formulas will illustrate the value of this brief method of expression. (1) The Formula for Simple Interest. By arithmetic : The interest (I) on a given principal at a given rate for a given time equals principal x rate x time. Ex- pressed as a formula: If p = the principal expressed in dollars, r = the rate of interest expressed decimally, t = the time expressed in years, Then / = prt is the general formula for simple interest. (2) The Formula for Compound Interest. (Interest com- pounded annually.) If p = the principal expressed in dollars, r = the rate of interest expressed decimally, n = the number of years in the interest period, Then /=p(l +r) n — p is the general formula for interest com- pounded annually. (3) The Formula for the Transformation of Temperatures. Both the standard thermometers, the Fahrenheit and the Centi- grade, are in everyday use in physical investigations, and the formula given below is used to change Fahrenheit readings to the Centigrade scale. 88 SUBSTITUTION If F = the given reading from a Fahrenheit scale, C = the required equivalent reading on the Centigrade scale, Then C sa $ (F — 32) is the formula for temperature transformation. Exercise 20 1. What is the area of a triangle having a base of 14 inches and an altitude of 9 inches ? 2. Find the simple interest on $ 1500 for 12 years at 5 per cent. 3. What is the interest on $6200 for 3 years 7 months 10 days at 4.5 per cent ? 4. Find the interest, compounded annually, on $ 1200 for 4 years at 5 per cent. 5. On a Fahrenheit thermometer a reading of 50° is taken. What is the equivalent reading on a Centigrade thermometer? 6. Find the circumference (C) of a circle, the radius (R) being 5 feet, and the constant (it) in the formula 0=2 ttR being 3.1416 + . 7. Find the area (S) of a circle whose radius (R) is 4 feet, the constant (71-) in the formula S = irR 2 being 3.1416 +. 8. With the formula of Example 7, find the area of a cir- cular pond whose diameter (D) equals 100 feet. (D — 2 R.) 9. If the diameter (D) of a sphere is 2 feet, what is the volume (V) of the sphere from the formula, V— ? (tt = 3.1416 +.) 6 10. Find the last term (Z) in a series of numerical terms of which the first term (a) is 3, the number of terms (n) is 8, and the difference between the terms (d) is 2, the formula be- ing l = a+ (n — l)d. 11. If a = 3, r = 2, and n m 5, what is the value of I in the expression I = ar n_1 ? 12. If r = 5, s = 4, and n = 6, find the value of a in the ex- pression a = (r — l)s -f- r n_1 . CHAPTER VIII SPECIAL CASES IN MULTIPLICATION AND DIVISION MULTIPLICATION The product of simple forms of binomials may often be written without actual multiplication, the result being obtained by observing certain laws seen to exist in the process of actual multiplication. Three common cases are : The square of the sum of two quan- tities, a + b a + b a 2 + ab + ab + b 2 II The square of the difference of two quantities, a — b a — b a 2 - ab - ab + b 2 III The product of the sum and difference of two quantities. a + b a — b a 2 + ab -ab-b 2 a 2 F a 2 + 2 ab + b 2 a 2 - 2 ab + b 2 Therefore, we may state, from (I), (II), and (III), respectively : 111. The square of the sum of two quantities equals the square of the first, jplus twice the product of the first by the second, plus the square of the second. 112. The square of the difference of two quantities equals the square of the first, minus twice the product of the first by the second, plus the square of the second. 113. The product of the sum and difference of two quantities equals the difference of their squares. Ability to apply these principles rapidly is essential in all later practice. 89 90 SPECIAL CASES IN MULTIPLICATION AND DIVISION Oral Drill Give orally the products of : 1. (a + m) 2 . 4. (a^+4) 2 . 7. (3a + 2) 2 . 2. (x + z) 2 . 5. (a + 3m) 2 . 8. (4a + 5) 2 . 3. (a; + 3) 2 . 6. (c+5d) 2 . 9. (5c+7) 2 . 10. (c-z) 2 . 13. (c-8) 2 . 16. (cd-4) 2 . 11. (&-4) 2 . 14. (3a-5) 2 . 17. (a 2 6 2 -6) 2 . 12. (m-5) 2 . 15. (7a-3) 2 . 18. (cd-Scx) 2 . 19. (a + a;)(a-aj). 27. (3 mn + 5 ma) 2 . 20. (m + y)(m-y). 28. (4c 2 d 2 -5) 2 . 21. (a + 4) (a- 4). 29. (2m 3 + 9)(2ra 3 -9). 22. (2a + l)(2a-l). 30. (5 xyz + 7 y 2 z 2 ) 2 . 23. (3a + 5)(3a-5). 31. (3a 4 -13)(3a; 4 + 13). 24. (7a + 10)(7a — 10). 32. {am 2 + xyz)(am 2 -xyz). 25. (5a-8 2/)(5a; + 8 2/). 33. (3c 5 + ll) 2 . 26. (3 a 2 + 5) (3 a 2 -5). 34. (7 m 6 -9 a 7 ) 2 . IV. THE DIFFERENCE OF TWO SQUARES OBTAINED FROM TRINOMIALS 114. Many products of two trinomials may be so written as to come under the binomial case of Art. 113. In such multiplications we group two of the three terms in each quantity so as to produce the same binomial expressions in each, the pa- renthesis being treated as a single term. Three different cases may occur : (1) (2) (3) (a + 6 + c)(a + b-c) (a + b + c)(a- b + c) (a + 6-c)(a -b + c) The terms inclosed in parentheses must form the same binomial in each expression. THE DIFFERENCE OF TWO SQUARES 91 From 1. (a + 6 + c)(a + b - c) = [(a + b) + e][(a + 6) - c] =s (a + &) 2 - c 2 = a 2 + 2 a& + 6 2 - c 2 . Result. From 2. (a + b + c) (a - b + c) = [(a + c) + 6] [(a + c) - 6] = (a + c) 2 - 6 2 = a 2 + 2 ac + c 2 - 6 2 . Result. From 3. In this case only one term has the same sign in each ex- pression, the a-term. Hence the last two terms of each expression ara inclosed in parentheses, the parenthesis in one case being preceded by the minus sign. (a + b - c)(a - b + c) = [a + (6 - c)][a - (6 - c)] = a 2 - (6 - c) 2 = a 2 - (6 2 - 2 6c + c 2 ) = a 2 - 6 2 + 2 be - c 2 . Result. Exercise 21 Write by inspection the following products : 1. [(a + 0) + 4][(a+«)— 4]. 8. (ra + a + 2)(ra + a; — 2) 2. [(ra + a;) + 2][(m + a)— 2]. 9. (m — w + c)(ra — n — c). 3. [(c-2) + m][(c-2)-m]. 10. ( c 2 + c + l)(c 2 + c-l). 4. [(a 2 + l) + a][(a 2 + l)-a]. 11. (c-d + 2)(c-d-2). 5. [a -r-(d + a>)] [a — (d + a;)]. 12. (m + w — 4)(m — n + 4), 6. [d+(y-«)][ (x*j tX -2){x i -x + 2). 15. (aj 4 + ^ + l)(» 4 -^+l). 16. (m 4 -2m 2 + l)(m 4 + 2m 2 + l). 17. (x 4 -x 2 -6)(x A + x 2 -6). 18. [( c + d) + (m + l)][(c + d)-(m + l)]. 19. [0 + x) - (y - z)] [(« + <*) + &- *)J 20. (t^-^-c-lX^ + ^ + c-l). 92 SPECIAL CASES IN MULTIPLICATION AND DIVISION V. THE PRODUCT OF TWO BINOMIALS HAVING A COMMON LITERAL TERM By actual multiplication : a + 7 a + 5 a 2 + 7 a + 5a + 35 a 2 + 12 a + 35 Hence, in the product : a? = a x a, the product of the given first terms. + 12a=(+7 + 5)a the product of the common literal term by the sum of given last terms. + 35 = (+ 6) (+ 7) the product of the given last terms. In like manner : (1 ) (x - 4) (x - 9) = z 2 - 13 x + 36. x* = x-x. -13z=(-4-9)z. + 36=(-4)(-9). (2) (e + 9)(z-3)=z 2 + 6z-27. x 2 = x-x. +6x=(+9-3)x. -27=(+9)(-3). (3) (w-15)(m+7)=m 2 -8m-105. m a = OT-m. -8w=(-15 + 7)to. -105=(- 15)(+7) In general : (x + a) (x + b) = « 2 + (a + b)x + ab. Therefore, in the product of two binomials having a com- mon literal term: 115. The first term is the product of the given first terms. The second term is the product of the common literal term by the algebraic sum of the given second terms. The third term is the product of the given second terms. Oral Drill Give orally the following products : 1. (a> + 3)(a> + ±). 6. (c + l2)(c + l). 2. (« + 4)(o; + 5). 7. (tt + ll)(n + 12). 3. (a> + 5)(» + 7). 8. (m + 3)(m + 20). 4. (m + 6)(m + 4). 9. (d + 12)(d + 16). 5. (y + 9)0/ + 4). 10. (c + 15)(c + 16). THE PRODUCT OF ANY TWO BINOMIALS 93 11. (6-4)(6-7). 15. (xz-9)(xz-l). 12. (n-ll)(n-10). 16. (c 2 - 3) (c 2 - 4). 13. (ax-3)(ax — 5). 17. (x 2 - xy)(x> - 2 xy). 14. (cd— T)(cd — 3). 18. (mn- 3n)(mw- 7 w). 19. («-3)(sc + 5). 24. (a-13)(a+10). 20. (c-3)(c + 8). 25. (a 2 + 5)(a 2 -3). 21. (x-9)(x + 7). 26. (m 3 4-8)(m 3 -3). 22. (a?4-9)(a?-10). 27. (a& - 9)(ao + 12). 23. (m + H)(m-12). 28. (cx 2 -7)(cx 2 + S). VI. THE PRODUCT OF ANY TWO BINOMIALS By actual multiplication 2a +5 3a +4 6 a 2 + 15 a + 8 a -f 20 6 a 2 + 23 a + 20 The two multiplications resulting in the terms + 15 a and + 8 a are cross products. It will greatly assist the beginner to imagine that the terms entering the cross product have this connection : I r=i 1 (2a+5)(3a+4) And, by inspection, the middle term results from (+2a)(+4)+ (+3a)(+5) = + 8 a + 15 a = + 23 a. Considering other possible cases : (1) In (4 a — 7) (3 a — 5) the middle term in the product will be (+4a)(-5) + (-f3a)(-7) = _ 20 a -21 a = - 41 a. (2) In (2 a + 5) (3 a — 4) the middle term in the product will be (4- 2 a)(- 4) + (+ 3 «)(+ 5) = - 8 a + 15 a = + 7 a. (3) In (3 a — 2) (9 a + 4) the middle term in the product will be (+3a)(-4) + (+9a)(-2) = + 12 a - 18 a = - 6 a. 94 SPECIAL CASES IN MULTIPLICATION AND DIVISION In general : (ax + b) (ex + d) = acx 2 + (acd + bc)x + bd. Therefore, in the product of any two binomials : 116. The first term is the product of the given first terms. TJie second term is the algebraic sum of the cross products of the given terms. The third term is the product of the given second terms. Exercise 22 Write by inspection the following products : 1. (3» + 2)(aJ + l). 13. (7aj-2)(3o?-7). 2. (2 a -fl) (3 a + 2). 14. (10 c- 11) (3 c- 7). 3. (3m+2)(2m + 3). 15. (7 n -9) (8 n- 5). 4. (4a? + JL)(a + 3). 16. (5mw — 1)(3 mn-4). 5. (5c + 4)(2c + 3). 17. (4a -f 7m)(3a - 2 m) 6. (66 + 5)(26 + 3). 18. (7 c - Sd)(Sc + 3d). 7. (5z + 7)(3z + 4). 19. (2 ac - 3 a>)(ac + 11 x) 8. (7m + 2)(2m + 9). 20. (14m — 5 no;) (2m + nx). 9. (3a-7)(2a-5). 21. (11 x + 9y)(9x - 2 y). 10. (3m-l)(2m-3). 22. (3 c - 13 mn) (4 c + 5 mn). 11. (5c-2)(3c-l). 23. (mna-ll)(2mna; + 3). 12. (6y-l)(2y-3). 24. (13 ac - 5 x)(3ac +2 x). VII. THE SQUARE OF ANY POLYNOMIAL By actual multiplication : (a + b + c) 2 = a 2 + b 2 + c 2 + 2 ab + 2 ac + 2 6c. ( a + & _ c _ + 3). "■■ (100 -8 la 4 ) -(10 -9 a 2 ). 5. (c 2 -36)^-(c-6). 12. (rfy 2 - 121) + (xy + 11). 6. (4m 2 -l)-f-(2m-l). 13. (81mW-169c 4 )-s-(9mw-13c*). 7. (9d 2 -25)-j-(3d+5). 14. (196c 6 -81d 4 )-(14c 3 + 9d 2 ). II. THE DIFFERENCE OF TWO CUBES By actual division : qB ~ &B = a 2 + ab + & 2 . a—b From the form of the quotient and its relation to the divisor we state: 119. The difference of the cubes of two quantities may be divided by the difference of the quantities. The quotient is the square of the first quantity, plus the product of the two quantities, plus the square of the second quantity. Oral Drill Give orally the quotients of: 1. (a 3 -c 3 )-5-(a-c). 3. (or* - z 3 ) -*- (x - *). 2. (m 8 -a?)-*-(m-a;). 4. (x*-l)-*-(x-l). THE SUM OF TWO CUBES 97 5. (c8_8) + (c-2). 9. (c 3 d 3 - 125) + (cd- 5). 6 . (^_27)_^(d_3). io. (8 ar>- 27 m 3 ) + (2 x -3 m). 7. (64 -a?) -s- (4 -x). 11. (512c 3 d 3 -729)-=-(8cd-9). 8. (ary-l)-r-(a;?/-l). 12. (216 afy 8 - 1000 z 3 )^ (6ay -10 2). m. THE SUM OF TWO CUBES By actual division : q3 + b * = a?-ab + 6 2 . a-tb From the form of the quotient and its relation to the divisor, we state: 120. The sum of the cubes of two quantities may be divided by the sum of the quantities. The quotient is the square of the first quantity, minus the product of the two quantities, plus the square of the second quantity. Oral Drill Give orally the quotients of : 1. (a* +c *)+ a 6 + 5 a 3 + 10 + 2 a 2 . 7. a2/-4a + 52/-20. 16 x 4 + i4 1 _2x-7 3?. 8. afoc — 2 a& -f ca> — 2 c. 17. — arfcc -f 3 d — 3 ex -f- aca^. 9. 2rana;-5a;-6mw+l5. 18. 3 d- 10 d 2 -15 + 2 d*. (b) The Grouping of Terms to form the Difference of Two Squares Type Form ... x 2 + 2 xy +/ - z 2 . The Origin : By multiplication, (x + y + z)(x + y-z) = [(a + y) + *][(*+» - z] = ( X + t/) 2 - z 2 = a; 2 + 2 a;y + y 2 - s 2 . 112 FACTORING The product is the difference of a trinomial square and a monomial square. Therefore, an expression of this type is factored by grouping three of the terms that will together form a perfect trinomial square ; the fourth term being a per- fect monomial square. The result is a difference of two squares. Illustrations : 1. a 2 + 2 ab + b 2 - c 2 = (a 2 + 2 ab + 6 2 ) - c 2 st (a + by - c 2 = (a + b + c)(a + b-c). 2. x 2 - y 2 - 2 2/0 - z 2 = x 2 _ (y2 + 2 ^ + s*) = x 2 - (y + zY «[•+ ft + ! «)K^- &■+"#)] = (» + y + «)(&-?-«)• 3. a 2 - 6 ax + 9 at 2 - 4 m 2 - 12 m - 9 = (a 2 - 6 ax + 9 x 2 ) - (4 m* + 12 m + 9) = (a-3x) 2 - (2m + 3) 2 = [(a - 3 x) + (2 m + 8>] [(a -3 x) - (2 m -f 3)] = (a - 3 x 4- 2 m-\- 3) (a - 3 x - 2 m - 3). The process consists mainly in finding three terms that, when grouped, form a perfect trinomial square. The key to the group- ing is the given term that is not a perfect monomial square. Or, considering the given square terms, we may state : 137. (1) When only one given square term is plus, it is written first, and the other three terms are inclosed in a negative paren- thesis. (2) When only one given square term' is minus, it is written last, and the other three terms are written first in a positive parenthesis. Exercise 33 Write the factors of : 1. a 2 + 2ax + a?-m 2 . 3. c 2 + d 2 -y 2 -2 cd. 2. tf-2yz + z*-4:. 4. l-c 2 -2cd-tf. REPEATED FACTORING 113 5. v?-m 2 -l + 2m. 10. c 2 - 10 ex + 25 x 2 - 49 m 2 . 6 . 4 c 2_ X1> a ^-4a;2/-9» 2 2/ 2 4-4?/ 2 . 7. 9-m 2 -2m2/-2/ 2 . 12. 2/ 2 + 16z 2 - 16 + 8 yz. 8. 4c 2 + c 4 -4c 3 -4. 13. ary-a 2 & 2 + 16a&-64. 9. 12x + ±x 2 -9y 2 + 9. 14. 490^ + 10 c 2 ^ 2 -.^ 4 - 25. 15. 81 -100 x*y* + 60 x 4 y 2 z -9 z 2 . 16. 4iB 2 -4» + l-9m 2 4-6wi?i-7* 2 . 17. c? — a? + x 2 — y 2 — 2cx — 2 ay. 18. c 2 -d 2 -x 2 + m 2 — 2cm + 2dx. 19. 36c 4 4-l-49ic 6 -92/ 2 -12c 2 -42of l y. REPEATED FACTORING 138. Any process in factoring may result in factors that | may still be resolved into other factors. The review examples following frequently combine two or more types already con- \ sidered. The following hints on factoring in general will i assist. 1. Remove monomial factors common to all terms. 2. What is the number of terms in the expression to be fac- tored ? (a) If two terms, which of the three types ? (b) If three terms, which of the three types ? (c) If four terms, how shall they be grouped ? 3. Continue the processes until the resulting factors are prime. Illustrations : 1. Sa 7 -3ax* = 3a(a?-x 6 ) = 3 a (a 3 + x 8 ) (a 8 - x 3 ) = 3 a (a + x) (a 2 - ax + x 2 ) (a - x) (a 2 + ax + x 2 ). 2. 12x5-75x 8 + 108x=3x(4x 4 -25x 2 + 36) = 3 x (4 x 2 - 9) (x 2 - 4) = 3x(2x + 3)(2x-3)(x + 2)(x-2). SOM. EL. ALG. 8 J 114 FACTORING 3. 8 a 8 x - 2 ax + 8 a 5 x - 2 a 4 x = 8 a 8 x + 8 a 6 x - 2 a 4 x - 2 as = 2 ax (4 a 7 + 4 a 4 - a 8 - 1) • = 2 ax [(4 a 7 + 4 a 4 ) - (a 8 + 1)] = 2 ax [4 a 4 (a 3 + 1) - (a 8 + 1)] = 2ax(4a 4 -l)(a 3 +l) = 2ax(2a 2 +l)(2a 2 -l)(a+l)(a 2 -a + l). MISCELLANEOUS FACTORING Exercise 34 Write the factors of : 1. 5a? + 25x + 20. 21 75 a 3 -90 a 2 -{-27 a. 2. 5 0^-45 a 3 . 22. 6a 2 x+9abx-8a 2 y-12aby. 3. 8 a 2 - 24 a + 18. 23. 5 m 3 + 135. 4. a 4 -8a. 24. 15 c 3 +33 c 2 + 6c. 5. 2am + 2ma;-4a-4aj. 25. 242 ar*- 98 a. 6. 2-2c 2 -4cd-2d 2 . 26. 12 m 4 + 10 m 3 -8 m 2 . 7. 3 m 4 + 81 m. 27. 160 a,- 5 + 20 x 2 . 8. 3a 2 -9a-84. 28. 3 ar 5 - 15 a 3 + 12 a;. 9. 2c 5 - 128c 2 . 29. 98 ar> + 18 a*/ 2 - 84 afy. 10. 16 m V + 8 mnx + x 2 . 30. m 4 - 2 m 3 + 8m- 16. 11. 15 - 60 xy + 60 x*y 2 . 31. 15 a 3 - 25 a 2 a - 10 ax 2 . 12. 7 aV - 175 ax. 32. 81 afy 4 - 3 a#. 13. c 4 +c 2 -3c 3 -3c. 33. 4cd + 2c 2 d 2 + 2-2ar J . 14. Ux*-S2x 2 -12x. 34. 250c 4 -16c. 15. 49m 3 -84m 2 + 36m. 35. 8mV+4m¥-112mw. 16. 4o 3 -36ar J + 56a;. 36. 5 c 5 - 10 c 3 - 315 c. 17. 7 a 7 - 7x. 37. 21 ar> + 77 a 2 -140 a;. 18. 4 a 3 + 16 a 2 + 15 a. 38. 512 - 32 m 4 n\ 19. 8a 2 -18c 2 + 24a& + 186 2 . 39. 27^+215^-8. 20. 24^-55^-24^ 40. 3 x» -51 a^ + 48 x. SUPPLEMENTARY FACTORING 115 41. 686 x* - 2 xtf. 45. 242 a 4 - 748 a 2 + 578. 42. 8m 5 + 8m 2 -18m 3 -18. 46. 2x 5 +64y 2 -$x s tf-16x 2 . 43. 8a 3 + 2a 5 -8a 4 -2a. 47. 64a 6 + 729a 2 . 44. 32 a 4 - 2ajy. *»• 8 a* - 10 a 8 - 432 a. SUPPLEMENTARY FACTORING (a) Compound Expressions in Binomial and in Trinomial Forms Comparative illustrations with corresponding processes : 1. z 2 + 12z+35 = (x + 5)(z + 7). Similarly, (a _ &)2 + 12 (a - 6) + 35 = (a - 6 + 5)(a -6 + 7). 2. 3£ 2 + 10zy + 32/2 = (3s + y)(z + 3y). Similarly, 3 (a - b) 2 + 10 (a - 6) (c - d) + 3 (c - d) 2 = [8(a-&) + (c- 2 -l)-6(a-l) 2 . . (6) The Difference of Two Squares obtained by Addition and Subtraction of a Monomial Perfect Square 140. No general statement of this process can be given in a simple form. Illustration : Factor 9^+6^ + 49. The first and third terms are positive perfect squares. Hence, if the middle term were twice the product of their square roots, the expression would be a perfect trinomial square. For a perfect trinomial square the middle term should be 2 (3 2^(7) = 42 x 2 . Adding + 36 x 2 to the expression, we obtain the perfect trinomial square required, and subtracting the same square, + 36 x 2 , the expression is unchanged in value but is now the difference of two perfect squares. Therefore : 9z 4 + 6z 2 + 49 = 9a 4 + 6 x 2 +49 + 36 x 2 - 36 x 2 = 9 & + 42 x 2 + 49 - 36 x 2 = (3x 2 + 7) 2 -36s 2 = (Sx 2 + 7+6x)(3x 2 + 7-6x). It is important to note that a positive perfect square only can be added. SUPPLEMENTARY FACTORING 117 Exercise 36 Write the factors of: 1. a^ + ^ + l. 7. 25^-51^ + 25. 2. a 4 + 3a 2 + 4. 8. 81 m 4 + 45 ra 2 + 49. 3. n 4 -7n» + l. 9. 36c 4 -61c 2 m 2 + 25m 4 . 4. (j* -28 c 2 + 16. 10. 64 a 4 + 79^ + 100 a 4 . 5. 9c 4 + 5cV + 25a; 4 . 11. 64 m 4 + 76 m 2 n 2 + 49 n 4 . 6. 9 d 4 - 55 d?x 2 + 25 a; 4 . 12. 81 x 4 - 169 zV + 64 z\ (c) The Sum and the Difference of Equal Odd Powers By actual division : 2 x 2 + ax 3 + x 4 . a — x Hence, for the factors : a 5 + x 5 = (a + x) (a 4 - a 3 x + a 2 x 2 - ax 3 + x 4 ). a 5 - x 6 = (a - x) (a 4 + a 3 x + a 2 x 2 + ax 3 + x 4 ). By Art. 121, the factors of any similar cases may be found. In general, therefore : 141. (1) One of the factors of the sum of equal odd powers is the sum of the quantities. The other factor is the quotient obtained by using the binomial as a divisor. (2) One of the factors of the difference of equal odd powers is the difference of the quantities. The other factor is the quotient obtained by using the binomial as a divisor. Write the factors of : Exercise 37 - 1. ra 5 -l. 5. ra 5 + n 5 . 9. 32(^+243. 2. tf + x*. 6. 32 - n 5 . 10. a¥-32. 3. m 7 — n 7 . 7. 128 +c 7 . 11. m 11 + w 11 . 4. a 7 + x 7 . ■ 8. a?Y- 343. 12. a io c iD_ a jy. 118 FACTORING MISCELLANEOUS FACTORING Exercise 38 1. 3a*- 54^ + 243a. 14. 72 m 8 + 94 m 4 + 128. 2. ( c 3 + 8) + 2c(c + 2). 15 . 2a 2 (a+2)-5a 2 -8a+4. 3. 4a 2 + 9ar>-*-(12aa; + 16). 16. (c 2 + a 2 ) 2 - 4 c'x 2 . 4. 5(c? + l)-15c-15. 17. (4ar 5 +32)-5a 2 -17a-14. 5. (m 2 -12) 2 -m 2 . 18. 98 a 5 - 16ajV + 8aw*. 6. (a + 2) 2 -7(a + 2) + 12. 19. m 6 -3m 4 + 3m 2 - 1. 7. 4 + 2 a? — (am + 05) am. 20. (a^ + a- l) 2 -(a 2 -a-l) 2 . 8. 3(a + l) 2 -19(a + l)+6. 21. 2(ar 3 -l) + 7(a 2 -l). 9. (m 2 -12) 2 -(m 2 -6) 2 . 22. m 4 -9a 4 + m 2 + 3a 2 . 10. 2 a 6 + 38 a 3 -432. 23. 12m 2 -m(w-l)-(n-l) 2 . 11. (a 2 + 4) 3 -125a 3 . 24. 2 d 10 - 1024 a\ 12. (m+fl?) 2 — 7— 3(m+a;+l). 25. cx+my—mx—cz—cy-\-mz. 13. (c 2 + 4) 3 -16c 4 -64c 2 . 26. 4 mV - (m 2 + n 2 - 1) 2 . 27. (c + d)(m 2 -l)-(m + l)(c 2 -c? 2 ). 28. 126 2 -18ma + 12a6-3a 2 + 3a 2 -27m 2 . 29. 7m 2 — 7x 2 + 7n 2 — 14(az — mri) — 7 z 2 . 30. (m — l)(a 2 — asc) + (l — m)(ax — x 2 ). 31. (a-3) 3 -2(a-3) 2 -15(a;-3). 32. (a 2 + 4a + 3) 2 -23(a 2 + 4a + 3) + 120. 33. (x-2)(x-3)(x-4:)-(x-2) + (x-2)(x-3). 34. a?b + b 2 c + ac 2 - a?c - a& 2 - 6c 2 . 35. (3a + 2)(9a 2 + 2a; + 12)~(27aj 3 + 8). 36. 8(a-a) 2 + 5a 2 -5a 2 -3(a + a) 2 . 37. (c-l)(c 2 -9)-(c + 5)(c- 3)- 3(^ + 9 0. 38. (x + 2y) 2 -3(x + 2y + l)-W. 39. m 2 (x — 1) — 2 mx — m + x 2 (m — 1) — x. 40. ^(a-5) 2 + 2a(a 2 -a;-20) + (a; + 4) 2 . CHAPTER X HIGHEST COMMON FACTOR 142. A common factor of two or more algebraic expressions is an expression that divides each of them without a re- mainder. Thus : a s b s , a 3 5 4 , and a?b 5 may each be divided by ab. Therefore, ab is a common factor of a 3 6 3 , a 3 6 4 , and a 2 b b . In this definition the algebraic expressions are understood to include only rational and integral expressions. 143. Expressions having no common factor except 1 are said to be prime to each other. Thus : 3 abx and 7 cmn are prime to each other. 144. The highest common factor of two or more algebraic ex- pressions is the expression of highest degree that divides each of them without a remainder. Thus, {a 3 6 3 3 4 a 2 6 3 is the expression of highest degree that will ... divide each of the three without a remainder. a 2 b 5 That is, a 2 b 3 is the highest common factor of a 3 & 3 , a 3 6 4 , and a?b & . 145. The highest common factor of two or more expressions is the product of the lowest powers of the factors common to the given expressions. The abbreviation " H. C. F." is commonly used in practice, 119 120 HIGHEST COMMON FACTOR THE H. C. F. OF MONOMIALS 146. The H. C. F. of monomials is readily found by inspection. Oral Drill Give orally the H. C. F. of : 1. a 3 6 4 anda 2 6 5 . 8. 15 mn z and 20 m 2 x. 2. m 2 nx 2 and mn 2 x. 9. 12 m s n 7 and 15 m 4 n 9 . 3. 2m¥and4m¥. 10. 8 c z d z m and 12 -14z + 12, 8^-32 a + 30, 18 a 2 - 69 x 4- 63. 27. a 2 + ax — 2 a — 2 #, am 4- a 4- m# 4- as, a 3 — ax 2 . 28. 6a 2 x + 6a 2 -6x 2 -6x,4:a 2 m-4:a 2 -±mx+4:X,2a 2 x-2x 2 -2a 2 -\-2x. 29. 3a 2 + 6a2/4-3aa;4-6iB2/, 6a 2 + 6ai/ + 6aa; + 6a^,9a 2 -9a2/ 4- 9 ax — 9 #?/. 130. am + an + a + mx + nx + x, m 2 — n 2 + m — n, m 2 + m 4- w 4- 2 mn 4- n 2 . CHAPTER XI FRACTIONS. TRANSFORMATIONS 148. An algebraic fraction is an indicated quotient of two algebraic expressions. Thus : 2, ^, a2 + a6 + 62 are algebraic fractions. b x a-2b 149. The numerator of a fraction is the dividend; the denomi- nator, the divisor. The numerator and the denominator are the terms of a fraction : The following principle is of importance in processes with fractions : Since, by definition, a fraction is an indicated quotient, we may let the quotient of a divided by b be represented by x. Then b The dividend being equal to the product of the divisor by the quotient, a = bx. Multiplying by m, am = bmx. (Ax. 3) Considering am a dividend, bm a divisor, and x a corresponding quotient, a ^ = x. bm Hence, ^m = a (Ax. 4) bm b 150. That is : The value of a fraction is unchanged when both numerator and denominator are multiplied or divided by the same quantity. 122 THE SIGNS OP A FRACTION 123 THE SIGNS OF A FRACTION 151. Three signs are considered in determining the quality of a fraction. From the law of signs, a _ — a _ _ — a _ _ a b~ -b~ b -j> ' Since each fraction has the same value, -, o From the second fraction : If the signs of both numerator and denominator of a fraction are changed, the value of the fraction is not changed. From the third and fourth fractions : If the sign of either numerator or denominator and the sign of the fraction are changed, the value of the fraction is not changed. 152. Consider the signs of the factors of the terms of a fraction : ■ (+<0 =| (-a) ■ (-a) ■ (+o) m (+&X+C) (+&)(- <0 (-&)(+e) (-&)(_<>) " KJ Note that in each fraction two signs are changed and that the sign of the fraction is not changed. + (+«) = _ (-«) - _ (+«) _. _ (-«) (2 \ (+*)(+•) C+^X+c) (-6)(+c) (-ft)C-c)' W Note that in these fractions one sign is changed or three signs are changed, and that the sign of the fraction is changed. (1) We may change the signs of an even number of factors in either numerator or denominator of a fraction without changing the sign before the fraction. (2) We may change the signs of an odd number of factors in either numerator or denominator of a fraction if we change the sign before the fraction. The following is a common application of these principles : , etc. (w — n) (x — y) (in — ri) (x — y) (n — m) (x — y) 124 FRACTIONS. TRANSFORMATIONS TRANSFORMATIONS OF FRACTIONS To reduce a Fraction to its Lowest Terms. 153. A fraction is in its lowest terms when- the numerator and denominator have no common factor. Illustrations : ! 45 aWc _ 3 x 3 x 5 aWc = 3 ab 2 Regult 60 a 2 b 2 c 4x3x5 a 2 b 2 c 4 2 a* — ax* _ a(a - x) (a 2 + ax + x 2 ) _ a 2 + ax + x 2 R esu i t a 8 — ax 2 a(a — x) (a + x) a + x 3 ab + bm — am — m 2 _ (a + m) (b — m) _ — (a + m) (m — b) m 2 - b 2 ~ (m + b) (m - b) ~ (m + b) (m - b) a + m Result. m + b For (b - m)= ( - m + 6) = - (w - 6) (Art. 152). In general, therefore, to reduce a fraction to an equivalent fraction in its lowest terms : 154. Factor both numerator and denominator. Cancel the factors common to both. To cancel is to divide both numerator and denominator by a factor common to both. The expression "cancel" cannot be applied to any other operation in algebra. The terms of an expression in a numerator cannot be canceled with like terms in a denominator, for such an operation is not division. Oral Drill Eeduce orally to lowest terms : 8 m 2 35afy AZrfyz 48 c 3 fe 2 ' 12m* ' 2Sxtf ' 28xyz' ' 72 c 2 d 4 x 2 15 a& 39 m 2 n 2 19 a 2 fz A . lUa s x 2 z 20ac* ' 65m 3 n* ' 76a 3 y 2 z' ' 95a 3 x'z' _ 14m 2 yi 24 c 3 d 2 q 69 m 2 ny 12 81 a 2 mn 3 x 21 mn ' 36 c A d' ' 46 m 2 n 2 y 2 ' ' 135 m 2 naT TRANSFORMATIONS OF FRACTIONS 125 Exercise 40 Eeduce to lowest terms : 4a 2 -8a 10 m 2 -(x + l)\ 4a 2 + 20a* ' (m-xf-1 3m 2 -3n 2 ■ (^-(c + l) 2 6m 3 -6r/ ' (c + l) 2 -^ 1. 3. 2a*4-2a 2 + 2a 9-(a-2) 2 a 2 + a + l ' ' (3-a) 2 -4* 4 8c 3 -8 13 ar i + (a + c)a; + ac '4c2_4 as 2 -|- (-J7L -f- c)x + cm 2a 3 -18a 4 s»-4a? + 4--tf. 14. tf(y-z\y(tf-#),y + z. 15. 3(aj» + ajy), 8(^- t/ 2 ), 12 (x 2 -i/ 2 ). 16. 2c 3 -c 2 -c, 2c 3 -3c 2 -2c. 17. 5« 4 -5^, 4^-8^ + 40?. 18. 7a 3 -7a, 4a(a-l) 2 , 3a 3 + 6a 2 + 3a. 19. m 3 + l, 3(2 + 3ra + m 2 ), 4ra-4m 2 + 4m 8 . 20. 18-2^,3^-81, 4a; 4 -324. 21. am-f-a+ m + 1, am 2 + am — m 2 — m. 22 . 5 c (c - 2) 2 , 4 ca> + 12 c - 24 - 8 x, 24 c 2 - 3 c 5 . 23. (a + xf-1, a 2 -(x-l) 2 , (a-lf-x 2 . 24. 4a 2 -14a; + l2, 8 »* - 32 a + 30, 18 as" - 69 » + 63. 25. 6 a 2 A5 + 6 a 2 — 6 x 2 — 6 a, 4 a 2 m — 4 a 2 — 4 m# + 4 as, 2a 2 2. Change — and x „ to equivalent fractions having af + x xr — x a common denominator. The L. G. M. of x 2 + x and x 2 - x is a: (jc + 1) (x — 1). Hence L. C. D. = x (x + 1) (x - 1). Dividing each denominator into the L. C. D., we have : x(x + l)(x~l) _ x 1 . a (g + 1) (a - 1) = g | x X 2 + X x 2 - x Multiplying both numerator and denominator of each fraction by the corresponding quotients obtained : £rl x *=± = I^zDL. Result. X 2 + X X-l x(x' 2 -l) x + l x x+l = (x+l) 2 . Regult x 2 -x x + 1 x(x 2 -l) In general, to change two or more fractions to equivalent fractions having a common denominator: THE LOWEST COMMON DENOMINATOR 131 163. If necessary, reduce the fractions to their lowest terms. Find the lowest common multiple of the given denominators, for the common denominator. Divide each given denominator into the common denominator. Multiply both numerator and denominator of each given fraction by the respective quotients obtained. Exercise 43 Change to equivalent fractions having a common denominator : 3m 2m 5ra 2 3 l " T' 3 ' 2 ' 7 ' 5a 3a a 2 - T' T' S' 8 * AAA 2a 4a'3a 9. 4. A, A, 1. 10 ax 1 bic 9 ab 5 ± 2- ± 11 ' a°b' aV aV 6. -i-. -i-, JU. 12 c+1' c- m 1 m m + n n c — m c _ _ 3_ 8 a 7 x 15. i-^7 134 FRACTIONS Exercise 44 Simplify : i. ~+ x 4 1 - i2. -^-2 • a a + 1 a — 1 2 a + x ~ 3 _l 1 ~ a + x , 13 3 4 a 2 + a a 2 — a 3 £±I_1_^+A 14 3c 4c 2 6 ' 2c-2 3c + 3* 4 2m-l m-1 7m+l 12 5 5 3 10 " 7^-7 3a; 2 + 3* 5 c ~ - 2 c ~ ^ - 16 m + 3 m + 1 12 6 ' m + 4 m + 2* a 2 4- 2 a — 1 a + l < m + 9 m — 9 4 a 2 ""3a 2a* * m-9 m+9' a—b.b—c.c—a ■ _ 5 a + 7 5 # — 7 7. — r~ + "T 1 18 - i s*— •= =• a& 6c ca 5-2 16. 3tf2_a;_2' 3a 4 -3a; 10-10 ' 3a*+3aj»" m 3 2 c 2 - 9 c-3 (ft + 2) 2 . a? -4 c 2 + 2c ' c 2 + 3c + 2* ' a?-5 * as 8 -125' 4- (m 2 - 1). 142 FRACTIONS k c 2 — ccc— 6ar* . c+3x x 2 — 4a; + 3 t x — 1 (P-9CX 2 ' c+2x 3-x ' 2 — x n a? + 6x-7 . x 2 -3x+2 aj"-l. 1-x x*-±x-2l x* + x-6 1-x 3 ar' + l+a; ^_8aj + 15 lx-10-x 2 9. 10. 6 — 50? + ^ ' x 2 — 4a;+4 (a.-l)2_ a 2 ^_(i_ a )2 (a._ a )2_i ' ^_( a _i)2' X1 a 2 -3a x ' \x J\m 2 -2mx-3x 2 J\3 J \a + z J\a — z J ar — z 2 (2. + « + D(8.-7 + ?) + (.-g. +2 y\ ( c + 2 y-(P / 2d Y (c + d + 2f \ ^c + 2-d) 10. / r 2a? + 5 + ?V3a-7 + - 11. (c + 2y-(c-2)' ; 8c ' 3m + 3 14. r * 4 i^-iu^+i+^j. LC^-ox 2 ) 2 ]^ 6 ) W a 1 ) Vm+i^ J\m-i A™- 1 A^+i / [2Va + 2 a-2y/ 2a 2 +4aJ V ; 144 FRACTIONS [_m J n mn J |__ftMr mn m'nj " [(•^)("8(- + ;)> ' w+ * w ' +1 ' ti4 ! a-\-xj c(acx + a + x) /'/7_1^2 ,v7_/*_9/7_i_9 \ cy \acx -f- c 4_2c 3 -c + 2 (d-1) 2 . cd-c-2d + 2 ±-d 2 ' c 2 -c ' d + 1 FRACTIONS IN THE COMPLEX FORM 172. A. complex fraction is a fraction whose numerator or denominator, or both, are fractions. The order of processes used in simplifying complex fractions varies with different types, and no general statement will cover all possible cases that arise. The student will easily under- stand the following Illustrations : 1. Simplify f ■ x * 1_1 c 3 + d 3 d c c 2 + cP c 2 + d? - cd d y c*-cP _ d x c 2 -d* 1_1 c 8 + # c-d c 3 + d 3 d c cd tf-cd+d* cd ' (c + d)(c-d) _ c d c-d (c + d)(c 2 -cd + d 2 ) x 2. Simplify * + 2 x ~ 2 , x-2 x+2 The L. C. D. of both numerator and denominator is (x + 2)(x Multiplying both numerator and denominator by (x + 2) (a; — 2), FRACTIONS IN THE COMPLEX FORM 145 x . x x + 2 x-2 _ x(x-2) + x(x + 2) _ x 2 -2x + x 2 + 2x _2x 2 _x ~lc ~ x(x + 2) -x(x-2) x 2 + 2x-x 2 + 2x 4x 2* x - 2 x + 2 Result. 3. Simplify 1 + ^ (The work begins by first sim- ■j . 2 plifying the lowest fraction, etc.) 1 — x i+ 1 \ = i+ t-V= i+ 7to =1+ ^ t , 2 3-s 3-x 3-x 1-x 1-x = 1 +f_=^ = _i_. Result. 1+x l+x Exercise 49 Simplify : 6 . ^-1- "- 1 . 1. •*■■■ M . ■' s x — o . 3 \ x — o x + 3 2 2 x + 3 1 1 2 z + 3 c-1 2o- c -±| c + 1 x*-tf-l ^^ 2a . 4 o Multiplying both members by 12, 12 (x + 6) 12 (x + 10) t 4 6 Reducing, 3 (* + 6) = 2 (x + 10). And the original equation is merely changed in form and is free from fractions. In general, to clear an equation of fractions : 174. Multiply both members of the equation by the L. C. D. of the fractions, remembering that the sign of each fraction becomes the sign of its numerator. Solve the resulting integral equation. Illustrations : # I. Solve^±l_^z2 = 6- £ . 3 ' 5 2 The L. C. D. = 30. Multiplying both members by 30, 10(2x + l)-6(3x-2)=15(6-x). 20 x + 10 - 18 x + 12 = 90 - 15x. 20x - 18x + 15x = - 10 - 12 + 90. 17x = 68. x = 4. Result. 147 148 FRACTIONAL AND LITERAL LINEAR EQUATIONS x-1 x + 2 „ 2x? 2. Solve 2- x + l x — 1 x 2 — l Multiplying both members by the L. C. D., (% + l)(x - 1), !)•-(* +!)(* + 2) = 2(x 2 - 1) x 2 -2a; + l From which x 2 -Sz-2 = 2x 2 1 2 a*. -2x2. x = — Result. 5 Solve : x 4-1 ,a5 — l 3 *-l 1. 3. 4. 4- 4 * + l 2 a; + 1 3 x-1 = 4. 3. Exercise 50 ' 2x-l A-x 2x+l o. 6. 3 2*4-1 5 3*4-1 14 15' 7. = 4:' 8. 3 5 10 4 »-l Q 2*4-1 3 J_ 5 3 x + 2 3*4-1 2*4-1 4 3 2 3 a; 4-2 *-6 1-2* = 0. 4-1. 9. i.(*_l)_f(*4-l)4-2=0. 10. f(2*-l)- T V = -f(5 + 3*). 11. (*4-|)(^-i) = * 2 . 12. (x-l)(x + i) = (x-l)(x-i). 13. i(*_2)-|( a; - r -l)-i(*-2) = 0. 14. 2f *-fl 2x- 15. (*4-l) 2 ' (*4-2) 2 ^ 14-* ! rA 16. 17. 18. 3 x — 6 #4-5 *+3 2*-3 4*-5 3* + 4 3*-2 6*-7 2*-5 6* — 1 4*4-1 = 0. 19. 20. 21. 4*-l 3*-l a? 4-1 " x-1 4 * * 1. * — 5 x 5*-l 3 = 0. *4-3 4-2 7*-3 *4-l = 0. SPECIAL FORMS OF FRACTIONAL LINEAR EQUATIONS 149 00 2x Sx x . 2x 5 Z6. — -J- — — — -+- 23. » + l a? + l 2a; + 2 3a? + 3 6 a^ + 2a? + 4 2 a; ^-2^ + 4 a + 2 4-a 2 x-2 •24. _6 2 _ 6. 25. 26. 27. 2a; + 3 3-2a 4a 2 -9 3a;-4 18g 2 + a? = 3a? + 4 3a> + 4 9^-16 4-3z* 9a^ + 3a? + l 6 a;== 9a;2 - 3a? + 1 3a; + l l-3a; 2 3 4 A o t _ 56 the i ar g er part required. 2-7 + 1 And, verifying, 56 +■ 2(4) = 7. GENERAL STATEMENT OF A PROBLEM 163 Exercise 55 1. Divide the number c into two parts such that m times the larger part shall equal n times the smaller part. 2. Divide a into two parts such that the sum of -th of the n larger part and -th of the smaller part shall be q. r 3. The sum of two numbers is s, and if the greater number, g, is divided by the less number, the quotient is q and the remainder r. Find the numbers. 4. If A and B can together mow a field in t days and B alone can mow the same field in b days, find the number of days that A working alone will require to do the work. 5. Show that the difference of the squares of any two con- secutive numbers is 1 more than double the smaller number. 6. m times a certain number is as much above k as d is above c times the same number. Find the number. 7. A and B are m miles apart, and start to travel toward each other. If they start at the same time and A goes at a rate of & miles an hour while B goes at the rate of s miles, how far will each have gone when they meet ? 8. When a certain number is divided by a, the quotient is c and the remainder m. Find the number. 9. The front wheel of a wagon is m feet in circumference, and the rear wheel n feet in circumference. How far has the wagon gone when the rear wheel has made r revolutions less than the front wheel ? 10. A and B can together build a barn in r days, B and C the same barn in s days, and A and C the same in t days. In how many days can each alone build it ? 164 PRACTICAL APPLICATIONS OF GENERAL SYMBOLS THE USE OF THE LABORATORY FORMS OF PHYSICS 179. Density is defined as mass per unit of volume, and may M be calculated by the formula D—^, in which D is the density of the body under examination, M its mass, and Fits volume. Volume is another name for cubical contents. The mass of a body may be found by weighing it, and the formula for density W may be written D = — , in which W is the weight of the body to be considered. Exercise 56 1. A block of iron 3 feet 6 inches long, 2 feet wide, and 6 inches thick weighs 1706.25 pounds. Calculate its density in pounds per cubic foot. 2. The density of water is approximately 62.5 pounds per cubic foot. What is the volume of a ton of water ? 3. A cubical block of wood 9 centimeters on an edge has a mass of 371.79 grams. Calculate its density in grams per cubic centimeter. 4. A block of iron having an irregular cavity weighs 3.265 kilograms. When the cavity is filled with mercury, the whole weighs 3997 grams. The density of mercury being 13.6 grams per cubic centimeter, calculate the volume of the cavity. 5. The density of a certain substance being a grams per cubic centimeter, calculate the mass of this substance necessary to fill a vessel of 1000 cubic centimeters' capacity. 180. To change a reading on a Centigrade thermometer to a corresponding reading on a Fahrenheit thermometer use is made of the formula _F=.f 0+ 32, In which F is the reading in degrees on the Fahrenheit scale, and C the reading in degrees on the Centigrade scale. To LABORATORY FORMULAS OF PHYSICS 165 change Fahrenheit readings to Centigrade readings we use the formula C = f(F- 32). Exercise 57 1. Change the following Centigrade readings to correspond- ing Fahrenheit readings : 20°; 40°; 0°; -15°; -273°; -180°. 2. Change the following Fahrenheit readings to correspond- ing Centigrade readings : 212°; 32°; 70°; -10°. 3. At what temperature would the reading on a Centigrade thermometer be the same as the reading on a Fahrenheit thermometer? 4. The Centigrade scale is marked 0° at the freezing point of water, and 100° at the boiling point of water. The Reaumur thermometer is marked 0° at the freezing point of water and 80° at the boiling point. Prepare (1) a formula for changing Reau- mur readings to Centigrade readings, and (2) a formula for changing Centigrade readings to Reaumur readings. 181. In the figure, we have a straight bar whose weight is __ ,m m to be neglected. The bar is sup- ——d- \~jf~ ! ported at the point C arid is 1 acted upon by the several forces, ,«// f C t» /? /'? /"> etc v * n tne directions indicated by the arrows. These forces tend to cause rotation of the bar about the axis C. The tendency of a force to produce rotation is called its moment, and the moment is calculated by multiplying the magnitude of the force by the distance of its point of application from the axis. Moments tending to rotate a body clockwise are given a positive sign ; those tending to produce rotation in the opposite direction are given a negative sign. In order that a body under the influence of moments may be in equilibrium, i.e. stationary, the algebraic sum of the 166 PRACTICAL APPLICATIONS OF GENERAL SYMBOLS moments acting on it must be 0. Thus in the figure, if the bar is in equilibrium, f'd" + f'"d'" -fd -f'd' -f""d"" = 0, or, better : (f'd" + /'"<*'") - (fd +fd' +f""d"") = 0. (/'»/"»/"'» etc., are read "/prime," "/second," "/third" etc.) Exercise 58 1. Two weights of 4 and 12 pounds respectively are bal- anced on a bar, the distance from the support to the further weight being 6 feet. What is the distance from the support to the nearer weight ? 2. A bar 11 feet long is in equilibrium when weights of 15 and 18 pounds are hung at its ends. Find the distance of the point of support from each end. 3. If weights are distrib- , — v: — ,o-- ED EI uted upon a bar as in the figure, where must a weight of 40 pounds be placed to keep the bar in equilibrium ? 4. A 10-pound weight hangs at one end of *a 12-foot bar, and a 15-pound weight hangs at the same side of the support- ing point, but 2 feet nearer it. If a 40-pound weight at the other end keeps the bar in equilibrium, at what distances from the ends is the point of support located ? THE VALUE OF ANY ONE ELEMENT OF A FORMULA IN TERMS OF THE OTHER ELEMENTS 182. The statement of a mathematical law by means of a formula always gives an expression for the value of the par- ticular element to which the law refers. By transpositions and divisions we are able to derive from any formula another expression or formula for any oue of the other elements. TRANSFORMATION OF FORMULAS 167 Illustrations : 1. Given the formula, R = & s . Derive a formula for s. 9 + s Clearing of fractions, (g + s) B = gs. Multiplying, Bg+ Bs = gs. Transposing, Bs — gs = — Bg. Dividing by — 1, gs — Bs = Bg. Collecting coefficients, (g — B)s = Bg. Dividing by (g — B) , s = — 2-, the required formula for s. g — B 2. Given the formula, I = a + (n — 1) d. Find an expression for n. * l = a+ (n-l)d. Multiplying, I = a + nd — d. Transposing, I — a + d = nd. Dividing by d, n= ~ a + , the required formula for n. d THE TRANSFORMATION OF FORMULAS Exercise 59 PHYSICAL FORMULAS 1. Given v = at, find the value of t in terms of a and v. 2. Given S = % g (2 1 — 1), find a formula for t in terms of s and g. 3. Given C = , derive a formula for S in terms of E, f*9 b, P, C, and R. 4. Giv p, and p' 4. Given -=--{-—, find an expression for each element, /, 168 PRACTICAL APPLICATIONS OF GENERAL SYMBOLS 5. Given F = f C + 32, derive a formula for C in terms of F. 6. Given F = — , v = gt, and w = .Fs, find a value for to in terms of m, g, and s. 7. Given ^1 = ^M, derive expressions for 2\ and v 2 - MISCELLANEOUS FORMULAS 8. Given 0=2 irR, obtain a formula for M. 9. Given / = a -f (n — 1) d, derive an expression for a in terms of I, n, and d. 10. Given 8 = ^ (a ■+■ Q> find the value of a in terms of Z, n, and & 11. Given T=nR(R + L), find £ in terms of the elements involved. 12. Given S = r ~ a , find each element in terms of the r- others. 13. Given d=-±- - 1 , find a formula for S. n(n — l) 14. Given a=p-\-prt, find each element in terms of the others. GENERAL REVIEW Exercise 60 1. Simplify 2a- [3-2{a-4(a-a + l)}]. 2. Solve (a>-l)(« + 3)-2(«-l)(3»+l) = (3-«)(2+5a). 3. Showthatfl-f-4^— ^-Y-^- i_+lVl=0. \ a + 1 a-f-3/\a — 3 a — 1 / GENERAL REVIEW 169 8 or a r l 1 a 2 M la 2) a a z J 4. Simplify 5. Factor a 5 -S(a 2 -x 2 )-a 3 x i . 6. What number added to the numerators of the fractions - and -, respectively, will make the results equal? Is there b d an impossible case ? 7. Factor 6 x 3 + 6 a 2 x - 37 ax 2 . 9. Solve + l-2a; l-4a; l-3z 10. Factor (a 2 - 9) (a + 2) - a - (4 + a) (3+ a) - 3. 11. Solve ^±^ S_ = ^-m 3 a + ra 3 12. By three different methods factor (sc 2 — 6) 2 — a£ 5 13. Solve JtlJfl-lL 2,-1 4^,-1 37 6(1-2/) 14. For what value of aiscc 4 — 3a^ + 2a^ + 21a; + 3a divis- ible by x 2 + x- 3 ? 15. The sum of the numerator and the denominator of a certain fraction is 39. If 3 is subtracted from both numerator and denominator, the result is \. Find the original fraction 16. Factor 225 - 4 x 2 (9 + x) (9 + x). 17. Find the H.C.F. and the L. CM. of x 4 - ax* - 2 a 2 x>, 2ar 5 -2a 2 a;, and 3 ar> + 12 aa 2 + 3 a 2 x. 170 PRACTICAL APPLICATIONS OF GENERAL SYMBOLS 18. Simplify ( C -±±- C -=i) + ( c ±A + c ^±). ■ 19. Solve and verify 2 -^=± + ^ = *JL=1 . 3 7 2s + 2 28 20. Prove that -i- 1 — = cd- 1+i 1-1 od-I cd cd cd 21. Simplify (a-l) 2 -4^r_3 21 (a-3} p J (a+2) 2 -l [a + 3 a-3j a-U 22. Solve 2 15 (M-.l a + 2 a + 3 23. A boy has a dollar, and his sister has 28 cents. He spends three times as much as she spends, but has left four times as much as she has left. How much did each spend ? 24. What is the value of m if 3m + 5n = 1 when n = - 1. m — n ■2 S_ x _7 25. Solve £_ + -* -— 4— J- f(a^l) 1(1+,) 1-1 26. Prove that the sum of any five consecutive numbers equals 5 times the middle one. 27. Factor (a 2 -{-5 a- 10) 2 + 2(a 2 + 5 a- 10)a - 8 a 2 . 28. Find three consecutive numbers such that if the second and third are taken in order as the digits of a number, this number will be 7 more than four times the sum of the 1st and 3d given numbers. 29. If a = 2, b = 3, and c = — 4, find the value of (a 2 + c*)(3 a - c)V(7a + c)(4 6-a). 30. Simplify - x GENERAL REVIEW 1 171 1 Sx2 -* i 5 1 + 3 3a + l I J — 2x 2x 2x +3 31. At what time between 8 and 9 o'clock are the hands of a watch 5 minutes apart ? 32. Find the H. C. F. and the L. C. M. of aj»-4, 8-a? 8 , -4+4a-a 2 , (2-oj)*, and (a?-2)(l-a?). />»2 ^yj ~, 33. Solve and verify -^ (- - 4^ — m 4 2a? + m 34. Simplify g + i+^)+j a?+l . x(x-l)+l ~\ . a 2 cc 2 » 3 — 1 a^— 1 35. What value of x will make (4 x + 3) (3 x — 1) equal to (6o> + 5)(2»-l)? 36. Simplify (6-c)(c-a)(a-5) 5-c c-a g-6, J (&4-c)(c + a)(a + &) & + c c + a a +6 37. By what must x 4 -}-2x s -\-Sx 2 -\-4:X-{-5 be divided to give a quotient of ar* + 4 a? -}- 14 and a remainder of 44 a? + 47 ? 8^-1 38. Simplify _. 1 1- l+» 39. Show that if a = - 1, i(l-"H 1('-)K 1+ -) SFFi " JF+3R 40. Factor mV - mV - 27 »V + 27. 41. Solve fg- > + c ) 2 - CT = a- 3ca; . n a 42. Factor 6 a 2 — 6np — (4n — 9p)a. 172 PRACTICAL APPLICATIONS OF GENERAL SYMBOLS 43. The head of a certain fish weighs h pounds, the tail weighs as much as the head and -J the body, and the body as much as the head and tail. What is the weight of the fish in terms of h ? 3a 2 -3al 44. Simplify I 6a?-6 4a a + 1 48^~ 45. In how many years will s dollars amount to a dollars at r per cent, simple interest ?• 46. Solve for m : (2m + 2)(3 b — c)+2ac = 2c(a — m). 47. If a certain number of wagons is sold at $80 each, the same amount is received as when 10 less are sold at $ 100 each. How many are sold in each case ? 48. Find the value of \a? a J a? cr + 1 a 2 — 1 m 49. Factor a 2 + a 2 -2(1 -ax)— (x + a). 50. Two bills were paid with a 10-dollar bank note. One bill was 25 % more than the other, and the change received was -i-f the smaller bill. Find the amount of each bill. 51. Solve for s: (s - a) (s + b) - (s + a)(s - b) - 2(a - b) = 0. 52. Factor aJ 6 - X s — 2 (3 a? + 4). 53. Solve for n, I = (n — 1) (a — I) + a. 54. An orderly is dispatched with an order, and 3 hours after he leaves a second orderly is sent after him with instruc- tions to overtake the first in 6 hours. To do this he must travel 4 miles an hour faster than the first traveled. How many miles an hour does each travel ? 55. Factor 50 x + 2 x 5 - 38 X s . CHAPTER XVI SIMULTANEOUS LINEAR EQUATIONS. PROBLEMS 183. If x and y are two unknown quantities and their sum equals 7, we may write x + y = 7. Clearly, an unlimited number of values of x and y will satisfy this equation. For example : If x = 2, y = 5. If* = l, y = 6. If x = 0, y = 7. • 11*= - 1, 2/ = 8. Iix= - 2, y = 9, etc. 184. Such an equation in two unknown quantities, satisfied by an unlimited number of values for the unknown quantities, is an indeterminate equation. 185. If, however, we have with this equation a second equa- tion stating a different relation between x and y } as x - y = 3, then the pair of equations, x + y = 7, x - y = 3, is such that each is satisfied only when x = 5 and y = 2. For x + y = 5 + 2 = 7, And x-y = 5-2 = 3. 173 174 SIMULTANEOUS LINEAR EQUATIONS No other values of x and y will satisfy this pair of equations. Hence, 186. Simultaneous equations are equations in which the same unknown quantity has the same value. 187. A group of two or more simultaneous equations is a system of equations. 188. Two possible cases of equations that the beginner may confuse with simultaneous equations must be carefully noted. (a) Inconsistent Equations. Given : x + y = 9, x + y = 2. It is manifestly impossible to find a set of values for x and y that shall satisfy both given equations. The equations are inconsistent. (b) Equivalent Equations. Given : x + y — 4, 3a + 3^ = 12. If the first equation is multiplied by 3, it becomes the same as the second equation, and every set of values that satisfies the second satis- fies the first as well. The equations are equivalent. 189. For a definite solution of a pair of simultaneous equa- tions we must have a different relation between the unknown quantities expressed by the given equations. Equations that express different relations are independent equations. • 190. Simultaneous equations are solved by obtaining from the given equations a single equation with but one unknown quantity. This process is elimination. Each of the three methods of elimination in common use should be thoroughly mastered. ELIMINATION BY SUBSTITUTION 175 ELIMINATION BY SUBSTITUTION Illustration : (5x + 2y = ll, (1) Solve the equations : { ^ . ' H {3x + ±y = l. (2) l-4y 3 From (2), Substituting in ( 1 ) , 5 1 1-4y > \ + 2 y = 11. From which 5-20j + 2 y = n. o Clearing, 5 - 20 y + 6 y = 33. - 14 y = 28. y=-2. Substituting in (1), 5x + 2(- 2) = 11. 5x-4 = ll. x= 3. l Result. y = - 2. J Check : Substituting in (1), 5 (3) + 2 (- 2) = 15 - 4 = 11. Substituting in (2), 3 (3) + 4 (- 2) = 9 - 8 = 1. From the illustration we have the general process for elimi- nation by substitution : 191. From one of the given equations obtain a value for one of the unknown quantities in terms of the other unknown quan- tity. Substitute this value in the other equation and solve. The method of substitution is of decided advantage in the solution of those systems in which the coefficients of one equa- tion are small numbers. In later algebra a knowledge of this method is indispensable. In applying this process of elimination care should be taken that the expression for substitution is obtained from the equa- tion whose coefficients are smallest. The resulting derived equation will usually be free from large numbers. 176 SIMULTANEOUS LINEAR EQUATIONS ELIMINATION BY COMPARISON Illustration : 5x + 2y = 9, (1) Solve the equations : , '2a5 + 32/ = 8. (2) From(l), x= 9 ~ 2y ' From (2), x-*^^- 2 By Ax. 5, 9-2 1 = 8-3_ y< 6 2 Clearing, 2 (9 - 2 y) = 5 (8 - 3 y). 18 - 4 y = 40 - 15 y. 11 y = 22. 2/ = 2. Substituting in (1), 5 x + 2 (2) = 9. Hence, *, _ X = 1,1 2/ = 2. J Check : Substituting in (1), 5 (1) + 2 (2) = 5 + 4 = 9. Substituting in (2), 2 (1) + 3 (2) = 2 + 9 = 8. In general, to eliminate by comparison : 192. From each equation obtain the value of the same unknown quantity in terms of the other unknown quantity. Place these values equal to each other and solve. The method of comparison is particularly adapted to those systems of simultaneous equations in which the coefficients are literal quantities. ELIMINATION BY ADDITION OR SUBTRACTION Illustrations : 1. Solve the equations : \ Sx -±y= 5 > H l5a> + 3y = 18. (2) Choosing the terms containing y for elimination, we seek to make the coefficients of y in both equations equal ; for, if these coefficients were ELIMINATION BY ADDITION OR SUBTRACTION 177 equal, adding the equations would cause y to disappear. The L. C. M. of the coefficients of y, 3 and 4, is 12. Dividing each coefficient into 12, we obtain the multipliers for the respective equations that will make the coefficients of y the same in both. Multiplying (1) by 3, 9 x - 12 y = 15 (3) Multiplying (2) by 4, 20 x + 12 y = 72 (4) Adding (3) and (4), 29 x = 87 x =3. Substituting in (1), 3 (3) - 4 y = 5. T 9-4, = 5. Hence,* = 3,l Regult , = 1. y = 1 *i It is to be noted that the signs of y, the eliminated letter, being unlike, the process of addition causes the y-term to disappear. 2. Solve the equations I2x-3y = l, (1) \3x + 7y = 13. (2) Multiplying (1) by 3, Multiplying (2) by 2, Subtracting (4) from (3), 6 x - 9 y = 3. (3) 6 x + 14 y = 26. (4) - 23 y = - 23. 2/ = l. 2s-3(l)=l. tt 2a- 3 = 1. Hence '* = M Result. x=2, y =1 -i Substituting in (1), In this example since the signs of the sc-term are like, the equations are subtracted to eliminate the sc-term having the same coefficients. In general to eliminate by addition or subtraction : 193. Multiply one or both given equations by the smallest num- bers that will make the coefficients of one unknown quantity equal. If the signs of the coefficients of the term to be eliminated are unlike, add the equations ; if like, subtract them. The necessary multipliers for the elimination of a letter may be found by dividing each coefficient of that letter into the lowest common multiple of the given coefficients of that letter. The method of addition or subtraction is of advantage when it is desirable to avoid fractions, and is used in the solutions of systems involving more than two unknown quantities. SOM. EL. ALG. 12 178 SIMULTANEOUS LINEAR EQUATIONS Exercise 61 Solve: 1. x + y = 3, 10. 3s + 7y = -8, 3x + 2y = 7. s + y = 0. 2. a;-2/ = 4, 11. 5x + 3y = 10, 2x + 3y = 13. Sx-5y = 16. 3. 4 x + 3 y = 7, 12. 7# — 3 = 4y, 3 = 22, 3m — 5n4-7p = — 1, 8m-9n-5jp = 21. 9. a-hz=10, 2/-z = 2, «-2/ = 2. 10. 2 # + 2/ =3, 3z-x = -3, ±y-3z = -12. 11. a?4-2/ + z = 0, 5x-3y = 50, 2z + y = -20. 12. y = 5 x + 17, 3^-22/4-37 = 0. 71 = - 7 * 4- 2 x. 13. a?4-2/ + z=24, x + y + u = 25, x + z + u = 26, y + z+u = 27. 14. a?4-y=18, 2/ 4- * = 14, z + w = 10, w 4- w = 6, a 4- u = 12. FRACTIONAL EQUATIONS 181 FRACTIONAL FORMS OF SIMULTANEOUS LINEAR EQUATIONS (a) When the Unknown Quantities occur in the Numera- tors of the Fractions 197. Systems of simultaneous linear equations in which the fractions have unknown quantities in the numerators only, are solved by first clearing of fractions and then eliminating by either of the three methods. Illustration: [ V - * 1_ ^~ 4 m ~2~ 3 ' W Solve the equations : 3 4 w From (1), By -Bx-6 = 2x-S. Simplifying, 5 x - 3 y = 2. (3) From (2), 4 (2y + 1) - 3 (a + 3) = 0. Simplifying, 3 x — 8 y = — 5. (4) From (3), x = ?JL±2. From (4), x = ^^ . o 3 Therefore, Sy±2 = Sy-5 t (5) 5 3 Solving (5), y = l,l Regult> Substituting in (1), x = 1> J 198. Extraneous Roots. In fractional simultaneous equa- tions we may reject any solution that does not satisfy both given equations. Exercise 63 Solve: 1. * , y - 3^4 6, 4^~2 7. 2. X 5~ y _ 3 •■ — X 4 .y - 7 :2. 3, 3x 2 T 3 " = 17, 2x 3 ■3 = 13. 4 2^fl + 3^1 = 6> 10 ?_4 = |+6, 3a; — 1 2y — l _g a; + y a; — 2 y = 35 a; 2 + 5 ' 10 *" 8 2 6* 5y-2 a;-9 = 5 3s + l 4y + l = 1 3 2 2' 3 2 2' 3a?~l y + 4 1 =0 2a?-l 4y-l ^l 2 3 6' 2 32* 2 3 ~4' iL_?.4-5 5-5 a2 J -l_y+l == _5. ' 3 + 4 ~8 ? 3 2 6* 5 (x +2/) = a; — y. 2y-a;-3 y-2x-3 _, 13. j - -4, 4y-3a;-3 2y-4a; + 9 _ 14. * + 2>+* = 10, 15. ^-^ + ^ = 0, 2 T 3 4 3 2 6' » y_z_ 8 4a; . 2y 3z_l 3 2 8~ ' 3 3 2 2' a:_y 2_o 5a; 5y 4z_7 6 5 + 4~ ' 16. | + | = 2, 17. 2 3 2>+*=4. 3 2 4 6 • y__ * 2 3 12' 18 a; z_ 1 3 4~24' 2/ *_ * . 2 3 12 a; 3" y_ 11 2 72 2/ 4 2 1 "3 48' a; 2 z _ 5 "5 24* FRACTIONAL EQUATIONS 183 (b) When the Unknown Quantities occur in the Denom- inators of the Fractions, giving Simultaneous Linear Equations in - and — x y 199. A solution of this type of simultaneous equations is obtained by considering the unknown quantities to be - and -, x y and the process of elimination is carried through without clear- ing of fractions. Much difficulty is avoided by this method. Illustrations : f3 2 == _31 x y 40 ' 5_10 = 11 # x y 8 1. Solve the equations : I Multiplying (1) by 5, Adding (2), Dividing by 20, Clearing of fractions, Substituting in (2) , - - - — = — 16 , 10 31 V — — — - — X y 8 5_ 10 11 X y 8 20 20 X 8 1 1 X Y X =- -8. 11 10 16 (1) (2) (3) 8' - so = 16 y< y=- 5 - Hence, x y ::;:) Result. 200. If the coefficients of x or y are greater than 1, the equations may be changed in form by multiplying each equa- tion by the L. C. M. of those coefficients. The resulting equa- tions will be of the same form as the equations just solved. 2. Solve the equations : . 2a; 3^ 4a; 6y (i) (2) Multiplying (1) by 6 i£ + l? = _12. - Multiplying (2) by 12, ?! + — = 12. x y From the system (3) and (4), jc = - and y = — 1 2 * 3' Exercise 64 Solve: , 1 , 3 o 3 5_7 5 2 _11 1. - + - = 2, x y 6. x y 6' 11. 2a; Sy 6' ?-§ = 2. 5_6 = 7 9 2/ — 8 x = xy. x y a; 2/ 2 12. 1,1 17 1 ~ TrT J A 2,3 ., 1 1 _ 5 a: y z 12 2. - + - = 1, x y 7. 2a; Sy 12' 1 1,1 5 1 — -i r>> 4_3_1 1 1 ___ 1. a; 2/ z ** a 2/ 2 3a; 2y 12* x y z 12 „ 3.5 o 1 1.7 3. - + - = 2, a ' 2/ 8. 3a; 2y 18' 13. ?-?+!=? 3_10 = 1 x y 1 \- 1 - 1 . a? y z 5 2a; 32/ 2* 2,3 2_17 x y z 30 4. ?_9 = _ 2 , 2 3 35 4 4 3_83 ( 9. — — SB • a; y 3a; 2y 6 a; 2/ z 60 1-1 = 0. •Li-ik 3 3 _3 a; y 2a; 32/ 14. ^""^""i' 5. i- 3 -=-2, 10. ^_ _5_ = _4 2 3 __1 x y 3a; 22/ 3* 3a; 2y 6' 5 2 13 6 3 _19 # 2 3 19 i i"'8 f 5a; 2y 5* 3y 2% 36* (3) (4) LITERAL SIMULTANEOUS LINEAR EQUATIONS 185 LITERAL SIMULTANEOUS LINEAR EQUATIONS Illustration : Solve the equations : \ r ~ ' )J {mx + ny=d. (2) From (1), x = l^M. .From (2), x = y - 2 18. 19. x — y+ n y+x-m =3 y — x— n ax _ .. my (a -f- m)x .. (ra ~ a)y by ax 2, by + cz ^ 2 a cz + ax 21. 22. 2. ay , bx _ cy dx 6 6 ~9 9 — a 52 3 ' a-1 a+1 a — c x-1 1-c ^=i, + a-1 y + i a PROBLEMS — SIMULTANEOUS LINEAR EQUATIONS 187 PROBLEMS PRODUCING SIMULTANEOUS LINEAR EQUATIONS WITH TWO OR MORE UNKNOWN QUANTITIES 202. A problem may be readily solved by means of a state- ment involving two or more unknown quantities provided that (1) there are as many given conditions as there are required unknown numbers, and (2) there are as many equations as there are required un- known numbers. Exercise 66 1. If a certain number increased by 3 is multiplied by another number decreased by 2, the product is 9 more than that obtained when the first number is multiplied by 1 less than the second number. The sum of the first number and twice the second number is 30. Find the numbers. Let x = the first number, and y = the second number. Then (a + 3) (y-2) = the product of (the 1st no. +3) by (the 2d no.- 2). Also, x{y — 1) = the product of (the 1st no.) by (the 2d no. — 1). Therefore, from the condition, (x + S)(y-2)-9 = x(y-l). (1) Also we have, x + 2 y = 30. (2) From (1), -x + 3y = 15. (3) From (2) and (3) , y = 9, the second number. x = 12, the first number. Verifying in (1) : (12 + 3) (9 - 2) - 9 = 12(9 - 1). (15) (7) -9 = 12(8). 105 - 9 = 96. 96 = 96. 2. If the larger of two numbers is divided by the smaller increased by 5, the remainder is 1 and the quotient 3; but if their product decreased by 43 is divided by 1 less than the larger number, the quotient is 1 more than the smaller number. Find the numbers. 188 SIMULTANEOUS LINEAR EQUATIONS 3. If A gives B $ 30, each will have the same amount; but if B gives A $ 30, the quotient obtained by dividing the number of dollars A has by the number of dollars B has will be -J. Find the amount each has. 4. If 1 is added to both the numerator and the denominator of a certain fraction, the result is f ; but if 2 is subtracted from the numerator, and 2 is added to the denominator, the fraction becomes £. Find the fraction. Let x = the numerator of the fraction, y = the denominator of the fraction. Then, By the first condition, By the second condition, Solving (1) and (2), The required fraction is, therefore, -. - = the required fraction. y x + 1 8 y + i 9 (1) x-2 1 «/ + 2 2* (2) x = 7, and y = 8. 5. A certain fraction becomes -§- when 3 is subtracted from its numerator and 4 is added to its denominator. The same fraction is increased by T ^- if -J- is added to its numerator only. Find the fraction. 6. If a certain fraction is divided by 3, and the result is increased by 2, a new fraction, |-£, is obtained ; but if the original fraction is multiplied by 2, and then both numerator and denominator are decreased by 2, the result is f . Find the fraction. 7. A certain number is made up of three digits. The hundreds' digit equals the sum of the units' digit and the tens' digit; the units' digit is 3 more than the tens' digit, and the sum of the three digits is 14. Find the number. PROBLEMS — SIMULTANEOUS LINEAR EQUATIONS 189 Let x = the digit in the hundreds' place, y = the digit in the tens' place, z = the digit in the units' place. Then 100 x + 10 y + z = the number. From the 1st condition, x =y + z. (1) From the 2d condition, z — y = 3. (2) From the 3d condition, x + y + z = 14. (3) Solving the system (1), (2), and (3), x = 7, y=2, z =6. Therefore, the required number is 725. 8. The sum of the two digits of a certain number is 13, and if 45 were added to the number, the digits would be reversed. Find the number. 9. If a certain number is divided by the difference of its two digits, the remainder is 1 and the quotient 18 ; but if the digits are interchanged and the new number is divided by the sum of the digits, the quotient is 3 and the remainder is 7. Find the number. 10. A certain sum of money placed at simple interest amounted to $ 1400 in 3 years, and to $ 1500 in 5 years. What was the sum at interest and what was the rate of interest ? Let x = the number of dollars in the principal, y ss the rate of interest. The interest for 1 year = -^-ths of the principal, = ^- dollars. 100 100 Therefore, ^-^ = the interest for 3 years, 100 — ^ = the interest for 5 years. 100 J Hence, 3 4.^= 1400. (1) 100 x+^=1500. (2) 100 # Vf From (1) , 100 x + 3 xy = 140000. 190 SIMULTANEOUS LINEAR EQUATIONS From (2), 100 x + 5xy = 150000. 500 x + 15 xy = 700000. (Multiplying (1) by 5.) 300 x + 15 xy = 450000. (Multiplying (2) by 3.) 200 x = 250000. x = 1250. Substituting in (1), y = 4. Principal = f 1250 1 R esu it Rate =4% J 11. A sum of money at simple interest amounted to $ 336.96 in 8 months, and to $348.30 in 1 year and 3 months. Find the sum at interest and the rate of interest. 12. A banker loaned $15000, receiving 5% interest on a portion of the amount, and 4 % on the remainder. The income from the 5 % loan was $ 60 a year less than that from the 4 % loan. What was the sum in each of the loans ? 13. Two loans aggregating $ 6000 pay 3 % and 4 % respec- tively. If the first paid 4 % and the second paid 3 °/ the total income from the loans would be $ 12 more each year. What is the amount of each loan ? 14. There are two numbers whose sum is 10, and if their difference is divided by their sum the quotient is $•. Find the numbers. 15. The two digits of a certain number are reversed, and the quotient of the new number divided by the original number is If, The tens' digit of the original number is 4 less than the units' digit. Find the number. 16. If the greater of two numbers is divided by the less, the quotient is 2 and the remainder, 1. If the smaller number is increased by 20 and then divided by the larger number de- creased by 3, the quotient is 2. Find the numbers. 17. Find two numbers such that the' first shall exceed the second by m ; and the quotient jof the greater by the sum of the two shall be s. PROBLEMS — SIMULTANEOUS LINEAR EQUATIONS 191 18. A certain number of two digits is 9 more than four times the sum of the digits. If the digits are reversed, the resulting number exceeds the original number by 18. Find the number. 19. A rectangular field has the same area as another field 6 rods longer and 3 rods less in width, and also has the same area as a third field that is 3 rods shorter and 2 rods wider. Find the length and width of the first field. 20. Two automobilists travel toward each other over a dis- tance of 60 miles. A leaves at 8 a.m., 1 hour before B starts to meet him, and they meet at 11 a.m. If each had started at 8.30, they would have met at 11 also. Find the rate at which each traveled. 21. Three pipes enter a tank, the first and second together being able to fill the tank in 3 hours, the second and third together in 4 hours, and the first and third together in 5 hours. How long would it require for all three running together to fill the tank? 22. The numerator of a certain fraction is the number com- posed by reversing the digits of the denominator. If the denominator is divided by the numerator, the quotient is 2 and the remainder, 5. If the numerator is increased by 18, the value of the fraction becomes 1. Find the fraction. 23. A man seeks to purchase two different grades of sheep, the whole to cost $210. If he buys 12 of the first grade and 13 of the second grade, he lacks $ 3 of the amount necessary to buy them. If he buys 13 of the first grade and 12 of the sec- ond grade, he still lacks $ 2 of the necessary amount. What is the cost of each grade per head ? 24. A tailor bought a quantity of cloth. If he had bought 5 yards more for the same money, the cloth would have cost f 1 less per yard. If he had bought 3 yards less for the same money, the cost would have been $1 more per yard. How many yards did he buy and at what price per yard ? 192 SIMULTANEOUS LINEAR EQUATIONS 25. An automobile travels over a certain distance in 5 hours If it had run 5 miles an hour faster, the run would have been completed in 1 hour less time. How far did it run and at what rate ? 1 1 26. If the length of a certain field were increased by 6 rods, and the breadth by 2 rods, the area would be increased by 102 square rods. If the length were decreased by 7 rods, and the breadth increased by 4 rods, the area would be unchanged. Find the length and breadth of the field. 27. A company of men rent a boat, each paying the same amount toward the rental. If there had been 3 more men, each would have paid $1 less; and if there had been 2 less men, each would have paid $1 more. How many men rented it, and how much did each pay ? 28. A boat runs 12 miles an hour along a river with the cur- rent. It takes three times as long to go a certain distance against the current as it does with it. Find the rate of the current and the rate of the boat in still water. 29. 144 voters attend a meeting and a certain measure is passed. If the number voting for it had been ^ as large, and the number voting against it had been f as large, the vote would have been a tie. How many voted for and how many against the measure ? 30. The total weight of three men is 510 pounds. The first and third together weigh 340 pounds, and the second and third together 350 pounds, and the first and second 330 pounds. Find the weight of each of the three. 31. The sum of the three angles of a triangle is 180°. The angle at A is £ of the angle at B\ and the angle at C is f the sum of the angles at A and B. Find the number of degrees in each angle. DISCUSSION OF A PROBLEM 193 THE DISCUSSION OF A PROBLEM 203. Since a problem is solved by means of equations based upon given conditions, it follows that a true solution can result from possible conditions only. Briefly, (a) Impossible conditions result in no solution ; or, (&) An impossible answer indicates impossible conditions. Some important cases are considered under the following heads : (a) A Negative Result may indicate an Impossible Problem 1. If a dealer doubles the number of horses he owns and also buys 9 additional head, he will then have -J- the number he might have possessed by adding 20 head to 5 times the original number. Find the number he originally possessed. Let x = the original number of horses. 2 x + 9 = the number under the first condition named. x + 20 = the number under the second condition. Then 2x + 9 = 6 -*±™. O 6x + 27 = 5z + 20. 6x- 5z = 20-27. x=-7. And the negative result indicates an impossible problem, due to a fault in the statement of the conditions. If the statement had been "and then sells 9 head" instead of "and also buys 9 additional head " the problem would have been possible. For 2x-9 = 6x + 20 , 3 6x- 27 = 5z + 20. x = 47. Result. (b) A Statement in a Problem may be Reversed 2. A certain man 42 years of age has a son 18 years old. How many years ago was the father twice as old as the son ? SOM. EL. ALG. 13 194 SIMULTANEOUS LINEAR EQUATIONS Let x = the number of years since the father was twice as ! old as the son. Then 18 — x = the son's age x years ago. 42 — x = the father's age x years ago. From the condition, 42 -x = 2(18 -x). 42 - x = 36 - 2 x. 2sc-x = 36-42. x — — 6. Result. The result indicates that 6 years have still to elapse before the condi- tion named will hold. (That is, in 6 years the father will be 48 and the son 24 years of age.) If the problem were changed to read " In how many years will the father be twice as old as the son ? " the solution would be possible. (c) A Fractional Result may indicate an Impossible Problem 3. If the number of boys in a certain schoolroom is de- creased by 5, there will be left 2 more than one third the original number. How many boys were there at first ? Let x = the number of boys at first. From the given condition, X -5 = |+2. Sx- -15 = a + 6, Sx — X = 6 + 15. 2x = 21. X = 10*. Clearly, the fractional result indicates an impossible condition. (d) Possible Discussions of Given Values and their Relations to each Other A general problem admits of discussion by assuming differ- ent relations between given values. DISCUSSION OF A PROBLEM 195 4. Two trains, an express and a mail, pass along the same railroad in the same direction, the express train traveling m miles per hour, and the mail n miles per hour. At 12 o'clock the mail is k miles ahead of the express. In how many hours will the two trains be together ? We may assume that they are together x hours after 12 o'clock, the express having traveled mx miles, and the mail nx miles. But, by the con- ditions, the express has traveled k miles more than the mail at 12 o'clock ; hence, mx — nx = k. k m — n Discussion : 1. Suppose m is greater than n. The value of x will be positive. The express will overtake the mail after 12 o'clock. 2. Suppose m less than n. The value of x will be negative. (For, if n is greater than w, the rate of the mail was the faster rate.) The trains were together before 12 o'clock. This assumption, therefore, is impossible, for we are going contrary to the condition that the trains are to be together after 12 o'clock. 3. Suppose m equal to n. The value of x will become _. If the rate m equals the rate n, the trains have not been together, are moving at the same constant distance apart, and will never be together. In this case, therefore, the supposition has led to the impossible con- k dition denoted by the symbol -. This symbol is usually denoted by oo, and is read " infinity." 4. Suppose m equal to w, and k equal to 0. If k equals 0, the mail and the express started together ; and since m equals n, the trains have been, and will continue to be, together. Expressing this final condition in symbols, we have, x = — - — = - = any finite number. m — n • 196 SIMULTANEOUS LINEAR EQUATIONS That is, there is an infinitely great number of points at which the two trains are together. In this case, therefore, - is the symbol of inde- terminate value. In general, therefore : From (2), A negative result indicates an error in statement. From (3), A result x = — = co indicates no possible solution. From (4), A result x = — indicates an infinitely great number of solu- tions. Exercise 67 Discuss the following problems and interpret the solution for each : 1. A is 12 years old, and B is 17 years old. In how many years will B be twice as old as A ? 2. The total number of boys and girls in a certain school is 32, and four times the number of boys plus twice the num- ber of girls equals 95. How many boys and how many girls are there in the school ? 3. A and B can together paint a sign in 8 hours, and B alone can paint the same sign in 5 hours. How many hours will A require if working alone ? 4. A boy has 45 coins, the value being in all 78 cents. A portion of the number consists of nickels, and the remain- der of cents. How many are there of each kind ? 5. A group of b boys bought a boat, agreeing to pay d dollars each, but / of the boys failed to pay their shares, and each remaining boy had to pay e dollars more than he had agreed. Find the cost of the boat in terms of b, e, and / Ans. ( 6 ~/> 6e . / CHAPTER XVII THE GRAPHICAL REPRESENTATION OF LINEAR EQUATIONS THE GRAPH OF A POINT In the figure two straight lines, XX' and YY', intersect at right angles at the point 0, the origin. These lines, XX' and YY'j are axes of reference. From O measure on OX the distance, OA = a. From measure on OF the dis- tance, OB — b. Through A draw a line parallel to Y, and through B draw a line parallel to OX. These lines intersect at P. And P is the graph of a point plotted by means of the measure- ments on OX and OY. 204. OA and OB, or their equals, a and 6, are the rectangular coordinates of point P. 205. A coordinate parallel to the XX' axis is called an abscissa. A coordinate parallel to the YY f axis is called an or- dinate. 206. Abscissas measured to the right of are positive, to the left of 0, negative. 197 Y B P /> x a X jA Y 198 GRAPHS OF LINEAR EQUATIONS Y P M X X O T N Y Ordinates measured upward from are positive, downward from 0, negative. In the accompanying diagram it will be seen that The abscissa of P is + 3 ; the ordidate, + 4. The abscissa of M is — 4 ; the ordinate, + 3. The abscissa of N is — 4 ; the ordinate, — 5. The abscissa of T is + 2 ; the ordinate, — 3. 207. It will be seen that the axes of reference di- vide the plane of the axes into four parts or quad- rants, and these quadrants are named as indicated, I, II, III and IV. From the principle governing the signs (Art. 206), , we observe that in Quadrant I : abscissa + ; ordi- nate + • Quadrant II : abscissa — ; ordi- nate + . Quadrant III : abscissa — ; ordi- nate — . Quadrant IV : abscissa + ; ordinate — . In naming the coordi- nates of a point, the ab- scissa is always named first, and the ordinate last. In the illustrations of this chapter the scale of the graphs is so chosen that one unit of division on the axes corresponds to one numerical unit in a solution. In later chapters the student will easily apply other scales as occasion requires. Y It X" _ x — -g II IE 7^ THE GRAPH OF A POINT 199 In the diagram let the student name the location of each of the several points, making a table in which the position of each is re- corded. 4= (2, 4), **< ), B= (-5,1), 0=( )i C=(-3,-5), R={ ), #=(7, -7), K={ ), fc=( ), £ = ( )• Y H A K B X X G O L F E C D Y Exercise 68 On properly ruled paper plot the following points : I 1. (2,5). 6. (-3, 4). 11. (4,0). 2. (3,4). 7. (5, -2). 12. (0,4). 3. (7,1). 8. (3,3). 13. (-5,0). 4. (1,7). 9. (-2,-1). 14. (0,-5). 15. (0, 0). 5)? 5. (-2,5). 10. (-4,-7). 16. In what quadrant does (- 3, 5) lie ? (2, -4)? (-3, (4, 7)? 17. In what quadrant does (5.5,. 7.5) lie ? (- 4.8, - 9.75) ? (- 5.4, 8.7) ? (- 14.75, - 6.1) ? 18. Using two divisions on your paper for one given unit of measure, plot the points ; (4,3), (2 f 5) f (-2,3), (3,-2), (0, 4), and (7, 0). 19. Using one division on your paper for two given units of measure, plot the points; (4, 6), (10, 18), (-8, 14), (20,-16), (-18,-18), (0,-24), (10,15), (-17,-21), (-3,25), (25, -3), and (-19, -12). 200 GRAPHS OF LINEAR EQUATIONS THE GRAPH OF A LINEAR EQUATION IN TWO UNKNOWN NUMBERS 208. A variable is a number that, during the same discus- sion, may have an indefi- X A J>r- \B s n 5s Sn S? _ Ss - X"~ m SEL - x 5 — ~S C s V S^ ^ _ B 7^ nitely great values. number of If 2/=0, *=4, P=(4,0). y=l, x=3, P 1= (3, 1). y=2, a>=2, P 2 =(2,2). 2/=3, a=l, P 3 =(l,3). 209. A constant is a number that, during the same discussion, has one and only one value. If, in the given linear equation, jc-f y = 4, we as- sume a succession of differ- ent values for y, the corre- sponding values of x are as follows: 2/=4, 0=0, P 4 =(0,4). y=5, a=-l,P 5 =(-l,5). y=6, x=-2,P 6 =(-2,6). etc., indefinitely. Kegarding the respective pairs of values as the coordinates of points which we may designate by P, P lt P 2 , etc., we plot these points and draw through them the line AB. Therefore, the line AB in the figure is the graph of the linear equation, x + y = 4. An indefinite number of points might have been obtained by assuming fractional values for y, and finding the corresponding values of x. Exercise 69 Plot the graphs of the linear equations : 1. x + 5y = 7. 3. x + 3y = ±. 5. x+y = 0. 2. 3x + y = 2. 4. 2x + 7y = 3. 6. #-4?/ = 0. LINEAR EQUATIONS IN TWO UNKNOWNS 201 A Shorter Method for Obtaining the Graph op a Linear Equation in Two Variables 210. It can be proved that the graph of every linear equa- tion in two variables is a straight line. (This fact justifies the use of the word "linear" in naming so-called equations of the first degree.) Now a straight line is determined by two points, hence the graph of a linear equation should be deter- mined by the location of any two points that lie in its graph. The two points most easily determined are those where the graph cuts the axes. Therefore, (1) Find the point where the graph cuts OX by placing y = 0, and calculating x. (2) Find the point where the graph cuts OFby -placing x — 0, and calculating y. Illustration: Plot the graph of Sx- If t/ = 0, x = 2. Plot Pi (2,0). If x = 0, y = - 3. PlotP 2 (0-3). Join PiP 2 . AB is the required graph of 3cc-2y=6. 2y = 6. Y Q-/k ™~" mmmm b ZZZ—Z Exceptions to the Shorter Method (a) A Linear Equation whose Graph passes through the Origin. 211. If a given linear equation has the form of ax = by, it is evident that when x = 0, y = also. That is, a graph of such an equation passes through the origin. Therefore, to plot the graph of an equation in this form, at least one point not on either axis of reference must be determined. 202 GRAPHS OF LINEAR EQUATIONS (b) A Linear Equation whose Graph is Parallel to Either Axis. 212. If a given linear equation is in form of #=7, it is evident that the value of a? is a constant. That is, the abscissa of every point in the graph is 7. Therefore, the graph of this equation is a straight line parallel to the axis, YY' and 7 units to the right of it. Exercise 70 Plot the graphs of the following : 1. x+2y = 5. 5. a? = 4. 9. Sx — 5y = — 3. 2. y = 3x + 4:. 6. y + 5 = 0. 10. 5x + 3y = 0. 3. x = 2y + 3. 7. 3x-2y = -10. 11. 12 a; -17 2/ =15. 4. x + ±y = 5. 8. y = 7 x. 12. 4?/ = 16 — 18 a. THE GRAPHS OF SIMULTANEOUS LINEAR EQUATIONS IN TWO UN- KNOWN NUMBERS (a) Independent Equations In the figure, the line AB is the graph of the linear equation, x + y = 5. The line CD is the graph of the linear equation, 4 a?— 3y =6. It will be seen that the graphs intersect at the point P, (3, 2). That is, the point, (3, 2), is common to both graphs. Solving the given linear equations, x + y = 5 and 3 x — 4 y = 1, we obtain as a result, x = 3, - s i * ^ y ss 2. And, clearly, the values obtained for a; and 2/ are the coordinates of the intersection of the graphs. 213. The coordinates of the point of intersection of the graphs of two simultaneous linear equations form a solution of the two equations represented by the graphs. SIMULTANEOUS LINEAR EQUATIONS 203 Exercise 71 Solve the following simultaneous linear equations and verify the principle of Art. 213 by plotting their graphs : 1. 5x — 3y = l, Sx + 5y = 21. 2. 5a -3?/ = 36, 7x— 5y = 56. 3. 8a + 3^ = 12, 12 a + 5y = 16. 4. 4x + 6y = -3, 2-10y = 6. 7. 2a> — 3y = l, 16 oj - 24 y = 8. 8. a?+% = 19, lly-3a> = 19. 9. 4#-32/ = 8, 16-8a; = -6y. 10. 3a; — 5*s%, 9z = 15 + 18y. CHAPTER XVIII INVOLUTION AND EVOLUTION INVOLUTION 214. Involution is the operation of raising a given expression to any required power. All cases of involution are multiplica- tions, the factors in each case being equal. In all elementary work the exponents of the powers are positive and integral. THE GENERAL PRINCIPLES FOR INVOLUTION (a) The Power of a Power When m and n are both positive integers : By Art. 61, a m = a x a x a ... to m factors. Therefore, (a m ) n = (a x a x a ... to m factors) (a x a x a ... to m factors) ... to n groups of factors, = a x a x a ... to mn factors, as a mn . The Third Index Law. Hence, to obtain any required integral power of a given integral power : 215. Multiply the exponent of the given power by the exponent of the required power. (b) The Power of a Product When n is a positive integer : (a&) n = ab x ab x ab ••• to n factors, = (a x a x a ••• to n factors) (b x b x b ••• to n factors), = (a«)(6»), = a n b n . The Fourth Index Law. 205 206 INVOLUTION AND EVOLUTION Hence, to obtain any required power of a product : 216. Multiply the factors of the required product, first raising each factor to the required power. (c) The Power of a Fraction When n is a positive integer : (a\ n a /. a „ a - . [ - =7 x T x 7-ton factors, \bj b b o — ( a x a x a '" to n ^ actors ) (b x b X b ••• to n factors) ~b» Hence, to obtain any required power of a fraction : 217. Divide the required power of the numerator by the same I required power of the denominator. These laws are general for unlimited repetitions of the powers or of the factors. Thus : C(o n ) m ] p = a mnp , etc., (abc ••• w) m = a m b m c m ••• n m , etc. THE SIGNS OF POWERS 218. All even powers of any quantity, positive or negative, are positive. Thus, (+a) 2 = + a 2 . (+ a) 4 = + a 4 . (-a) 2 =+a 2 . (- a) 4 = + a 4 . 219. j£B odd powers of any quantity, positive or negative, have the same sign as the given quantity. Thus, (+ a) 8 = + a 3 . (+a) 6 =+a 5 . (-a) 8 =-a 8 . (-a) 5 =~a 5 , (a) The Involution of Monomials Illustration : 1. I -1*LY '..(«*^'.-«^ ■ Result. V 3m»y' (b) The Involution of Binomials The process of raising a binomial to a required power, or, the expansion of a binomial, is best shown by comparative illustra- tions. By actual multiplication : (a + 6) 2 = a 2 + 2 ab + 6 2 . (a + 6)3 = a 3 + 3 a 2 6 + 3 a62 + 6 3. (a + 6)4 = a 4 + 4 a t 6 + 6 a i b 2 + 4 a?> 3 + 6 4. (a + 6) 5 = a 5 + 5 a 4 6 + 10 efib* + 10 a 2 ^ + 5 ab* + & 5 . 208 INVOLUTION AND EVOLUTION * 221. From a study of the form of each product we note : 1. The number of terms in the expansion exceeds by 1 the ex- ponent of the binomial. 2. The exponent of a in the first term is the same as the expo- nent of the power to which the binomial is raised, and it decreases by 1 in each succeeding term. 3. b first appears in the second term with an exponent 1, and this exponent increases by 1 in each succeeding term until it is the same as the exponent of the binomial. 4. TJie coefficient of the first term is 1, and of the second term the same as the exponent of the binomial. 5. The coefficient of each succeeding term is obtained from the term preceding it, by multiplying the coefficient of that term by the exponent of a, and dividing the product by the exponent of b increased by 1. 6. If the second term of the given binomial is positive, the sign of every term of the expansion will be positive; and if the sign of | the second term of the given binomial is negative, the signs of the terms of the expansion will be alternately positive and negative. Illustrations : 1. Expand (a — #) 4 . For the a-factor : a 4 a 3 a 2 a x-f actor: x x 2 X s x* coefficients: 4 6 4 signs : - + - + Combining, a 4 — 4 a 8 x + 6 a 2 x 2 — 4 ax* + x*. Result. 2. Expand (3 a -|Y. By observing the formation of the expansion of (a — &) 6 , we write : (a - h)'° = a 6 - 5 a 4 6 + 10 a 8 6 2 - 10 a 2 6 3 + 5 a& 4 - b & . (3a-^-(3a)5_5(3a) 4 ^+10(3a)3^ 2 _10(3a) 2 (|y EVOLUTION 209 Simplifying, 2 4 16 32 A polynomial is raised to a power by the same process, if the terms are so grouped as to express the polynomial in the form of a binomial. 3. Expand (1-a + ar 2 ) 3 . (1 - X + Z 2 ) 3 =[(1 - x) + *2]3 = (1 - X) 3 + 3(1 - X )*X* + 3(1 _ s)(z2)2 +(>2)8 = (1 _ Sx + 3x 2 - x* + 3x 2 - 6x 8 + 3x* + 3x 4 - 3x6 + 36) = (1- 3x + 6x 2 - 7x 8 + 6x* - 3x 6 + x 6 ). Result. Exercise 73 Expand : 1. (m + w) 3 . 8. (3a-l) 4 . 13. (*-i)r- 2. (m + ny. 9. (ax-2yy. 3. (m-*) 6 - i . (c 2 -^) 5 . 14 - (^ + ^ + l) 2 - 4. (c + 3) 4 . / 3.N3 15. (m 2 -2m + 3) 2 . 5. (a-2) 5 . n ' ( C + 2y* 16- C^ + c + l) 3 . 6. (3a-2y)\* / _2V 17 « (rf-x + lf- 7. (2c 2 -3) 3 . I 37 18. (2a J -aj 2 + 2) 3 . EVOLUTION 222. Evolution is the process of finding a required root of a given expression. 223. The symbol for a required or expressed root is the radical sign, ^/. Thus : Va = the square root of a ; y/a = the cube root of a. ; etc. 224. The number written in the radical sign and indicating the root required is the index of the radical. SOM. EL. ALG. — 14 210 INVOLUTION AND EVOLUTION Thus : 3 is the index of the indicated cube root above. It will be noted that in the case of a square root the index is not usually written, the absence of an index being an " understood " square root. The vinculum is commonly used to inclose the expression affected by a radical. ' 225. From the definition of root (Art. 60) it is clear that y/a means " Required : One of the two equal factors of a." v/& means " Required : One of the three equal factors of &." y/c means " Required : One of the four equal factors of c." {Vx means " Required : One of the n equal factors of x," etc. THE GENERAL PRINCIPLES OF EVOLUTION In the discussion of these principles both m and n are posi- tive and integral numbers. (a) The Root of a Power By Art. 215, (a m ) n = a mn . Therefore, by definition (Art. 60), a m is the nth root of a mn . For a m is one of the n equal factors of (a m ) n . That is, a m = {Var™. The Fifth Index Law. The conclusion is a direct result of a division of the exponent of the given quantity by the index of the required root. Or, mn y/a^n- — a/" = a m . Hence : 226. Any required root of a power is obtained by dividing the exponent of the power by the index of the required root. (5) The Root of a Root By Art. 226, ( m KVa) mn = a. Extracting the nth. root, ("\/a) m = Va. Extracting the with root, ( w -v/a) = Zjtya. GENERAL PRINCIPLES OF EVOLUTION 211 Hence : 227. The mnth root of an expression is equal to the mth root of the nth root of the expression. (c) The Root of a Product By Art. 226, ({Vab) n = ab. Therefore, (Va x {Vb) n = (Va) n x ({Vb) n . Or, (y/aby=(Vax y/iy. Hence, {Vab={Vax Vb. Therefore : 228. Any required root of a product of two or more factors is equal to the product of the like roots of the factors. w The Root of a Fraction By Art. 217, /a\ n _a n \b) b» Therefore, >&» b Hence : 229. Any required root of a fraction is obtained by finding the like roots of its numerator and denominator. THE SIGNS OF ROOTS (a) Positive Even Powers By Art. 218, (+ a)(+ a) = + a 2 and (- a)(- a) = + a 2 Therefore, V+a 2 = + a or — a. Hence : 230. Every positive number has two square roots whose abso- lute value is the same, but whose signs are opposite in kind. The double sign, ± , is used to indicate two roots. Thus, Va 2 = ±a. The sign ± is read, " plus or minus." 212 INVOLUTION AND EVOLUTION (b) Negative Even Powers Since — a 2 = (+ a)(— a), we have a product of unequal factors. Hence : 231. It is impossible to obtain an even root of a negative number. (c) Positive and Negative Odd Powers ByArt.219, (+a)( + a)(+a) = -f a 3 . Also, (_a)(_ a )(-a) = -a 8 , etc. Therefore, V+ a 3 = -f a, V— a 8 = — a, etc. Hence : 232. !Z7ie odd roots of a positive quantity are positive, and the odd roots of a negative quantity are negative; or, briefly, the odd I roots of a quantity bear the same sign as the given quantity. THE EVOLUTION OF MONOMIALS 233. Illustrations : 1. Eequired the cube root of 8 a% 3 c 9 . The root is odd ; the given quantity, positive ; the sign of the result, +. Dividing each exponent by the index of the root, \/S a*b*c* = \/2*cfib*

3a; a 2 12 a' 9 9 ■'■ T 4 T 9 ^16^ 4 Find three terms of the square root of : 19. 1+9 x. 21. ar 2 -}-^. 23. 9^ — 1. 20. a?-S. 22. 4 0^-5 0;. 24. 36 -12a;. THE SQUARE ROOT OF ARITHMETICAL NUMBERS 237. Since an arithmetical square integer is the result of the multiplication of some integer by itself, we are assisted in obtaining arithmetical square roots by noting a certain relation that exists between such numbers and their squares. I 2 = 11 A number of one place has not more than two 9 2 = 81 J places in its square. 10 2 = 100 1 A number of two places has not more than four 99 2 = 9801 J places in its square. 100 2 = 10000 1 A number of three places has not more than six 999 2 = 998001 J places in its square. Conversely, therefore : If an integral square number has two figures, its square root has one figure. If an integral square number has four figures, its square root has two figures. If an integral square number has six figures, its square root has three figures. a + 2 ab + b' 2 \a + b 1 2 302 = a + b 1296 180 + 6 Result. 900 396 396 __ + 2 ab + 6 2 2 (30) = 60 + 2ab + b 2 (60 + 6) (6) = 218 INVOLUTION AND EVOLUTION Hence : 238. Separate any integral square number into groups of two figures each, and the number of groups obtained is the same as the number of figures in its square root. Illustrations : 1. Find the square root of 1296. Beginning at the decimal point, separate into periods of two figures Parallel Algebraic Process each. (*)* = 2(a) = 2 a (2 a + 6) (6) In the square root of 1296 : The greatest square in 1296 is 900. The square root of 900 is 30. The trial divisor is 2 (30) = 60. The second term of the root is (396 -f- 60 =6). For the completion, (60 + 6) (6) = 396. The process is repeated in the same order if the given integer is of a higher order. 2. Eind the square root of 541,696. The following process is given in a form commonly used in practice. Separating into periods of two figures each : , • • • -n, i ,. 541696 /So Explanation : The greatest square contained in the first period (54) is 49. The square root of 49 is 7. 7 is, there- fore, the first figure of the root. Subtracting 49 from 54, and bringing down the two figures of the next period, we have 516, the remainder. Annex- ing a to the first figure of the root, 7, our trial divisor is 2 (70) = 140. Dividing 516 by 140, we obtain 3, the second figure of the root. (140 + 3) 3 = 429, which product is subtracted from 516. With the remainder (87), we bring down the last two figures (96), and the new remainder is 8796. Annexing a to the figures of the root already found, our trial divisor is 2 (730) = 1460. Dividing 8796 by 1460, we obtain 6, 541696 140 + 3 49 1516 3 [429 1460 + 6 8796 6 8796 SQUARE ROOT OF ARITHMETICAL NUMBERS 219 the third figure of the root. (1460 + 6)6=8796, which product, sub- tracted from 8796, gives a remainder of 0, and the square is completed. The addition of the to the figures of the root already obtained gives a trial divisor of the same order as the remainder, or of the next lower order. The process gives fewer figures and less likelihood of error. If a given square number has decimal places, we point off by beginning at the decimal point, first separating the whole number as before, finally separating the decimal into periods from left to right. Ciphers may be annexed, if necessary, to complete any period. If a given number is not a perfect square, its approximate square root can be found to any desired number of places. The square root of a common fraction is best found by chang- ing the fraction to a decimal and extracting the square root of the decimal to the required number of places. Illustrations : 1. Find the square root of 2. Find, to three decimal 19920.4996. places, Vl2f. 20 + 4 4 280 + 1 1 2820 + 1 1 28220 + 4 4 19920.4996| 1 99 96 320 281 3949 2821 141.14 f = .375. 12| = 12.375. Result. The required three decimal places necessitate six decimal figures in the square. Hence, with three ciphers annexed, we have to obtain the square root of 12.375000 13.517+.-. 9 Result. 112896 112896 The decimal point in the result is located easily by noting between which periods of the given example the given decimal point lies. In the example above there are two periods to the right of the decimal point given ; therefore, there will be two decimal places in the root obtained. 60 + 5 5 337 325 700 + 1 1 1250 701 7020 + 7 7 54900 49189 220 INVOLUTION AND EVOLUTION Exercise 75 Find the square root of : 1. 9216. 5. 186624. 9. .717409. 13. .00002209. 2. 67081. 6. 4202500. 10. 9617.7249. 14. .0001752976. 3. 32761. 7. 49.434961. 11. 44994.8944. 15. .009409. 4. 182329. 8. 9486.76. 12. .00119716. 16. .0000879844. Find, to three decimal places, the square root of : 17. 3. 19. 7. 21. .5. 23. f 25. .037. 18. 5. 20. 10. 22. .05. 24. f. 26. .0037. Find, to three decimal places, the value of : 27. 5 + 3 V'2~. 29. 2V3-V5T 31. Ve + VK 28. V7+-VI0. 30. 3V7-7V3. 32. ^VS-Vf. 33. How many rods in the side of a square field whose area is 2,722,500 square feet? 34. Simplify and extract the square root of ^(10^ + 13)-2(2a 4 + 3 + 7^)a; + (ar J + l) (a 4 - a 2 + 1). 35. Simplify and extract the square root of (x 2 - 2x - 3) (x 2 - x - 6) (x 2 + Sx + 2). 36. Show that the required square root in the preceding example can be readily obtained by factoring and inspection. 37. If a, b, and c are the sides of a right triangle, and a lies opposite the right angle, we may prove by geometry that a 2 = b 2 + c 2 . Find the length of a in a right triangle in which b and c are 210 feet and 350 respectively. 38. Applying the principle given in the preceding example, find, to three decimal places, the distance from the " home 1 plate to the second base ; the four base lines of a regulation baseball diamond forming a square 90 feet on each side. CHAPTER XIX THEORY OF EXPONENTS 239. The Index Laws for Positive Integral Values of m and n have been established : (1) For Multiplication: a m x a n = a m + n (61) (2) For Division (m greater than n) : a m ■* a n = a m ~ n (79) (3) For a Power of a Power : (a m ) n = a mn (215) (4) For a Power of a Product : (ab) n = a n b n (216) (5) For a Root of a Power : Va"*™ = am ( 2 ~ 6 ) 240. The extension of the practice of algebra requires that these laws be also extended to include values of m and n other than positive integral values only; and the purpose of this chapter is to so extend those laws. We shall, therefore : 1. Assume that the first index law (a m x a n = a m+n ) is true for all values of m and n. 2. Define the meaning of the new forms that result under this assumption, m or n, or both, being negative or fractional. 3. Show that the laws already established still hold true with our new and broader values for m and n. THE ZERO EXPONENT If m and n may have any values, let n = 0. Then, a m x a° = a TO +° (61) = a m d m Dividing by a m , a = — That is, a° = 1. Hence, we define a as equal to 1. Or : 221 222 THEORY OF EXPONENTS 241. Any quantity with the exponent equals 1. Illustrations : 1. s° = 1. 2. (wm)° = 1. 3. 3 m° = 3-1 4. 4aP + (2y)° = 4.1 + l=5. Oral Drill Simplify orally : 1. a°x. 4. 4a°6°c. 7. 5 a? -5. 10. 3a + 3 + a°6 . 2. 2aa°. 5. 3a° + l. 8. 2a°+3a°. 11. 2a° + (a°-l). 3. 3aV- 6- 4a°-3. 9. 4a; - 52/°. 12. 4?x + (a + 3 a?) . THE NEGATIVE EXPONENT If m and w may have any values, let n be less than and equal to — m. (61) (241) Then, a m x a -7 " = a mr ~ m lence, a m x a- m = 1. Dividing by a~ m , 1 Dividing by a m , 1 a _m = — • a™ Hence, we define ar™ as 1 divided by a m . From this definition we obtain an important principle of con- stant use in practice : 242. Any factor of the numerator of a fraction may be trans- ferred to the denominator, or any factor of the denominator may be transferred to the numerator, if the sign of the exponent of the transferred factor is changed. Illustrations : 1 <,&-*-« 2 ar^s-i-JL. <* 2- l mx-* _3mn FRACTIONAL FORM OF THE EXPONENT 228 Oral Drill Transfer to denominators all factors having negative exponents : 1. ax~ 2 . 4. 2a 2 ~b~ 2 . 7. crtc" 3 . 10. 2 ar^ar 1 . 2. crf-V" 3 . 5. 2- 1 ar 2 b- s . 8. x~ l y-h-\ 11. 5 ary 3 * 4 - 3. 2ab~ 2 . 6. Stf-Vr 1 - 9 - S-'a-^c- 1 . 12. iHar^Mr^. Read the following without denominators : a 2 cd Sa c 2 1 13. -• 15. 17. ~- 19. -^r- . 21. b 2 mn a?y 2 ' ar 2 b~ 2 2~ 2 x~ 2 f xyz ax m 5 x AA - 2 a ab 14. - 2 -. 16. — • 18. — • 20. -rt. 22. 53 2 m z J a l m 6 n t —3 "era Read the following with all exponents positive : 23. 5a -1 &c. 27. = ; , - 31. 2ar>" 1 2- 1 m 2 w- 1 2 24. 12 a- 1 ^. 28. 5 5-5-5* 32. r — rr— j. 25. -5«*V. 29. — 53. 33. 3^. 26. 3a-Vd 30. |S^C 3 4. 3-V^z - 3 m- 1 ^-^ y- 1 ^ 2 THE FRACTIONAL FORM OF THE EXPONENT ■ If the expression a 2 can be shown to conform' to the first index law, we may find a definition for the fractional form of exponents. By the first index law, (a*) 2 = at x J = a? + % = a. Hence the meaning of a% is established, and the exponent in this form still agrees with the fundamental index law. That is: 224 THEORY OF EXPONENTS a? = Va is one of the two equal factors of a. Similarly, (a^) 3 = eft x eft X eft = cft + * + $ = eft = a 2 . 2 That is, a 3 is one of the three equal factors of a 2 . In like manner, a? = Va = one of the three equal factors of a. 3 4 . a* = va 3 = three of the four equal factors of a. Therefore : 243. In the fractional form of an exponent we may define the denominator as indicating a required root, and the numerator as indicating a required power. Illustration : Vs 2 = 8» = 2 2 = 4. And 4 contains two of the three equal factors of 8. \/8P = 81^=3 8 =27. And 27 contains three of the four equal factors of 81. In general : m m m m m m a n • a n • a n ••• to n factors — a n n n ••• to n terms mn = a m . m That is, a n is an expression whose nth power is a m . m And a n is the nth root of a m - m Or, as above, a n is one of the equal factors of a m . % It is understood that while a* must equal either + Va or — Va, we i r - consider the positive value only ; and a 2 is denned as -f V a, the principal square root of a. In future operations we may apply the definition of Art. 243 to expressions given in radical forms, observing that 244. The index of a radical may be made the denominator of an exponent in the fractional form, the given exponent of the power of the quantity becoming the numerator of the fractional form. FRACTIONAL FORM OF THE EXPONENT 225 Illustrations : 1. Vrn* = wA 2. Vtf = x\ 3. y/b^ = b^. And in the converse operation of the principle of Art. 244 : 4. **=v^. 5. a$=Va?. 6. c^=«v^ In the reduction of numerical forms : 7. v / l66 = (\/l6)5=(2)5=32. 8. v^8* = (\/=8)6 = (-2*) = -32. It is to be noted that the root is extracted first. Oral Drill Express with radical signs : 1. eft. 4. cfix?. 7. y*z\ 10. x$y%z$. 2. x\ 5. m^/*. 8. cW. 11. rftp^q*. 3. c*. 6. «&*. 9. e^mi 12. ir&rfix?. Express with exponents in fractional form : 13. Va?. 16. Vz\ 19. y/a • Vc 3 . 22. V4~a* • V^" 8 . 14. Vm 5 . 17. -v^S*. 20. V^-^c 3 . 23. ^^-V367 4 . 15. Va 7 . 18. <^. 21. ^/m^.-^/n 3 . 24. VlScSJ* • \/8?. Give the numerical value of: 25. 4* 28. 64*. 31. 4*. 34. 81*. 37. 32^. 26. 9i 29. 4*. 32. 8*. 35. 25*. 38. 64*. 27. 27*. 30. 8*. 33. 27*. 36. 49*. 39. 128* Having established a meaning for the new forms of ex- m ponents, a , a -1 , and an, we must show that the index laws hold true for these new forms; thus fulfilling the third and final clause of our agreement in Art. 239. SOM. EL. ALG. 15 226 THEORY OF EXPONENTS Proof of the Index Laws for Negative, Fractional, and Negative and Fractional Values of m and n 245. In the following proofs, m and n are rational integers or rational fractions. The Law a m xa n = a m+n . 1. When m and n are negative and integral. a~ m x a~ n — — x — = = a-*-*. a m 2. When w is jpo^iYwe and fractional. Let w =^, p and g being positive and integral. Q P mp m p (a m ) n = (a wl ) < * = V(a m )p = Va m » = a « — a '«, APPLICATIONS OF PKINCIPLES OF EXPONENTS 227 3. When n is negative and fractional. Let n = — ■?, p and q being positive and integral. 0. = — =a « =a v « y (a m ) n =(a m ) * = (a m ) * « The Law (a6) w = a n b n . 1. When n is negative and it Let n — — p, p being positive and integral. (ab) m = (db)-p = — — = — ?— = a-vb-P. (aby aPbP 2. When n is positive and fractional. Let n = -, p and &p = i'a*' • VbP = a* 6« 3. When w is negative and fractional. Let n = — -, p and q being positive and integral. -E i i _£ _£. («&)"»=(«&) « = ; = "T7 = a ?6 q (ab)i aib* Let the student discuss this law (1) when m and n are both positive and fractional, and (2) when m and n are both negative and fractional. APPLICATIONS OF THE PRINCIPLES OF EXPONENTS (a) Simple Forms involving Integral Exponents 246. In processes with exponents no particular order of method can be said to apply generally. Experience with dif- ferent types will familiarize the student with those steps that ordinarily produce the clearest and best solutions. As a rule, results are considered in their simplest form when written with positive exponents. 228 THEORY OF EXPONENTS 247. Illustrations : 1. Simplify x~ A • x 5 • x~ 7 . x-* • sc 5 - x- 7 = z-4+6-7 = a;-«= — . Result. x 6 2. Simplify m 3 n~ 2 • m~ 2 n~ 5 m 8 n- 2 • m _2 n -6 = wi 3_ 3. Simplify ^^ - ^Vl ar 1 (ma; -1 ) ~ J m 8 n- 2 • m- 2 n- 5 = wi 3 -%- 2 " 5 = mn~ 7 = S . Result. ML 7 ^m-gft 4 (a" 2 ) 3 _ a 2 m~ 3 y>cg- 6 X" 1 (jnix- 1 )- 2 x-im-W = a 2-6 m~ 5+2^4+1-2 = a^w-ix 3 - -_^!_. Result. a 4 m Simplify : Exercise 76 1. a 5 x a -3 . 5. m 2 • m~ 3 • m 5 . 9. x- 3 + x- 2 . 2. c~ 3 x c 7 . 6. x* • ic° • or 4 . 10. m~ 4 -r- m~ 7 . 3. ra" 2 X m 3 . 7. a-V- 1 ^- 8 ^ 11. a~ s m~ 2 -r- a 2 m~ 3 . 4. a; -4 • x 3 . 8. c^d-cM" 4 . 12. ar 1 ^" 5 + x~ 5 y~ l . 13. C" 2 dV 3 'd-'z 5 x®y~ l m 3 n~ 9 m 5 n~ 7 x°y- 8 14. a a ~ 2 b- ~ 2 b " 2 ^°. 16. a?m~hr 3 x~ 5 a~ 5 m~ 3 n 5 x 7 17. (a" 3 ) 2 . 22. (»-^l|)" 1 . 27. (a° + ay)- 3 . 18. (c- 3 )- 2 . 19. (a)" 5 . 23. (a-V 2 ) 2 . 24. (a- 1 ^- 2 ) 3 . 28. (mV)°x(3a; ) 2 . 20. (a" 1 ) 5 . 21. (a- l b)\ 25. (a 2 a°a- 3 )- 2 . 26. (c- 1 ^ • c ^- 1 )- , >°-m'- APPLICATIONS OF PRINCIPLES OF EXPONENTS 229 (6) Types involving the Fractional Form 248. In the following illustrations attention is called to each important feature of the process, and the order of the principles that is emphasized in each is such as will, under similar conditions, produce the best form of solution. Illustrations : 1. Simplify (a^V 1 )" 2 . (asrtr 1 )" 2 = 25Vm/ / 9 v^F2 \-f __ p^ -f _ / 9 \"i = / 25 mM = 25? ro* _ 125 m^ V25Vm/ V 26m i/ \ 25m V V 9 / " 9* 27 Result. Note that in the third step inverting the fraction changes the sign of the exponent of the fraction. In general, ( -\ ~* = 225 = £ = (k\ x . 230 THEORY OF EXPONENTS V m f Vn~i _ j m Vn" 4 _ [" w^n""* _ / mn~ 2 T = [mJrT* -T- Vm%- 8 ] 6 b= (wi*n~^ -h m« _i ) 6 = (m^VM)* = (m^) 6 = m 2 . Result. Exercise 77 Simplify : 1. (8a-2a" 1 )i 6. (^Txyt+i-l/Slx 7 )- 1 . if 3/ =- 7. JaY(*X\Z 2. \(aV- 3 Vaary. Af * V2T7 2/ 3. [m^m-^m-?) 2 ] 6 . g ^ c~ V § J m* y/tfc _ r a~Wm m*a 16 Vc 4. [V a -i c i(c-W)^]" 8 . r 3/ \ 1-2 9. \ r- X = 5. L^ ax -Vax-U ' c*d- 2 cy/d 10 A c~Wx 4/ I6 n Jc^n* n As a ~s/x 2 + 27 xi Va-(r + (r} (rf) Types involving Literal Exponents 250. Illustration : m m _i 2m— 1 n* Simplify [a» -a" -^a^ 7 "] . p m m 2m— -i«x r- m+n m— n 2m l-.nx \_an + • a»~ ■+■ a «~ J = [_a « • a~^*~-r- a - ** - _ Cm+n m— n_2m— 1-ji a~n~ « «~J a*. Result. 232 THEORY OF EXPONENTS Exercise 79 Simplify : 1. a 2m+n • a m+n -!-a? m+n . 7 f *+i . 1 \^ 2. (m*) x+1 + m x+2 . K ^ J 3. («*)•*■ + (<**•). 8. ^_ 2n+2 x8r 4. (m* - ")' (m y ~*y -*- m _(ae+2)y . 16" 5. (03& 2C6)ax. 9 - [^'(^t^)^} a x+2y a y+z a x+3a 4~ T~ (e) Miscellaneous Processes involving the Principles op Exponents 251. Illustrations : 1. Multiply a - 3 J - 2 a$ by 2 - a* + 3 a~i g — 3 gf — 2 g^ 2- - a "i + 3 g~$ 2a- - 6 g$ - 4 g$ - gt + 3gi + 2 + 3 a \ _ 9 _ -6a -1 2 a - 7 of + 2 rf - 7 - 6 g~£ Result. 2. Divide 9 a* - 3 a* + 1 + 7 * = a 8 - 8 a 4 + 24 - 32 a~* + 16 Result. 7. If ar 3 = 2/ -2 , and 2/ -1 = —8, what is the value of x ? We first require the value of y~ 2 from the second equation. Hence, if y~ 1 = -S, Tnen > x" 3 = (-8) 2 . (y-i) 2 zz(-8) 2 , * = [(-8) 2 r* (-8) 2 . = -8-1 = J. Result. -2m 234 THEORY OF EXPONENTS Exercise 80 Multiply : 1. a;" 2 + 3 a;" 1 -2 by a;- 2 -2 a;- 1 -4. 2 . a* + 2a±&* + 6*by a*-2a*&* + &i 3. 4ar 3m + 6ar 2m — 5ar OT -3 by 3 x~ 2m + 2 ar TO - 1. 4. a*-9a* + 27a~*-27a"Hy a^-6 + 9a"i 5. va^ ^ + 2v^-2V^by -^ + Va + -^. W Va 3 Va Divide : 6 . c -4_ c -3_ 8c -2 + llc -i_ 31:?v c -2 + 2c -i_3. 7. 2a-a* + 4a* + 4a*-3by a*-a* + 3. 8. 35+4a- w -16a- 2m +19a- 3w -6a- 4m by 7+5or w -3cr 9. a 2 + 2a;*-7ar*-8af£ + 12ar 3 by a? - 3 oT* + 2 af*. 10 . a-M? + ^- 8 K + H 7 V-a-^+±. ■y/c Vc Vc 3 c Vc Vc Simplify : 11. [( x + 3)(x-3)- 1 -(x-3)(x+3)- 1 ] + [l-(x i +9)(x+3)- 2 ]. 1 2 . [(o^-SXc-^- ^^^^] x [Sc^-O)-}- 1 Expand : / 2V 17, I — = F/ 14. (V^+2^)\ 16 " ( 3ViF -aJ' W ° 2V ^ Extract the square root of : 18. 4 ar 4 + 12 ar 3 + ar 2 - 12 a;- 1 + 4. 19. 9a- 2 " l -6a- w -ll + 4a m + 4a 2w . APPLICATIONS OF PRINCIPLES OF EXPONENTS 235 20 . x~% — 4 x ^ + 8 x %y* — 8 x~*yl + 4 y^. * -y/x 3 \y \x Find the value of x in : 22. a£==3. 25. y~* = 4. 28. #"*=£. 23. a;* = 2. 26. 2"* = 9. 29. x» =2. 24. a>* = 25. '27. af* = — 8. 30. x m = 3 n . Multiply by inspection : Divide by inspection : 31. (3 a" 3 - 2 a 3 ) 2 . 35. (a" 3 - 9) by (a* + 3). 32. (2 or 1 + 3) (3 ar 1 + 7). 36. (a - 81) by "(a* + 3). 33. (3a- 1 -2a)(5a" 1 -3a). 37. (a- 3 -86" 8 )by (a- 1 -26~ 1 ). 34. (2a- 2 + a~ 1 H-3) 2 . 38. (27a"*+125)by(3a~ J +5). Find the value of a; in each of the following : 39. x' 1 = y, and y 2 = 9. 42. a£ = y _1 , and y = 3. 40. sb = 2/ -1 > and y* = 3. 43. a£ = 2/^, and y» = 9. 41. x~ 3 = ?/, and y~ 2 = 2. 44. aT* = IT 3 , and y~* = 9. Simplify : 45. (a" 1 - or 1 ) 2 - (a" 1 + ar 1 ) (a" 1 - x- 1 ). 46. (a- ffl + 3a m ) 2 -(a- m -3a w ) 2 . 47. (a- 3 + l)a- 3 -(a- 3 -l) 2 + (l-a- 3 ). V d»-c*A c^ + dU J___L 49 f ^ + 13 ajy* a^ + 4yH . |" 5^ + 2^ "| L^4-2a-3^ — 3^ a* + 3^J L*t— 9yt.J CHAPTER XX RADICALS. IMAGINARY NUMBERS. REVIEW 252. A radical expression is an indicated root of a number j or expression. Thus : V2, fyl, VlO, and Vx + 1 are radical expressions. 253. Any expression in the form -\/# is a radical expression, or radical. The number indicating the required root is the index of the radical, and the quantity under the radical is the radicand. In \/7, the index is 3, and the radicand, 7. 254. A surd is an indicated root that cannot be exactly ob- tained. 255. A radical is rational if its root can be exactly obtained, irrational if its root cannot be exactly obtained. Thus : V25 is a rational expression ; VlO is an irrational expression. 256. A mixed surd is an indicated product of a rational factor and a surd factor. Thus : 3 V6, 4V7x, abVa + b are mixed surds. 257. In a mixed surd the rational factor is the coefficient of the surd. Thus : In 4VEx, 4 is the coefficient of the surd. 258. A surd having no rational factor greater than 1 is an entire surd. Thus : Vbac is an entire surd. 236 TRANSFORMATION OF RADICALS 237 259. The order of a surd is denoted by the index of the required root. Thus : . a/5 is a surd of the second order, or a quadratic surd. \/7 is a surd of the third order, or a cubic surd. 260. The principal root. Since (+ a) 2 = -f- a 2 and (— a) 2 = -f a 2 , we have V+a 2 = ± a. That is, any positive perfect square has two roots, one + and the other — , but in elementary algebra only the -f value, or principal root, is considered in even roots. THE TRANSFORMATION OF RADICALS TO REDUCE A RADICAL TO ITS SIMPLEST FORM 261. A surd is considered to be in its simplest form when the radicand is an integral expression having no factor whose ; power is the same as the given index. There are three com- mon cases of reduction of surds. (a) WJien a given radicand is a power whose exponent has a factor in common with the given index. By Art. 244, tfp = a* = a% - Va. Hence, to reduce a radical to a radical of simpler index : 262. Divide the exponents of the factors of the radicand by the index of the radical, and write the result with the radical sign. Illustration : V%ahfi = y/WahP = 2 y a M a Wcfix- -Result. Exercise 81 Simplify : 1. -v/aV. 5. VW&. 9. a/243. 2. -Vote*. 6. VWm*. 10. J/WaW. 3. vW. 7. ^36. 11. \/ffiS. 4. tyrftf*. 8. -y/lSS. 12. ^216 aV. 238 RADICALS. IMAGINARY NUMBERS. REVIEW (b) When a given radicand has a factor that is a perfect power whose exponent is of the same degree as the index. By Art. 244, Vcfib = (a 2 &)* = aM = a$ = aVb. Hence, to remove from a radicand a factor of trie same power as the given index : 263. Separate the radicand into two factors, one factor the product of powers whose highest exponents are multiples of the given index. Extract the required root of the first factor and write the result as the coefficient of the indicated root of the second factor. Illustrations : 1. vW = V4 a 2 x 3 a = 2 aVSa. Result. 2. 2V72 aWy 1 - 2 V36 a 2 x*y« -2 ay = 2(6 ax 2 y s ) y/2 ay = 12 ax 2 y* y/2ay. Result. Exercise 82 Simplify : 1. V28". 8. VI§2. 14. |V*p; 2. vs: 9 2 * /mu «• 1^375- 3 - ^* io. 3VI62: 16 - *W 4. V98T 17. V9 a 3 . 11. 3^80. 5. -4 W! 10. M ,4 11. 4—- *a§ 12. J*. -4 13. mM >-4 14 c 4 /^\ SC Ac 3 -4 K 2c s/T 15. — a • 3 V'2c •■# - 6 *VS- 17. m 3/54 m 4 18. J-1% *# + 1 19. (l_a)^f-— 20 . bi^SS a + 1 *# — 1 21 1 3/ (m-iy(m + 2) 6 m + 2\ (m + 1) 2 22 m * p ^-12a; + 38 ^ — ft \ a? — o * fir 23. (a-2) x f^± a — 1 w J^+2^+1 TO CHANGE A MIXED SURD TO AN ENTIRE SURD The process is the reverse of that of Article 261, Section (a). In general : ay/x = a • x% = a% 2 = Va 2 #. Hence, to change a mixed surd to. an entire surd : 265. liaise the coefficient of the surd to the same power as the degree of the radical, and multiply the radicand by the result. TJie indicated root of the product is the required entire surd. Illustrations : 1. 3 V5 = VPT5 = \/9^5 = V45. Result. 2. 2 VI = \f¥Ti = %%7i = ^32- Result. TRANSFORMATION OF RADICALS 241 Exercise 84 Change the following to entire surds : 1. 2V3. 2. 5V2. 3. 3^/4. u 3™ /~8^ »X 2 4.2^2. 2a;V27TO 5. 3^4. 12. i^Vg. a *ar 6. 2xV2x. , 7. fcrfffii 13 ' ( a + 1 )\(^i)^- 8. So 2 ^*;. 14. g±Sj ^- 1 » — 1 ^ar J + 4a;H 9. 3a-\/3a; 2 . ' a-1 \ar 2 + 4a; + 4* TO CHAISE RADICALS OF DIFFERENT INDICES TO EQUIVALENT RADICALS HAVING THE SAME INDEX We have Va = a? and -\/a = a 5 . Expressing the exponents as equivalent fractions having a common denominator, a%=a? = -\/a 3 , • a* = a* = -v'a*. Hence, to change radicals of different indices to equivalent radicals having the same index : 266. Express the given radicals with exponents in fractional form and change these fractions to equivalent fractions having a common denominator. Rewrite the results in radical form. Illustrations : 1. Change V5 and VlO to radicals having the same index. V5 = 5^ = 5* = #5* = #125. #10 = 10^ = 10^ = #10 2 = #100. SOM. EL. ALG. 16 Result. 242 RADICALS. IMAGINARY NUMBERS. REVIEW 2. Arrange ^11, V5, and V90 in order of magnitude. vTl = 113 = ni = ^n-2 = ^121. V5 = 5* = 5^ = \/p = vT25. ^90 = 90^ = #90. The order of magnitude is, therefore : V5, vTI, #90. Result. Exercise 85 Change to equivalent radicals of the same index : 1. V2, VS. 4. -y/n, -v/6. 7. V2, #3, ^5. 2. V3, -Vl. 5. ^6, ^/8. 8. V5 ^^5. 9. y/l,/6)(2 Vl5) = (3 • 2)( V6 . Vl5) = 6V90 = 18\/l0. Result. 2. Multiply V28 X V42 X Vl5. Expressing each radical in prime factors, V28)(V42)(Vl5) = V(7.22)(7-2.3)(5.3) = V(7 2 • 22 . 3 2 ) (5 . 2) = (7-2.3) VTO = 42 VlO. Result. 3. Multiply 2-v/4 by 3 + l). DIVISION OF RADICALS 272. As in multiplication of radicals we may divide any two radical expressions of the same index. Illustrations : 1. Divide -^96 by y/2. M = ^.= m = 2Vz. Result. 2. Divide y/l by V6. n = w = yii = ^i^ Result V6 W >216 *27 3 246 RADICALS. IMAGINARY NUMBERS. REVIEW 3. Divide yfiff by >^f. Exercise 89 Divide : 1. V8by Vl2. 5. 6V5by2Vl5. . 9. ^36 by -y/U. 2. V6by V18. 6. 4-^4 by 3^12. 10. Result> . 2V3 2V3 V3 2.3 6 3 2. Divide 2 -tys by -y/L 2jft = 2V8xtt = 2( l Vy^ ) = 2WW) = 2 i^ Regult The process of rationalization of the divisor simplifies numerical calculations with radical expressions. 3. Find, to the three decimal places, the value of " Y, ' Vl2 3V6 3V6xV3 3V18 9V2 3 _ S„„ A N mmm £ —=z = —= 7= = — =■ = -7— = - v5 = o (1.414+-..) = 2.121+.... Vl2 Vl2 x V3 V36 6 2 2 ^ > OPERATIONS WITH RADICALS 247 Exercise 90 Eationalize the denomi- Find, to three decimal places, nators of : the value of : 1. — -• . 4. -• 7. 10. V2 . SV2 V5 V75 & 5. A. 8 . ju ii. b* V6 V2 3V2 3V2 -*-. 6 . _L. 9. ?V5. 12 A. 2V3 a/9 3V3 ^i (5) When Either or Both of the Terms of a Binomial Divisor are Quadratic Surds. By Art. 104, (« + 6) (a - b) = a 2 - 6*. Similarly, ( Va + V&)( Va - Vb) = a - b. 274. Two binomial quadratic surds differing only in sign are called conjugate surds. ( Va + Vb) and ( Va — Vb) are conjugate surds. From the multiplication above we may conclude : 275. The product of two conjugate surds is rational. Illustrations : 2V3-3V2 1. Rationalize the denominator of 2 V3 - V2 2V3-3v^ _ 2V3 -3\/2 x 2\/3+ V2 2V3-V2 2V3-V2 2V3 + V2 _ 6-4V6 _, 2(3-2V6) ^ 3-.2V6 Result 10 10 5 Two successive multiplications will rationalize a trinomial denominator in which two quadratic surds are involved. 248 RADICALS. IMAGINARY NUMBERS. REVIEW 2. Eationalize the denominator of — ^t— • V3+ V2-1 V3_ V2+ 1 = V3-(V2-1) V3-(V2-1) V3+V2-1 V3+(V2-1) V3-(V2-1) = 6-2V6 + 2a/3-2V2* 2V2 3 - V6 4- V3 - V2 V2 3_V6+V3-V2 x V2_ 3\^-2V3+ V6 - 2 V2 V2 2 Exercise 91 Eationalize the denominators of: 4 V2+ V3 7> 2V2-4 2 + V2 V2-V3 3V2+2 3 5 5-V2 g 4V3-3V2 V2-2* ' 2 + V2* ' 2V3 + 3V 2 ' 5_ 3V2 + 1 9 5V6-2V3 V7-V2 V2-1 3V2 + 3 10 V^+V^ - 15. Vs — 2 — Va^ Va— V# V# — 2 + Va; „ 2Vm+Vn . -„ Va + 1 + V2 a — 1 11« = r* ■*■*>. — rzz= z' 3 Vm — Vw Va + 1 — V2 a — 1 2Va + V2a , 17. V3 + V2-1 , 3Va-V2^ V3-V2 + 1 13 qVft + cVa? 3- V2 + V3 aVaj — cVa 3 + V2 — V3 V^fl+2 4 - V 3 + V5 14. — ■ ly- — -=• Va + 1-2 4_VB-V5 12. INVOLUTION AND EVOLUTION OF RADICALS 249 INVOLUTION AND EVOLUTION OF RADICALS 276. By the aid of the principles governing exponents, we may obtain any power or any root of a radical expression. Illustrations : 1. Find the value of (2 Va 5 ) 2 . (2^)2= (2 afy = W = 4 a? = 4 Vol Result. 2. Find the cube root of VaV. VVaW sk VaW = a?x* = Vatf = xVax. Result. Exercise 92 3/ — — \ Find the value of : 1. (Va) 6 . • 5. y/^/S. 9. V^a + fc) 2 *. 2. (-^y) 3 . 6. + 3 = Vaj" + 15. X + 3 = Vx 2 + 15. Squaring, x 2 + 6 x + 9 = x 2 + 15. Transposing, x 2 — x 2 + 6 x = — 9 + 15. Uniting, 6 x = 6. Whence, , x = l. Substituting 1 in the original equation, and taking -the positive value of the square root, we have, 1 + 3 = vT+To. 4=4. Therefore, the solution, x = 1, is a correct solution. 2. Solve VaJ — 2 + Vx + 5 = 7. y/x-2 + Vx+5 ss 7. Transposing, Vx — 2 = 7 - Vx + 5. Squaring, x-2 =49-14V» + 5 + X + 5. Whence, 14 Vx + 5 = 56. Dividing by 14, Vx + 5 = 4. Squaring, x + 5 = 16. EQUATIONS INVOLVING IRRATIONAL EXPRESSIONS 253 And, a; = 11. In the original equation, Vll — 2 + Vl 1 + 5 = 7. V9+ Vl6 = 7. 7 = 7. Hence, x = 11, is a solution of the given equation. 3. Solve VaT+3 — 5 = Vx — 2. Vx + 3 _ 5 = Vx - 2. Squaring, x + 3 - 10 Vx + 3 + 25 = x - 2. Transposing, x — x — 10 Vx + 3 = - 3 — 25 — 2. Collecting, - 10 Vx + 3 = - 30. Dividing by - 10, Vx + 3 = 3 . Squaring, x + 3 = 9. Whence, x = 6. Substituting in the original equation, V6 + 3 - 5 = V6 - 2. V9 - 5 = VI. 3-5=2. Since the original equation is not satisfied by the solution x = 6, this solution must be rejected. It will be found by trial that the solution, x =6, satisfies the equation 5 - Vx + 3 = Vx^2 . Exercise 94 Solve and test the solutions of : 1. Va;+1 = 2. 8. V7+a;-7 + VaJ = 0. 2. v^i=3. 9 - Vt+s-vs+4 3. V ^T = ,-1. 10 - 2(V^5)(V^-5)=-35. 11. Vx — 1— Va = 2. 4. 2VS-3-V5+2. 12 V -_ [=V5 _ TI _ V ^ 5. ^a?-l = V6. 13 3V^+V^ = 2= V4»-l. 6. 2VaJ+3=VaJ + 2. 14. V# + a + Va 4- 2 a = V4 a; — a. 7. V« 2 + 3 = a;-l. 15. yz-5-V4a;-2+- Va + 3 = 0. 16. V4 a + c = V25 x — 3 c — V9 d?-±^-a 2 3 10. cV-l-f- V-16c 2 + V-c 2 c 11. (2 + 5V^) + (7-2V^)-(8-2V^T). 12. (x + y^/^) + (x-z^~^) + (y-x->J^l). MULTIPLICATION OF IMAGINARY NUMBERS 295. The positive integral powers of the imaginary unit, By Art. 292, (V^T) 2 =- 1. Therefore, (V^H) 3 = (\/^T) 2 (\/^T) =(- 1) (V^I) = - \T^\. ( V31 ) 4 =( V31 ) 2 ( V31)2=(_1)(-1) = +1^ (V^T)^(v r ^l)HV^l) = (+i)(V^I)-v r -i. That is, the first power of V— 1 = V— 1. the second power of V— 1 = — 1. the third power of V— 1 = — V— 1. the fourth power of V— 1 = 1. And this succession repeats itself in order for the following higher powers. OPERATIONS WITH IMAGINARY NUMBERS 257 In the multiplication of imaginary numbers it is helpful to remember that 296. The product of two minus signs under a radical is a minus sign outside the radical. Illustration : 1. Multiply V^3 by V^6. VZ3 by V^6 = V(-3)(-6)= V(18)(-l) 2 = - Vl8= -3 V2. Result. Exercise 96 Multiply : 1. V^ by V" 2. V^G by V- 3. V- 12 by V- 3. 4. V^36by -V^16. 5. _V-18by -V-54. 6. -yf^-2 by V^^T by V- 7. -yf^l by V^9 by V- = »)( 28. 16. (• V- 25) (V-ioo). 9. (V3^)(_V-4c 2 )(- V-16 6 2 ). 10. (-V^^C-aV^H-aV^^. 11. (-V^l) (2V^1) (-3V=i). 12. (-2V" = l8)(-3V" :r 8)(5V^ : 32). 13. (V^aV) (- -V^^x) ( - V - c 2 x). 14. (a V" 11 ^) (a" 1 -/-!) ( - a-V^l). 15. (— Vm — ») ( vn - to). 16. (V-^-2x-l) (V^I) (VT=^). MULTIPLICATION OF COMPLEX NUMBERS 297. We assume that the principles underlying the multipli- cation of real numbers still hold true in the multiplication of complex numbers. SOM. EL. ALG. 17 258 RADICALS. IMAGINARY NUMBERS. REVIEW Illustrations : 1. Multiply 3 + V^2 by 2 - V^2. 2 - Bv^-2 6 + 4V^2 - 15V^2-10(~2) 6-llV^2 + 20 = 26- 11V^2. Result. 2. Expand (3 - V^) 4 . In processes involving frequent repetitions of the imaginary unit, V— 1, it is convenient to substitute t in place of V— 1, replacing the imaginary unit after the simplification is com plete. Let (3 - v/^T)* = (3 - iy. Then : (3 - iy = 81 - 108 i + 54 & - 12 + & = 81 - 108V^T + 54(-l) - 12C-V~T) + (1) (Art. 292) = 81 - 108v^I - 54 + 12V^1 + l = 28 - 96 V^l. Result. Simplify Exercise 97 3). 1. (2 + V- 1) (3 + V- 1). 2. (3-V^l)(4-V^l). 3. (2 + 2 V^T) (3 + 2V" =: T). 4. (V^+V^)(V^T+V- 5. (2V ::: l-3V :r 3)(2V" :r T- 6. (3 V2 + 2 V^5) (2 V2 - 3V^ 7. (V— a — x) (V— a + a). 8. (-V^l + 2 V^2 + 3 V"^) 2 . 9. (2 + 3V^I) 8 . 11. (3V2 - 2V = T) 8 . 10. (3 + 2V-T)« 12. (aV^T- bV^iy -2). 4V" rg). OPERATIONS WITH IMAGINARY NUMBERS 259 DIVISION OF IMAGINARY AND COMPLEX NUMBERS BY RATIONALIZATION OF THE DIVISOR 298. Illustrations: 1. Divide V- 28 by V^8. ■ = ) . {} -= =± = \T = \T = 5 VIl. Result. 2. Divide V- 12 by V3. yrrT2 = (V—2)xVs = V^ == 6V=T =2Vzr - v Regult V3 V3xV3 3 3 3. Divide 2 + 2V :r 3 by 2 -2V :r 3. 2 + 2V^3 2 + 2 V^l? = (2 + 2\Z^3)« _ 4 + 8 V^3 - 12 2 + 2V-3 2 2 -(2V-3) 2 = 8V~3-8 _- 16 4 + "3-1 Result. Exercise 98 Simplify 5 2. V^2 V18 V^3* V-12 4. 13 V-12 2V^3* 2V^3- 5. 3V- x + 3+V- «-V-l V-27 2V^~3 2V^~4 -SV^ -14 V^2 -4V20 V-5 15. 16. 10. 11. 12. 5-V-2 2+V- 3+V" 3-V-3 V2+V^T V2-V^l* V^2~+V^3 V-2-V-3 Vc + 3V-2c Vc— V— c 2 m - 3 # V^^T 2m -1 260 RADICALS. IMAGINARY NUMBERS. REVIEW 299. Conjugate Imaginaries. Two complex numbers differing only in the sign of the term containing the imaginary are con- jugate imaginary numbers. Thus, a +bV— 1 and a — by/— 1 are conjugate imaginaries. By addition, (a + b v^T) + (a - 6 V^l) = 2 a. By multiplication, (a + b l)(a 1) = a 2 + & 2 . 300. T7ie sum and the product of two conjugate complex num- bers are real. MISCELLANEOUS PROCESSES WITH IMAGINARY NUMBERS Exercise 99 Simplify : i. (V"^i) 5 + (V=T) 3 . 4. (_V^I) 5 + (-V^i) 4 . 2 . (_v^T) 3 -(-V"=T) 2 . 5. (_V^2) 2 -(-V^2) 4 . 3 . (_2V^2) 2 + 3(V^3) 3 . 6. (V^I-1) 2 -(1 + V^T) S 7. (V"=3-l) 4 . 8. (1 + V"^T) 3 + (l - V^T) 2 - (1 - V^l). 9. 3(V^) 3 -3(V"^)(V^-1) 2 . 10. (m + w V— l)(m — nV- 1). 11. 12. 13. V- -1+1 1 v: -i-i =1-1 1 _ ri+i (V :r 2) 3 + (V- ^y. 14. (2 + V-l) 2 + (2- 1£ 3V-1 15 a + #V— 1 a — x 1 2V-2 + V-1 (V3i)4, ( V=2)4 yzri_V-2 a — xV—l a + x-V— 1 i6. ft^i-ff-v=i ; 17. V-i ( V3l)6 _ ( V3T)5 + ( V3l)4 V^2 GENERAL REVIEW 261 GENERAL REVIEW Exercise 100 1. Expand (xk - 2 -s/xf. 2. Find the square root of (x 2 + 3 x + 2)(x* — l)(a? + x — 2). 3. Simplify Vf + V^f + V| - VJ - V8. 5. Factor 27" V - 8" 1 . 6. Find the approximate numerical value of 2V2 + 1 . 3V2-1 V2-1 I V2 + 1 ' [ Simplify (^-y(^)« 1 8. So i ve f^-^ = w2 -^ [ my — %# = ra J — w . 9. Find the square root of a* + a* — 4 a 6 -f- 4 a + 2 a« — 4 al 10. | si mp iif y (V^W^) 2 11. Simplify (- V^) 3 - (- V^T) 4 - (- V- l) 5 . 12. How may the square root of (x 2 — x— 2)(x 2 — 4)(ar J +3a;+2) be found without multiplying ? i(* + 2/)-z = -3, 13. Solve J 2x + (z-y) = 9, _z + \(x + y + z) = 2. 14. Factor a; (x + 1)+ a (2 a? + 1) + a 2 . 15 - Simplify , ^C^%#£±gg • 1 J (x-l)- 1 -(x-2)- 1 -(x-3)- 1 262 RADICALS. IMAGINARY NUMBERS. REVIEW 1 16. Simplify • — 17. Simplify(V^-V^-V^3)(V^I+V^2-V^3). 18. Solve x — 2y = 3 f 3y + z = ll, 2z — 3x=l. 19. Simplify [^.^B 4 . L ar 1 V a -1 a" 2 V ar 3 J 20. Draw the graphs of 2 a? + 3 y = 13 and 3 a? — y = 3, and check by solving the system. 21. Solve 5x + 2y — z = 3x + 3y + z = 7x + 4:y — z = 3. 22. Factor ar 4 — 4ar 2 — 21. 23. What is the square root of x (x 3 - 4) + 1 - 2 x 2 (2 x - 3) ? 24. Find the cube of - 2 V^l + (V^I)" 1 . rixl6«x2*l-i; 25. Simplify Li___J_^. 26. Simplify cc + [x + a? (as + ar 1 )" 1 ]- 1 . 27. Simplify (- 2 V^ - V"^) (3 V^J.+ V-2). 28. Expand (2 a 2 -3 a;) 5 . «~ o i J? j ra — n + 1 2 „ , 3m+n- 1 r 29. Solve for m and n : ! — - = — and — =$ 2m + n—l 13 m — ti + 2 30. Simplify g^+g^i; 31. Factor c(<* -f » + 2 c) + c(a? + 1) + (» + 1)* ««ai 3 4 1 4 . 2 06,8 1 32. Solve = 1, - + -=3, - -f - = 1. x y x z z y 33. Explain the meaning of a~ n — — • GENERAL REVIEW 263 34. Simplify — - 3 x° + 27"^ - 1 3 *. 8"~* 35. Simplify 16c4 + 8c3 - 2c - M" 2 -l .fl L^\ J l-9a 2 4c 2 + 2c ^ 1_2-^V # 36. Solve for s : V 4 s — 9 — Vs + 1= Vs — 4. 37. Find two numbers whose sum is m and whose differ- ence is n. 38. Simplify |^ + flz£ . (c-Q(l + c-i) T 39. Find to three decimal places the value f 2 ^+3V2 t 3V3-2V2 40. Solve 2x = 5y, y- X -2z = l, 12y-3 + z = 4:X. 41. Simplify - ^-l-f. 2+V-l 5 2-V-l 42. Find the square root of 1 ~^( 2 ~^) 1£ 43. Is the expression a 4x + 2 a 3 * + a 2x + 2 a* -{- 2 + a" 2 * a per feet square ? 44. Simplify 2m ( m ~ 1 ) + L_ . (l_ m 2)f (l- m 2)i ^ 45. Factor 1 + 2 ( 3 + *> + ?*±i. (a 2 ^^)- 1 ^ a- 1 ^ 2- 1 af) 46. Solve -i + -i = m,—-i = n. • ex ay dx cy 47. If a~% = c" 1 , and c* = f, find the numerical value of a. 48. Simplify Y— L^-f ^Y— -i L^ V3+V-1 3-V-1A3+V-1 3-v"^iy 49. Collect 3Vl|-4V3-6V5| + 9V8j. 264 RADICALS. IMAGINARY NUMBERS. REVIEW 50. Under what condition is a solution of mx + ny = a and kx + ly = b impossible ? 51. What must be the values of m and n in order that the division of x 4 + x 8 + mx 2 + nx — 3bj x 2 — x + 1 may be exact ? 52. Extract the square root of 4 a 2n - 24 a~ n + 9 a~ 2n - 16 a n + 28. 53. Simplify V(a + l)(a>*- 1) - Va^-ic 2 - VaT^l. 54. A discount of 10 % was given on a bill of rubbers, and one of 5 % on a bill of shoes. The amount of both bills was $ 250, but, with discounts off, the merchant paid both with a check for $230. What was the amount of each bill ? 55. What is the value of 2 a°[2°+(2 a )°] ? 56. Simplify [ab-h- 1 + a-'bc- 1 + a^b^c) ' \ (a + b + c)~ 2 J ' 57. Expand |> + x l — 2] 3 . 58. Draw the graphs of x-\-3y — 12 and 3x — 4?/ = 10. Checlf by a solution. 59. Calculate to three decimal places the value of (1 - V3) + (1 + V3). 60. Simplify °-°~ 1 a + a_1 a* — a » a* + a"J 61. Solve (« + l)(»-l)->- — — 1— -— = 8ar>. 62. Introduce the binomial coefficient under the radical and simplify : (V3 - 2)V7 + 4V3. ' GENERAL REVIEW 265 63. Simplify 1«- - 3 aP+ 3 (3 x)° - 1° + (1 + a°)(l - a ). 64 - tt * ^M-.^ +»-W 65. Find the value of ( a _ V6)(a-f V6)(a— V^l>)(a+ V^6) when a = 2 and 6=3. 66. Solve and verify \ I a; -f 2 y — m 2 + mw -f ™ 2 » 67. Simplify T^_^!l ( e \ . 68. Show that (m 2 ^-?! 2 )?!" 1 — m m 2 — n 2 _ m 2 -mw w _1 — »i _1 m 3 + w 3 7i — m 69. Solve and test the solution of ^/5~x~^a — V5 a; + . : = . V5a; — a 70. Draw the graphs of 2sc + 3?/ = 7 and 4 - 20. 7. 3^ + 1 = 0. 8. ±ax 2 -l6a 2 = 0. 9. 7/ = 35a 4 . 12. 3a 2 + 18 - a; 2 - 5 1 1 17. 18. 19. 20. 21. 9). x 2 -\-x — 6 07+3 X + 1 X 0. a — 3 a + 4 or* + a; — 1 x 2 — x — x+2 ~x—2 a^ + a ?-2 = a3 2 + a?-l + 7a;. 10. 5a^-8a;. 3. ^-16x. • 7. tf-Sx. 11. 2a^-3aj. 4. ^ + 24^. 8 . 2/ 2 +ll2/. 12. 6^-17 x. The principle of completing the square is applied to the solu- tion of affected quadratic equations in the following : 270 QUADRATIC EQUATIONS Illustrations : 1. Solve a 2 + 12 a =13. Adding to both members the square of one half the coefficient of x, x 2 + 12 x + (6) 2 = 13 + 36 = 49. Extracting square root, x+ 6 = ± 7. Whence, x = ± 7 - 6. That is, x= + 7-6orx=-7-6. From which, x = 1 or x = — 13. Result. Both values of x are roots of and satisfy the given equation. Verification : Ifx = l: l 2 .+ 12(l) = 13, 1+12 = 13. Ifx=-13: (-13) 2 + 12(-13)=13, 169-156 = 13. 2. Solve 2z 2 -3a;-20 = 0. 2x 2 -3x-20 = 0. Transposing, 2 x 2 - 3 x = 20. Dividing by 2, x 2 — | x = 10. Completing the square, x 2 — f x + (f ) 2 = 10 + ^ Extracting square root, x — f = ± l ¥ 3 -. Whence, x = \ 6 -, or - J^. Or, x = 4, or — f . Result. Verification : If X = 4 : 2 (4) 2 - 3 (4) - 20 = 0. 32 - 12 - 20 = 0. Ifx=-f: 2(-f) 2 -3(-|) -20=0. ^ + Y__ 2 0=0. « o i x* — x-\-l q? + x — 1 ., 3. Solve J J — - — =1. x—2 x+2 Clearing of fractions, (x 2 - x + l)(x + 2) - (x 2 + x - l)(x - 2) = (x + 2)(x - 2). x 8 + x 2 - x + 2 - x 3 + x 2 + 3 x - 2 = x 2 - 4. Whence, x 2 + 2 x = - 4. Completing the square, x 2 + 2 x + 1 = — 4 + 1 = -3. Extracting square root, x + 1 = ± V— 3. From which, x = — 1 ± V— 3. Result. AFFECTED QUADRATIC EQUATIONS 271 Irrational roots result when, after completing the square, the right member of an equation is irrational. The original equation will be satisfied by either (— 1 + V— 3) or (-l-v^3). From these illustrations we may state the general process for completing the square in the solution of affected quadratic equations: 308. Simplify the given equation and reduce to the form x 2 -j- bx = c. Add to each member the square of one half the coefficient of x. Extract the square root of each member, and solve the two resulting simple equations. Exercise 102 Solve and verify : 1. ^-4a?==12. 11. 15x* + l±x=-S. 2. a? + 2x=z35. 12. 14 a 2 - 5 a; = 24. 3. ar J -6a? = 27. 13. 20x*-lx = Q. 17. x 2 -\-5x-S = 0, 8. 3^ + 17 a;+20 = 0. 18. 2ar s -8a = l. 9. 4:X 2 -5x — 6 = 0. 19. x 2 -4:X=-9. 10. 6x 2 -x = 2. 20. 3a^-a; + l=0. 21. 2x 2 -{-Sx= -10. 22. (3a + l) 2 -(4« + l)(2a;-l) = ll. 23. (x 4 -l)-(x 2 + 2)(x 2 -3)-(x + 5)=0. 24. (2x + 3)(x-2)-(3x-iy = x(x-3)+l. 25. (2 z + 1) (3a -2)-(z + l) (2z -!) = (» + l)(3a?-l). 272 QUADRATIC EQUATIONS 26. x 2 -2(x-l) + x(x-T)-(x-l)(x-2) = 0. 27. x(x 2 -x-3) + (x + 3)(x + ll) = (x — 1) (x 2 + 1). 28 2 't 3 -1. 34. ^±±Z^«^±£±2. ' # + 1 x — 1 ^ + # + 2 a^ + x — 3 89 . «±! + i«4. 35. E±| + l = ^3. oj + 3 a; 3 x — 3 # + 3 30. i±i + ^L-^l. 36. -A- =6+1- 05 + 1 a? — 4 a? + 4 x' 2 — 16 31 a? + 1 I Ez^ asl 37 - I X == a?2 + 5a; + 8 , ' x — 1 a + 2 " ' a + 2 a + 5 a 2 + 7 a + 10* 32 g-1 , g-r2^3 a+3 g-2 = 2(s-5) ' x -2 ai-3 4* x 2 a 2 + l x 4 + a^ ' 33 ?* + l ,3* + 2 = 39 2^ + 1 3a?-l = 5a?-2 2x-l 'Sx-2 ' ' a? + l x-1 " a+2 3z + 4 2 x 2 + x-l 40. ^ + 2^ + 4 2 — aj a^-8 The Quadratic Formula 309. Every affected quadratic equation may be reduced to the form ax 2 + bx + c — 0, in which, form the coefficients, a, b, and c, represent numbers, positive or negative, integral or fractional. 310. Solving this affected quadratic equation by completing the square, we have, ax 2 + bx = — c. a a a \2a] a 4a 2 = 6 2 - 4 ac 4 a 2 AFFECTED QUADRATIC EQUATIONS 273 x + b 2a X Vb 2 - 4 ac 4 a 2 b V6 2 - 4 ac 2a 2a _ -b±Vb 2 -,4ac 2a This value of x from the general equation, ax?-{- 6a; + c = 0, serves as a formula for the solution of affected quadratic equa- tions. The formula is expressed in terms of the coefficients of the given general equation, and by substitution of particular values for a, b, and c from a given equation we obtain the roots of that equation. The formula is the most practical of the many methods of solution and it should be memorized. To obtain the solution of an affected quadratic equation by means of the general formula : 311. Transpose all terms of the given equation to the left member, and reduce to the form ax 2 +- bx + c = 0. In the formula substitute the coefficient of the given x 2 for a, the coefficient of the given x for b, and the given constant for c. Simplify the resulting expression. Illustrations : 1. Solve by the formula, 2 x 2 + 5 x = 12. Transposing, 2 x 2 + 5 x - - 12 = 0. For the formula : Then in a = 2, 6 = 5, c = - 12. x _ - 6 ± V6 2 - 4 ac 2a -(5)±V(5) 2 -4(2)(-12) 2(2) _ - 5 ± V25~+ 96 4 _ -5 ±11 4 = f , or — 4. Result. SOM. EL. ALG. — 18 274 QUADRATIC EQUATIONS 2. Solve by the formula, x 2 — ±g. x = -§-. Transposing and clearing of fractions, 3x 2 -10z-8 = 0. For the formula, a = 3, b = — 10, c = — 8. Substituting, ^ = -(-10)±V(-10) 2 -4(3)(-8). 6 x ='4, or — f . Result. Exercise 103 Solve by the formula : 1. x? -11 x + 24 = 0. 2. o 2 -9# = 22. 3. ^ + 9 # + 14 = 0. 4. aj*-lla>=-28. 5. ^ + 9z = 52. 6. a^-a;-72 = 0. 7. o*-21 z = 46. 8. 6o 2 -a = 2. 9. 2ar>+7a; = l5. 10. 8x 2 + 2x = 3. 11. 15^ + 44^ = 20. The Solution by Factoring 312. Many affected quadratic equations may be solved by an application of factoring. This method is based upon the principle that : 313. The product of two or more factors is zero when one of the factors is equal to zero. 12. 21 a* + 29^-10 = 0. 13. 14^ + 53^ + 14 = 0. 14. 30 a 2 -11 a = 30. 15. 42 <& — m so = - 20. 16. a; 2 --u- # + 2 = 0. 17. ^ + -V 3 ^ = f 18. x 2 -- = 2. 6 19. 4 8. 20. ^_*? = o. 30 3 AFFECTED QUADRATIC EQUATIONS 275 To solve an affected quadratic equation by factoring : 314. Reduce the given equation to the general quadratic form, ax 2 + bx -f- c = 0, and factor the resulting trinomial. Assume that each factor in turn equals zero, and solve the other factor for the value of the unknown quantity. Illustrations : 1. Solve by factoring, x 2 + 6 x = 7. Transposing, x 2 -f 6 x — 7=0. Factoring, (x + 7) (x - 1) = 0. From which, x + 7 = 0, or x — 1 = 0. Solving, x = - 7 and x = 1. Result. 2. Solve by factoring, 2x 2 + 1 ^x=%. Simplifying, 10 x 2 + 11 x - 6 = 0. Factoring, (2 re + 3) (5 x - 2) = 0. Whence, 2 as + 3 = and 5 x - 2 = 0. And x = -f, or x = % . Result. Exercise 104 Solve by factoring: 1. a^-5a; = 14. 10. 8a*-38a; = -35. 2. a?-8x + 15 = 0. 11# 15 x 2 - 77 a + 10 = 0. 3. « 2 -3a; = 4. 4. z 2 - 13 a + 12 = 0. 5. a^-12a;-13 = 0. 6. ar J -19 = 3»-ff. 276 QUADRATIC EQUATIONS LITERAL AFFECTED QUADRATIC EQUATIONS 315. Any one of the three methods given applies readily to literal affected quadratic equations. . Illustrations : 1. Solvec?(x i -l)=x(x + 2c). Clearing of parentheses, c 2 x 2 — c 2 = x 2 + 2 ex. Transposing, c 2 x 2 — x' 1 — 2cx = c 2 . Uniting coefficients of x 2 , (c 2 — 1) x 2 — 2 ex = c 2 . Dividing by (c 2 - 1) , x 2 - (j^) * = j£j ■ c 2 Completing the square, x 2 - -^- + ( —^—r) = -^—- + — c 2 — 1 \c 2 — 1/ c 2 — 1 (c 2 - l) 2 = c 2 (c 2 -l) + c 2 (C 2 - l) 2 (c 2 -!) 2 ' Extracting square root, x -—r— - = ± c 2 — 1 ± * C 2 _ 1 - C 2 _ X From which, x = — ^— orx= — • Result. c- 1 c + 1 2. Solve *+»•-*£• Transposing and clearing of fractions, n 2 x 2 + mnx - 2 m 2 = 0. For the formula : a = n 2 , b = ran, c = — 2 m 2 . x _ -mn± V (m») 2 - 4 (n 2 ) (-2 m 2 ) 2(w 2 ) _ — mn ± V9 m 2 n 2 In 2 _ — mn ± 3 mn 2n 2 = », or- 2 -^. Result. DISCUSSION OF AFFECTED QUADRATIC EQUATION 277 Exercise 105 Solve: 1. x 2 -2ax = 3a\ 8. a?-2ax + a* = l. 2. x 2 = 5 ax — 6 a 2 . 9. cV + c(m — ri)x = mn. 3. x 2 + ax = 2a 2 . 10. mx 2 — (m + l)x + l = 0. 4. 3 aV + acx — 2 c 2 . 11. acx 2 + anx = cmx + mn. 5. IOcAb 2 - 21 c«a> + 9 = 0. 12. cV-c(a + l)« + a = (). 6. 35c 6 x 2 — c 3 x = 6. 13. (2*+a)(*— c)=(aj— a)(»+c). 7. cc 2 + 4a» = 4a + l. 14. (as + a -f- mf = (a — m) 2 . 15. « 2 -2aa;-26x = c 2 -a 2 -6(2a + 6). 1iS a;-i-a . a 5 -„ 1 1 _ 1 1 lb. 1 ; — — • 17. — • a x + a 2 m — x m c—x c ma x-2a . x + 2a 2x 2 + 3a 2 lo. 1 -— " • • cc + 3 a x — 3 a 0^ + 9 a 2 DISCUSSION OF THE AFFECTED QUADRATIC EQUATION Character of the Roots 316. If the two roots of the affected quadratic equation, ax 2 + bx -+- c = 0, be denoted by t-j and r 2 , we have (Art. 310), — b + V& 2 — 4 ac — 6 — V & 2 — 4 etc 2 a 2a The character of the result in each root depends directly upon the value of the radical expression V& 2 — 4ac, for the radicand, b 2 — 4 ac, may be positive, zero, or negative. (I) When b 2 — kac is positive. (a) The roots are real, for the square root of a positive quan- tity may be obtained exactly or approximately. (6) The roots are unequal, for Vb 2 — 4 ac is -f in r x and — in r 2 . (c) The roots are rational if 6 2 — 4ac is a perfect square, irrational if not. ■ 278 QUADRATIC EQUATIONS Illustrations : (1) In2sc 2 + lla; + 12=0, a=2, 6 = 11, c=12. 6 2 -4ac=121-96=25. Hence, the roots of 2 0, the roots are real and unequal. II. If b 2 — 4 ac = 0, the roots are real and equal. III. If b 2 — 4 ac < 0, the roots are imaginary. Illustrations : 1. Determine the character of the roots of x* — 5 x = 6. a a - 5z-6 = 0. a = l, 6 = -5, c = -6. 62 _ 4 ac = [(-5)2- 4 (1)(- 6)] = (25 + 24) = 49. Therefore, the roots are real, unequal, and rational. (316, 1.) 2. Show that the roots ofar 2 — 4sc+5=0 are imaginary. 6 2 -4ac=[(-4) 2 -4(l)(5)] = (16-20) = -4. Hence, the roots are imaginary. DISCUSSION OF AFFECTED QUADRATIC EQUATION 279 3. Determine the value of m for which the roots of 4or J + 10a; + m=0 are equal. By Art. 316, II, the discriminant must equal 0. Hence, (10) 2 - 4 (4) (m) = 0, 100 - 16 m = 0, m = ^. That is, the roots of 4 z 2 + 10 * + ^ = are equal. 4. For what value of m will the roots of (m + l)a; 2 + (6m + 2)a; + 7m + 4 = 0be equal ? The discriminant must equal 0. a = (m + 1), 6 = (6m + 2),c = (7m + 4). Then (6 m + 2) 2 - 4 (m + 1) (7 m + 4) = 8 m 2 - 20 m - 12. Solving, 8 m 2 - 20 w - 12 = 0, m = 3 and - J. Result. By substituting these values, 3 and — §, in the given equation, two dif- ferent equations result, both of which have equal roots. Exercise 106 By inspection of the discriminant, determine the character of the roots of : 1. a? + 7x=&. 7. 2 x* + l a + 3 = 0. 2. a 2 -6a; = 40. 8. 3 ar 2 - 5 x= - 12. 3. tf 2 + 5a-84 = 0. 9. a? + 10« + l = 0. 4 . ^-3^ + 54 = 0. 10. 3a^-12a;=-17. 5. x* + 12x = -36. 11. 5ar> + 40a; = l. 6. 3a 2 + 8a; + 4 = 0. 12. 7 ar> + 11 a? + 12 = 0. Determine the values of m for which the two roots of each of the following equations are equal : 13. 4x 2 + 20a + ra = 0. 16. a 2 -}- (m -f 5) a; + 5m+l f=0. 14. 9x* + mx + 25 = 0. 17. (m-f-l)a5 2 -(m-2)a;-hl = 0. 15. 2m^- 30 # — 15 = 0. 18. 2ma^+7ma!=^+5a;-5m. 19. (m + l)a; 2 + ma;=-9(a; + l). 20. m — 7 + mo 2 — m# = — 2 a; — x 2 . 280 QUADRATIC EQUATIONS Relation of the Roots and Coefficients 318. If the roots of ax 2 + bx + c = Oorx 2 + -x+- = a a are obtained by Art. 310, and are denoted by r x and r 2 respec- tively, we have 2a 2a 1*^ «aam^ „ _l v ~ b + V6 2 - 4 ac - b - V& 2 - 4 ac By addition, n + r 2 = ! 2a Or, ri + r2= _&. a By multiplication, nn = (- » + V5' - 4«c) (- 5 - Vt° -4 ac) 4 a 2 _ ft 2 - ft 2 + 4 qc 4 a 2 Or, nr 2 =-- a Hence, we may state : 319. In any affected quadratic equation of the form a a (a) The sum of the roots equals the coefficient of x with its sign changed. (b) The product of the roots is equal to the constant term. Illustrations : 1. Find by inspection the sum and the product of the roots of6x 2 + 5x = 6. Changing to the required form, x 2 + f x — 1 = 0. From Art. 319 : Sum of the roots = — f . Product of the roots = — 1. Result. 2. One root of 2 x 2 + 5 x = 12 is — -§. Find the other root. Transposing and dividing, x 2 + f x — 6 =0. Dividing the product of the roots, - 6, by the known root, - f, we have, (-6)-(-f)=+(6x|)=4, Result. DISCUSSION OF AFFECTED QUADRATIC EQUATION 281 Exercise 107 Find by inspection the sum and the product of the roots of : 1. x> + 9x + U = 0. 5. 3ar>-10a; + 3 = 0. 2. x 2 -21x = ±6. 6. 2ar> + 3a = 2. 3. 2 a^+7 »«.!«. 7. 7^ + 9^-10 = 0. 4. 6x 2 -x = 12. 8. 15^ + 14^ + 3 = 0. 9: One root of 5 x 2 — 26 x + 5 = is 5. Find the other root. 10. One root of 8 x 2 = 15 x + 2 is — -|-. Find the other root. 11. With the values of r, and r 2 from the equation ax 2 + foe + c = 0, find the value of rs—r 1 . 1 n + r 2 Also the value of — + Formation of an Affected Quadratic Equation with Given Roots b c Consider again the form x 2 + -x + - = 0. (1) a a If, as before, i\ and r 2 denote the roots of this equation, we have (Art. 319) : 7*1 + 7*2= a Whence, - = - r s - r 2 (2). Also (319), -= r x r 2 (3) Ct CL Substituting in (1) the values found in (2) and (3), x 2 + ( — r x — r 2 )x + r x r 2 = 0. x 2 — r x x — r 2 a* + r^ = 0. (x 2 — r&) — (i\x — rfo) = 0. x(x — r 2 ) — r x (x — r 2 ) = 0. (»-r 1 )(a;-r 2 )=0. 282 QUADRATIC EQUATIONS Therefore, to form a quadratic equation that shall have any given roots : 320. Subtract each root from x, and equate the product of the resulting expressions to 0. Illustrations : 1. Form the equation whose roots shall be 3 and 7. By Art. 320, (* - 3) (x - 7) = 0. Or, x 2 - 10 x + 21 = 0. Result. 2. Form the equation whose roots shall be f and — -|. By Art. 320, (z_f)( x _(_|)) = o ; (x-f)(x + §) = 0; l5x 2 + x-6 = Q. Result. Exercise 108 Form the equations whose roots shall be : 1. 2,5. 5. -2* 8. a — 1, a -f- 1. ! 3 9. a — 1, 2 a. 2. 3, -8. 6 5 7 6 * 2' ~8 10. 2±V=3. 3. -4,-7. 11. V2, -1. 4. 3,-L 7. -?,-3. 12. Va + 1 Vtt— 1 ' 2 3' 4 2 ' 2 GRAPH OF A QUADRATIC EQUATION IN ONE VARIABLE 321. The graph of a quadratic equation is obtained by application of the principles governing the graphs of linear equations. The given equation is written in the typical form, ax 2 + bx + c — O, and the left member is equated to y. By as- suming successive values for sc, the corresponding values of y are obtained as before. GRAPHS OF QUADRATIC EQUATIONS GRAPHS OF QUADRATIC EQUATIONS HAVING UNEQUAL ROOTS 283 322. Plot the graph of a? + 2x = 8, Assume y = x 2 + 2 x — 8. In the figure, If x = 3, y = 7. P. X= 2,y = 0. Pi. x = 1, y = - 5. P 2 . X = 0, y = - 8. P 8 . x = - 1, y = - 9. P 4 . x= -2,y= -8. P 6 . x = - 3, y = - 5. P 6 . x = - 4, y = 0. P 7 . x = - 5, y = 7. P 8 . etc. Bsag i The curve representing the equation, ^ + 2^ — 8 = 0, might be indefinitely extended by choosing further values of x. In- termediate points on the curve may be obtained by assuming fractional values of x and obtaining corresponding values of y. 323. The lowest point of the graph of a quadratic equation in one variable may, in general, be obtained by completing the square. Thus, x 2 + 2x + 1=8+ 1; (x + l) 2 = 9; (x+l) 2 -9 = 0. Now (x + l) 2 — 9 has its greatest negative value when x = — 1. Hence, the coordinates of the lowest point of the curve are (— 1, — 9). Plot the graph of 2 x 2 + 7 x - 4 = 0. Let y = 2 x 2 + 7 x - 4. Then in the first figure on p. 284 : (The student will note that, in the figure, the scale of the graph is so chosen that one unit of division on the axis corresponds to two units from the solutions.) 284 QUADRATIC EQUATIONS If x = 2, y = 18. x — 1,2/ = 5. x = 0, y = - 4. s= — 1, 2/ =-9. P. Pi. P 2 P s 2 [T P f i i . 1 . -1 -it f X' ( )t x in: X t- T^ t. ll| fL J I B ~Y" a; =-2, y =-10. x=-S,y=-7. x=-4,y = 0. x=-b, y = 11. P 4 . Ps- P 6 . P 7 . On completing the square the lowest point in the curve is (— |, - - 8 ^). Exercise 109 Plot the graphs of: 1. x>-§x + 5z=0. 2. x 2 + 6x + 8 = 0. 3. a^ + a; — 12 = 0. 4. 4 x 2 — 5 x = 0. 5. 2^-5x = -3. 6. 8a 2 + 2z-3 = 0. GRAPHS OF QUADRATIC EQUATIONS HAVING EQUAL ROOTS 324. Plot the graph of x>- -6x + 9 = 0. Let y : = x*- -6s + 9. the figure Ifx = 0,2/ = 9. P. « = 1*9 = 4. Pi. x = 2,2/ = 1. P 2 . X = 3,2/ = 0. P 3 . X = 4, y = 1. P 4 . X = 5, y = 4. Ps. X = 6,2/ = 9. P 6 . In « PV 45 Now, ce 2 — 6se + 9 = 0, may- be written (x — 3) 2 = ; and x in this equation can never be 0. Hence, the curve cannot cut the XX 1 axis, but is tangent to it at (3, 0), the lowest point of the graph. GRAPHS OF QUADRATIC EQUATIONS 285 GRAPHS OF QUADRATIC EQUATIONS HAVING IMAGINARY ROOTS 325. Plot the graph of x 2 - x + 2 = 0. Let y = x' 2 — x + 2. In the figure : x = 3, y = $. x = —l, y = 4. x = - 2, y = 8. lix = 0, y = 2. P. x - 1, y = 3. Pi. X - 2, y = 4. P 2 . x = in Completing the square x 2 -x + 2 = 0: X 2 - x+2 = (z 2 - = (x- -i) 2 + |. i + 2 p 3 . p 4 . p 5 . - 3, y = 14. P 6 . Pif 1 EEEEEEEEK2EEEEEEE FTP And this expression not being or negative for any value of x, it follows that y cannot be or nega- tive ; that is, the graph cannot touch the XX' axis. Important Conclusion. The graph of any quadratic equation in one variable and in the form ax 2 + bx + c = 0, intersects the XX' axis at two points whose abcissas are roots of the equation, provided that the given equation has real and unequal roots ; has one point in the XX' axis if the roots are equal ; and does not cut the XX' axis if the roots are imaginary. Exercise 110 Plot the graphs of : 1. a 2 -4# + 4 = 0. 2. 4ar>-|-4:r + l=0. 4. x 2 + 2x + S=0. 5. x? + x+6 = 0. 6. 2x 2 -6x + ll = 0. CHAPTER XXII THE QUADRATIC FORM. HIGHER EQUATIONS. IRRATIONAL EQUATIONS EQUATIONS IN THE QUADRATIC FORM 326. An equation in the quadratic form is an equation having three terms, two of which contain the unknown num- ber ; the exponent of the unknown number in one term being twice the exponent of the unknown number in the other term. Thus: HIGHER EQUATIONS SOLVED BY QUADRATIC METHODS 327. It will be seen at once that many equations in the quadratic form must be of a higher degree than the second. The method of factoring permits the solution of many such equations, and is generally employed in elementary algebra. 328. If quadratic factors result from the application of factoring to higher forms of equations, they may ordinarily be solved by completing the square or by the quadratic formula. Such factors most frequently occur in connection with binomial equations. Illustrations : 1. Solve x 4 — 13 x 2 + 36 = 0. Factoring, And, + 2) (i "Whence, (z 2 c - 2) (x - 4) («2 _ 9) = + 3) (x - 3) = x = 286 = 0. = 0. -2,2, 3, and 3. Result. EQUATIONS IN THE QUADRATIC FORM 287 2. Solve a? + 7a?-8 = 0. Factoring, (x 3 + 8) (x 3 - 1) = 0. And, (as + 2) (x 2 - 2x + 4) (x - 1) (x 2 + x + 1) = 0. By inspection, x = — 2 and 1. Solving, x 2 -2x + 4=0. x = 1 ± V^3. Solving, x 2 + x + 1 = 0. Therefore, x = - 2, 1 ± V^3, 1, and ~ + ^ ~ ■ Result. 3. Solve ^? - 4a/^ = 12. Expressed with fractional exponents and in the transposed form, we have, a;f _4»i_ 12 = 0. Factoring, (x* - 6) (x^ + 2) = 0. Whence, x^ = 6, or x* = - 2. "From which, x = 216, or — 8. Result. When tested both roots prove to be solutions. 4. Solve x* - 12 x~l = - 1. Jk 12 _ i With positive exponents, x 7 — * x* Clearing of fractions, x^ — 12 = — x*» Transposing, x^ + x* — 12 = 0. Factoring, (x^ + 4) (x* - 3) = 0. Whence, x* = — 4, or x* = 3. And, x s 256, or 81. The equation is satisfied by the negative value of v / 256 and by the positive value of VH. 329. For convenience of solution a single letter may be substituted for a compound expression in equations in the quadratic form. Care should be taken that no solution is lost in the final substitutions. 288 THE QUADRATIC FORM. HIGHER EQUATIONS Illustrations : 1. Solve (a 2 - 4) 2 - (a 2 - 4) = 20. Transposing, (x 2 - 4) 2 - (x 2 -4) - 20 = 0. Let (x 2 - 4) = y. Then, y 2 -y- 20 = 0. Factoring, (y - 5) (y + 4) =0- Whence, y = 5, or — 4. If y = 5, If*/ = -4, x 2 - 4 = 5, X 2 - 4 = - 4, x 2 = 9, x 2 = 0, x = ± 3. x = 0. Therefore, x = ± 3 and Result. 2. Solve Let X + IV , » + 1 X x + \) + = y- x-l Then, y 2 + y - 6 = 0. Factoring, (j/ + 3) (y - 2) = 0- And, y = — 3 and 2. If 2/ = - 3, x+1 x-l Hy = 2, x + l . x-l" = -3, x + l = -3(x-l), 4x = 2, r Result. 2, x + 1 = 2(x - 1), - x = - 3, x = 3. 3. Solve 4a^ + 36 a;- 2 = 25. Transposing, 4 x 2 + 36 x~ 2 - 25 = 0. Then, 4z 2 +!^_25 x 2 Whence, 4 x 4 + 36 - 25 x 2 s= 0. Or, 4x 4 -25x 2 + 36 =0. Factoring, (x 2 - 4) (4 x 2 - 9) = 0. Or, (x + 2) (x - 2) (2 x + 3) (2x - 3) = 0. And, x = - 2, 2, - } , f. Result. EQUATIONS IN THE QUADRATIC FORM 289 Exercise 111 Solve : 1. a; 4 -5ar> + 4 = 0. 2. z 4 -8ar 9 = 9. 3. a;- 2 - 3 a;- 1 = 10. 4. x 9 + 9a? + $ = 0. 5. a 6 -7ar* = 8. 6. ar* + 7a;- 2 = 144. 7. »- 1 + ^ = 20. 8. 2ar* + 5ar* + 2 = 0. 9. 3 a + \/3 = 2. 10. or 1 — 3ar* = 4. 11. 6aBr* + 13«r* + 6=aO. 12. 8ar°-3ar 2 = -10. 13. 2^ + ^ = 7. 14. + 3 a; + 6)* - 8 = 0. 27. a^ + a;-f-2 = 7vV + a; + 2-10. 28. (x + — Y_3(W— ^ + 2 = 0. ar> + 2 29. a^-2 a^ + 2 4-8-6^ 30. SOM. EL. ALG ?-2 aT3T + °(a?T2j =6 19 290 THE QUADRATIC FORM. HIGHER EQUATIONS EQUIVALENT EQUATIONS AND THE REJECTION OF ROOTS 330. If, in two equations involving the same unknown number, the solutions of each include all the solutions of the other, the equations are equivalent. Thus, 3 x X 2 = x + 6. (1) Whence 2 x = 4. (2) Equations (1) and (2) are equivalent, for x — 2 is a solution j of each. 331. In the solution of fractional equations, and in the solution of irrational equations also, it must be remembered that the processes of simplification may introduce roots that | will not satisfy the given equation. That is, equivalent equations do not always follow in the successive steps of a process. (a) Processes that do not change the Roots op an Equation I 6 Given the equation x + 3 x-2 Clearing, (x + 3) (x - 2) = 6. From which, x 2 + x - 12 = 0. And, x = 3, or — 4. By trial both 3 and — 4 are found to be roots of the given equation. Hence, multiplying by x — 2, the lowest common denominator, intro- duced no new root. The only root that could be introduced by the multiplier, x — 2, would be the root of the equation, x - 2 = 0, or x = 2. And this root has not been introduced. In general : 332. The roots of a fractional equation are unchanged if: (1) the given fractions having a common denominator are com- bined, and (2) the equation is multiplied by the loivest common multiple of their denominators. EQUIVALENT EQUATIONS 291 (b) Processes that change the Roots of an Equation Given the equation x + 5 = 0. Multiplying by as - 1, (se - 1) (x + 5) = 0. From which, x = — 5, or 1. The solution x = 1 fails on trial to satisfy the given equation. In general : 333. A root is introduced if both members of an integral equation are multiplied by an expression involving the unknown number in the equation. Given the equation x — 2 = 0. Transposing, x = 2. Squaring, x 2 = 4. Whence, a 2 - 4 = 0. Factoring, (x + 2) (x - 2) = 0. Whence, x = - 2, or 2. Of these two solutions only the solution, « = 2, satisfies the given equation, the solution, x = — 2 being introduced by the process of squar- ing. In general : 334. A root is introduced if both members of an equation are raised to the same power. The Rejection of Roots. Since it is clear that certain processes may affect the solu- tions of involved types of equations, we make the following conclusion : 335. Before accepting all the solutions of a given equation as roots of that equation it is necessary that each solution be tested; and all solutions not satisfying the given equation must be rejected. 292 THE QUADRATIC FORM. HIGHER EQUATIONS IRRATIONAL EQUATIONS INVOLVING QUADRATIC FORMS 336. Illustration: Solve V2 x + 7 — V#— 5 = VS. Squaring, 2 x + 7 - 2 V(2x + 7)(x - 5) + x - 5 = x. Transposing, — 2 V2 x 2 — 3 a; — 35 = — 2 a; — 2. Dividing by - 2, V2 x 2 - 3 x - 35 = x + 1. Squaring, 2 x 2 - 3 x - 35 = x 2 + 2 35 + 1. From which, x 2 - 5 x - 36 = 0. And, x = 9, or — 4. Result The solution, x = 9, satisfies the given equation, but the solution, x = — 4, does not satisfy, and is rejected. Note. If the solution, x = — 4, is tested and, in extracting the square root of the right member the negative value of the root is taken, we have : i V2(-4)+ 7-V(-4)-6= V-4. t/Zi - V^9 = */-!. V^T-8 V^T = -2 But this process is at variance with the accepted condition that positive square roots are to be consistently taken through the practice of ele- mentary algebra. Exercise 112 Solve and test the solutions of : 1. aj — l = V* + l. 4. x + V2 = Va + 2 x V2 + 4. 2. Vz 2 + « + l = V3a + 4. 5. V2(aj»-4) = (aj-2). 3. x + Va + l = 19. 6. ar 2 + a-3 = Va 4 -r-2ar J +l0. 7. Vx-f-2- Va-3 = V .*• 8. (a: + 3)(a>-l) = Va>(a 8 -7) + 10 + 2a. Vg + l--v^-l =!i 3 a . 4 . 2t Va + 1 + Vx — 1 FORMULAS INVOLVING QUADRATIC EQUATIONS 293 PHYSICAL FORMULAS INVOLVING QUADRATIC EQUATIONS Exercise 113 1. From the formula, E = -—, obtain an expression for v. 2. From the formula, H= .24<7V£, obtain an expression for C. 3. Find expressions for I and g when t = it -i/- . 4. Find an expression for t from the formula S = \gt 2 . Wv 2 5. Given, J57 = -— — ; find an expression for v. 6. Given the formulas, V=gt, and S=^gt 2 . Find an expres- sion for the value of t in terms of V and S. 7. Given the formula, F = — - — . Find a formula for t in t • terms of F, m, n, and r. 8. Given the formula, t = v\j- . Find the value of I when Li. v » 9. From the formula, S = V t + igt 2 , obtain the value of t in terms of V , g, and S. 10. Given the formulas, E=Fs, F=ma, and v= V2as; obtain a formula for i? in terms of m and v. v 2 11. If a = — , and t*=2 wr, find a formula for a in terms of r 7r, r, and £. 12. E represents the energy of a moving body, m its mass, 7YIV and v its velocity, in the formula E= -— • What is the rela- z tion of the energies of two bodies if one has twice the mass but only two thirds the velocity of the other? CHAPTER XXIII SIMULTANEOUS QUADRATIC EQUATIONS. PROBLEMS i 337. In the solution of simultaneous quadratic equations we have particular methods for dealing with three common types ; but no general method for all possible cases can be given. SOLUTION BY SUBSTITUTION 1. Solve 338. When one equation is of the first degree and the other q the second degree. '2x*-3xy-y* = l, (1) _5x+y=3. (2) From (2), # y = S-6x. (3) Substituting in (1), 2 x 2 - 3 x (3 - 5 x) - (3 - 5 a) 2 = 1. 2a; 2 - 9x + 15 x 2 - 9 + 30x - 25x 2 = 1. 8x 2 -21x + 10 = 0. (4) Factoring, (8 x - 5) (x - 2) = 0. From which, x = f , or 2. Substituting in (3), Ifx = f, y = 3-5(|) = -f Ifx = 2, */ = 3 -5 (2) = -7. Hence, the corresponding values of x and y are , **» • 1 Result. Corresponding values must be clearly understood as to mean- ing. From (4) in the above solution two values of x result. Each value of x is substituted in (3), and two values for y result. Therefore, we associate a value of x with that value of y result- ing from its use in substitution. 294 SOLUTION BY COMPARISON AND FACTORING 295 Exercise 114 Solve : 1. x + 2y = 5, 6. « 2 + an/ + 2/ 2 = 19, x* + 3xy = 7. x + y = 5. 2. x -3y=2, 7. x*-3xy -y 2 + 3 = 0, xy + y 2 = ti. 3x + 2y — 5 = 0. 3. 2x + 3y = 12, 8. 3x + 2y = 5, xy-6 = 0. 2-xy-y 2 = 0. 4. 3x + 2y = 9, 9. x? + y 2 + x +y = 6, xy — x — 2. x — 4 y = 6. 5. a; + 22/ + l = 0, 10. 2 z + 3 # + 7 = 0, 3ay-2/ 2 = -4, 3aj*+2y*-a> + 4y=:12. SOLUTION BY COMPARISON AND FACTORING 339. When both equations are of the second degree and both are homogeneous. Solve (**-*-* (*) 13^-^ = 11. (2) A factorable expression in x and y results if the constant terms are eliminated from (1) and (2) by comparison. Multiplying (1) by 11, 22 x 2 - 11 xy = 110. (3) Multiplying (2) by 10, 30 x 2 - 10 y 2 = 110. (4) Equating left members, 30 x 2 -10y 2 = 22 x* - -11 xy. From which, 8x 2 + llxy - 10y 2 = 0. (5) Factoring (5), (x + 2y)(8x-5y) =0. Therefore, ,-f- (6) And, Sx V ~ 6 * • (7) 296 6 SIMULTANEOUS QUADRATIC EQUATIONS By substitution in (1): Again in (1) : From (6) : If y = - 1, From (7): tfys**, 5 »*-*(-§) = 10. 2z 2 -x(^Wo. 2x 2 + — = 10. 2i 2x2-^=10. 5 5a; 2 = 20. 2 x 2 = 50. x 2 = 4. x 2 = 25. x =±2. Hence, in (6), x—± 5. Hence, in (7), "=-ih-2 1(±2)=T1 - * = if=!<:k6)=±& o Therefore, x = ± 2, y==Fl, x 2/ = ±5;1 f Result. = ±8.J Note. The sign T is the result of a subtraction of a quantity having the sign ±. Thus, — (± a) = — (+ a) or — (- a) = — a or + a, = T a. Solve : Exercise 115 1. x 2 -\-xy=6, xy—y 2 = l. 6. tf-xy + y 2 = %, x* + xy + y 2 = %. 2. x 2 — xy — 4, 7. x 2 — 3xy + y 2 = — 5, 2ic 2 +a*/ -2?/ 2 = -4. 3. x 2 -2xy = -8, y 2 -3xy = -9. 8. 0^ + ^ — 4^ = 18, a? 2 — a??/ — 3 y 2 = — 9. 4. x 2 + 2«2/ = 9, 3 xy — i/ 2 = — 4. 9. 10« 2 + 7«2/-2/ 2 = -ll, 12x 2 -^-9xy-y 2 = -15. 5. ar 2 + 5 xy — — 6, ar> + 2/ 2 = 5. 10. 2x 2 + 4 : xy + 7y 2 -15 = 0, 5^ + 3x2/-32/ 2 -15 = 0. SOLUTION OF SYMMETRICAL TYPES 340. When the given equations are symmetrical with respect to x and y ; that is, when x and y may be interchanged without changing the equations. SOLUTION OF SYxMMETRICAL TYPES 297 1. Solve (x + y = 7, [xy = 10. (1) xy = 10. (2) Squaring (1), x 2 + 2 xy + y 2 = 49. Multiplying (2) by 4, 4 xy = 40. By subtraction, x 2 — 2 xy + y 2 = 9. Extracting square root, x — y — ± 3. (3) Adding (1), x + y = 7. 2x=+3+7, or-3 + 7. 2 x = 10, or 4. x = 5, or2;l Regult Subtracting (1) from (3), y = 2, or 5. J 2. Solve (^= 12 > ft Multiply (1) by 2, 2 set/ = 24. (3) Add (2) and (3), x 2 + 2 xy 4 y 2 = 49. (4) Subtract (3) from (2), x 2 - 2 xy + y 2 = 1 . (5) Extract square root of (4), x 4 y = ± 7. (6) Extract square root of (5), x — y = ± 1. (7) Four pairs of equations result in (6) and (7), viz. : x + y=7 x + y= 7 x + y = -l x + y = -7 x — y = 1 x - y =— 1 x- y = 1 x — y = — 1 x=4 x=3 x=-3 x = — 4 y =3 y= 4 y =-4 y =-3 Hence, the corresponding values for x and y are x = ±4, y=±3,i x = ±3; y=±4.j Result. Solve : Exercise 116 1. a + y = 6, 3. ^ + 2/ 2 = 5, ojy = 6. as + y = 3. 2. x + y = 5, 4. a#— 3=0, zy-4 = 0. a^4-2/ 2 = 10. 298 SIMULTANEOUS QUADRATIC EQUATIONS 5. x 2 + xy + y 2 = 7, 7. x 2 + y 2 =5, x + y = 3. i (x + y) 2 = 9. 6. x + y-5 = 0, 8. o 2 + 3a?/ + 2/ 2 = 59, x 2 -xy + y 2 = 7. x 2 -f o?y + y 2 = 39. SOLUTIONS OP MISCELLANEOUS TYPES 341. Systems of simultaneous quadratic equations not con- forming to the three types already considered are readily recognized, and the student will gradually gain the experience necessary to properly solve such systems. No general method for these types can be given. It is frequently possible to obtain solutions of those sys- tems in which a given equation is of a degree higher than the second, derived equations of the second degree resulting from divisions and multiplications. . Illustrations : l. Solve j^ + 3* + 3, = 28, (1) lajy-6 = 0. (2) Transposing and multiplying (2) by 2, 2 xy = 12. (3) Adding (1) and (3), x 2 + 2 xy + y 2 + 3 x + 3 y = 40. Whence, (x + y)* + 3 (x + y) - 40 = 0. Factoring, (x + y + 8) (x + y — 5) = 0. Hence, x + y =— 8, or x + y = 5. Combining each of these results with (2), we have two systems : (a) x + y = - 8, (6) x + y = 5, xy - 6 = 0. xy - 6 = 0. The two systems (a) and (&) may be readily solved by the principle of Art. 340. *. Solve mf**-* ■ I [x — y = l. (*) From (1), xiy* + 5 xy - 14 = 0, (xy + 7 j (xy - 2) = 0. SOLUTIONS OF MISCELLANEOUS TYPES 299 Whence, xy = 2, or xy = — 7. Combining each of these derived equations with (2), we have, (a) xy = 2, (6) xy=-l, x — y = \. x — y = l. Solving these two systems, we find from (a), x = 2, or — 1, y = 1, or - 2 ; from (&), _ lj-3 V33 ^ *- 2 _l±3v3 y 2 3. Solve J fl + l = ? 1 1 = 3 a y 2 Squaring (2), Dividing (1) by (2), Subtracting (4) from (3), Whence, x 2 xy y 2 4 xy y* _3 = 6 xy 4' J L = 1 xy 2 Subtracting (5) from (4) , — - — + - o = 1 . x 2 scy y 2 4 x y 2 Extracting square root of (6) , x Combining (2) and (7), we obtain, x = *' or J ; 1 y = 2, or 1. J Result. 4. Solve *« + y* = 82, se+y=4 Let x = w + v, and y = u — v. Then in (1), (u + t>) 4 + (« - O 4 = 82. (1) (2) (3) (4) (5) (6) (7) (1) (2) 800 SIMULTANEOUS QUADRATIC EQUATIONS Or, m 4 + 4m 3 h 6 u 2 v 2 + 4 uv s + v 4 ' + u l — 4 w 3 t? + 6 rt'V 2 — 4 2fy 3 + w 4 2 w 4 +12 ztV 2 + 2 1? 4 = 82 Whence, w 4 + 6 w 2 v 2 + v 4 = 41. Substituting in (2), w + v + w ~ v = 4. w = 2. Substituting (5) in (4), (2) 4 + 6 (2)V + v i = 41 . Or, tf + 24 1>2 _ 25 = 0. Whence, v = ± 1, or ±5 V- 1. Therefore, if u = 2 and t> ==± J, And, if u = 2 and v = ±5 y/~— 1, x = w-fv = 2±l=3orl, x=u+v=2±5 V^T, y — u — v = 2 + l=lor3. 2/ = w-y-l, ■ ^ + / = 8^ 2 -3^-3. 22 « ^-^2/ + 2/ 2 = 3, 8. xy + x-y = b, fcy(a - 2/) = 6 - 9. ^ 2 + 2/ 2 4-^ + 2/ = 2, a*/-2 = 0. 10. IB 2 — 42/ 2 = a; + 2y, x + 4 ?/ = 7. 11. 3*4-^=^+10, x—y=xy+2. 12. x 4 +ary + 2/ 4 = 19, ar 2 — a*/ 4- 2/ 2 = 7 - 13. x 4 + 2/ 4 -17=0, a;4-y--3 = 0. 14. (z-?/) 2 -a; 2 ?/ 2 -5 = 0, 15 3a; 2 +2«?/-2?/ 2 =14(a;-2/), 2a? + xy-3y 2 = 7(x-y). x — xy+y = l. 23. it 12 6 y .-i-l-o. o, 24. x 2 y 2 16 25. 1 1_7 x* f $' 1_1_1 x y 2' 1 1 1 + ^ = o, 26. x + 2/ » — y 2 3_12 x y xy' + 5 ' 302 SIMULTANEOUS QUADRATIC EQUATIONS GRAPHS OF QUADRATIC EQUATIONS IN TWO VARIABLES TYPE FORMS OF EQUATIONS AND CORRESPONDING GRAPHS (I) Type Form 7 3 J£Be ^•jT ^ -£ tn 2 -\ ZZZZt J )r„± .x -±- — 5 r 1 t V ^E S^ -.^5 _5ta;~*2B jegff K jr 2 +/ = c. Illustration : Plot the graph of x 2 + y 2 = 25. From the equation we have If y = ± V25 - P- jc=0, y= ±' V25, or 5. P. «s=X, y=±V24, or ±4.89+. Pi. jc=2, y=± V2l", or ±4.06+. P 2 . a; =3, y= ± Vie, or ±4. P 3 . »=4, y=±V9, or ±3. etc. Plotting these points, we ob* tain a circle as the graph of the equation, x 2 + y 2 = 25. It will be seen that the coordinates of any point (x, y) are legs of a right triangle whose hypotenuse is the dis- tance from the origin to the point (x, y). That is, for any point on the curve we have (see figure), ^ + 2/ 2 =5 2 , or, x 2 + y 2 = 25. In general, therefore: 342. Tfie graph of any equation in two variables in the form x 2 + y 2 = c is a circle. GRAPHS OF QUADRATIC EQUATIONS 303 (II) Type Form ••• f = ax + c. Illustration : Plot the graph of If x=0, y = ± V8, or ±2.8+. P 2 . x= l,y = ±V&, or ±3.4+. P 3 . x=-l, y = ±V4, or ±2. Pl 35= -2, y=0; etc. P. (Note the enlarged scale.) Plotting these points, we ob- tain a parabola as the graph of the equation, y 2 = 4 x + 8. ¥ ^' *'% IS «3 X P 6^" 2L y £ x ir x PI o V \ V S^vp 1 ^s s 5 ^N V It will be seen that if x is less than — 2, y is imaginary ; hence, no point in the curve lies to the left of P. From this type form and graph we have, in general : 343. The graph of any equation in two variables in the form y 2 = ax + c is a parabola. (Ill) Type Form ••• ax 2 + by 2 = c. Illustration : Plot the graph of ±x 2 + 9y 2 = 36. If y = 0, x 2 = 9, x = ±3. A. x = 0, y 2 = 4, y = ±2. B. Hence, x= + 3 and — 3, the points where the graph cuts XX ; and, y = + 2 and — 2, the points where the graph cuts YY'. 9 zn g B P ^ j^ -7 V +- t \ xi : o i * A^ JA \ Z -S« ^ ^._ -^ B B p Y^ 304 SIMULTANEOUS QUADRATIC EQUATIONS For any other points let x = ± 1. Then, 9 y 2 = 32, y = ± f i/% ttf f =s± 1 .9+. Hence, for P, P if P 2 , and P 8 , we have (1, 1.9), (1, - 1.9), (- 1, 1.9), (— 1, — 1.9) respectively. Plotting these points, we obtain an ellipse as the graph of the equation 4x 3 + 9y 2 = 36. By assuming values sufficiently greater or less than those by which the points above are obtained, it can be shown that no points in the graph can lie to the right or teft, above or below, the intersections with the axes of reference. From this type form and graph we have, in general : 344. The graph of any equation in two variables in the form ax* -\-by 2 = c is an ellipse. (IV) Type Forms • • • ax 2 — by 2 = c and xy = c. Illustrations : 1. Plot the graph of x 2 -4y 2 = l. From the given equation X s. ^^** ^£ P + ' r -S ta -«£_ Hi 5I--A2 x /- Q *s >' s ^ +' £ _ U s s *~ Y*~ Iix = ±l,y 2 = 0,y=0. x = ±2,y*=%,y=±iy/3, or ±.86+. For the values of x we plot A, (1, 0), and 5, (-1, 0). AlsoP, (2, .86); P u (-2, .86); P 2 ,(-2,-.86);P„(2,-.86). With these points we obtain an hyperbola as the graph of the equation, x 2 — 4 y 2 = 1. It will be found by trial that any value of x between -f- 1 and — 1 gives an imaginary value for y, hence no part of the curve can lie between the points A and B. GRAPHS OF QUADRATIC EQUATIONS 305 2. Plot the graph of If x = -4, ••• ••• + 4, etc., y = _l, ••• ±qo ». + l, etc. Plotting the points, (—4, — 1), etc., we obtain an hyperbola whose branches lie in the angles FOX and VOX*. From the two cases : 345. TJie graph of any equation in two variables in the form ax* — by 2 — c, or in the form xy = c, is an hyperbola. Exercise 118 = || | ||| *i till lllilll HH-Iy- 1 1 1 1 1 1 - Plot the graph of : 1. x 2 + y 2 = 16. 2. s 2 + y 2 = 49. 3. y 2 = 4a?4-4. 5. ^ ii-zfiiiifc::::::: i£^l&^ J! -*>— ^PF-s- E^S o, By measurement of the graphs we find the coordinates of P % and P 2 to be (4, 3) and (— -J-|, — -f-f ). Solving the equations, we obtain x — 4, y = 3 ; or x = — -}-|, y = — ^|. The accuracy of this and of subsequent cases may be increased by plotting on a larger scale. 2. Given x 2 + y 2 + 9x + U y 2 =4:X + 16. From the intersections of the graphs of the given equations obtain the roots, and check the results by solving the equations. In the figure the graph of x 2 + y 2 + 9 x + 14 = is a circle and the graph of 2/ 2 = 4# + 16isa parabola. By measurement of the graphs we find the coordinates of P x and P 2 to be (— 3, 2) and (—3,-2) respectively, and these coordinates correspond to the real roots obtained from the solution. For the solution of the system gives x = — 3, y = ±2, or x = — 10, y = ± V— 24. We cannot find a point corresponding to the imaginary root, and we make the important conclusion that : 346. Since there can be no point having one or both coordinates imaginary, the graphs of two equations can have no intersections corresponding to imaginary roots. 3. Given x 2 + 4 y 2 = 17, xy = 2. From the intersections of the graphs of the given equations obtain the roots, and check the results by solving the equations. GRAPHS OF QUADRATIC EQUATIONS 307 Iii the figure the graph of a? 2 + 4 ?/ 2 = 17 is an ellipse, and the graph of xy = 2 is an hyperbola. By measurement of the graphs we find the co- ordinates of P lt P 2 , P 3 , and P 4 tobe'(4, £), (1,2), (-4, -h), (-1, -2) respectively. The solu- tion of the equations gives x= ±4, y= ±i, or x= ±1, |= ±2. It will be noted that this last case is the first in which both equations are homogeneous and of the second degree, and that the graphs serve to emphasize once more the importance that attaches to the association of corresponding values of x and y in the solution of a system of simultaneous quadratic equations. 4. Given 2a,- 2 + 5y 2 =125, 5 x + 2 y = 10. From the intersections of the graphs of the given equations obtain the roots, and check the results by solving the equations. In the figure the graph of 2^+22/2 = 125 is an ellipse, and the graph of 5 x + 2 y = 10 is a straight line. 308 SIMULTANEOUS QUADRATIC EQUATIONS Y "^w ■\ ^< "" m "™ ^ < P z ~ qL 2 §\ T ^ J-^Sr x: o s x ^ I 4 s V -,£ ^ ^ v -*». — tf^ SI By measurement of the graphs we find the coordinates of Pj and 1\ to be (0, 5) and (3.7+, 4.4-) respectively. The solution of the equations gives x = 0, y = 5, or 3 = 3.75+, ?/ =4.38+. 5. Given, a; 2 + f = 100, 3a; + 4?/ = 50. From the intersections of the graphs of the given equations obtain the roots, and check the results by solving the equations. In the figure the graph of x 2 + y 2 = 100 is a circle, and the graph of 3 x + 4 y = 50 is a straight line. Solving the system, we find that the equation resulting from substitution gives equal roots for y. Hence, for x we find but one value, the solution of the system being x = 6, y = 8. It will be seen from the ' graph that the circle and straight line have one point only in common ; that is, the line is tangent to the circle. The coordinates of the point of tangency are found to be (6, 8) or P. In general : 347. If, in the solution of a system of equations, a derived equation has equal roots, the graphs of the equations are tangent to eadi other. It will be found to assist the student in the exercise follow- ing if the type forms of equations and the corresponding graphs are remembered. 1 . ax + by — c, The Straight Line. 2. jr 2 +/ = c, The Circle. 3. / 2 = ax 4- c, The Parabola. GRAPHS OF QUADRATIC EQUATIONS 309 4. ax 2 + by 2 = c, The Ellipse. 5. ax 2 -b/=c,ov\ Th e Hyperbola. 6. xy — c, J Exercise 119 Plot the graphs of the following, and check the roots deter- mined by a solution of each system : 1. x 2 +y 2 = 25, 5. x 2 + y 2 + x + y = 18, x — y=l. xy + x + y = 11. 2. ^+2/2=74, 6. r = 4a + 8, xy = 35. « 4- y — 6 = 0. 3. a; — y = 6. 7. 9 x 2 - 16y 2 = 144, ^4-^ = 36. 4. a? + y = 12, a*/ = 32. 8. 0^+4^ = 4, x 2 + tf '■ = 17. 9. a 2 a 2 4-2/ 2 = -xy = 32, + 2/ 2 = 28. 10. Solve the system, x + y = and a^/ = 4, and determine if the solution, alone will show that the graphs intersect, or do not intersect. 11. How do the graphs of xy = 4 and x l — y 2 — 16 differ in their positions relative to the axes of reference ? 12. Can you describe the position of the graph of the equa- tion, x 2 + y 2 — 2 x = 0, without plotting it ? 13. How do the graphs of the equations, x 2 -f 2y 2 = 32 and 4 # 2 + y 2 = 16, differ in their relative positions ? 14. Plot the graphs of x 2 — xy 4- y> = 28 and x — y = 0, and show that their intersections check the roots found by solution. 15. In how many points may the graphs of x 2 + y 2 = 25 and a,- 2 4- 2 y 2 = 64 intersect? Prove your answer by plotting. 810 SIMULTANEOUS QUADRATIC EQUATIONS 16. Show the positions of the graphs of x 2 + y 2 . = 100 and x 2 + 4?/ 2 = 100, and determine the number of real roots. 17. Show by solution that y 2 — 4 x = 12 and x 2 + y 2 -f 3 x = can have but one point in common, and prove your answer by a graph of the system. PROBLEMS PRODUCING QUADRATIC EQUATIONS 348. In the solution of a problem from which a quadratic equation or equations result we retain only the solution that satisfies the given conditions. As a rule negative results will not ordinarily satisfy the conditions even if they satisfy the equations. Illustrations : 1. If 6 times the number of laborers in a field are increased by the square of the number at work, there will be in all 55 men. How many laborers are there in the field ? Let x t= the number of laborers in the field. From the given conditions, 6cc + « 2 = 55. From which, x 2 + 6 x — 55 = 0. Solving, x as 5, or x = — 11. Clearly the positive result only is retained. Hence, 5 laborers. Result. 2. A company of boys bought a boat, agreeing to pay for it the sum of $ 60. Three of the boys failed to pay as agreed, so each of the others was compelled to pay $1 more than he had promised. How many boys actually paid for the boat? Let x = the number of boys actually paying for the boat. x + 3 = the number of boys first agreeing to share its cost. — = the number of dollars paid by each boy. = the number of dollars each had expected to pay. x + S PROBLEMS PRODUCING QUADRATIC EQUATIONS 311 Then, ■ S>— S»-«i x x + 3 From which, x 2 + 3 x - 180 = 0, and x = 12, or - 15. That is, 12 boys actually shared in the cost of the boat. Many of the problems in the following exercise must be stated by the use of two unknown quantities, and simulta- neous quadratic equations will result. But as far as pos- sible the student should attempt to state most of the earlier examples by means of one unknown number only. Exercise 120 1. Find those two consecutive odd numbers the sum of whose squares is 290. 2. Find a number that, added to 7 times its reciprocal, equals 8. 3. Two factors of 48 are such that one exceeds the other by 2. What are the factors ? 4. The sum of two numbers is 10, and the sum of their squares is 52. Find the numbers. 5. Find two numbers such that their sum and the difference of their squares are each 13. 6. Find the two factors of 600 whose sum is 49. 7. Find two numbers whose sum is 11, and whose product is 17 less than 15 times their difference. 8. The sum of the cubes of two numbers is 126, and the sum of the two equals 6. Find the numbers. 9. Find two numbers the sum of whose squares is 130 and whose product is 63. 312 SIMULTANEOUS QUADRATIC EQUATIONS 10. How many yards of picture molding will be required for a room whose ceiling area is 1200 square feet, the diagonal of the ceiling being 50 feet ? 11. One of two numbers exceeds 30 by as much as the other number is less than 30, and the product of the numbers is 875. Ifind them. 12. The sum of the squares of two numbers equals 13 times the smaller number, and the sum of the numbers is 10. What are the numbers ? 13. If twice the product of the ages of two children is added to the sum of their ages, the result is 13 years. One child is 3 years older than the other. Find the age of each. 14. The diagonal of a rectangle is 100 feet, and the longer side is 80 feet. Find the area of the rectangle. 15. Find three consecutive numbers such that the sum of their squares shall be 194. 16. The simple interest on $600 for a certain number of years and at a certain rate is $ 120. If the time were two years shorter and the rate 2 % more, the interest would be $108. Find the time and the rate of interest. 17. The sum of the squares of two numbers is 2 a 2 + 2, and the sum of the numbers is 2 a. What are the numbers ? 18. The combined capacity of two cubical tanks is 637 cubic feet, and an edge of the one added to an edge of the other equals 13 feet. Find the length of a diagonal on any one face of each cube. 19. The product of two numbers is 15 greater than 5 times the larger number, and is 6 less than 16 times the smaller number. Find the numbers. 20. If a number of two digits is multiplied by the tens' digit, the product is 96; and if the number is multiplied by the units' digit, the product is 64. Find the number. PROBLEMS PRODUCING QUADRATIC EQUATIONS 313 21. Divide 15 into two parts such that their product shall equal 10 times their difference. 22. The difference of two numbers is 1, and the sum of the numbers plus the product is 19. Find the numbers. 23. If the sum of two numbers is multiplied by the less, the product is 5 ; and if the difference of the numbers is multi- plied by the greater, the product is 12. Find the numbers. 24. A garden 40 feet long and 28 feet wide has around it a path of uniform width. If the area of the path is 960 square feet, what is its width ? 25. A dealer would have received $ 2 more for each sheep in a drove if he had sold 6 less for $ 240. How many were there in the drove, and at what price was each sold ? 26. In a number of two digits the units' digit is 3 times the tens' digit, and if the number is multiplied by the sum of the digits, the product is '208. Find the number. 27. A bicyclist starts on a 12-mile trip, intending to arrive at a certain time. After going 3 miles he is delayed 15 minutes and he finds he must travel the remainder of the journey at a rate 3 miles an hour faster in order to arrive at his destination on time. Find his original rate of speed. 28. The sum of the squares of the two digits of a number is 13, and if the square of the units' digit is subtracted from the square of the tens' digit and the remainder is divided by the sum of the digits, the quotient is 1. Find the number. 29. The difference between the numerator and the denomina- tor of a certain improper fraction is 2, and if both terms of the fraction are increased by 3, the value of the fraction will be decreased by -£$. Find the fraction. 30. From the formula, t = vxj- 9 find the length of a pen- es' dulum that vibrates once a second at a point where g = 32.16 feet. 314 SIMULTANEOUS QUADRATIC EQUATIONS 31. A body falls through a space of 3216 feet at a point where g equals' 32.16 feet. From the formula, S = \ gt 2 , determine the number of seconds required for the fall. 32. If an automobile traveled 3 miles an hour faster, it would require 2 hours less time in which to cover a distance of 120 miles. What is the present rate of the automobile in miles per hour ? 33. A certain floor having an area of 50 square feet can be covered with 360 rectangular tiles of a certain size ; but if the masons use a tile 1 inch longer and 1 inch wider, the floor can be covered with 240 tiles. Find the sizes of the different tiles. 34. One leg of a right triangle exceeds the other leg by 2 feet, and the length of the hypotenuse is 10 feet. Find the length of the legs of the triangle. 35. 168 feet of fence inclose a rectangular plot of land, and the area inclosed is 1440 square feet. Find the dimensions of the field. 36. The sum of the squares of two numbers is increased by the sum of the numbers, and the result is 18. The difference of the squares of the numbers is increased by the difference of the numbers, and the result is 6. Find the numbers. 37. If the difference of the squares of two numbers is divided by the smaller number, the remainder is 4 and the quotient 4. If the difference of the squares of the numbers is divided by the greater number, the remainder is 3 and the quotient 3. What are the numbers ? 38. If the length of a certain rectangle is increased by 2 feet, and the width is decreased by 1 foot, the area of the rect- angle will be unchanged. The area of the rectangle is the same as the area of a square whose side is 3 feet greater than one side of the rectangle. What are the dimensions of the rectangle ? CHAPTER XXIV RATIO. PROPORTION. VARIATION. RATIO 349. If a and b are the measures of two magnitudes of the same kind, then the quotient of a divided by b is the ratio of a to b. Ratios are expressed in the fractional form, -, or with the b colon, a : b. Each form is read " a is to b." 350. In the ratio, a : b, the first term, a, is the antecedent, and the second term, b, is the consequent. THE PROPERTIES OF RATIOS 351. The properties of ratios are the properties of fractions; for the ratios, -, m:n, (#,+ y):(x — y), etc., are fractions. b (a) The Multiplication and the Division of the Terms of Ratios 352. The value of a ratio is unchanged if both its terms are multiplied or divided by the same number. 353. A ratio is multiplied if its antecedent is multiplied, or if Us consequent is divided, by a given number. 354. A ratio is divided if its antecedent is divided, or if its consequent is multiplied, by a given number. 315 316 RATIO. PROPORTION. VARIATION (6) Increasing or Decreasing the Terms of a Ratio 355. If a, b, and x are positive, and a is less than b, the ratio a : b is increased when x is added to both a and b. For a + x a = x Q*- a ) ' b + x b b(b + x) And, since a < b, the resulting fraction is positive and the given ratio, - , is increased accordingly. 356. If a, b, and x are positive, and a is greater than b, the ratio a:b is decreased when x is added to both a and b. For the resulting fraction in Art. 355 is negative when a > b. 357. An inverse ratio is a ratio obtained by interchanging the antecedent and the consequent. Thus, the inverse ratio of m : n is the ratio n ; m. 358. A compound ratio is a ratio obtained by taking the product of the corresponding terms of two or more ratios. Thus, mx : ny is a ratio compounded from the ratios, m : n and x : y. 359. A duplicate ratio is a ratio formed by compounding a given ratio with itself. Thus : a 2 : 6 2 is the duplicate ratio of a i b. In like manner, a z : 6 3 is the triplicate ratio of a : b. Exercise 121 1. Write the inverse ratio of a : x ; of m : n ; of 7 : 12 ; of 3x:5x; of (2a + l) : (2a- 1); of (a? + xy + y 2 ) : (x 2 -xy + y 2 ), 2. Arrange in order of magnitude the ratios 2:5, 3:7, 4 : 9, 5 : 8, 10 : 17, 12 : 19, 21 : 27, 32 : 39, and 40 : 51. 3. Compound the ratios 3 : 7 and 10 : 17. 4. Find the ratio compounded of 3 : 8, 4 : 9, and 6 : 11. PROPORTION 317 5. Compound the ratios (x 2 — 9) : (ic 3 + 8) and (x+2) : (x—3). 6. What is the ratio compounded from the duplicate of 2 : 3 and the triplicate of 3 : 2 ? 7. Find the value of the ratio (x + 6) : (x 2 + 7 x + 6). 8. Two numbers are in the ratio of 4:7, but if 3 is added to each number, the sums will be in the ratio of 5 : 8. Find the numbers. 9. The ratio of a father's age to his son's is 16 : 3, and the father is 39 years older than the son. Find the age of each. 10. In a certain factory 5 men and 4 boys receive the same amount for a day's work as would be paid if 3 men and 12 boys were engaged for the same time. What is the ratio of the wages paid the men and the boys individually? PROPORTION 360. A proportion is an equation whose members are equal ratios. Thus, the four numbers, a, 6, c, and d, are in proportion if - = -. b d 361. A proportion may be written in three ways : (1) 2-£. (2) a:b = c:d. (3) a:b::C:d. b d Each form is read " a is to b as c is to d." We understand the mean- ing of a proportion to be that the quotient of a ■*■ b is the same in value as the quotient of c -=- d. 362. The extremes of a proportion are the first and fourth terms. 363. The means of a proportion are the second and third terms. 318 RATIO. PROPORTION. VARIATION 364. The antecedents of a proportion are the first and third terms, and the consequents the second and fourth terms. In the proportion a:b = c:d a and d are the extremes, a and c are the antecedents, b and c are the means ; b and d are the consequents. 365. If the means of a proportion are equal, either mean is a mean proportional between the first and fourth terms. Thus : in a : b = b : c, b is a mean proportional between a and c. 366. The last term of a proportion whose second and third terms are equal is a third proportional to the other two terms. Thus, in a:b = b :c, cisa third proportional to a and b. 367. A fourth proportional to three numbers is the fourth term of a proportion whose first three terms are the three given numbers taken in order. Thus, in a : b = c : d, diss, fourth proportional to a, 6, and c. 368. If, in a series of equal ratios, each consequent is the same as the next antecedent, the ratios are said to be in con- tinued proportion. Thus : a :b = b:c = c:d=d:e = etc. 369. In the treatment of proportions certain relations are conveniently discussed if we recall that - = -=?*. Whence J b d a = br, and c = dr. Substitutions of these values for a and c will be of frequent service in practice. 370. In order that four quantities, a, b, c, and d, may be in proportion, a and b must be of the same kind, and c and d of the same kind. However, c and d need not be of the same kind as a and b. PROPERTIES OF PROPORTION 319 PROPERTIES OF PROPORTION 371 . Given a:b = c:d. Tlien ad = be. Proof : b~ d Multiplying by bd, ad = be. That is: In any proportion, the product of the means equals the product of the extremes. 372. Givena:b = c:d. Then a = — , b = — , etc. d c From the equation ad = be we obtain by division „ be d _bc — » a o = — , c = c ad b That is : Either extreme of a proportion equals the product of the means divided by the other extreme; and either mean of a proportion equals the product of the extremes divided by the other mean. 373. Given a:b = b:c. Then b = -Vac. Proof : a : b = b : c. By Art. 371, b 2 = ac. Extracting square root, b = y/ac. That is: The mean proportional between two numbers is equal to the square root of their product. 374. Given ad = be. Then a : b = c : d. Proof : ad = bc. Dividing by bd, 9l = 9:. b d Or, a:b = c:d. 320 RATIO. PROPORTION. VARIATION That is : If the product of two numbers is equal to the product of two other numbers, one pair may be made the extremes, and the other pair the means, of a proportion. In like manner we may obtain a proportion, a:c=b:d, etc. 375. Given a 6 = : c : d. Then b:a = d:c. Proof : a _c b~ d Then, b d Whence, b_d a c Or, b:a = d:c. . That is : If four numbers are in proportion, they are in proportion by inversion. 376. Given a:b = c:d. Then a:c = b:d. Proof: °l = c -. b d Multiplying by 5, «& = &£. c be cd Whence, ^ = -. c d Or, a : c = b : d. That is : If four 7iumbers are in proportion, they are in proportion by alternation. In applying alternation all four quantities considered must be like in kind. 377. Given a:b = c:d. Then a+b:b = c +d:d. Proof: 2«'5. b d PROPERTIES OF PROPORTION 321 Adding 1 to both members, - + 1 = - + 1. b d Whence, a + V _c + d b d Or, a + b :b = c + d:d. That is : If four numbers are in proportion; they are in proportion by composition. 378. Given a:b = c:d. Then a — b:b = c — d:d. Proof : a _ c b~ d Subtracting 1, |- 1 = ^-1. b d Whence, a — b c — d b d Or, a — b.b = c — d:d. That is : If four numbers are in proportion, they are in proportion by division. 379. Given a:b = c:d. Then a-\-b:a—b = c + d:c — d. Proof: a±b^c±d, (1) b d \„a a — b c — d /ON And, — = — • (2) Dividing (1) by (2), £±| = c - ± ^,- a — b c — a Or, a + b :a — b = c + d:c — d. That is: If four numbers are in proportion, they are in proportion by composition and division. 80M. EL. ALG. 21 322 RATIO. PROPORTION. VARIATION 380. Given a:b = c:d. Then a n :b n = c n : d n . Proof: Sfil 6 d Raising both members to the nth power, 9- — SH. b n ~ d n ' Or, a n :b n = c n : d n . That is : Like powers of the terms of a proportion are in proportion. 381. Given a:b = c:d = e:f= •••. T7iew (a + c -+- e + •••) : (b + d + f+.>.)=a:b. Proof: Since ® = -? = * = ..., b d f ' then, - = r, - = r, - = r, etc. 6 d / Whence, a = &r, c = dr, e =^fr, etc. Adding, a -f c + e + ••• = br + dr +fr + •••. Whence, a + c + e + ••• = (& + d + / + •••)»*. And, q + c + e +- = r=g. Or, ( + c+e + -) :(& + <* +/+•••) = a:&. That is : Jn a series of equal ratios, the sum of the antecedents is to the sum of the consequents as any antecedent is to its consequent. 382. Given a :& = = b c. Then a :c—a 2 : b 2 . Proof : Since a b 6"c' it follows that, « v b _a a b c~b b Whence, a_a 2 c & 2 ' Or, a : c = a 2 : b 2 . APPLICATIONS OF THE PROPERTIES OF PROPORTION 323 That is : If three numbers are in continued proportion, the first is to the third as the square of the first is to the square of the second. 383. Given a:b = b:c = c:d. Then a:d = a 8 :b s . Proof : Since it follows that, « x ^x- = ?x?x a b~ _b_ c c = ~d' c d = <** b a b d~ ~b*' a d = = a* b* b c d b b b Whence, Or, That is : If four numbers are in continued proportion, the first is to the fourth as the cube of the first is to the cube of the second. 384. Given a:b = c: d, e:f=g:h, k:l = m:n. Then aek : bfl = cgm : dhn. Proof: Since §«.£, 5 = 2, *«»« b d f h I n Multiplying, J7i=T- bfl dhn Or, aek : bfl = cgm : dhn. That is : The products of the corresponding terms of two or more pro- portions are in proportion. APPLICATIONS OF THE PROPERTIES OP PROPORTION 385. Of the important properties of proportion in common use in the solution of problems involving certain relations, we may note briefly the following : (1) Since, in any proportion, the product of the means equals the product of the extremes (Art. 371), we may readily find any one term of a proportion when three terms are known. 324 RATIO. PROPORTION. VARIATION (2) Since — = - = r, we have a — br, and c= dr ; and the substitution b d of these values for a and c will be of frequent service in reductions. (3) Composition and division (Art. 379) are of frequent use in mini- mizing the work necessary for the solution of certain types of equations. (4) An assumed identity involving any four numbers, a, 6, c, and d, is shown to be true if, by transformations, we obtain a proportion, a:b = c:d. Illustrations : 1. Find the ratio of x to y when 3a? + 2y = - . In the form of a proportion we have Zx + 2y.4:X-3y = 5:6. By Art. 371, 6 (3* 4 2 y) = 5 (4 x - 3 y) . Whence, 18x4l2*/ = 20x- 15 y. 2x = 27 y. Therefore (Art. 374), x : y = 27 : 2. Result. 2. If a: 6 = c: d, show that a + 3c: 6 + 3<2 = 2a+c : 26+riL Since a : & = c : d, we nave - = - — r. Whence, a = br, and c = dr. 6 d Reducing each ratio separately, a 4 3 c _ (6r) + 3 (dr) = r (b 4 3 y : 6^. 3. 3:5* = 6:10x. 6 ' *>£-*^J? Determine whether the following are true proportions : 7. 34:53 = 19:39. 9. 12.1 : 4.4 = 2.2 : .8. 8. 4£:4 = 8f:8J. 10. ar>-l : x + 1 = x-l : x. Find the value of the unknown in each of the following : 11. 8:10 = 12:». 13. 32:12 = x:6. 12. 4: z = 16: 15. 14. y: 8 = 12: 4. 18. 4c:a = 3sc:6a. 19. 2 am : 3 = mx : 15. 1 20. a-1 326 RATIO. PROPORTION. VARIATION 15. 5: z = 10.5:. 3. 13. 2a:5 = 6:3. 17. 3. 41. 2:8 and 3:5. n 4 n 2 APPLICATIONS OF THE PROPERTIES OF PROPORTION 327 Find a fourth proportional to : 45. 4, 7, and 9. 48. x 2 — 1, x + 1, and x — 1. 46. 3a;, 2 y, and 62. 49. c 3 - d 3 , c 2 - a* 2 , and c 2 -f cd + cP. 47. |, |, and |. 50. 1 + V2, 2 + V2, and 2 - V2. Change the form of each of the following proportions so that the unknown quantity shall occur in but one term : 51. 2:5 = 3 — a?: a;. 56. m: 7i=p — z:z. 52. 5 : 3 = 4 — x : x. 57. 3 : 2 = x + 1 : x. 53. 6 : 7 = 12 — y : y. 58. 15 : 7 = x + 1 : a;. 54. 4a : 3a = 10 — a;: #. 59. 3a : 5 c = x 4- 1 : x — 1. 55. a 3 : a 2 c = c — a;: x. 60. a 2 +l : a 2 — l = #-}-l : &— 1. 61. c-\-d:c — d = c + y:c — y. 62. m 2 4-m + l:m 2 — m — l = a; — 2:a; — 8. 63. Find two numbers in the ratio of 2 : 3, such that if each is decreased by 1 their ratio becomes as 3 : 5. 64. Find two numbers in the ratio of 4 : 7, such that if each number is increased by 2 the ratio becomes as 5 : 8. 65. If ad = bc, write all the possible proportions whose terms are a, b, c, and d. 66. Find two numbers whose sum is to their product as 8 : 15, and whose sum is to their difference as 4:1. 67. What number must be added to each of the numbers, 4, 14, 10, and 30, that the resulting sums may be in proportion ? 68. Separate 10 into two parts such that their product shall be to the difference of their squares as 6 : 5. 69. Two rectangles have equal areas, and their bases are to each other as 5 : 16. What is the ratio of their altitudes ? 70. The lengths of the sides of three squares are in the ratio of 2, 3, and 4. Find a side of each if the sum of the three areas is 725 square units of area. 328 RATIO. PROPORTION. VARIATION 71. If a : b = 2 : 3, and b : c = 3 : 5, find the ratio of a : c. 72. If x : y = 3 : 4, and y : z = 5 : 6, find the ratio of a? : z. 73. If m : n = 3 : k, and n : p = k : 4, what is the ratio of m : p? 74. If c : d = 3.2 : 4.8, and d : fc = 3.2 : 8, find the ratio of c : k. 75. If a : b = ft : c = 2 a; : 3 y, find the ratio of a to c in terms of x and ?/. 76. If a : ft = 3 : 4, find (3a + 2&) : (2a + 36). 77. If m : w = 5 : 2, find (2 m + n) : (2 m — ?i). 78. If a : b = 2x : Sy, find (2 a + #) : (2 a — #) in terms of b and y. 79. If a + b : b = x-\-y : y, find (# + 2/) : (®—y) in terms of a; and y. 80. If a; -|- a : # — a = y + x : y — z, find (# -f ?/) : (x — y) in terms of a and z. If a : b — c : d prove that : 81. ab:cd = b 2 : d\ 82. a + c : b + d = c : d. 83. a 2 d:b = be 2 : d. 84. 3a 2 -26 2 :26 2 = 3ac-26d: 2bd. 85. Va 2 -6 2 : Vc 2 -d 2 = 6 : d 86. a 2 + 2a& :a& = c 2 + 2cc?: cd If a : b = b : c, prove that : 87. ab-b 2 : bc-c? = b 2 : c 2 . 88. a — b : 6 — c = 6 : c. 89. a\a + 6) : c(b + c) = a 2 (a - 6) : c(b- c) . 90. a + b : & + c=Va 2 -& 2 : V& -c 2 . APPLICATIONS OF THE PROPERTIES OF PROPORTION 329 Solve the equations : 91. 2a + 3 : 2x + 5=3a+2 : 3a; + 4. 92. 2^ + 2^ + 1 : 2a 2 -2# + l:=a + l : x-1. 93. x + m : x — m = m-\-n : m — n. 94. x 2 -l:x + l=x 2 + l:x-l. 95. x 2 + 2x + 4 : : x*-x-l = x + 2 : x-2. a? + x-l x?-x + 2 96. 97. 98. 99. 100. a^-z + a ^ + x _2 a; + n + 1 _ 3#— n+1 ' x + n — 1 3 a; — n — 1 s 2 + 2a? + 3 == s 2 + 3a;-4 aj 2 + 3a; + 2 z 2 + 4a;-2' 3s s + o; + l = 3s 2 — a;— 1 4:X? + X+1 4ar*— a;— 1 5_Va7+2 6- Vaj + 1 4 + Va; + 2 3+Va + l 101. Separate 100 into three parts which shall be in the Tatio of 2 : 3 : 5. 102. Three angles of a certain triangle are in the ratio of 1:2:3. If the sum of the angles of a triangle is 180°, find the number of degrees in each of the angles. 103. The sum of three sides of a triangle is 240 feet, and the ratio of the sides is as 3:4: 5. Find the length of each side. 104. For what value of a will the quantity a — 1 be a mean proportional between the quantities a + 3 and a + 2 ? 105. If 2 is subtracted from the smaller of two numbers and 4 is added to the larger, the ratio is as 3 : 5, but if 1 is subtracted from the greater while 1 is added to the smaller, the ratio is as 10 : 9. Find the numbers. 330 RATIO. PROPORTION. VARIATION 106. Two trains approach each other between two points 100 miles apart, and their rates of traveling are as 4 : 5. How many miles will each have traveled when they meet ? 107. Find two numbers whose sum is 10, such that the ratio of the sum of their squares to the square of their sum is 13:25. 108. Find the length of the two parts into which a line I inches long is divided if the parts are in the ratio of a : b. 109. What must be the value of x in order that 2x — 1, 2 x -f 1, 2 x + 5, and 2 (2 x + 1) may be in proportion. 110. Find each side of a triangle whose perimeter is n inches, the ratio of the sides being as a : b : c. . VARIATION 386. A variable is a quantity that, under the conditions of a problem, may have many different values. 387. A constant is a quantity that, in the same problem, has a single fixed value. 388. If the ratio of any two values of a given variable equals the ratio of any two corresponding values of a second variable, then the first quantity is said to vary as the second quantity. Illustration : Suppose an automobile is running at a speed of 10 miles per hour. The total distance traveled at the end of any hour depends upon the total number of hours that have elapsed since the start was made. If it runs 6 hours, the distance covered is 60 miles ; If it runs 9 hours, the distance covered is 90 miles. Clearly, therefore, the ratio of the two periods of time is the same as the ratio of the two distances covered. That is : 6 : 9 = 60 : 90. We may say, therefore, that the distance (d) varies as the time ($)» or d cc t. The symbol cc is read "varies as." VARIATIONS UNDER DIFFERING CONDITIONS 331 389. If xccy, then x equals y multiplied by some constant quantity. Proof : Let a and b denote any one pair of corresponding values of x and y. Then by definition, - = -. y b From which, % = - y • b Denoting the constant ratio, -» by c, x = cy. b In general : We may change any variation to an equation by the introduc- tion of the constant factor or ratio. 390. If one pair of corresponding values for the variables in a given variation is known, the constant ratio is readily obtained. Illustration : If x varies as y, and x = 12 when y = 3, find the value of x when y = 10. We have x — cy. Substituting, 12 = c. x 8. Whence, c = 4. Hence, when y = 10, x = 4 y = 4x10 = 40. Result. VARIATIONS UNDER DIFFERING CONDITIONS (a) Direct Variation 391. The simple form, xccy, is a direct variation. Illustration : The distance covered by a train moving at a uniform rate of speed varies directly as the time elapsed. That is : d x t or d = ct, where c is the constant ratio. 832 RATIO. PROPORTION. VARIATION (6) Inverse Variation 392. A quantity is said to vary inversely as another quantity when the first varies as the reciprocal of the second. ■ Illustration: The time (£) required by an automobile going from New York to Phila- delphia varies inversely as the speed (s) ; for if the speed is doubled, the time required will be but one half the former time. That is : 1 c tec- or t = -, where c is the constant ratio. s s I (c) Joint Variation 393. If a quantity varies as the product of two quantities, the first quantity is said to vary jointly as the other two quantities. Illustration: The number of dollars (N) paid to a motorman for a certain number of trips (£) varies jointly as the number of dollars paid to him for one trip (w) and the number of trips made. That is : Nccnt or ST as cnt, where c is the constant ratio. (d) Direct and Inverse Variation 394. One quantity may vary directly as a second quantity, and, also, inversely as a third quantity. In such a case the quantities are said to be in direct and inverse variation. Illustration : In mowing a field, the time required for the work (t) varies directly as the number of acres in the field (A), but inversely as the number of men engaged at the task (n) . That is : A cA t oc — or t = — , where c is the constant ratio. n n VARIATIONS UNDER DIFFERING CONDITIONS 333 (e) Compound Variation 395. The result obtained by taking the sum or the differ- ence of two or more variations is a compound variation. Illustration : If y equals the sum of two quantities, a and b, and if a varies directly as x 8 while b varies inversely as x 2 , then c' a = ex 8 and b = — x 2 c' Adding (since y = a + 6), y = ex 8 + — . x' 2 * It is to be noted that in such cases tw% different factors are necessary. (/) Important Principle 396. If x depends only upon y and z, and if x varies as y ivhen z is constant, and x varies as z when y is constant, then x varies as yz when both y and z vary. Let x, y, z (1), Xi, yi, z (2), and x 2 , jfi, z\ (3) be three sets of corre- sponding values. Then, Since z has the same value in (1) and (2), Since y x has the same value in (2) and (3), Multiplying (I) by (II), Or, That is : The ratio of any two values of x equals the ratio of the corre- sponding values of yz, and, by definition, x varies as yz. Illustration : The area (A) of a rectangle varies as the base (B) when the height (H) is constant, and the area varies as the height when the base is constant. Therefore, when both the base and height vary, the area will vary as the base and height jointly. *i y\ (I) X\ _ z X 2 9% (II) x _ yz x 2 y\z\ x-.x 2 = yz: y x z\. 334 KATIO. PROPORTION. VARIATION I II III IV 8 10 8 . 12 5 10 5 IS A = BH = 105 = 50 A =[BH = 12.5 = 60 A = BH = 10-8 = 80 A = BH = 12-8 = 96. In II. B changes, H constant, A changes (AccB, H constant). In III. H changes, B constant, A changes (AxH, B constant). In IV. B changes, H constant, A changes (AccBH, Band H vary). APPLICATIONS OP VARIATIONS 397. Illustrations: 1. If x varies as y 2 , and x = 2 when y — 4, find x when y = 16. Since xocy 2 , we have a; = my 2 . And, if x = 2 and y = = 4, 2 = m x 4 2 . Whence, m = £. Then, when y = 16, x = K16) 2 . » = H*. x = 32. Result. 2. If x varies inversely as y*, and x = 2 when y = 4, find a; when y =2. Since xoci-, y 8 we have Whence, 2 =P' and m = 128. Then, when y = 2, 23 x = 16. Result. APPLICATIONS OF VARIATIONS 335 3. If s is the sum of two quantities, one of which varies directly as x 2 and the other inversely as x, and if s = 6 when x = 2, and s = 2 when x — — 2, find s when a; = — 1. Let w and v represent the two quantities. Then & = w + t>; uccx 2 ; floe-* sacs 2 +-- X From which, s = mx 2 + -• (1) Substituting : Ifs = 2, 2 zJ (2) If x = - 2, 2 = m(-2)2+i = 4m- — 2 n 2 (8) From (2) and (3), at =s 1, fi s 4. If x = — 1, in (1), S = l(_l)2 + _£_ = 1-4 = - 3- Result. 4. The volume of a sphere varies as the cube of its diame- ter. If three metal spheres whose diameters are 6, 8, and 10 inches, respectively, are melted and recast into a single sphere, what is its diameter ? Let V denote the volume of the required sphere, and D its diameter. Then Foe 2)3. Whence, V= mD 8 . (1) Denote the volumes of the three given spheres by V u V 2 , and Vs. We have, therefore, Vi = wi(6) 8 = 216 m, F 2 = m(8) 8 = 512m, F 3 = m(10) 8 = 1000 m, Whence, by addition, V\ + V 2 + V s - 1728 m. But, by the conditions, V x + V 2 + V* = V= m 7) 3 . (From 1.) 336 RATIO. PROPORTION. VARIATION Hence, mi) 3 =1728 m, D 3 = 1728, D =12. Result. That is, the diameter of the sphere obtained from the given spheres is 12 inches. Exercise 123 1. If a; varies as ?/, and x = 10 when y — 2, find, x when y = 5. 2. If a; varies as y, and # = 3.2 when y = 0.8, find a; when y = 5.6. 3. If x + 1 varies as y — 1, and cc = 6 when y = 4, find 03 when y = 7. 4. If 2 a; — 3 varies as 3 2/ + 2, and ?/ = 2 when x — 0.2, find y when a; = 1.5. 5. If x 2 varies as y 2 , and x = 3 when y = 2, find y when a; = 4. 6. If x varies inversely as y, and x = 2 when 2/ = 4, find a; when y = S. 7. If a; varies inversely as y 2 , and x = 2 when y = ^, find y when a; = 1£. 8. If x varies jointly as y and z, and x = 3 when y = 4 and 2! = 2, find x when ?/ = 5 and 2 = 4. 9. If a; varies inversely as y 2 — 1, and a; = 4 when y =5, find a; when ?/ = 15. 10. If a; varies as -, and y = — 2 when a? = 7, find the equa- tion joining aj and y. 11. If the square of x varies as the cube of y, and if x— 6 when 2/ = 4, find the value of jf when x = 30. 12. If s is the sum of two quantities, one of which varies as x while the other varies inversely as x ; and if s = 2 when a? = £ and s = 2 when a; = — 1, find the equation between s and x. APPLICATIONS OF VARIATIONS 337 13. If w varies as the sum of x, y, and z, and w = 4 when x = 2, y = — 2, and 2 = 5, find a; if w = — 3, 2/ = 2, and 2= -6. 14. Given that s = the sum of three quantities that vary as x, x 2 , and se 3 , respectively. If x = 1, s = 3 ; if a; = 2, s = 6 ; and if x = 4, s = 16. Express the value of s in terms of #. 15. The area of a circle varies as the square of its diameter. Find the diameter of a circle whose area shall be equivalent to the sum of the areas of two circles whose diameters are 6 and 8 inches respectively. 16. The intensity of light varies inversely as the square of the distance from the source. How far from a lamp is a certain point that receives just half as much light as a point 25 feet distant from the lamp ? 17. The volume of a sphere varies as the cube of its diameter. If three spheres whose diameters are 3, 4, and 5 inches, respec- tively, are melted and recast into a single sphere, what is the diameter of the new sphere ? 18. The volume of a rectangular solid varies jointly as the length, width, and height. If a cube of steel 8 inches on an edge is rolled into a bar whose width is 6 inches and depth 2 inches, what will be the length of the bar in feet ? 19. If the amount earned while erecting a certain wall varies jointly as the number of men engaged and the number of days they work, how many days will it take 4 men to earn $ 100 when 6 men working 9 days earn $ 135 ? 20. The pressure of the wind on a plane surface varies jointly as the area of the surface and the square of the wind's velocity. The pressure on a square foot is one pound when the wind is blowing at a rate of 15 miles an hour. What will be the velocity of a wind whose pressure on a square yard is 81 pounds ? SOM. EL. ALG. — 22 CHAPTER XXV • PROGRESSION ARITHMETICAL PROGRESSION 398. A series is a succession of terms formed in accordance with a fixed law. 399. An arithmetical progression is a series in which each term, after the first, is greater or less than the preceding term by a constant quantity. This constant quantity is the common difference. 400. We may regard each term of an arithmetical progres- sion as being obtained by the addition of the common difference to the preceding term; hence, An increasing arithmetical progression results from a positive common difference, and a decreasing arithmetical progression re- sults from a negative common difference. Thus : 1, 5, 9, 13, . . . etc., is an increasing series in which the common differ- ence is 4. 7, 4, 1, —2, . . . etc., is a decreasing series in which the common differ- ence is —3. 401. In general, if a is the first term, and d the common difference, a, a + d, a + 2 d, a 4- 3 d, a + 4 d, . . . etc., is the general form of an arithmetical progression. 338 ARITHMETICAL PROGRESSION 339 402. The nth Term of an Arithmetical Progression. In the general form, a,a + cf.fl + 2d,a + 3d,a + 4d ) . . . etc., it will be seen that the coefficient of d in any term is less by 1 than the number of the term. Hence, the coefficient of d in the nth term is n — 1. Therefore, if a = the first term of an arithmetical progression, d = the common difference, n = the number of terms in the series, I = the last term (that is, any required term), then, 1 = a + (n - l)d. (I) 403. The Sum of the Terms of an Arithmetical Progression. If s denotes the sum of the terms of an arithmetical progression, we may write the following identities, the second being the first written in reverse order : s = a + (a + d) + (a + 2 d) + + (I - 2d) + (* - &) + I 8 = 1 + (l-d) + (l-2d) + • + (a + 2d) + (a + d) + a. 2 s = (a + + (a + I) + (a + I) + .-• + (a + 1) + (a + I) + (a + I). Whence, 2s = n(a + l)> Or, s = 5(a + l). (II) 404. Combining the two formulas, (I) and (II), we obtain the following convenient formula for finding the sum of an arithmetical progression when the first term, the common difference, and the number of terms are known. That is : s = 5[2a + (n-l)d]. (Ill) z 405. The first term (a), the common difference (d), the number of terms (n), the last term (I), and the sum of the terms (s), are the elements of an arithmetical progression. APPLICATION OF THE FORMULAS FOR ARITHMETICAL PROGRESSION 406. Illustrations : 1. Find the 10th term and the sum of 10 terms of the arithmetical progression, 1, 4, 7, 10, . . . 340 PROGRESSIONS We have given, a — 1, d = 3, n = 10. In the formula, I = a + (n — l)d. Substituting, I = 1 + (10 - 1) (3) = 28, the required 10th term. In the formula, s = - (a + Z)- 2 Substituting, s = — (1 + 28) z = 5 x29 = 145, the required sum of 10 terms. 2. The first term of an arithmetical progression is 3, the last term, 38, and the sum of the terms, 164. Find the series. We have given, a = 3, I = 38, s = 164. Then, s = - (a + Z) or 164 = | (3 + 38), whence n = 8. Also, Z = a + (w - 1) d, or 38 = 3 + (8 - 1) d, whence d = 5. Therefore, 3, 8, 13, 18, 23, 28, 33, 38, is the required series. 3. How many terms of the series, 2, 5, 8, •••, will be re- quired in order to give a sum of 126 ? We have a - 2, d = 3, s = 126. Then Z = a+(n-l)d = 2 + (»-l)3 = 2 + 3n-3 = 3n-l. This value for Z in terms of n is substituted in the formula S =|(a + Z). Whence, 126 = 2 (2 + 3 n - 1), or 3 n 2 + w = 252. 2 Solving this quadratic equation in n, we have (using the quadratic formula), n = -l ± VrT3024 - n = -L±JL5, M = 9)0 r-9|. 6 6 Therefore, the required number of terms for the given sum is 9. Formula III (§ 404) may be used here without finding Z. ARITHMETICAL PROGRESSION 341 Exercise 124 1. Find the 16th term of 7, 10, 13, .... 2. Find the 10th term of 2, - 1, - 4, .... 3. Find the 12th term of 4, 3.2, 2.4, •••. 4. Find the 10th term of J, f, J, .... 5. Find the -20th term of f, f, |j ..... 6. Find the 10th term of - 12, -9.5, - 7, -4.5, .... 7. Find the 14th term of x + 1, x + 3, x + 5, •••. 8. Find the 11th term of x — 5 a, x — 4 a, #— 3 a, •••. 9. Find the 10th term of 4 + V2, 3 + 2^2, 2 + 3V2, .... 10. Find the 358th term of .0075, .01, .0125, .015, .... Find the sum of : 11. 3, 8, 13, 18, ... to 24 terms. 12. -2, -9, -16, ... to 12 terms. 13. .25, .3, .35, ... to 40 terms. 14 - i>lbih -to 16 terms. 15. 20V2-10V3, 18V2-9V3, 16V2-8V3, *.- to 21 terms. Find the first term and the common difference when : 16. s = 297, w = 9, 1=16. 19. Z = 0, s = 100, n = 25. 17. s = 294, 7i = 12, 1 = 41. 20. n = 6, s = 20 A, I = 4.9. 18. 7i = 13, s = 260, Z = -f. 21. s = 0, Z = 34.5, w=24. Find the common difference when: 22. a = 4, ? = 40, n = 13. 24. I = 2, ti = 7, s = 19.25. 23. s = -27.5, = 4.5,71=11. 25. Z = .97, a = .8, s = 35.4. 26. a = .08, s = — 25, 7* = 25. 27. 7i = 30, a=-13V2, Z = 16V2. 342 PROGRESSIONS Find the first term when : 28. n == 12, I = 10, s = 60. 29. d = - 1, n = 10, s = - 100. 30. d = - .6, n = 14, Z = - .83. Find the number of terms when: 31. a = 7, d = -3, Z = -23. 32. Z = f, a = }, s = 17. 33. a = -|,Z = -A,d = -^. How many consecutive terms of 34. 2, - 3, - 8, •-. will give a sum of - 205 ? 35. 1, 1J, If, ••• will give a sum of 831 ? 36. — 3, - 3 }, — 4, • • • will give a sum of — 97£ ? 37. 0.36, 0.32, 0.28, ••• will give a sum of -.4 ? 38. 12V3, 10V3, 8V3, will give a sum of 0? THE DERIVATION OF GENERAL FORMULAS FROM THE FUNDAMENTAL FORMULAS 407. From the fundamental formulas (I) and (II) in Arts 402 and 403 we may derive a formula for any desired element in terms of any other three elements. Illustrations : 1. Given a, w, and s; derive a formula for d. From the fundamental formulas, l = a+(n — l)d and s=-(a + J), Ss we must eliminate the element I ; and by substituting the value of I from the first formula for I in the second formula, we have s =-[a + a + (n- l)d]. ARITHMETICAL PROGRESSION 343 From which, 2 s = ?i[2 a + (n — l)d], 2 s = 2 a?i + w(w — l)a*, 2 s — 2 an = n(» — l)d", d _ 2(g-an)^ the required formula for d. n(n — 1) 2. Given d, I, and s, find a formula for n. From (I), Art. 402, I = a + (n - l)rZ. Hence, a = Z — (ft — l)a\ Substituting this value for I in (II), Art 403, we have, « = |[2 a+ (n- l)d] * =^[2Z-2(n-l)d+(n-l)d]. 2 2s = 2 »Z-n(n-l)d. on 2 - (2Z + d>-f 2s = 0. Simplifying, Whence, Solving for n, 2 Z + d ± V(2 Z + d) 2 - 8 ds. _ 1. Given 2. Given 3. Given 4. Given 5. Given 6. Given 7. Given 8. Given 9. Given 10. Given 11. Given 12. Given Exercise 125 dj ft, and I ; derive a formula for a. n, Z, and s a, ft, and s a, I, and n a, ft, and s a, I, and s dj I, and n n, ?, and s a, Z, and s dj ft, and s a, c?, and s a, d, and s derive a formula for a. derive a formula for I. derive a formula for d. derive a formula for d. derive a formula for d. derive a formula for s. derive a formula for d. derive a formula for n. derive a formula for a. derive a formula for I. derive a formula for ft. 344 PROGRESSIONS ARITHMETICAL MEANS 408. If we know a and b, the first and last terms respec- tively in an arithmetical progression, we may form an arith- metical progression of m + 2 terms by inserting m arithmetical means between a and b. Illustration : Insert 7 arithmetic means between 3 and 43. We seek an arithmetical progression of (m + 2) = (7 + 2) = 9 terms. (For two terms, the first and the last, are given.) Therefore, a = 3, I = 43, n = 9. It remains to find d. In the formula, 1= a+ (n — l)d. Substituting, 43 = 3 + (9 — 1) — => — —•> ••• . J V2 V2 2V2 20. Insert 3 geometrical means between a~* and a 4 , and find the sum of the resulting series. o x 21. Insert 5 geometrical means between — 3 and — and find the sum of the series. * 22. What must be added to each of the numbers, 5, 11, 23, that the resulting numbers may be in geometrical progression ? 23. The sum of the first 3 terms of a decreasing geomet- rical progression is to the sum to infinity as 7 : 8. Find the common ratio. 24. If the numbers, x — 2, 2x — l, and 5 x + 2, are in geo- metrical progression, what is the common ratio of the series ? 25. The sum of the first 8 terms of a geometrical pro- gression is equal to 17 times the sum of the first 4 terms. Find the common ratio. 26. The population of a certain city is 312,500, and it has increased uniformly by 25 % every 3 years for a period of 12 years. What was the population 12 years ago ? GEOMETRICAL PROGRESSION 357 27. A ball on falling to the pavement rebounds ^ of the height from which it was dropped, and it continues to succes- sively rebound ^ of each preceding distance until it is at rest. If the height from which it originally fell was 60 feet, through how great a distance does it pass in falling and rebounding ? 28. What is the condition necessary that a + 1, a-j-S, a+ 7, and a + 15 shall be in geometrical progression? For what value of a is this condition true ? 29. Show that if 4 numbers, m, n, x, and y, are in geo- metrical progression, then m + n, n + a?, and x + y are also in geometrical progression. 30. A sum of money invested at 6 % compound interest will double itself in 12 years. What will be the amount of $10 invested at compound interest at the end of 60 years ? 31. If 4 numbers are in geometrical progression, the sum of the 2d and 4th divided by the sum of the 1st and 3d is equal to the common ratio. 32. Find 3 numbers in geometrical progression whose sum is 21, and the sum of whose squares is 189. 33. Find an arithmetical progression whose first term is 2, and whose 1st, 3d, and 7th terms are in geometrical pro- gression. 34. If the alternate terms of a geometrical progression are removed, the remaining terms are in geometrical progression. 35. Find to infinity the sum of the series X + 1 \X + 1J \x + 1 36. Find the 4th term of an infinite decreasing geometric series the sum of whose terms is ff> and whose first term is .25. CHAPTER XXVI THE BINOMIAL THEOREM. POSITIVE INTEGRAL EXPONENT 423. A finite series is a series having a limited number of terms. 424. The binomial theorem is a formula by means of which any power of a binomial may be expanded into a series. 425. By actual multiplication we may obtain : (a + by = a 2 + 2 ab + 6 2 . (a + 6)3 = a 3 + 3 a 2 6 + 3 a& 2 + b s . (a + by = a* + 4 a 3 & + 6 a 2 6 2 + 4 a& 3 + 6* ; etc. In the products we observe : 1. The number of terms exceeds by 1 the exponent of the binomial. 2. The exponent of a in the first term is the same as the exponent of the binomial, and decreases by 1 in each succeed- ing term. 3. The exponent of b in the second term is 1, and increases by 1 in each succeeding term until it is the same as the expo- nent of the binomial. 4. The coefficient of the first term is 1, the coefficient of the second term is the same as the exponent of the binomial. 5. If the coefficient of any term is multiplied by the exponent of a in that term, and the product is divided by the exponent of b in that term increased by 1, the result is the coefficient of the next following term. 358 THE BINOMIAL THEOREM 359 426. By observing these laws we may write the expansion of (a + 6) 4 thus: / , ni i ( .-ijn'*'*Ji'j 4.3-2^,3 . 4.3.2.1 M (a + 6)4 = a 4 + 4 «3 6 + __ a 2 6 2 + ^^a&B + j^-M In like manner, we may write the expansion of (a + b) n in the form : (a + &)• = a» + na-ift + n (" - 1) a n-2 6 2 + n(n-l)(n- 2) g ^ ft , + ... 1 «2 1 • 2« o This expression is the binomial formula, and we will now prove that it is a general expression for any power of (a -+- 6), for positive integral values of n. PROOF OP THE BINOMIAL THEOREM FOR POSITIVE INTEGRAL EXPONENTS 427. We have shown by actual multiplication that the laws governing the successive expansions of (a-f b) are true up to and including the fourth power. If, now, we assume that the laws of Art. 425 are true for any power, as the nth power, and if, furthermore, we show the laws to hold for the (n -f- l)th power, then the truth of Art. 425 for all positive integral values of n is established. This method of proof is known as mathematical induction. Both members of the formula (1) below are multiplied by (a + 5). (a + by = a» + nfl»-*6+ n(n " ^ «-%* 4- n < n ~ &g ~ 2) a»-'6»+ - (1) 1 • 2 1 * 2 • o (a + 6) (a + 6) (a + &)•+! = *h-i + na»6 + "("-^ o-iy + ^- 1 )(^- 2 ) a a-2 & 3 + ... 1 • 2 1 « 2 • o + a*6 + wa"- 1 ^ 2 + w(n-l) a«- 2 6 3 + - 1*2 (a + 6)«+i = tfH-i + (n + l)a»6 + f" 71 ^- 1 ) + »1 a »-i&» r n (n-l)(n-2) n(n -1)"1 w _ 2&8 , ... ^L 1-2-3 1-2 J 360 THE BINOMIAL THEOREM m tfH-1 + (n + l )a - 6 + QL±^a»-W + (» + 1 M;- 1) ^268. + .... (2 ) 1 • 2 1 • 2 * O It will be observed that (2) is identical with (1) excepting that every n of (1) is replaced by n -j- 1 in (2). That is, we assumed the laws of 425 to be true for n, and have shown them to be true for n + 1. Similarly, we might show the laws to be true for n -f- 2, and so on, indefinitely. Hence, the laws of Art. 425 being true for the 4th power may be shown to hold true for the 5th power ; and holding for the 5th power, may be shown to hold for the 6th power. Therefore, for any positive integral values of n : (a + b)» = a« + na*-*b + HCEzJQ a «-2 b 2 + n(n-l)(n-2) ari _ 31)8 1-2 1-2-3 428. The Factorial Denominator. In practice it is convenient to write (_3_ f or 1 • 2 • 3, [6_ for 1 . 2 • 3 • 4 • 5 • 6, etc. We read [3 as "factorial 3," [6_ as "factorial 6," etc. In general, \n_ means the product of the natural numbers 1 . 2 • 3-4-5 • • • n inclusive. 429. An expansion of a binomial is a finite series when n is a positive integer. For, in the coefficients, we finally reach a factor, n — n, or 0. And the term containing this factor disappears. Moreover, every following term contains this factor, hence each term following disappears. 430. The Signs of the Terms in an Expansion. If both a and b are positive in (a + 6) n , the signs of all the terms in the expansion are positive. If b is negative, that is, given (a— b), all terms involving even powers of — b are positive, while all terms involving odd powers of — b are negative. Therefore, the signs of the terms of the expansion of (a— b) n are alternately + and — . APPLICATIONS OF THE BINOMIAL FORMULA 361 APPLICATIONS OF THE BINOMIAL FORMULA 431. Illustrations : 1. Expand (2a + 3 6) 5 . (2 a + 3 by = (2 ay +5(2 a)* (8 &) + ^J (2 a )«(8 6) 2 + f^| (2 a)«(S 6)3 1 • J 1 • 1 • o 1.2.8*4 N yv 1.2.3.4.5^ ' = 32 a 5 + 240 a 4 6 + 720 a 8 6 2 + 1080 a*& 4- 810 a6* + 243 b 5 . Result. 2. Expand /^-^Y. Changing to a form best suited to the binomial formula, (2 er*-af)«=(2 a-!) 6 -6(2 a-*)*(rft) + 15(2 a- 1 )*(a*) 2 -20(2 a- 1 ) 8 ^)* + 15 (2 a-i)2( a t)4 - 6 (2 a- 1 ) (a*) 5 + (a§)« = 64 a" 6 - 192 a- 5 at + 240 a-*a% - 160 a~ 3 a 2 + 60 a~H% - 12 a-^r+a* = 64_192 + 240_160 + 60a |_ 12a | + a4 ^ a + x- 1 ) 10 . 32. X s in (^ + 3 aj"Y- 30. a 12 in (2 *" -jy ' 33. ar 16 in A* or 2 - |Y° • 31. xmfex- ?Y 3 • 34. x~» in f 3 a;" 1 - ~r J • CHAPTER XXVIT LOGARITHMS 433. By means of the exponents, 2, 3, 4, etc., we may express certain numbers as exact powers of 10. Thus, 100 = 10 2 , 1000 = 103, 10000 = 10*, etc. 434. Suppose, now, that 10 is given any real exponent, as x. Then some positive number, N } results as the #th power of 10. That is, N=ltf. 435. If, therefore, we knew the necessary approximate values for x, we might express any number as an approximate power of 10. 436. By a method of advanced algebra, these approximate values for x have been obtained. For example, it has been found that 180 = io 2 - 2663 , 4500 = 103-6532, 19600 = lO*^ 23 , etc. 437. These exponents are called the logarithms of the num- bers they produce. 438. The exponent that must be given 10 in order to produce a required number, JV", is called the logarithm of N to the base 10. The expression 10* = N is usually written x = logic -ZV» and is read, M x is the logarithm of N to the base 10." 364 THE PARTS OF A LOGARITHM 365 The object and use of logarithms is to simplify numerical work in the processes of multiplication, division, involution, and evolution. 439. Logarithms to the base 10 are known as common log- arithms, and are in universal use for numerical operations. Unless otherwise stated, the discussions of this and subsequent chap- ters refer to common logarithms. Any positive number, except unity, may be taken as the base of a system of logarithms. A negative number is not considered as having a logarithm. THE PARTS OF A LOGARITHM 440. Consider the results in the following : IO 3 = 1000, log 1000 = 3, IO 2 = 100, log 100 = 2, 101 = io, log 10 =1, 10° =1, logl =0, 10-i = .1, log.i = _l, IO" 2 = .01, log .01 = - 2, IO" 8 = .001, log .001 = - 3 ; etc. From these results it is evident that (1) The common logarithm of a number greater than 1 is positive. (2) The common logarithm of a number between and 1 is negative. » (3) The common logarithm of an integral or a mixed number between 1 and 10 is + a decimal, between 10 and 100 is 1 + a decimal, between 100 and 1000 is 2 + a decimal, etc. (4) The common logarithm of a decimal number between 1 and 0.1 is — 1 -f- a decimal, between 0.1 and 0.01 is — 2 + a decimal, between 0.01 and 0.001 is — 3 4- a decimal, etc. 366 LOGARITHMS 441. The integral part of a logarithm is the characteristic. The decimal part is the mantissa. In log 352 = 2.5465, 2 is the characteristic and .5465 is the mantissa. I. To obtain the characteristic of a logarithm. (a) When the given number is integral or mixed. 442. By Art. 440 (3), the characteristic of the logarithm of a number having one digit to the left of the decimal point is 0, of a number having two digits to the left of the decimal point is 1, and of a number having three digits to the left of the decimal point is 2. In general, The characteristic of the logarithm of a number greater than unity is 1 less than the number of digits to the left of the decimal point. (b) When the given number is a decimal. 443. By Art. 440 (4), the characteristic of the logarithm of a decimal having no cipher between its decimal point and its first significant figure is — 1, of a decimal having one cipher between its decimal point and its first significant figure is — 2, and of a decimal having two ciphers between its decimal point and its first significant figure is — 3. In order to avoid writing these negative characteristics — 1, — 2, — 3, etc., it is customary to consider that - 1 = 9 - 10, - 2 = 8 - 10, - 3 = 7 - 10, etc. With the negative results written in this form, we have, in general, The characteristic of the logarithm of a number less than unity is obtained by subtracting from 9 the number of ciphers between its decimal point and the first significant figure, annexing — 10 after the mantissa. USE OF FOUR-PLACE TABLES 367 II. To obtain the mantissa of a logarithm. (a) Important principle governing the finding of all mantissas. 444. It has been computed that the logarithm of 35,200 is 4.5465, and from our discussion we have seen that the logarithm of 100 is 2. We may write, therefore, 104.5465 = 35200 (1) and 10 2 = 100. (2) These logarithms being, by definition, exponents, we may treat them as such in the following operation. Dividing (1) by (2), we have : 104.5466 35200 10 2 100 From which, 102.5465 = 352. That is, log 352 = 2.5465. Clearly, therefore, the mantissas of the logarithms of 352 and 35,200 are equal. In like manner, we may show that In general : log 35.2 = 1.5465, log 3.52 = 0.5465, log .352 = 9.5465 - 10, etc. If two numbers differ only in the position of their decimal points, their logarithms have the same mantissas. THE USE OF THE FOUR-PLACE TABLE 445. The table of logarithms is used for two distinct and opposite operations. (1) Given a number, to find the corresponding logarithm. (2) Given a logarithm, to find the corresponding number. 368 LOGARITHMS I. To find the logarithm of a given number. 446. (a) Numbers having three figures. Illustrations : 1. What is the logarithm of 247 ? On page 370, in the column headed "N," we find "24," the first two figures of the given number. In the same horizontal line with 24, under the heading corresponding to the last figure of the given number, "7," we find the mantissa, 3927. Since the given number has three figures to the left of the decimal point, the required characteristic is 2 (Art. 442). Therefore, log 247 = 2.3927. Result. 2. What is the logarithm of .0562 ? Opposite "56" of the "N" column (p. 371), and under "2," we find the mantissa 7497. Since the given number is a decimal and has one cipher between its decimal point and its first significant figure, we subtract 1 from 9, and annex — 10 to the mantissa (Art. 443). Therefore, log .0562 = 8.7497 - 10. Result. (b) Numbers having two figures. Illustrations : 1. What is the logarithm of 76 ? Opposite "76" of the "N" column, and under "0," we find the mantissa 8808. The characteristic is 1 (Art. 442). Therefore, log 76 = 1.8808. Result. 2. What is the logarithm of .0027 ? Opposite " 27 " of the " N " column, and under "0," we find the man- tissa 4314. The characteristic is (9—2), or 7, with —10 annexed (Art. 443). Therefore, log .0027 = 7.4314 - 10. Result. INTERPOLATION 369 (c) Numbers having onejlgure. Illustrations : 1. What is the logarithm of 7 ? Opposite "70" of the"N" column, and under "0," we find the mantissa 8451. The characteristic is (Art. 442). Therefore, log 7 = .8451. Result. 2. What is the logarithm of .00008 ? Opposite "80" of the "N" column, and under "0," we find the mantissa 9031. The characteristic is (9 — 4), or 5, with — 10 annexed (Art. 443). Therefore, log .00008 = 5.9031 - 10. Result. 447. It is evident that the logarithms of numbers having one or two figures have mantissas from the " " column of the table. Exercise 132 K nd the logarithms of the following numbers : l. 124. 7. 84.2. 13. 6.73. 19. .0642. 2. 283. 8. 39.6. 14. .829. 20. .0006. 3. 589. 9. 2.85. 15. .342. 21. .00016. 4. 676. 10. 6.76. 16. .676. 22. .0676. 5. 643. 11. 3.70. 17. .037. 23. .00809. 6. 540. 12. 5.89. 18. .0681. 24. .00000734. INTERPOLATION 448. Interpolation is based upon the assumption that the dif- ferences of logarithms are proportional to the differences of their corresponding numbers. While the assumption is not absolutely correct, the results obtained are exceedingly close approximations. The process is necessary in obtaining the logarithms of numbers having four places by means of the four-place table. SOM. EL. ALG. — 21 370 LOGARITHMS N 1 2 3 4 5 6 7 8 9 10 0000 0043 0086 0128 0170 0212 0253 0294 0334 0374 11 0414 0453 0492 0531 0569 0607 0645 0682 0719 0755 12 0792 0828 0864 0899 0934 0969 1004 1038 1072 1106 13 1139 1173 1206 1239 1271 1303 1335 1367 1399 1430 14 15 1461 1492 1523 1553 1584 1614 1644 1673 1703 1732 1761 1790 1818 1847 1875 1903 1931 1959 1987 2014 16 2041 2068 2095 2122 2148 2175 2201 2227 2253 2279 17 2304 2330 2355 2380 2405 2430 2455 2480 2504 2529 18 2553 2577 2601 2625 2648 2672 2695 2718 2742 2765 19 20 2788 2810 2833 2856 2878 2900 2923 2945 2967 2989 3010 3032 3054 3075 3096 3118 3139 3160 3181 3201 21 3222 3243 3263 3284 3304 3324 3345 3365 3385 3404 22 3424 3444 3464 3483 3502 3522 3541 3560 3579 3598 23 3617 3636 3655 3674 3692 3711 3729 3747 3766 3784 24 25 3802 3820 3838 3856 3874 3892 3909 3927 3945 3962 3979 3997 4014 4031 4048 4065 4082 4099 4116 4133 26 4150 4166 4183 .4200 4216 4232 4249 4265 4281 4298 27 4314 4330 4346 4362 4378 4393 4409 4425 4440 4456 28 4472 4487 4502 4518 4533 4548 4564 4579 4594 4609 29 30 4624 4639 4654 4669 4683 4698 4713 4728 4742 4757 4771 4786 4800 4814 4829 4843 4857 4871 4886 4900 31 4914 4928 4942 4955 4969 4983 4997 5011 5024 5038 32 5051 5065 5079 5092 5105 5119 5132 5145 5159 5172 33 5185 5198 5211 5224 5237 5250 5263 5276 5289 5302 34 35 5315 5328 5340 5353 5366 5378 5391 5403 5416 5428 5441 5453 5465 5478 5490 5502 5514 5527 5539 5551 36 5563 5575 5587 5599 5611 5623 5635 5647 5658 5670 37 5682 5694 5705 5717 5729 5740 5752 5763 5775 5786 38 5798 5809 5821 5832 5843 5855 5866 5877 5888 5899 39 40 5911 5922 5933 5944 5955 5966 5977 5988 5999 6010 6021 6031 6042 6053 6064 6075 6085 6096 6107 6117 41 6128 6138 6149 6160 6170 6180 6191 6201 6212 6222 42 6232 6243 6253 6263 6274 6284 6294 6304 6314 6325 43 6335 6345 6355 6365 6375 6385 6395 6405 6415 6425 44 45 6435 6444 6454 6464 6474 6484 6493 6503 6513 6522 6532 6542 6551 6561 6571 6580 6590 6599 6609 6618 46 6628 6637 6646 6656 6665 6675 6684 6693 6702 6712 47 6721 6730 6739 6749 6758 6767 6776 6785 6794 6803 48 6812 6821 6830 6839 6848 6857 6866 6875 6884 6893 49 50 6902 6911 6920 6928 6937 6946 6955 6964 6972 6981 6990 6998 7007 7016 7024 7033 7042 7050 7059 7067 51 7076 7084 7093 7101 7110 7118 7126 7135 7143 7152 52 7160 7168 7177 7185 7193 7202 7210 7218 7226 7235 53 7243 7251 7259 7267 7275 7284 7292 7300 7308 7316 54 7324 7332 7340 7348 7356 7364 7372 7380 7388 7396 1 2 3 4 5 6 7 8 9 FOUR-PLACE TABLE 371 N 1 2 3 4 5 6 7 8 9 55 7404 7412 7419 7427 7435 7443 7451 7459 7466 7474 50 7482 7490 7497 7505 7513 7520 7528 7536 7543 7551 57 7559 7566 7574 7582 7589 7597 7604 7612 7619 7627 58 7634 7642 7649 7657 7664 7672 7679 7686 7694 7701 59 60 7709 7716 7723 7731 7738 7745 7752 7760 7767 7774 7782 7789 7796 7803 7810 7818 7825 7832 7839 7846 61 7853 7860 7868 7875 7882 7889 7896 7903 7910 7917 62 7924 7931 7938 7945 7952 7959 7966 7973 7980 7987 63 7993 8000 8007 8014 8021 8028 8035 8041 8048 8055 64 65 8062 8069 8075 8082 8089 8096 8102 8109 8116 8122 8129 8136 8142 8149 8156 8162 8169 8176 8182 8189 66 8195 8202 8209 8215 8222 8228 8235 8241 8248 8254 67 8261 8267 8274 8280 8287 8293 8299 8306 8312 8319 68 8325 8331 8338 8344 8351 8357 8363 8370 8376 8382 69 70 8388 8395 8401 8407 8414 8420 8426 8432 8439 8445 8451 8457 8463 8470 8476 8482 8488 8494 8500 8506 71 8513 8519 8525 8531 8537 8543 8549 8555 8561 8567 72 8573 8579 8585 8591 8597 8603 8609 8615 8621 8627 73 8633 8639 8645 8651 8657 8663 8669 8675 8681 8686 74 75 8692 8698 8704 8710 8716 8722 8727 8733 8739 8745 8751 8756 8762 8768 8774 8779 8785 8791 8797 8802 76 8808 8814 8820 8825 8831 8837 8842 8848 8854 8859 77 8865 8871 8876 8882 8887 8893 8899 8904 8910 8915 78 8921 8927 8932 8938 8943 8949 8954 8960 8965 8971 79 80 8976 8982 8987 8993 8998 9004 9009 9015 9020 9025 9031 9036 9042 9047 9053 9058 9063 9069 9074 9079 81 9085 9090 9096 9101 9106 9112 9117 9122 9128 9133 82 9138 9143 9149 9154 9159 91(55 9170 9175 9180 9186 83 9191 9196 9201 9206 9212 9217 9222 9227 9232 9238 84 85 9243 9248 9253 9258 9263 9269 9274 9279 9284 9289 9294 9299 9304 9309 9315 9320 9325 9330 9335 9340 86 9345 9350 9355 9360 9365 9370 9375 9380 9385 9390 87 9395 9400 9405 9410 9415 9420 9425 9430 9435 9440 88 9445 9450 9455 9460 9465 9469 9474 9479 9484 9489 89 90 9494 9499 9504 9509 9513 9518 9523 9528 9533 9538 9542 9547 9552 9557 9562 9566 9571 9576 9581 9586 91 9590 9595 9600 9605 9609 9614 9619 9624 9628 9633 92 9638 9643 9647 9652 9657 9661 9666 9671 9675 9680 93 9685 9689 9694 9699 9703 9708 9713 9717 9722 9727 94 95 9731 9736 9741 9745 9750 9754 9759 9763 9768 9773 9777 9782 9786 9791 9795 9800 9805 9809 9814 9818 96 9823 9827 9832 9836 9841 9845 9850 9854 9859 9863 97 9868 9872 9877 9881 9886 9890 9894 9899 9903 9908 98 9912 9917 9921 9926 9930 9934 9939 9943 9948 9952 99 9956 9961 9965 9969 9974 9978 9983 9987 9991 9996 N 1 o 1 2 3 4 5 6 7 8 9 372 LOGARITHMS Illustrations : 1. What is the logarithm of 6874 ? Since the table gives mantissas for numbers having but three places, we will consider that we are seeking the logarithm of 687.4. (See Art. 444.) From the table we find the mantissa of 687 = .8370 the mantissa of 688 = .8376 Difference = .0006 That is, an increase of 1 in the number causes an increase of .0006 in the mantissa. Therefore, an increase of .4 in the number (for 687.4 is .4 greater than 687) will cause an approximate increase in the mantissa of .0006 x .4 = .0002. (The fifth decimal place being disregarded if less than .5 of the fourth.) Hence, for the mantissa of the logarithm of 687.4, we have .8370 + .0002 = .8372. Therefore, with the proper characteristic, log 6874 = 3.8372. Result. 2. What is the logarithm of 23.436? Consider the number to be 234.36. From the table we find the mantissa of 234 = .3692 the mantissa of 235 = .3711 Difference = .0019 Then, as above, .36 x .0019 = .000684, or, approximately, = .0007. Therefore, the mantissa of the logarithm of 234.36 = .3692 + .0007 = .3699. Whence, log 23.436 = 1.3699. Result. Exercise 133 Eind the logarithms of the following numbers : 1. 2762. 5. 8625. 9. 9824. 13. 68.741. 2. 3894. 6. 8.642. 10. .06421. 14. 430.05. 3. 2007. 7. 726.4. 11. .005432. 15. .0006941. 4. 6492. 8. 54.29. 12. 44.212. 16. .0004682. THE ANTILOGARITHM 373 II. To find the number corresponding to a given logarithm. 449. The number to which a given logarithm corresponds is called its antilogarithm. 450. When the mantissa of a given logarithm is found in the four-place table, the antilogarithm is readily found, and if the mantissa given is not found in the table, a close approxi- mation for the required antilogarithm is obtained by a process of interpolation. Illustrations : 1. What is the number whose logarithm is 1.4669 ? Look in the table for the mantissa .4669. In the same horizontal line with this mantissa and in the " N " column, we find "29," the first two figures of the required number. The mantissa is found under the "3." Therefore, the required sequence of figures is 293. Now the given logarithm has a characteristic, 1. Therefore (Art. 442), there must be two figures to the left of the decimal point in the required number. Hence, the antilogarithm of 1.4669 = 29.3. Result. 2. What is the antilogarithm of 7.9112 -10 ? The mantissa, .9112, lies in the horizontal line with " 81 " and under " 5." The sequence of figures is, therefore, 815. Since the characteristic is 7 — 10, the number has two ciphers between the decimal point and the first significant figure (Art. 443). Therefore, the antilog of 7.9112—10 = .00815. Result. 3. What is the antilogarithm of 2.8828 ? The mantissa, .8828, is not found in the table, but lies between the mantissa of log 763 = .8825 and the mantissa of log 764 = .8831 The difference of these mantissas = .0006 That is, an increase of .0006 in the mantissas causes an increase of 1 in their antilogarithms. Now between the given mantissa and the next lowest mantissa there exists a difference of .8828 - .8825 = .0003- 374 LOGARITHMS Therefore, an increase of .0003 in the mantissas will cause an approxi- mate increase in the antilogarithms of .0003 _ 5 .0006""' ' Hence, .8828 is the mantissa of the log 763.5. Or, antilog 2.8828 = 763.5. Result. Exercise 134 Find the antilogarithms of the following logarithms : 1. 1.6085. 8. 0.1088. 15. 8.7219-10. 2. 1.7284. 9. 3.9799. 16. 7.9007-10. 3. 2.9345. 10. 2.6897. 17. 6.7370-10. 4. 3.8317. 11. 4.9733. 18. 9.6477-10. 5. 0.5509. 12. 9.8987-10. 19. 5.8566-10. 6. 3.5974. 13. 9.7306-10. 20. 9.9995-10. 7. 2.0095. 14. 8.9099-10. 21. 4.9543-10. THE PROPERTIES OF LOGARITHMS 451. In any system, the logarithm of 1 is 0. For, a = 1. (Art. 241) Whence, log 1=0. (Art. 438) 452. In any system, the logarithm of the base is 1. For, a 1 = a. Whence, log a = 1. (Art. 438) While, for convenience, the base of the common system is used in the following discussions, each theorem is a general one for any base, a, hence, for any system. 453. The logarithm of a product is equal to the sum of the logarithms of its factors. THE PROPERTIES OF LOGARITHMS 375 Let 10* = to (1) and 10^ = n. (2) Multiplying (1) by (2), \v x +v = mn. Whence, log mn = x + y. (3) (Art. 438) From (1) and (2), x = log to and y = log n. (Art. 438) Substituting in (3), log mn = log to + log n. 454. The logarithm of a quotient is equal to the logarithm of tlie dividend minus the logarithm of the divisor. Let 10* = m (1) and 10*/ = n. (2) Dividing (1) by (2), iv _ m 10» ~~ n That is, 10*-y = Tl. n Whence, log^ = x-y. (3) n (Art. 438) But, from (1) and (2), x = log m and y = log n. (Art. 438) Hence, in (3), log — = log m — log n. ft 455. The logarithm of a power of a number is equal to the logarithm of the number multiplied by the exponent of the power. Let 10* = m. (1) Raising both members of the equation to the pth. power, IOp* = mP. Whence, log mP = px. But, from (1), x = log m, (Art. 438) Hence, . log m" =p(log m). 456. TJie logarithm of a root of a number is equal to the logarithm of the number divided by the index of the root. Let 10* == w. (1) Raising both members to the power indicated by-, x 1 10* = TO*. 1 Whence, log m q = -. (Art. 438 ) But, from (1), x = log to, (Art. 438) - _ log to Therefore, log m q — — - — 376 LOGARITHMS 457. These demonstrations may be briefly summarized as follows : (1) To multiply numbers, add their logarithms. (2) To divide numbers, subtract the logarithm of the divisor from the logarithm of the dividend. (3) To raise a number to a power, multiply the logarithm of the number by the exponent of the required power. (4) To extract a root of a number, divide the logarithm of the number by the index of the required root. The antilogarithm of a result obtained by one of these processes is the required number. THE COLOGARITHM 458. The cologarithm of a number is the logarithm of the reciprocal of the number. By its use an appreciable saving of labor is made in computations with logarithms. By definition, I cologiV=log— 2v = log 1 — log N (Art. 454) = 0- log .2V (Art. 451) = -log2V. . To avoid this negative form we may write, colog JV= 10 - log 2V— 10. 459. In general, to obtain the cologarithm of a number : Subtract the logarithm from 10 — 10. In practice the subtraction is usually accomplished by be- ginning at the characteristic and subtracting each figure from 9, excepting the last significant figure, which is subtracted from 10. LOGARITHMS IN COMPUTATIONS 377 Illustration : Find the cologarithm of 16. The log of 1 is 0, which we may write as 10 — 10. Then, colog 16 = log 1 — log 16. log 1 = 10 - 10 log 16 = 1.2041 Subtracting, colog 16 = 8.7959 — 10. Result. 460. If the characteristic of a logarithm is greater than 10 but less than 20, we use in like manner, colog 2V = 20 - log N— 20. Similarly, 30 — 30, 40 — 40, etc. , may be used if necessary. 461. Negative factors in computations with logarithms. The logarithms of negative factors in any group are found without regard to the negative signs, and the result is positive or nega- tive according as the number of negative factors is even or odd. USE OP LOGARITHMS IN COMPUTATIONS 462. Illustrations : 1. Multiply 6.85 by 37.8. log 6.85 = 0.8357 log 37.8 = 1.5775 (Art. 453), 2.4132 = log of product, antilog 2.4132 = 258.9. Result. 2. Divide 30,400 by 1280. log 30400 = 4.4829 log 1280 = 3.1072 (Art. 454) , 1 . 3757 = log of quotient, antilog 1.3767 = 23.75. Result. 3. Divide 2640 by 36,900. log 2640 = 3.4216 (Art. 458), colog 36900 = 5.4330 - 10 8.8546 - 10 = log of quotient, antilog 8.8546 - 10 = 0.07155. Result. 378 LOGARITHMS 4. What is the value of ^846"? log 846 = 2.9274. (Art. 456) 3)2.9274 .9758 = log of cube root, antilog .9758 = 9.458. Result. 5. Find the value of ^.00276. log .00276 = 7.4409 - 10. When a negative logarithm occurs in obtaining a root, the characteris- tic must be written in such a form that the number subtracted from the logarithm shall be 10 times the index of the root. The divisor in this case being 6, we change the form of the logarithm by adding 50 — 50, and the subsequent division gives the required negative characteristic, log .00276 = 7.4409 - 10. Adding, 50 - 50 57.4409-60 (Art. 456) 6 )57.4409-60 9.5735 - 10 = log of sixth root, antilog 9.5735 - 10 = .3745. Result. 432 2 x27.1 3 6. Find the value of 35.1 4 (Art. 455) 2 log 432 = 2 (2.6355) = 5.2710 (Art. 455) 3 log 27 . 1 = 3 (1 .4330) =4 .2990 (Arts. 455, 458) 4 colog 35.1 = 4 (8.4547 -10) = 3.8188 - 10 3.3888 = log of result, antilog 3.3888 = 2448. Result. 7. Simplify V/ 2720* X V288 X 432* . .068 7 X V27.8 3 x 624 2 \ log 2720 a | (3.4346) = 1.7173 | log 288 z= i(2.4594) =1.2297 -flog 432 =|(2.6355) =3.5140 7 colog .068 = 7 (1.1675) =8.1725 f colog 27.8 = | (8.5560 - 10) = 7.8340 - 10 2 colog 624 = 2 (7.2048 -10 ) = 4.4096 - 10 6.8771 (Art. 456) 4)6.8771 1.7193, log of result, antilog 1.7193 = 52.40. Result. MISCELLANEOUS APPLICATIONS OF LOGARITHMS 379 1. 4600 x. 85. 2. 72x380. 3. .28 x. 00012. 4. .017 X .0062. 5. 4.96 x 58.4. 6. .00621 x .000621 7. 73400 x .00811. 8. .0293 x .000602. 9. 691 x .0000131. 624 x 372 x 891 19 20. 21. Exercise 135 e value of : 10. 1280 -r -.0064. 11. 68.5 + 6.12. 12. 2.741 - - .00822. 13. .00461 - .0931. 14. .07241 -r- .3623. 15. (2741 K 3.623) - -242. 16. (4.625 X .5821) - 2.067. 17. 34.74 - - (2.851 x 4.309). 18. 6.904- - (3.676 x .00275^ .0372 X .584 x . 30027 457 x 196 x 583 43.2 x 3.28 x .246 .537x3.41x56.8' 630 x 2100 x .007 23. 24. 3.25 X 472 x 6500 25. 4.52 2 . 28. ^377. 26. 2.74 5 . 29. VWM. 27. .0276 4 . 30. a/,00724. .273 x .00042 x .0121 .007 x .07 x .7 X 7 35.7 x 7.14 x .1428* .643 X. 0468x2760 346 x .0072 x .01 ' 31. 3.207f. 32. 5.602*. 33. .0007544. 34. 4 ,0027 3 x5.82 4 x762 3 3.207 4 x6.42 3 x26.7 2 35. 4 V.2809 x ^.003241 V^962X ^.000011 MISCELLANEOUS APPLICATIONS OF LOGARITHMS 463. I. Changing the base of a system of logarithms. Let a and b represent the bases of two systems of logarithms, and m the number under consideration. We are to show that log b m l0ga& 380 LOGAKITHMS Let a* = m (1) and bv = m. (2) Then, ' x = \og a m and y = log b m. (Art. 438) From (1) and. (2), & = W . X Extracting the yth root, a* = b. Therefore, log«& = - or y = — — V log a b That is, lo g6 m=^^. log a 6 Illustration: 1. Find the logarithm of 12 to the base 5. By Art. 463, log 6 12=^ " logi 5 = 1.0792 .6990 = 1.5439. Result. II. Equations involving logarithms. 464. An equation in which an unknown number appears as an exponent is called an exponential equation. Thus : a x = b is a general form for such equations. Illustrations : 1. Solve the equation, 4 X = 64. If, in an equation in the form of a x — b, b is an exact power of a, the solution is readily obtained by inspection. From 4* = 64, we have 4* = 4 3 . Therefore, x = 3. Result. 2. Find the value of x in 5* = 13. Using logarithms, * log 5 = log 13. x log 5 1.1139 .6990 1.5935. Result. MISCELLANEOUS APPLICATIONS OF LOGARITHMS 381 3. Solve the. equation 2 s/x = VS. 8 /)*=, when n is odd. Illustrations : 1. Divide a?° + y w by tf + tf. We may write x 10 + y™ = (x 2 ) 6 + (y 2 ) 5 . Then, (x 2 ) 6 + (y 2 ) 6 is divisible by a 2 + y 2 . (Art. 470, IV.) Whence, (S 2 )6+y) 6 =(a . 2)4 _ (a . 2) 3 (y2) + (^ )2(2/2) 2 _ (^)(y«)8 + (y2)4 a;-* + y-* = x 8 — xV -I- x 4 */ 4 - xV + f. Result. 2. Divide 64 a 12 - tf by 2 a? - */. We may write 64 x 12 — y* = (2 x 2 ) 6 — y*. Then, (2 x 2 )6 - y6 _ ^ + ^ + ^ ^^ + ^ ^ g + ^ x2)y4 + y6 2^ — y = 32 x 10 + 16 x 8 y + 8 xfy 2 + 4 x 4 y 3 + 2 xV + y 5 . Result. Exercise 138 Divide : 1. a? + 32y 5 byx + 2y. 2. a? 7 -128bya;-2. 3. 32^ + 243 by 2o; + 3. 4. ^-729 by --S. x 6 x 5. What are the exact binomial divisors of 81 — a* ? HIGHEST COMMON FACTOR 389 6. What are the exact binomial divisors of x 10 — y 10 ? 7. What are the exact binomial divisors of x 2 — 64 c 6 ? 8. Obtain the three factors of x 9 — y 9 . 9. Obtain the four factors of X s — y 8 . 10. . Obtain the six factors of x u — 4096. 11. Obtain the ten factors of a 12 — 64 a 6 . THE HIGHEST COMMON FACTOR OF EXPRESSIONS NOT READILY FACTORED 471. The method of obtaining the highest common factor of two expressions that cannot be factored by inspection is analo- gous to the process of division used in arithmetic for obtaining the greatest common divisor of two numbers. The principles involved in the algebraic process are estab- lished as follows : Let A and B represent two expressions, both arranged in the descend- ing powers of some common letter ; and let the degree of A be not higher than the degree of B. Let B be divided by A, the quotient being Q and the remainder B. Thus, A)B (Q A$ B It follows, therefore, that B = AQ + B, (1) and B = B-AQ. (2) Now a factor of each of the terms of an expression is a factor of the expression. We make, therefore, three important statements : (1) Any common factor of A and R in (1) is a factor of AQ + E, hence a factor of B. Or, A common factor of A and R is a common factor of B and A. (2) Any common factor of B and A in (2) is a factor of B— AQ, hence a factor of R. Or, A common factor of B and A is a common factor of A and R. 390 SUPPLEMENTARY TOPICS (3) Hence, the common factors of B and A are the same as the common factors of A and R. Or, The H. G. F. of A and B is the H. G. F. of A and R. Each succeeding step of the process may be proved in a similar manner. If, at any step, the remainder becomes 0, the divisor is a factor of the corresponding dividend and is, therefore, the H. G. F. of itself and the corresponding dividend. Or, The last exact divisor is the required highest common factor. 472. We may multiply or divide either given expression, or may multiply or divide any resulting expression, by a monomial that is not a common factor of both expressions. For the process refers to polynomial expressions only, and the introduction or the rejection of monomial factors not com- mon to both expressions cannot affect the common polynomial expression sought. Such factors are introduced or rejected merely for con- venience, as will be shown in the following illustrations. In finding the highest common factor of three or more expres- sions, as A, B, and G, we first find F 1} the highest common factor of A and B. It remains to find the highest common factor of F 1 and G. Let this result be F 2 . Then F 2 is the required highest common factor of A, B, and G. Illustrations : Find the H. C.F. of 2 a*-a*+ x -6 and 4 a*+2 a^-lO x-3. The following arrangement is universally accepted as most satisfactory : 2 as 8 — x 2 + x- 6 2 4x 3 + 2 a; 2 -10 x - 3 4 a; 8 - 2 x' 1 + 2 x - 12 X 4 a; 8 - 2x 2 + 2 a; -12 4 a 8 - 12 a; 2 + 9 a; + 4 x 2 - 12 x + 9 5 5x + 10 x 2 - 7 a; -12 + 10 a: 2 - 15 x 20 x a - 60 x + 45 20 a; 2 -14 a; -24 -23)- 46 a; + 69 H.C. F. 2x- 3 4 + 8 a; -12 + 8 a; -12 HIGHEST COMMON FACTOR 391 Explanation : 1. Dividing 4 a; 3 + 2 x 2 — 10 x - 3 by 2x* — x 2 + x — 6, we obtain a remainder, 4 x 2 — 12 x + 9, an expression lower in degree than the divisor just read. 2. Dividing 2x 3 - x 2 + x — 6 by 4 oj 2 — 12 x + 9, we obtain a remainder having such coefficients that the next succeeding division does not reduce the degree of the remainder. To obtain a division that will reduce the degree of the remainder we multiply 4 x 2 — 12 * + 9 by 5, and the product, 20 x 2 — 60 x + 45, contains 10 x 2 — 7 x — 12, with a remainder as desired, — 46 se + 69. From this remainder we may discard the factor, —23, since this factor is not common to both expressions at this point. • 3. The remainder, 2 x — 3, divides 10 x 2 — 7 x — 12 exactly. Therefore, 2 x — 3 is the required H. C. F. It may be noted that the process might have been completed by multi- plying 10 x 2 — 7 x — 12 by 2, the resulting product being used as a divi- dend with 4 x 2 — 12 x + 9 as a divisor. Division either way is possible when both expressions are like in degree. Exercise 139 Find the H.C.F. of: 1. ^-3^ + 4a-2and^- r -o 2 -4a; + 2. 2. a 3 -2a 2 -a + 2anda 3 -2a 2 + 3a-2. 3. c 3 + 5c 2 + 7c-f-3andc 3 + c 2 -5c + 3. 4. ^_^_5a;_|_2 and a^ + 4 » 2 -f 3 a; - 2. 5. m 3 + 3m 2 -f-5ra + 3 and m 3 + 2m 2 -2m-3. 6. a 3 -5a 2 + 7a-3anda 3 -a 2 -7a + 3. 7 . c 4 + c 2 + 2candc 4 -c 3 -3c 2 -c. 8. 2a 6 + 4ar i + 6a 4 -f-4ar J + 2a^ and 4a 5 - 4 a 3 -8^-40;. 9. m 3 + ra 2 — m + 15 and m 3 + 6 m 2 + 5 m — 12. 10. 2c 3 + 5c 2 + 5c + 6and3c 3 + 5c 2 -c + 2. 11. 6a; 4 + 24(» 3 + 30a! 2 -f-36a;and3a; 4 -f-6a5 3 -12ic 2 -9a;. 12. 2 a 4 + a 3 + 3 a 2 + a + 2 and a 4 + 2 a 3 +4 a 2 + 3 a + 2. 13. c 3 -3c 2 + 5c-3and2c 4 -3c 3 -3c 2 + 10c-6. 14. 2» 4 +a^-4^ + 3a;-2and3a; 4 + 4^-3ar 2 -4. 392 SUPPLEMENTARY TOPICS 15. 4a 4 -4 a 3 - 5 a 2 + 6a- land 6a 4 -5a 3 -6a 2 + 3a + 2. 16. I m 5 -3m 4 -2m 3 +5m 2 -2m and 3m 6 -7m 5 + 6m 4 - 3m 3 +m 2 . Reduce the following to lowest terms : 1? m 3 -4m 2 + 2m + l , 20 c 3 -c 2 -5c-3 m 3 -2m 2 + 3ra-2 c 3_4 c 2_n c _ 6 3-5a + 7a 2 + 3a 3 , gl 3 a 3 + 14 or 2 -5 a -56 6-7a + 3a 2 + 2a 3 6a 3 + 10 a 2 + 17 a + 88 10 4a 2 + 9a-9 00 2m 4 -2m 2 + m-l 4a 4 -f-10a 3 -7a 2 + 9 m *_ m 3 + 2 m 2 -m-l Find the H. C. F. of: 23. a 3 -2a + l, a 3 -2a 2 + l, and a 3 -2 a 2 + 2a-l. 24. a 3 + 4 a 2 +5 x+2, ^+3^+4^+4, and ^+3^+3 a +2. 25. 2a? 4 -2af'4-4a; 2 -4ic, 3 a; 4 -f- 9x 2 -12a;, and 4a 4 -8a; 3 -20« 2 + 24a-. 26. a 4 -4a 2 -a + 2, a 5 - 3a 3 + 3a 2 -l, and a 4 -f-3a-2. THE LOWEST COMMON MULTIPLE OP EXPRESSIONS NOT READILY FACTORED 473. In finding the lowest common multiple of two ex- pressions not readily factored we first find the highest common factor by Art. 471, after which the process is established by the following. Let A and B represent any two expressions whose highest common factor we find to be H. Dividing both A and B by H, we obtain ^ = a and ^ = 6. II H Then, A-Hxa, (1) B = Hxb. (2) Now, since H is the highest common factor of A and 2?, the two quo- tients, a and 6, can have no common factor. Hence, for the lowest com- mon multiple of A and B we write L.C.M.=Rxaxb. (3) Writing (3) thus, L. C. M.=5ax 6. LOWEST COMMON MULTIPLE 393 Multiplying and dividing by J9", L.C.M. -Hax^- H Substituting Ha = A from (1), and Hb — B from (2), L.C. M. = ^x— . H In general, therefore: To find the lowest common multiple of two expressions, divide one of the expressions by their highest common factor , and multi- ply the other expression by the quotient. The product of two expressions is equal to the product of their highest common factor by their lowest common multiple. For, Multiplying (1) by (2), we obtain AB = Hx ax Hxb = H(Hab). From (3) L. C. M. = Hab. Hence, AB = H(L. C. M.). In finding the lowest common multiple of three or more expressions as A, B, and C, we first find L\, the lowest common multiple of A and B. It remains to find the lowest common multiple of L\ and C. Let this result be'-L 2 . Then L 2 is the required lowest common multiple of A, B, and G. Illustrations : 1. Find the L. C. M. of : 2 ra 3 — ra 2 + m — 6 and 4 m 3 + 2 ra 2 — 10 m — 3. By the method of division we find the H.C.F.=2m-3. Then, L. C. M. = ( *™*-m* + m-Q \ (4 m s + 2 w2 _ 10 m _ 3) V 2i»-3 ) K J = (m? + m + 2)(4m 8 + 2m 2 - 10m - 3). Result. Exercise 140 Find the L. C. M. of: 1. ^ + 5^ + 705 + 3 and + 9. 2. a 3 + a 2 -10a + 8 and a 3 -4a 2 + 9a- 10. 3. 2m 3 + 5m 2 -f-2m — 1 and 3 m 3 + ra 2 — ra + 1. 4. 4c 3 -8c 2 -3c + 9and 6c 3 + c 2 -19c + 6. 5. 4ar 5 -7a + 3and2;c 3 + ar 2 -3a; + l. 394 SUPPLEMENTARY TOPICS 6. 2a 3 -7a 2 -16a + 5anda 3 -9a 2 + 18a + 10. 7. c 3 + 5c 2 -c-5 and c 4 -c 3 -f-c 2 + c-2. 8. a^ + 2^-a; + 6and2ar J -f-llar J + 16a; + 3. 9. m 4 -3m 2 + l andm 3 + 2m 2 -4m-3. 10. 4^-2a; 2 + 6a + 4and6a; 4 + 9x 3 -l5ar 2 -9a. 11. 2a 4 + 3a 3 + 4a 2 + 2a + l and 3a 4 -f-2a 3 + a 2 -2a-l. 12. m 5 - 3 m 4 + 3 m 3 -ll ra 2 +6 m and 2 m 4 -18 ra 2 +8 m-24. 13. 4a 4 -a 3 6 + 2a 2 6 2 + 2a6 3 + & 4 and3a 4 -5a 3 &+9a 2 6 2 -6a6 3 + 46 4 . 14. a^+2a; 2 4-2a;4-l, ^+2^+305+2, and a 3 -^ ^+4^+3. 15. 4 m 3 — 4 m 2 4- 3 m — 1, 6 m 3 — m 2 -f- m — 1, and 8 m 3 — 2 m 2 + m-l. THE CUBE ROOT OF POLYNOMIALS 474. If a binomial, (a+6), is cubed, we obtain (a 3 +3a 2 6 -f3a& 2 -f-& 3 ). We have in the following process a method for extracting the cube root, (a + b), of the given cube, (a 3 + 3 o?b + 3a& 2 + 6 3 ). a 3 + 3 a 2 b + 3 ab 2 + 6 8 la + 6 Cube Root. Trial Divisor Complete Divisor a? 3a 2 (3a 2 + 3ab + b 2 ) b +3a 2 6+3a& 2 +& 8 + 3a 2 &+3a6 2 +6 8 The first term of the root, a, is the cube root of the first term of the given expression, a 3 . Subtracting a 3 from the given expression, the re- mainder, 3 a 2 b + 3 ab 2 + ft 3 , results. The second term of the root, b, is obtained by dividing the first term of the remainder, 3 a 2 b, by three times the square of a. This divisor, 3 a 2 , is the Trial Divisor. The complete divisor, (3 a 2 + 3 ab + b 2 ) , consists of (the trial divisor) + 3 (the trial divisor) (the last quotient obtained) + (the square of the last quotient). The product of the complete divisor by the last quotient obtained, (3 a 2 + 3 ab + b 2 )b, completes the cube. The process may be repeated in the same order with any polynomial perfect cube whose root has more than two terms. CUBE ROOT OF POLYNOMIALS 395 Illustration : 1. Find the cube root of 8 a 6 -36 a 5 + 102 a 4 -171 a 3 + 204 a 2 -144a + 64. The polynomial is first arranged in descending order. |2q 2 -3q + 4 8a 6 -36a5+102a*-171a 3 +204«2_i44a+64 8«s 3(2a 2 ) 2 =12a 4 3(2 a 2 ) (-3 a) = -18 a 9 (-3q) 2 = +9a' (12a*-18a 3 +9« 2 )(-3q): -36 a* + 102 a*- 171 a 3 -36 a 5 + 54 a*- 27 a 3 3(2a 2 -3a) 2 = 12a 4 -36a 3 +27a 2 3(2a 2 -3a)(4) = 24a 2 -36a 4 2 = +16 (12 a*- 36 a 3 +51 a 2 - 36 a +16) (4) 48 a 4 - 144 a 3 + 204 a 2 - 144 a +64 48 a 4 -144 a 3 +204 a 2 -144 a + 64 475. The distribution of the volume of a cubic solid whose edge is a + 6, and the relation of the separate portions to the separate terms of the polynomial representing its cube, can be readily seen from the following illustrations. The planes cutting the cube pass at right angles to each other and parallel to the faces of the cube ; each at a distance of b units from the faces of the cube. I n m is? In the figure let the edge of the cube in (I) be a and the edge of the complete cube in (IV) be a + 6. I. a 3 . II. a 3 + 3a 2 6. III. a 3 + 3a 2 6 + 3a6 2 . IV. a 3 + 3a 2 6 + 3a6 2 +6 3 . 396 SUPPLEMENTARY TOPICS „ M THE CUBE ROOT OF ARITHMETICAL NUMBERS 476. It will be observed that l 8 = 1 A number of one figure has not more than three 9 s as 729 figures in its cube. 10 8 = 1000 A number of two figures has not more than six 99 8 = 970299, etc. figures in its cube. Conversely, therefore: If an integral cube has three figures, its cube root has one figure. If an integral cube has six figures, its cube root has two figures, etc. Hence : 477. Separate any integral number into groups of three figures each, beginning at the right, and the number of groups obtained is the same as the number of integral figures in its cube root. It is to be noted that the process of cube root of numbers as given in. arithmetical practice is based directly upon the al- gebraic principles learned in Art. 474. Illustration : 1. Find the cube root of 421875. Separating the number into groups of three figures each, we have 42i875. 42J875 |70 + 5 = 75. Result. a 8 = 70 8 = 343000 a + b 3(a) 2 = 3(70) 2 = 14700 3(a)(6) =3(70) (5)= 1050 (6)2 - (5)2 = 25 (15775) (5) = '8875 78875 The analogy between the formation of the cube of (a + 6) and the cube of 75, or (70 + 5), may be seen in the following : (a + b) s = a 3 + 3a 2 6 .+ 3«6 2 + 6 8 (70 + 5) 8 = 70 8 + 3(70)2(5) + 3(70) (5) 2 + 6 8 = 343000 + 73500 + 5250 + 125 = 421876. CUBE ROOT OF ARITHMETICAL NUMBERS 397 2^Find the cube root of 12812904. In practice the process is usually abbreviated as follows: . 12812904 |234. Result. 2» = 8 3(20)2 _ 1200 4812 3(20) (3) = 180 32 = 9 3(1389) = 4167 3(230) 2 = 3(230) (4) = 42 = 158700 2760 16 4(161476) 645904 645904. Exercise 141 Find the cube root of : 1. a 3 + 6a 2 + 12a+ 8. 2. 27 c 3 - 54c 2 + 36 c -8. 3. 8a 3 -36a 2 + 54a-27. 4. 27 a 6 - 135 a 4 + 225 a 2 -125. 5. a 6 + 6a 5 + 15a 4 + 20a 3 +15a 2 + 6a + l. 6 . ^-6^ + 21a 4 -44a 3 + 63a 2 -54a-f27. 7. a 6 + 9a 5 + 21a 4 -9a 3 -42a 2 + 36a-8. 8. 102a 4 + 204 x 2 - 171a 3 -144a + 64 -36a 5 + $x«. 9. 27m 10 - 35 m 6 + 12m 2 - 8 + 30m 4 - 45m 8 + 27m 12 . 10. c 9 -3c 8 + 6c 7 -4c 6 + 6c 4 -2c 3 + 3c + l. 11. 74088. 16. 3112.136. 12. 389017. 17. 1470612$. 13. 658503. 18. 48.228544. 14. 912673. 19. .559476224. 15. 1953125. 20. .000138991832. 398 GENERAL REVIEW GENERAL REVIEW - | Exercise 142 1. Find the numerical value of 81"*, 16^", (i)~ 4 , 5°, 25~t Define briefly the law that governs each reduction. 2. Solve x + V.25 jc + 0.06 = 0. 3. Form the equation whose roots shall be a — V — 1 and a + V—1. 4. Solve for x : 2(4 - 3a)^ - 12 = s& 5. Factor 125 ar 6 -Va7. 6. Solve a? -f 10 aT +7 = 0, and test the roots obtained. 7. By inspection determine the nature of the roots of Sx 2 -7x= -5. 8. Solve 4 x 2 — 4 ax + a 2 — c 2 = 0. 9. Solve and verify both solutions of (x + T)\x — 2) = a,(a>-3)(&-l)-4. 10. Solve ?— - = a?(aj + 2) + 1. 0* - !) 11. Find the square root of g*-s» + l 6(0* + 1) 15a? 20 . (a^ + 1)- 1 a;- 1 (a^ + l)" 1 12. Find the value of m for which the equation, 3x 2 — 6x + m = 0, has equal roots. 13. Factor 2[2 a 2 - (x - a) - 2 a(2 x - a)]. 14. Write the 8th term of the quotient of x l5 —y 15 divided by x-y. 15. Solve x* + 2xi- 3 = 0. 16. Solve *- 2m + 3m = * + « ■ c + d a? -j- 2m a? + 2m 17. Find the five roots ofar 5 -4ar J -a? 2 + 4 = 0. GENERAL REVIEW 399 18. Factor 15 x 15 — 15. 19. Solve x(x ~y) — 2, (x + y) 2 = 9. 20. Form that quadratic equation whose roots shall be * - i and - f 21. Obtain the factors of x 2 — 5# + 2 by solving the equa- tion, x 2 — 5 x + 2 = 0. 22. Divide 16 into 3 parts in geometrical progression, so that the sum of the 1st and 2d shall be to the difference of the 2d and 3d parts as 3 : 2. 23. Find log of 7, given log 5 = .6990 and log 14 = 1.1461. 24. Solve and test the solutions of V2 m + x — V2 m — x — V2# = 0. ,.-2 25. Factor 4 + 4i? + m + mx~ x p~ x 26. Form the quadratic equation whose roots shall be l+V^andl-V^T. 27. Solve x 2 + xy + y 2 = 7 x, x 2 — xy-{-y 2 — Sx. 28. What is the sum of the first n numbers divisible by 5 ? „ ^.(H+fXf-M)-*. 30. Solve 2 s " 1 x 2* = 40. 31. Givenl+V«:l + 3V^ = 2:5; find a. 32. Factor ar 4 - 2 -f a 4 . 33. If 7i is an odd integer, which of the following indicated divisions are possible ? x n + y n xF+y" x n — y n x n — y n x + y ' x — y ' x + i/ ' # — ?/ 34. Form the quadratic equation whose roots are 1 -+- a yi and 1 — a V#. 400 GENERAL REVIEW 35. Under what condition will the roots of ax 2 -+- bx -f- c = be imaginary ? Prove your answer. 36. If the length and breadth of a certain rectangle are each increased by 2 rods, the area will become 48 sq. rd. ; but if each dimension were decreased by 2 rods, the area would become 8 sq. rd. Find the dimensions of the rectangle. 37. Factor x* + 2x 2 -5x-6. 38. Write in logarithmic form 27 x = 81, and find x. 39. Solve for n : -^ 2 c = n n — 1 40. What is the price of candles per dozen when 3 less for 36 cents raises the price 12 cents per dozen ? 41. Show that the product of the roots of x 2 — 5x — 2 = is —4. 42. Prove that — -, 3, — - — , are in geometrical progression, and find the sum of 10 terms. 43. If m : x = n : y, show that mx — ny : mx -\-ny = m 2 —n 2 :m 2 + n 2 . 44. Find an expression for the nth odd number, and illus- trate your answer by a numerical substitution. 45. Find n in the formula I = ar n ~ l . 46. Solve for a and m : 2 a 2 — am — 12, 2 am — m 2 = 8. 47. Factor , A •*" ■ - - — ±— — • (V + 3)- 1 (1+3 a)" 1 48. Rationalize the denominator of — i L^l — . 2-V6 + 2V2 49. Insert 6 geometrical means between T \ and 12|. C a -cv a • xy/S -v/18 50. Find x in — ^— = ~ 12 y/i GENERAL REVIEW 401 51. Without solving, prove that the roots of 6 x 2 + 5 x — 21 are real and rational. 52. Solve m-f \Z3x-\-x 2 : m + l=m- V3 #-}-#* : m — 1. 53. The difference between the 5th and the 7th terms of an arithmetical progression is 6, and the sum of the first 14 terms is — 105. Find the first term and the common difference. 54. Without solving, determine the nature of the roots of 16ar> + l=8a;. 55. Find the 5th term of (x 2 - ar 2 ) 15 . 56. Show that the sum of the squares of the roots of »2-3iC + l = is 7. 57. If m 2 — n 2 varies as x 2 , and if x = 2 when ra = 5 and ra = 3, find the equation between m, n, and x. 58. Find x and y if 2 x+ » = 16 and S x ~ v = 9. 59. Find a 4th proportional to x* — 1, x 2 -f 1, and cc 2 — 1. 60. Construct the quadratic equation, the product of whose roots shall be twice the sum of the roots of x 2 — 7 sc -f- 12 = ; and the sum of whose roots shall be 3 times the product of the roots of x 2 -f 2 x = 3. 61. Factor 3 x*-2 + x 2 + x 4 -3 x. 62. Find by logarithms the value of -y/2 x (£)* X .01 x 3*. 63. The sum of two numbers is 20, and their geometrical mean increased by 2 equals their arithmetical mean. What are the numbers ? i _ 64. What is the interpretation of x 1 = Vsc? 65. Write the 5th term of (a -f b) m . 66. Solve c 2 O 2 + 1) = m 2 + 2 c 2 x. 67. The sum of the last 3 terms of an arithmetical progres- sion of 7 terms equals 3 times the sum of the first 3 terms. The sum of the 3d and 5th terms is 32. Find the 1st term and the common ratio. SOM. EL. ALG. — 26 402 GENERAL REVIEW 68. How many digits in 35 s5 ? 69. Expand and simplify (a _1 Va — aVa -1 ) 4 . 70. What must be the equation between m and n if the roots of mxP+nx+p are real ? if equal ? 71. Find the (r + l)th term of (1 -x) 20 . 72. Find two numbers in the ratio of 3:2, such that their sum has to the difference of their squares the ratio of 1 : 5. 73. What is the sum and product of the roots of ,> 5 x~ 2 x~ l 74. Solve x 2 + y 2 = 26, 5 x + y = 24. Plot the graphs of the system and verify the solutions. 75. Show that either root of x 2 — c = is a mean propor- tional between the roots of x 2 + bx + c = 0. 76. Solve x 2 — 2 x + 3 == V&- 2 — 2 a; + 5. Are all the solu- tions roots of the given equation ? 77. What is the value of the 6th term of f x ] when x = 2? 78. The intensity of light varies inversely as the square of the distance from its source. How far must an object that is 8 feet from a lamp be moved so that it may receive but \ as much light? 79. Find an expression for x in a 2x = 3 c. 80. Solve x — 1.3 = .3 x~\ 81. If a : b = c > : d, show that (ma — nb)(ma +nb)~ l _ (mc — nd)(mc + wcT)" 1 mn win 10 82. Solve V7 x - V3 a + 4 = V3a+4 GENERAL REVIEW 403 83. If r x and r 2 represent the roots of x 2 -f bx -f c = 0, find r* -f- r 2 2 an( i ^i 2 **2 2 » 84. Solve x~% + af* + 1 = 0. 85. Find the geometrical progression whose sum to infinity is 4^, and whose 2d term is .002. 86. Expand ( V2 - V^) 6 . 87. Solvefors: -2 - 2 s 2 + V7 +2 s + 4 s 2 - 2s = 2s 2 + 5. 88. Find the value of k in order that the equation (k + 6)x 2 -2k(x 2 -l)-2kx-3 = may have equal roots. 89. The floor area of a certain room is 320 sq. ft., each end wall 128 sq. ft., and each side wall 160 sq. ft. What are the dimensions of the room ? 90. For what values of m are the roots of the equation (m + 2)x 2 + 2ma;+l = equal? 91. Given a:b = c: dj prove that 3a + 2c:3a — 2c = 12 b + 8 d : 12 b - 8 d. 92. Plot the graph of 3 x 2 -f- 10 x = 12, and check the result by solving. 93. Find the ratio between the 5th term of f 1 + - ) and the 4th term of (l + |Y 2 - 94. Calculate by logarithms the fourth proportional to 3.84, 2.76 and 4.62, and, also, the mean proportional between -\/T2 and V12. 95. Solve for sand t: sP + P = 91, s = 7 — t. 96. Findnins = arW ~ a » 404 GENERAL REVIEW 97. The velocity of a body falling from rest varies directly as the time of falling. If the velocity of a ball is 160 feet after 5 seconds of fall, what will it be at the end of the 10th second ? 98. In an arithmetic progression, a— — V— 1, cZ=l+ V^T, n a* 20. Find I and s. 99. Write the (r + l)th term of (a + b) n . 100. Find the middle term of ( — X~3Y°. W-l ,* J 101. Plot the graphs of 4 x 2 + 9 f = 36, x + 2 y = 3. Check by solution. 102. Insert 4 geometrical means between V— 1 and — 32. 103. Prove the formula for I in each of the progressions. 104. Form the quadratic equation which has for one root the positive value of V J 4- 4V3, and for the other root the arithmetic mean between 4 — 2 V3 and zero. 105. If m : n — n : s = s : t, show that n + s is a mean pro- portional between m + n and t -\- s. 106. Solve xy = 2 m 2 + 5 m + 2, ar> + ?/ 2 = 5 ra 2 + 8 m + 5. 107. Solve and test the solution : V# 2 — mx -f» *- x*s m, f 2\ 12 108. Find the term of (x 4 J that does not contain x. 109. Solve for sand t: s 2 + st + 2t 2 = 46, 2s 2 -st + t 2 =29. 110. Plot the graphs of a^ + y 2 + x + y =34, x + y-7 = 0. 111. What is the value of 1.027027 •;• ? 112. Find x if 3 2 *~ 2 j== (9~ l y-\ 113. If a -f b — 61, and a^ — 6^ = 1, find the values of a and b. INDEX (Numbers refer to pages.) Abscissa, 197. Addends, 22. Addition, 22. Affected quadratic equation, 2 Aggregation, signs of, 29. Algebraic expression, 20. Algebraic fraction, 122. Algebraic number, 16. Alternation, 320. Annuity, 382. Antecedent, 315, 318. Antilogarithm, 373. Arithmetical means, 344. Arithmetical progression, 338. Ascending order, 44. Associative law, 22, 39. Axes of reference, 197. Axiom, 13, 65. Base of logarithm, 364. * Binomial, 21. Binomial formula, 359. Binomial theorem, 358. Brace, 29. Bracket, 29. Characteristic of logarithm, 3 Checking results, 25, 46, 59. Clearing of fractions, 147. Coefficient, 12.- compound, 35. detached, 51, 61. Cologarithm, 376. Collecting terms, 25. Common difference, 338. Common factor, 119. Common multiple, 127. Common ratio, 348. Commutative law, 22, 39. Complex fraction, 144. Complex number, 255. Composition, 321. Compound ratio, 348. Comoound variation, 333. Conditional equation, 64. Conjugate imaginary, 260. Consequent, 315, 318. Constant, 200, 300. Continued proportion, 318. Coordinates, rectilinear, 197. Cube root, 99, 394, 396. Definite number symbols, 9. Degree, of an expression, 44. of a term, 44. Denominator, 126. factorial, 360. lowest common, 129. Density, 164. Descending order, 44. Difference, 32. common, 338. Direct and inverse variation, 332. Direct variation, 331. Discriminant, 278. Discussion of a problem, 193. Distributive law, 39, 52. Dividend, 52. Division, 52, 321. Divisor, 52. Divisors, theory of, 387. Duplicate ratio, 316. Elements, of an arithmetical 'progres- sion, 339. of a geometrical progression, 349. Elimination, 165. Equality, 12. Equation, 64. affected quadratic, 268. complete quadratic, 268. conditional, 64. equivalent, 174, 290. exponential, 380. identical, 64. in the quadratic form , 286. incomplete quadratic, 266. inconsistent, 174. 405 406 INDEX Equation, independent, 173. irrational, 252. linear, 65. pure quadratic, 266. quadratic, 266. simple, 65. simultaneous, 174. solution of an, 64. systems of, 174. Equilibrium, 165. Equivalent equation, 174. Exponent, 41. in the fractional form, 223. negative, 222. zero, 221. Exponential equation, 380. Expression, algebraic, 20. homogeneous, 45. integral, 99. mixed, 125. prime, 100. rational, 99. Extraneous roots, 181. Extremes, of a proportion, 317. Factor, 12, 99. common, 119. highest common, 119. theorem, 385. Formula, 87. binomial, 359. physical, 164, 293. quadratic, 273. Fourth proportional, 318. Fourth root, 99. Fraction, algebraic, 122. complex, 144. terms of a, 122. Fractions, clearing of, 147. General number symbols, 9. Geometric means, 354. Geometric progression, 348. Graph, of a linear equation, 200. of a point, 197. of a quadratic equation in one vari- able, 282. of a quadratic equation in two vari ables, 302. Grouping, law of, 22, 39. Identical equation, 64. Identity, 64. Imaginaries, conjugate, 260. unit of, 254. Imaginary number, 254. pure, 254. Inconsistent equation, 174. Independent equation, 174. Indeterminate equation, 173. Index law, 42, 53, 205, 210. Indicated operations, 12. Integral expression, 99. Interpolation, 369, 373. Inverse ratio, 316. Inverse variation, 332. Inversion, 320. Involution, 205. Irrational equation, 252. Joint variation, 332. Law, associative, 22, 39. commutative, 22, 39. distributive, 39, 52. Linear equation, 65. Literal expression, 20. Literal number symbols, 9. Logarithm, 364. base of a, 364. characteristic of a, 366. common, 365. mantissa of a, 366. Lowest common denominator, 129. Lowest common multiple, 127. Mantissa of logarithm, 366. Mean proportional, 318. Means, arithmetical, 344. geometrical, 354. of a proportion, 317. Mixed expression, 125. Moment, 165. Monomial, 21. Multiple, common, 127. lowest common, 127. Multiplicand, 38. Multiplication, 38. Multiplier, 38. Negative exponents, 222. INDEX 407 Negative numbers, 14. Number, algebraic, 16. negative, 14. Numerator, 126. Numerical expression, 20. Order, ascending, 44. descending, 44. law of, 22, 38. Ordinate, 197. Origin, 197. Parenthesis, 13, 26. Physical formulas, 164, 293. Polynomial, 21. Power, 41. Prime expression, 100. Principal root, 237. Problem, 70. discussion of a, 193. solution of a, 71. Progression, arithmetical, 338. geometrical, 348. Proportion, 317. continued, 318. extremes of a, 317. means of a, 317. Pure imaginary, 254. Pure quadratic equation , 266. Quadrants, 198. Quadratic equation, 266. Quadratic form, 286. Quadratic formula, 273. Quality, signs of, 14. Quotient, 52. Radical, 236. index of a, 209, 236. sign of, 209. similar, 242. Radicand, 236. Ratio, common, 348. compound, 316. duplicate, 316. inverse, 316. Rational expression, 99. Rationalization, 246. Real number, 254. Reciprocal, 140. Rectilinear coordinates, 197. Remainder, 32. Root, 41, 65. cube, 99, 394, 396. square, 99. Roots, rejection of, 291. Series, finite, 258. infinite decreasing, 352. Signs, of aggregation, 29. of quality, 14. Similar terms, 20. Simple equation, 65. Simultaneous equations, 174. Solution of equations, 64. Square root, 99. Substitution, 84. Subtraction, 32. Subtrahend, 32. Sum, 22. Surd, 236. coefficient of a, 236. entire, 236. mixed, 236. Surds, conjugate, 247. Symbols of operation, 11. Systems of equations, 174. Term, 20. degree of a, 44. Terms, of a fraction, 122. similar, 20. Third proportional, 318. Transposition, 66. Trinomial, 21. Unit of imaginaries, 254. Variable, 200, 330. Variation, compound, 333. direct, 331. direct and inverse, 332. inverse, 332. joint, 332. Verification of a root, 67. Vinculum, 17. | Zero exponent, 221. ROBBINS'S GEOMETRY By EDWARD RUTLEDGE ROBBINS, A.B., Senior Mathematical Master, The William Benn Charter School, Philadelphia. Plane and Solid Geometry, $1.25 Plane Geometry . . . $0.75 Solid Geometry . . . #0.75 THIS text-book is intended to meet the needs of all secondary- schools and the requirements in geometry for entrance to all colleges and universities. It is clear, consistent, teachable, and sound. ^j So suggestively and comprehensively is the work outlined that the teacher is saved many explanations, while the pupil receives the help he desires wherever it is required. The preliminary matter is intentionally brief and simple, so that the theorems and their demonstrations may be reached as early as possible. Each theorem is employed in the demonstration of other theorems as promptly as is practicable and desirable, ^j The book has been so constructed that it meets the various degrees of intellectual capacity and maturity in all ordinary classes. The successive truths in a demonstration are stated, and the pupil is asked the reasons. But the latter is not left in ignorance if he is unable to perceive the correct reason, because the numbers of the paragraphs containing these desired truths are cited wherever experience has shown that the pupil is likely to require this assistance. ^j The conciseness and rigor of demonstration, the great wealth of original exercises, both classified and graded, the economy of arrangement, the full treatment of measurement, and the superior character of the diagrams distinguish the book. Pre- ceding the earlier collections of original exercises are summaries, while useful formulas are grouped both in the Plane and in the Solid. All the theorems in the Plane Geometry are stated at the beginning of the Solid. AMERICAN BOOK COMPANY C65) ELEMENTS OF TRIGONOMETRY By ANDREW W. PHILLIPS, Ph.D., and WENDELL M. STRONG, Ph.D., Professors in Yale University Plane and Spherical Trigonometry. With Tables $i-4Q The same. Without Tables 90 Logarithmic and Trigonometric Tables 1.00 IN this text-book full recognition is given to the rigorous ideas of modern mathematics in dealing with the funda- mental series of trigonometry. Both plane and spherical trigonometry are treated in a simple, direct manner, free from all needless details. The trigonometric functions are defined as ratios, but their representation by lines is also introduced at the beginning, because certain parts of the subject can be treated more simply by the line method, or by a combination of the two methods, than by the ratio method alone. ^J Many valuable features distinguish the work, but attention is called particularly to the graphical solution of spherical triangles, the natural treatment of the complex number, and the hyperbolic functions, the graphical representation of the trigonometric, inverse trigonometric, and hyperbolic functions, the emphasis given to the formulas essential to the solution of triangles, the close and rigorous treatment of imaginary quan- tities, the numerous cuts which simplify the subject, and the rigorous chapter on computation of tables, •fl" Carefully selected exercises are given at frequent intervals, affording adequate drill just where it is most needed. An exceptionally large number of miscellaneous exercises are included in a separate chapter. ^[ The tables include, besides the ordinary five-place tables, a complete set of four-place tables, a table of Naperian logarithms, tables of the exponential and hyperbolic functions, a table of constants, etc. AMERICAN BOOK COMPANY (70) ELEMENTS OF DESCRIPTIVE GEOMETRY By CHARLES E. FERRIS, Professor of Mechanical Engineering, University of Tennessee #1.25 THE leading engineers and draughtsmen, as investigation shows, do nearly all their work in the third quadrant or angle. It seems reasonable, therefore, that the subject of descriptive geometry should be taught in technical and scientific schools as it will be used by their graduates, •[f Many years of experience in teaching descriptive geometry have proved to the author that the student can learn to think with his problem below the horizontal, and behind the vertical and perpendicular planes, as well as above and in front of those planes. ^[ This volume forms an admirable presentation of the subject, treating of definitions and first principles ; problems on the point, line, and plane ; single curved surfaces ; double curved surfaces ; intersection of single and double curved surfaces by planes, and the development of surfaces ; intersection of solids ; warped surfaces ; shades and shadows ; and perspective. ^| Besides dealing with all its problems in the third angle instead of in the first, the book presents for each problem a typical problem with its typical solution, and then gives numerous examples, both to show variations in the data, and to secure adaptability in the student. In consequence, no sup- plementary book is necessary. *H To show the projections on the horizontal, and on the vertical planes, it uses v and h as exponents or subscripts instead of the usual method of prime, etc. ^[ In scope the treatment is sufficiently broad, and yet it is not so abstruse as to make the book difficult for the average college course. Both text and plates are bound together, thus being very convenient for use. There are 1 1 3 figures. AMERICAN BOOK COMPANY (63) PLANE SURVEYING #3.00 By WILLIAM G. RAYMOND, C. E., member Ameri- can Society of Civil Engineers, Professor of Geodesy, Road Engineering, and Topographical Drawing in Rensselaer Polytechnic Institute. IN this manual for the study and practice of surveying the subject is presented in a clear and thorough manner; the general method is given first and afterward the details. Special points of difficulty have been dwelt on wherever necessary. 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By ARTHUR MAYER WOLFSON, Ph.D., First Assistant in History, DeWitt Clinton High School, New York. ESSENTIALS IN MEDIEVAL AND MODERN HISTORY $1.50 From Charlemagne to the present day. By SAMUEL BANNISTER HARDING, Ph.D., Professor of Euro- pean History, Indiana University. ESSENTIALS IN ENGLISH HISTORY . . fi.50 From the earliest records to the present day. By ALBERT PERRY WALKER, A.M., Master in His- tory, English High School, Boston. ESSENTIALS IN AMERICAN HISTORY . $1.50 From the discovery to the present day. By ALBERT BUSHNELL HART, LL.D., Professor of History, Harvard University. THESE volumes correspond to the four subdivisions re- quired by the College Entrance Examination Board, and by the New York State Education Department. Each volume is designed for one year's work. 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HOADLEY, A.M., C.E., Professor of Physics, Swarthmore College A COURSE, containing a reasonable amount of work for an academic year, and covering the entrance require- ments of all of the colleges. It is made up of a reliable text, class demonstrations of stated laws, practical questions and problems on the application of these laws, and laboratory experiments to be performed by the students. ^J The text, which is accurate and systematically arranged, presents the essential facts and phenomena of physics clearly and concisely. While no division receives undue prominence, stress is laid on the mechanical principles which underlie the whole, the curve, electrical measurements, induced currents, the dynamo, and commercial applications of electricity. •J]" The illustrative experiments and laboratory work, intro- duced at intervals throughout the text, are unusually numerous, and can be performed with comparatively simple apparatus. 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This book is DUE on the last date stamped below. 12Nov'53Vl OCT 2 91953 RETURNED TO BECLCHl OCT 115 LD 21-100m-7,'52(A2528sl6)476 M306035 ^66 THE UNIVERSITY OF CALIFORNIA LIBRARY .y.n£;.Sf RKELE Y LIBRARIES 1 III ill! mSm iiii ■ 1! HI l ill I I