cca oe DARREN abe Rates MATTER AVN ma s: Sencar y o < Sie ae, baie teas eae e = Sarr en = it Se Fe Set a 2 LOOM POG pte Le i, e Ras melee 2 = 5 . , read is greater than, indicates that the num- ber which precedes the sign is greater than the number which follows it. Thus, 8+4-> 10 means that 8+ 4 is greater than 10. The sign <, read zs less than, indicates that the number which precedes the sign is less than the number which fol- lows it. Thus, 8+4<16 means that 8+4 is less than 16. 10. Signs of Deduction and of Continuation. The sign .. stands for the word “therefore” or “hence.” The sign stone or -———-— stands for the words ‘and so on.” 11. Number-Symbols in Algebra. Algebra, like Arith- metic, treats of numbers, and employs the letters of the alphabet in addition to the figures of Arithmetic to represent INTRODUCTION. 5 numbers. The letters of the alphabet are used as general symbols of numbers to which any particular values may be assigned. In any particular problem, however, a letter must be supposed to have the same particular value throughout the investigation or discussion of the problem. These general symbols are of great advantage in investi- gating and stating general laws; in exhibiting the actual method in which a number is made up; and in represent- ing unknown numbers which are to be discovered from their relations to known numbers. . The advantage of representing numbers by letters will be more clearly seen later on. For the present it will be sufficient for the beginner to understand that every letter, and every combination of letters, and every combination of figures and letters used in Algebra, represents some number. Thus, the number of dollars in a package of bank-notes can be represented by x; but if the package consists of ten- dollar bills, five-dollar bills, two-dollar bills, and one-dollar bills, and if we denote the number of ten-dollar bills by a, of five-dollar bills by 0, of two-dollar bills by ¢, and of one- dollar bills by d, the whole number of dollars in the package will be represented by 10a+ 56+ 2c+d. In this particular case and 10a+56+2ec+d both stand for the same number. 12. Substitution. It is obvious that the same operation on each of the above expressions will produce results that agree in value, and therefore that either may be substituted for the other at pleasure. In short, Every algebraic expression represents some number, and may be operated upon as uf it were a single symbol standing for the number which tt represents. 18, Factors. When a number consists of the product of two or more numbers, each of these numbers is called a factor of 6 SCHOOL ALGEBRA. the product. If these numbers are denoted by letters, the sign X is omitted. Thus, instead of a x b, we write ad. 14, Coefficients. A known factor prefixed to another factor to show how many times that factor is taken is called a coefficient. 15. Powers. A product consisting of two or more equal factors is called a power of that factor. The index or exponent of a power is a small figure placed at the right of a number, to show how many times the number is taken as a factor. Thus, 2‘ is written instead of 2xX2x2~x 2; a instead of aaa. The second power of a number is generally called the square of that number; the third power of a number, the cube of that number. 16. Roots. The root of a number is one of the equal fac- tors of that number; the sguare root of a number is one of the éwo equal factors of that number; the cube voot of a number is one of the three equal factors of that number ; and so on. The sign \/, called the radical sign, indicates that a root is to be found. Thus, 4, or V4, means that the square root of 4 is to be taken; 8 means that the cube root of 8 is to be taken; and so on. The figure written above the radical sign is called the index of the root. 17. An algebraic expression is a number written with alge- braic symbols; an algebraic expression consists of one sym- bol, or of several symbols connected by signs of operation. A term is an algebraic expression the parts of which are not separated by the sign of addition or subtraction. Thus, 3ab, 52x 4y, 8ab+4 zy are terms. INTRODUCTION. 3 A simple expression is an expression of one term. A compound expression is an expression of two or more terms. 18. Positive and Negative Terms. The terms of a com- pound expression preceded by the sign + are called posi- tive terms, and the terms preceded by the sign — are called negative terms. The sign ++ before the first term is omitted. 19, Parentheses. If a compound expression is to be treated as a whole it is enclosed in a parenthesis. Thus, 2x (10+ 5) means that we are to add 5 to 10 and multiply the result by 2; if we were to omit the parenthesis and write 2x 10+ 5, the meaning would be that we were to multiply 10 by 2 and add 5 to the result. Instead of parentheses, we use with the same meaning brackets [ ], braces {?, and a straight line called a vinculum. | Thus, (5+ 2), [5+ 2], {5+ 23, 5+2, ave) all mean that the expression 5-+ 2 is to be treated as the single symbol 7. 20. Rules for removing Parentheses. If a man has 10 dol- lars and afterwards collects 8 dollars and then 2 dollars, it makes no difference whether he adds the 8 dollars to his 10 dollars, and then the 2 dollars, or puts the 8 and 2 dollars together and adds their sum to his 10 dollars. The first process is represented by 10+ 3+ 2. The second process is represented by 10-+(8 + 2). Hence 10+(8+ 2)=10+ 3+ 2. (1) If a man has 10 dollars and afterwards collects 3 dol- lars and then pays a bill of 2 dollars, it makes no differ- ence whether he adds the 8 dollars collected to his 10 dollars and pays out of this sum his bill of 2 dollars, or pays the 2 dollars from the 3 dollars collected and adds the remainder to his 10 dollars. 8 SCHOOL ALGEBRA. The first process 1s represented by 10+ 3 — 2. The second process is represented by 10 + (8 — 2). Hence 10+(38—2)=10+3—2. (2) From (1) and (2) it follows that if a compound expres- sion is to be added, the parenthesis may be removed and each term in the parenthesis retain its prefixed sign. If a man has 10 dollars and has to pay two bills, one of 3 dollars and one of 2 dollars, it makes no difference whether — he takes 8 dollars and 2 dollars in succession, or takes the 3 and 2 dollars at one time, from his 10 dollars. The first process is represented by 10—3— 2. The second process is represented by 10 —(8+ 2). Hence 10—(8+ 2)=10—38 — 2. (3) If a man has 10 dollars consisting of 2 five-dollar bills, and has a debt of 3 dollars to pay, he can pay his debt by giving a five-dollar bill and receiving 2 dollars. This process is represented by 10 — 5 + 2. Since the debt paid is three dollars, that is, (6—2) dol- lars, the number of dollars he has left can evidently be expressed by 10 —(5~2) Hence 10 — (5 — 2) = 10—5+4 2. (4) From (8) and (4) it follows that if a compound expres- sion is to be subtracted, the parenthesis may be removed, provided the sign before each term within the parenthesis is changed, the sign + to —, and the sign — to +. Exercise 1. Perform the operations indicated, and simplify . 1 7+(8—2). 4. 5x(2+8). 7. (7—38)x (5—2). 2. 7T—(8--2). 5. (54+8)+2. 8. (8—2)+(5—2). 38. 7—(8+2). 6 5x(38—2). 9. 3xX(2—6—2). INTRODUCTION. 9 21, Fundamental Laws of Numbers. We are so occupied in Arithmetic with the application of numbers to the ordi- nary problems of every-day life that we pay little attention to the investigation of the fundamental laws of numbers. It is, however, very important that the beginner in Algebra should have clear ideas of these laws, and of the extended meaning which it is necessary to give in Algebra to cer- tain words and signs used in Arithmetic; and that he should see that every such extension of meaning is con- | sistent with the meaning previously attached to the word or sign, and with the general laws of numbers. We shall, therefore, give general definitions for the fundamental oper- ations upon numbers and then state the laws which apply to them. 22, Addition. ‘The process of finding the result when two or more numbers are taken together is called addition, and the result is called the sum. 28, Subtraction. The process of finding the result when one number is taken from another is called subtraction, and the result is called the difference or remainder. ‘The number taken away is called the subtrahend, and the number from which the subtrahend is taken is called the minuend. In practice the difference is found by discovering the number which must be added to the subtrahend to give the minuend. Ifthe subtrahend consists of two or more terms, - we add these terms and then determine the number which must be added to their sum to make it equal to the minu- end. Thus, if a clerk in a store sells articles for 10 cents, 15 cents, and 30 cents, and receives a dollar bill in pay- ment, he makes change by adding these items and then adding to their sum enough change to make a dollar. From the nature of this process it is obvious that the general laws of numbers which apply to addition apply 10 SCHOOL ALGEBRA. also to subtraction, and that we may take for the general definition of subtraction The operation of finding from two given numbers, called minuend and subtrahend, a third number, called difference, which added to the subtrahend will give the minuend. 24. Multiplication. The process of finding the result when a given number is taken as many times as there are units in another number is called multiplication, and the re- sult is called the product. This definition fails when the multiplier is a fraction, for we cannot take the multiplicand a fraction of a tume. We therefore consider what extension of the meaning of multi- plication can be made so as to cover the case in question. When we multiply by a fraction we divide the multiplicand into as many equal parts as there are units in the denomi- nator and take as many of these parts as there are units in the numerator. If, for instance, we multiply 8 by #, we divide 8 into four equal parts and take three of these parts, getting 6 for the product. We see that ¢ is ¢ of 1, and 6 is $ of 8; that is, the product 6 is obtained from the mul- tiplicand 8 precisely as the multiplier 2 is obtained from 1. The same is true when the multiplier is an integral number. Thus, in 6 X 8= 48, the multiplier 618 1-14 1-4-1 ae and the product 48 is 8+8+8+48+48+8. We may, therefore, take for the general definition of multiplication The operation of finding from two given numbers, called multiplicand and multepler, a third number called product, which is formed from the multiplicand as the multipher is formed from unity. INTRODUCTION. ‘ 11 25. Division. To divide 48 by 8 is to find the number of times it 1s necessary to take 8 to make 48. Here the product and one factor are given and the other factor is required. We may therefore take for the general definition of division The operation by which when the product and one factor are given the other factor is found. With reference to this operation the product is called the dividend, the given factor the divisor, and the required: factor the quotient. 26. The Commutative Law. If we have a group of 3 things and another group of 4 things, we shall have a group of 7 things, whether we put the 3 things with the 4 things or the 4 things with the 3 things; that is, 4+38=344. It is evident that the truth of the above statement does not depend upon the particular numbers 3 and 4, but that the statement is true for any two numbers whatever. Thus, in case of any two numbers we shall have First number + second number = second number + first number. If we let a stand for the first number and 6 for the second number, this statement may be written in the much shorter form atb=b-+a. This is the commutative law of addition, and may be stated as follows: Additions may be performed in any order. 27. Also, if we have 5 lines of dots with 10 dots in a line, the whole number of dots will be expressed by 5 x 10. 12 ; SCHOOL ALGEBRA. If we consider the dots as 10 columns with 5 dots in a column, the number will be expressed by 10 x 5. That 1s, DLO Sh eco Again, if we divide a given length into 6 equal parts, 3 eel 1 2 one-third of the line will contain 2 of these parts, and one- half the line will contain 8 of these parts. Now one-third of one-half will be 1 of these parts, and one-half of one- third will be 1 of these parts; that is, tXt=tXF. Therefore, if a and 6 stand for any two numbers, integral or fractional, we shall have ab = ba. This is the commutative law of multiplication, and may be stated as follows: Multipheations may be performed in any order. 28. The Distributive Law. The expression 4 x (5+ 38) means that we are to take the sum of the numbers 5 and 3 four times. ‘The process can be represented by placing five dots in a line, and a little to the right three more dots in the same line, and then placing a second, third, and fourth line of dots underneath the first line and exactly similar to it, eocee50 ° INTRODUCTION. 13 There are (5-++ 3) dots in each line, and 4 lines. The total number of dots, therefore, is 4 x (5 + 8). We see that in the left-hand group there are 4 x 5 dots, and in the right-hand group 4 x 3 dots. The sum of these two numbers (4 X 5)-+ (4 x 3) must be equal to the total number; that is, 4x (5+3)=(4x 5)+(4x 8). Again, the expression 4 x (8 — 3) means that 3 is to be taken from 8, and the remainder ‘to be multiplied by 4. The process can be represented by placing eight dots in a line and crossing the last three, and then placing a second, third, and fourth line of dots underneath the first line and exactly similar to it. e@ee8 @ eee @ eee ®@ e@e38 @ e@®ee @ ew ea ea eB ee Oe ee OB ew ee 8 OB The eile. number of dots not crossed in each line is evidently (8 —3), and the whole number of lines is 4. Therefore the total number of dots not crossed is 4x (8—8). The total number of dots (crossed and not crossed)-is (4 x 8), and the total number of dots crossed is (4 x 8). Therefore the total number of dots not crossed is Eee AO) (a8) ; that is, 4x (8—3)=(4 x 8)—(4 x 3). Hence, by the commutative law (8 —3)x4=(8 x 4)—(8 x 4). In like manner, if a, 6, ¢, and d stand for any numbers, we have ax(b+e—d)=ab+ace—ad, 14 SCHOOL ALGEBRA. This is the distributive law, and may be stated as follows: In multiplying a compound expression by a simple ex- pression the result is obtained by multiplying each term of the compound expression by the simple expression, and writ- mg down the successwe products with the same signs as those of the original terms. 29. The Associative Law. The terms of an expression may be grouped in any manner. For if we have several num- bers to be added, the result will evidently be the same, whether we add the numbers in succession or arrange them in groups and add the sums of these groups. Thus, at+btetdte =a-+(b+¢)+(d +e) =(a+8)+(et+dto) Likewise, if in the rectangular solid represented in the margin we suppose AB to contain 5 units of length, BC 3 units, and CD 7 units. The base may be divided into square units. There will be 3 rows of 5 square units each. Upon each square unit a cubic unit may be formed, and we shall have (8x5) cubic units. Upon these another tier of (3 x 5) cubic units may be formed, and then another tier of the same number, and the process continued until we have 7 tiers of (8 X 5) cubic units. Hence the number of cubic units in the solid will be represented by 7 x (8 x 5). Upon the right-hand square in the back row a pile of 7 cubic units may be formed, upon the next square to the left another pile of 7 cubic units may be formed, and upon the next square another, and the process continued until we have a pile of 7 cubic units on each square in the NENENE NENT EEN] INTRODUCTION. 15 back row. We shall then have (5 x 7) cubic units in the back tier, and as we can have 8 such tiers, the number of: cubic units in the solid will be represented by 3 x (5 x 7). Again, if we form a pile of 7 cubic units on the right-hand square of the back row, then another pile of 7 cubic units on the next square in front, another pile of 7 cubic units on the next square in front, we shall have a tier of (3 x 7) cubic units. We can have 5 such tiers, and the number of cubic units in the solid will now be represented by 5 x (8 X 7). It follows, therefore, that the total number of cubic units in the solid may be represented by 7X(8x 5), or by 3X(5~x 7), or by 5X (8X 7). It is obvious that no part of this proof depends upon the particular numbers 3, 5, and 7, but the law holds for any arithmetical numbers whatever, and may be expressed by Oxiax b)—aXx (bx ¢c)=5 X(aXc). This is called the associative law of addition and multi- plication, and may be stated as follows: The terms of an expression, or the factors of a product, may be grouped nm any manner. 30. The Index Law. Since a =aa, and a’ = aaa, oer ax a a a 6 NENT CEG TT cai | ee | haan If a stands for any number, and m and 2 for any integers, since a”= aaa: to m factors, and a” = aaa: to factors, a” X a" = (aaa ---- to m factors) X (aaa..... to n factors), = aaa: to (m+n) factors, 16 . SCHOOL ALGEBRA. Hence, the index law may be stated as follows: The index of the product of two powers of the same number as equal to the sum of the indices of the factors. 31. These four laws, the commutative, the distributive, the associative, and the index laws, are the fundamental laws of Arithmetic, and together with the daw of signs, which will be explained hereafter, they constitute the fundamental laws of Algebra. 32. Quantities Opposite in Kind. If a man gains 6 dollars and then loses 4 dollars, his actual gain, or, as we com- monly say, his net gain, is 2 dollars; that is, 4 dollars’ loss cancels 4 dollars of the 6 dollars’ gain and leaves 2 dollars’ gain. If he gained 6 dollars and then lost 6 dollars, the 6 dollars’ doss cancels the 6 dollars’ gain, and his net gaz is nothing. If he gained 6 dollars and then lost 9 dollars, the 6 dollars’ gain cancels 6 dollars of the 9 dollars’ loss, and his net loss is 8 dollars. In other words, loss and gain are quantities so related that one cancels the other wholly or in part. If the mercury in a thermometer rises 12 degrees and then falls 7 degrees, the fall of 7 degrees cancels 7 degrees of the vise, and the net rise is 5 degrees. If it rises 12 de- grees and then fal/s 12 degrees, the net rise is nothing. If it vuses 12 degrees and falls 15 degrees, there is a net fall of 3 degrees. In other words, rzse and fall are quan- tities so related that one cancels the other wholly or in part. An opposition of this kind also exists in motion forwards and motion backwards; in distances measured east and distances measured wes‘; in distances measured north and distances measured south; in assets and debts; in time be- fore and time after a fixed date; and so on. | INTRODUCTION. Ly 33. Algebraic Numbers. If we wish to add 4 to 3, we begin at 4 in the natural series of numbers, Dette 804. B68 8 count 3 units forwards, and arrive at 7,the sum sought. If we wish to subtract 3 from 7, we begin at 7 in the natural series of numbers, count 3 units backwards, and arrive at 4, the difference sought. If we wish to subtract 7 from 7, we begin at 7, count 7 units backwards, and arrive at 0. If we wish to subtract 7 from 4, we cannot do it, because when we have counted backwards as far as 0 the natural series of numbers comes to an end. In order to subtract a greater number from a smaller it is necessary to asswme a new series of numbers, beginning at zero and extending to the left of zero. The series to the left of zero must proceed from zero by the repetitions of the unit, precisely hke the natural series to the right of zero; and the opposition between the right-hand series and the left-hand series must be clearly marked. This opposition is indicated by calling every number in the right-hand series a positive number, and prefixing to it, when written, the sign +; and by calling every number in the left-hand series a negative number, and prefixing to it the sign —. The two series of numbers will be written thus: ae rem ee er PO Sap a8 4 Ae ALGEBRAIC SERIES OF NUMBERS. If, now, we wish to subtract 9 from 6, we begin at 6 in the positive series, count 9 units in the negative direction (to the left), and arrive at —3 in the negative series; that is, 6—9=— 8. The result obtained by subtracting a greater number from a less, when both are positive, is always a negatwe number, + 18 SCHOOL ALGEBRA. In general, if a and } represent any two numbers of the positive series, the expression a—6 will be a positive num- ber when a is greater than 6; will be zero when a is equal to 6; will be a negative number when a is less than 0. In counting from left to right in the algebraic series num- bers werease in magnitude; in counting from right to left numbers decrease in magnitude. Thus —38, —1, 0, +2, +4 are arranged in ascending order of magnitude. 34. We may illustrate the use of algebraic numbers as follows $ =Se 0 8 90 Spee Ee a= pes D A C Suppose a person starting at A walks 20 feet to the right of A, and then returns 12 feet, where will he be? Answer: At C,a point 8 feet to the right of A; that is, 20 feet —12 feet = 8 feet; or, 20—12=8. Again, suppose he walks from A to the right 20 feet, and then returns 20 feet, where will he be? Answer: At A, the point from which he started; that is, 20 —20=0. Again, suppose he walks from A to the right 20 feet, and then returns 25 feet, where will he now be? Answer: At D, a point 5 feet to the left of A; that is, 20 —25=— 5; and the phrase ‘‘5 feet to the left of A’ is now expressed by the negative quantity, — 5 feet. 35, Every algebraic number, as + 4 or —4, consists of a sign + or — and the absolute value of the number. The sign shows whether the number belongs to the positive or negative series of numbers; the absolute value shows what place the number has in the positive or negative series. When no sign stands before a number, the sign + is always understood. Thus 4 means the same as +4, a means the same as-+a. But the sign — is never omitted. INTRODUCTION. 19 36. Two algebraic numbers which have, one the sign + and the other the sign —, are said to have unlike signs. . Two algebraic numbers which have the same absolute values, but unlike signs, always cancel each other when combined. Thus +4—4=0, +a—a=0. 37. Double Meanings of the Signs + and —. The use of the signs + and — to indicate addition and subtraction must be carefully distinguished from the use of the signs ++ and — to indicate in which series, the positive or the nega- tive, a given number belongs. In the first sense they are signs af operations, and are common to Arithmetic and Algebra: in the second sense they are signs of opposition, atid are Beateeeds in Algebra alone. 38, Addition and Subtraction of Algebraic Numbers. An algebraic number which is to be added or subtracted is often inclosed in a parenthesis, in order that the signs ++ and —, which are used to distinguish positive and negative numbers, may not be confounded with the + and — signs that denote the operations of addition and subtraction. Thus + 4 -+ (— 8) expresses the sum, and + 4 — (— 8) ex- presses the difference, of the numbers -++ 4 and — 8. In order to add two algebraic numbers we begin at the place in the series which the first number occupies and count, in the direction indicated by the sign of the second number, as many units as there are in the absolute value of the second number. Thus the sum of + 4-+ (+8) is found by counting from +4 three units in the positive direction; that is, to the right, and is, therefore, ++ 7. The sum of + 4+ (— 8) is found by counting from + 4 three units in the negative direction; that is, to the left, and is, therefore, + 1. 920 SCHOOL ALGEBRA. The sum of —4-+(4 8) is found by counting from — 4 three units in the positive direction, and is, therefore, — 1. seeee aa ee a gems | 041442 +3 44 45 +6-— The sum of — 4-+ (— 3) is found by counting from — 4 three units in the negative direction, and is, therefore, — 7. Hence to add two or more algebraic numbers, we have the following rules: Case I. When the numbers have lke signs. Find the sum of thewr absolute values, and prefix the common sign to the result. Case II. When there are two numbers with wnlcke signs. Find the difference of their absolute values, and prefix to the result the sign of the greater number. Case III. When there are more than two numbers with ~ unlike signs. Combine the first two numbers and thas result with the third number, and so on; or, find the sum of the positive numbers and the sum of the negative numbers, take the difference between the absolute values of these two sums, and prefix to the result the sign of the greater sum. 39. The result is called the sum. It is often called the algebraic sum, to distinguish it from the arithmetical sum, that is, the sum of the absolute values of the numbers. 40, Subtraction. In order to subtract one algebraic num- ber from another, we begin at the place in the series which the minuend occupies and count in the direction opposite to that wndicated by the sign of the subtrahend as many units as there are in the absolute value of the subtrahend. Thus, the result of subtracting + 3 from + 4 is found by counting from +4 three units in the negatwe durection; that is, in the direction opposite to that indicated by the sign + before 8, and is, therefore, +1. INTRODUCTION. 21 The result of subtracting —3 from +4 is found by count- ing from +4 three units in the positive direction; that is, in the direction opposite to that indicated by the sign — be- fore 3, and is, therefore, -+- 7. The result of subtracting +3 from —4 is found by count- ing from —-4 three units in the negative direction, and is, therefore, — 7. The result of subtracting —-3 from —4 is found by count- ing from —4 three units in the positive direction, and is, . therefore, — 1. Collecting the results obtained in addition and subtrac- tion, we have ADDITION. SUBTRACTION. +4-+-(—3)=+4—3=-+1. peeeeete eee} eereidemecr Ly Pers t448=4+7. +4. (—8)=44438=+47. pee 48> 7 4 (4 8)a + 4 827. eee ye 43 i, | 4 (8) 44 81 No part of this proof depends upon the particular num- bers 4 and 3, and hence we may employ the general symbols a and 6 to represent the absolute values of any two ME braic numbers. We shall then have ADDITION. SUBTRACTION. +a+(—b)=+a—6. +-a—(+6)=+a—6. (1) +a+(+6)=+a-+0. +a—(—b)=+a+b6. (2) —a+(—b)=—a—b. —a—(+6)=—a—b. (8) —a+(+6)=—a+b. —a—(—b)=—a+b. (4) From (1) and (8), it is seen that subtracting a positwe wumber is equivalent to adding an equal negative number. From (2) and (4), it is seen that subtracting a negative number is equivalent to adding an equal positive number. pas SCHOOL ALGEBRA. To subtract one algebraic number from another, we have, therefore, the following rule : Change the sign of the subtrahend, and add the subtra- hend to the minuend. This rule is consistent with the definition of subtraction given in § 23; for, if we have to subtract — 4 from +3, we must add +4 to the subtrahend —4 to cancel it, and then add + 3 to obtain the minuend; that is, we must add +7 to the subtrahend to get the minuend, but +7 is obtained by changing the sign of the subtrahend —4, making it +4, and adding it to +3, the minuend. 41, The commutative law of addition applies to algebraic numbers, for +4-+(— 3)=—38-+(+4). In the first case we begin at +4 in the series, count three units to the left, and arrive at +1; in the second case we begin at —8 in the series, count four units to the right, and arrive at +1. The associative law, also, of addition is easily seen to apply to algebraic numbers. 42. Multiplication and Division of Algebraic Numbers. By the definition of multiplication, § 24. Since +38=+1+4+1+1; “3X (+ 8)=4+8+4848 =-+ 24, and 3x (—8)=—8—8—8 = — 24, Again, since —s=—1—-1-l1; “.(—38)X 8=—8—8-—8 = — 24, and = (= 8)x(—8)=—=(—8)—(- 8) -( 8) =+8+8+8 Sao. INTRODUCTION. ; 23 No part of this proof depends upon the particular num- bers 3 and 8. If we use a to represent the absolute value of any number, and 6 to represent the absolute value of any other number, we shall have (+a) x (+ 6)=+ ab. | (1) (+a) x (— 6) =— ab. (2) (=a) x (+ 6) = — ab. (3) (— a) X (—b)=+ ab. (4) » 43, Law of Signs in Multiplication. From these four cases it follows that, in finding the product of two algebraic numbers, Lake signs give +, and unlwke signs gwe —. 44, Law of Signs in Division. Since (+a) x(+6)=+ab, «. +ab+(+a)=+6. Since (+a) x (— 6) =—ab, «. —ab+(+a)=—8. Since (—a) X (+0)=—ab, .. —ab+(—a)=+0. Since (—a) x (—b)=+ ab, «. +ab-+(—a)=—0b. That is, if the dividend and divisor have like signs, the quotient has the sign-+; and if they have unlike signs, the quotient has. the sign —. Hence, in division, TInke signs gwe +; unlike signs gwe —. 45. From the four cases of multiplication that we have given in § 42 it will be seen that the absolute value of each product is zndependent of the signs, and that the signs are independent of the order of the factors. Hence the com- mutative and associative laws of multiplication hold for all algebraic numbers. 94 SCHOOL ALGEBRA. 46. The distributive law also holds; for, if a(b + c)=ab sen then —a(b+c)=—ab—ae, and (b +c) (— a) =b(—a) +e(—a). Therefore, for all values of a, 6, and e, a(b+c)=ab+ae. From the nature of division the distributive law which applies to multiplication applies also to division. 47. We have now considered the fundamental laws of Algebra, and for convenience of reference we formulate them below: a+t(b+c)=a+b+e a+(b—c)=a+b—e % : (1) a—(b+c)=a—b—e a—(b—c)=a—b+e | (+4) X (+ 6) =+ ab Ca) x (— 0) =) (—a) x (+ 6) =— ab | (— a) x (— 6) =+ ab The commutative law: Addition atb=b+a . (3) Multiplication ab = ba The associative law: | Addition ee = ae (4) Multiplication a(be) = (ab)ec=abe INTRODUCTION. Db) The distributive law: Multiplication a(b+c)=ab-+ ae rnees Tae Ae _ Division ecicte ita (5) a et The index law: Multipheation iat lia NOeeW nh a Semen mame a can 6) These laws are true for all values of the letters, but in (6) m and n are for the present restricted to positwe integral values. 48. Value of an Algebraic Expression. Every algebraic ex- pression stands for a number; and this number, obtained by putting for the several letters involved the numbers for which they stand, and performing the operations indicated by the signs, is called the value of the expression. In finding the values of algebraic expressions, the begin- ner must be careful to observe what operations are actually indicated. Thus, 4ameansat+ta+a-+a; that is, 4 xa. a*meansaxXaxXaxXa. Vabe means the square root of the product of a, }, and. /abe means the product of the square root of a by be. Nore. The radical sign V before a product, without a vinculum or a parenthesis, affects only the symbol immediately following it. Va-+ 6 means that d is to be added to the square root of a. Va-+6 means that b is to be added to a and the square root of the sum taken. 49, In finding the value of a compound expression the operations indicated for each term must be performed before the operation indicated by the sign prefixed to the term, 26 SCHOOL ALGEBRA. Indicated divisions should be written in the fractional form, and the sign X omitted between a figure and a letter, or between two letters, in accordance with algebraic usage. Thus, (6 — ec) 2 X e+ 26 should be written Notr. The line between the numerator and denominator of the fractions serves for a vinculum, and renders the parenthesis un necessary. If 6=4, and c= — 4, the numerical value is A eearac ria eee =") 4 96 = 9.4 26 =—1+4 26= 25, Spa aly a um cag YL in Exercise 2. Notz. When there is no sign expressed between single symbois or between compound expressions, it must be remembered that the sign understood is the sign of multiplication. If a=1, b= 2, and .¢=.3, find the valwer 1. Ta— be. 5. 2a—b+e. 9. /4abe. 2. ac+b. 6. ab+ be— ae. 10. V6abe. 3. 4ab—c. 7. P+a+e’. 11. &— 8. 4. 6ab—b—ec. 8. 2ab— 5d5be’. 12. Ve—ai. 13. a—2(b+c). 16. 66—10bc+12a+2e. 14. (a+ 6)+2(c—a). 17. 5¢e+(b—a)- (b+a). 15. V6bce—(b —c). 18. V62. If a=1, b=2, c=8, and d= 0, find the value cf 19. Ta—be+ 6d. 21. 4ab—cd—d. 20. ac+b—d, 22. 2a—bte, INTRODUCTION. yA 23. ab+ be — ad. 24. 2ab— d5bc. 25. W4abcd. 26. V4abcd. 21. 28. 29. 30. b—c+d. a—2(b+e). 2b (3 —5e)+(a— 2c). 2 (a+ by’. Exercise 3. Remove the parentheses (§ 20), and find the algebraic sum of (6—2)+(3+)). =o) (2—8). (—844)—(245). —6—(2—3-—1). 2—(5—7+8). 8 —(7—5+4). 10 —(5— 6— 7). 2—(8—38+4). (5 — 10) +(8 — 2). —7+(8+2-—4). ao =) fewest )) o= 2, and ¢=— 3, 21. at+b-+e. 22. a—b+te. 23. a—b—ce. 24. 25. 26. ah OR (HES Gy: Wee pe 5). Mees Ae 8), Ry by Beep |) Pe (be 10 = 38). = Heh eiea:) Fy Homepage dy 4. Cre ly aee tt 10), eal Seth 1}, find the value of a—(—b)+e. a—(—b)—e. (Sa)-P (By (=e): CHA Pi heghle ADDITION AND SUBTRACTION. INTEGRAL EXPRESSIONS. 50. If an algebraic expression contains only wltegral forms, that is, contains no J/edfer in the denominator of any of its terms, it is called an integral expression. Thus, e+ Tcxv’—c—5ex, 4ax—Ltbcy, are integral expressions, Wy 2 = a aad —s . ' . but As lt seis is a fractional expression. v—ab+ 6 An integral expression may have for some values of the letters a fractional value, and a fractional expression an integral value. If, for instance, a stands for $ and 6 for 1, the integral expression 2a — 506 stands for $—$=1; and the fractional expression - stands for 43+2=65., Integral and fractional expressions, therefore, are so named on account of the form of the expressions, and with no refer- ence whatever to the numerical value of the expressions when definite numbers are put in place of the letters. 51, A term may consist of a single symbol, as a, or may be the product of two or more factors, as 6a, ab, 5a7be. If one of the factors is an arithmetical symbol, as the factor 5 in 5a°be, this factor is usually written first, and is called the coefficient of the term; the other factors are called literal factors. Nore. By way of distinction, a factor expressed by an arithmeti- cal figure is called a numerical factor, and a factor expressed by a letter is called a literal factor. ADDITION AND SUBTRACTION. 29 52. Like Terms. ‘Terms which have the same combina- tion of “itera/ factors are called like or similar terms; terms which do not have the same combination of literal factors are called unlike or dissimilar terms. Thus, 5a°be, —7Ta’bc, abe, are like terms, but 5a%be, 5ab’c, 5abc’, are unlike terms. 53. A simple expression, that is, an expression of one term, is called a monomial. A compound expression, that is, an expression which contains two or more terms, is — called a polynomial. A polynomial which contains two terms is called a binomial, and a polynomial which contains three terms is called a trinomial. 54. A polynomial is said to be arranged according to the powers of some letter when the exponents of that letter either descend or ascend in the order of magnitude. Thus, daz — 4bz?— 6bax-+8b is arranged according to the de- scending powers of x, and 8b—6axr—462?+ 8az’* is arranged according to the ascending powers of wx. 55. Addition of Integral Expressions, The addition of two algebraic expressions can be represented by connecting the second expression with the first by the sign +. If there are no like terms in the two expressions, the operation is algebraically complete when the two expressions are thus connected. If, for example, it is required to add m+n—~p to a+b-+e, the result will ba+6+e+(m+n—~p); or, removing the parenthesis (§ 20),a+d+e+m+n—p. 56. If, however, there are like terms in the expressions to be added, the like terms can be collected; that is, every set of like terms can be replaced by a single term with a coefficient equal to the algebraic sum of the coefficients of the like terms. 30 SCHOOL ALGEBRA. If it is required to add 5a?+4a+3 to 2a@—8a—4, the result will be 2a’— 3a—4+(5a?+ 4a+8) =2¢—38a—4+5¢+4a+38 § 20 = 207+ 5¢—38a+4a—4+3 § 26 =Ta+a—l. This process is more conveniently represented by arrang- ing the terms in columns, so that like terms shall stand in the same column, as follows: 20 —8a—4 5a+4a+38 T+ a—-l The coefficient of a in the result will be 5+ 2, or 7; the coefficient of a will be —8+4, or 1; and the last term is —4+3, or —l. Notre. When the coefficient of a term is 1, it is not written, but understood ; conversely, when the coefficient of a term is not writ- ten, 1 is understood for its coefficient. If we are to find the sum of 2a°—8@b+4ae’?+ 6°, a+ 4a7b — Tab? — 26°, — 8a*® + ab — 8ab? — 486°, and 20+ 20¢b+ 6ab?— 30°, we write them in columns, as follows” 2a°— 80% +40bt+ B a + 407b — Tab? — 26° —8a°+ ab—3a’?—48° 2a°+ 2076 + 6ab? — 36° 20+ 407 — 86° The coefficient of a in the result will be 2+1—38-442, or +2; the coefficient of a7b will be —8+4+1-+2, or +4; the coefficient of ab? will be 4—7—8-+6, or 0; and the coefhcient of 6* will be 1—2—4-— 8, or —8. ADDITION AND SUBTRACTION. dl Exercise 4, Add Ta, 2a, —8a, and — 5a. — Tay, 2xy, —4axy, and — 5xy. 4a°b, —3a’b, and — 5a’d. day, 4ay, Tax, and — 3az. a+6 and a—b. xv? —x and 2*— 2’. 5a? +62—2 and 827—Tx+2. 3827°—2ry+y and 2’? —2ry4+3y’. ax’ + be —4, 3a27—2bx+4, and — 4a2? —2bxa+ 85. 5a+3y+z2, 8x+2y+ 382, and x—d8y—5z. 11. 8ab—2a2z*+ 3a'%z, 4ab—6a'?x+5a2’, ab+ aa— az’, and az’— 8ab—5a’x. 12. at —2a° + 8a?—a+7 g 2 at mens +2a¢—a+6, and —a —2¢04+2¢— 13. 38¢0—ab+ac— 30'?+4be—¢, —5a’?—ab—ac+5be, —4b6e+ 52+ 2ab, and —40?+ 8? —5bce4+ 2c’. amg oe? 4a ty Bat 4 O28 4+ 2? — 5a — 6, —4a'+32*-—382°+9x—2, and 22t—a2’?4+2?—2+1. 15. 3a7°—42°y+2', 5a°?—llay—1222, —Ty+2°y—22’, and — 4227+ 7?— 2° 16. a —2a°+3a7, dt+a’+a, 4at+5a', 2V7+3a—2, and — a’? —2a— 83. ay ie 17. #& +227 —axy—y’, 22° —382y — 42y — Ty’, and 2° — 8 ay"? — 7. ne SCHOOL ALGEBRA. 57. Subtraction of Integral Expressions. The subtraction of one expression from another, if none of the terms are alike, can be represented only by connecting the subtra- hend with the minuend by means of the sign —. If, for example, it is required to subtract a+6-+e¢ from m+n — p, the result will be represented by m+n—p—(a+b+e); or, removing the parenthesis, § 20, mtn—p—a—b—e. If, however, some of the terms in the two expressions are alike, we can replace two like terms by a single term. Thus, suppose it is required to subtract a’—2a’+2a—1 from 3a°—2a?+a—2; the result may be expressed as follows : 380° —2¢+a—2—(a— 20+ 2a—1); or, removing the parenthesis (§ 20), 8e0—2V+a—-2—a+2¢0—2a+1 =38ae—a—2¢4+2¢0+a—2a—-241 =2a—a—l. This process is more easily performed by writing the subtrahend below the minuend, mentally changing the sign of each term in the subtrahend, and adding the two expres- sions. Thus, the above example may be written 38a°—2a+ a—2 @&—2a?+2a—1 2a — a—l The coefficient of a® will be 8—1, or 2; the coefficient of a? will be —2+ 2, or 0, and therefore the term a? will not appear in the result; the coefficient of a will be 1— 2, or --1; the last term will be —2-+-1, or —1. ADDITION AND SUBTRACTION. 338 Again, suppose it 1s required to subtract a°+ 4 a*x’— 3 aa —4az* from a*2?+ 2a’2?—4azx‘. Here terms which are alike can be written in columns, as before: a’a? + 2 a?x*® — 4az* a + 4a°2 — 8a’2* — 4az* — o°— Baz’ + 5a’2’ There is no term a in the minuend, hence the coefficient of a in the result is O—1, or —1; the coefficient of a’a? . will be 1 —4, or —3; the coefficient of a’z* will be 2+8, or +5; the coefficient of ax* will be —4-+4, or 0, and therefore the term az* will not appear in the result. Exercise 5. 1. From 8a— 46 — 2c take 2a— 36 — 8c. 2. From 8a—46+ 8c take 2a— 8b—c—d. 8. From 7a?—9a—1 take 5a°—62—83. 4. From 22?— 2axv-+ a’ take 2? — ax — a’. 5. From 4a— 36 — 8c take 2a— 36+4e. 6. From 52°+ 72+4 take 32?— 7x4 2. 7. From 2ax+ 3 by +5 take 38ax — 3 by — 5. 8. From 4a?— 6a04+ 207 take 8¢’+ab-+ 6. 9. From 407) + 7ab?+ 9 take 8 — 3a0’. 10. From 5a’ + 6a7b — 8a’ take 6° + 6a’) — 5a’e. 11. From a — 2 take 0”. 13. From 0? take a? — 0. 12. From a? — 0’ take a’. 14. From a’ take a? — 6’, 15. From 2*+ 3axz* — 262? + 38cx—4d take 82*+ az®—40?+ 6er+d. 34 SCHOOL ALGEBRA. If A=3e—200+587, C=Te—8ab+58', B=9¢—50ab4+30?, D=11ld—38ab— 4B, find the expression for 146. 4A+C+ B+ D. 19. A+ C—B-D. 17. A—C— B+ D. 20. A—C+ B+ D. 18. C—A—B+D. 21. A+C—B+D. 58, Parentheses. From the laws of parentheses (§ 20), we have the following equivalent expressions : a+(6+ec)=atb+e, «. a+b+e=a+(b6+0e); a+(b—c)=a+b—c, ».a+b—c=a+(6—e); a—(b+ce)=a—b—c, «.a—b—c=a-—(b+e); a—(b—c)=a—b+e, .«.a—b+c=a—(b—-oc); that is, if a parenthesis is preceded by the sign +, the parenthesis may be removed without changing any of the signs of the terms within the parenthesis ; conversely, any number of terms may be enclosed within a parenthesis preceded by the sign +, without changing the sign of any term. If a parenthesis is preceded by the sign —, the paren- thesis may be removed, provided the sign of every term within the parenthesis 1s changed, namely, + to — and — to -+; conversely, any number of terms may be enclosed with- in a parenthesis preceded by the sign —, provided the sign of every term enclosed is changed. 09, Iixpressions may occur with more than one paren- thesis. In such cases parentheses of different shapes are used, and the beginner when he meets with a ( or a[ ora } must look carefully for the other part, whatever may in- tervene; and all that is included between the two parts of ADDITION AND SUBTRACTION. 35 each parenthesis must be treated as the sign before it directs, without regard to other parentheses. It is best to remove each parenthesis in succession, beginning with the innermost one. Thus, (1) a0 -=(e--.0) +e] =a—[b—c+d+e] =a—btc—d—e. (2) Pane sacra ti ey 7 |, =a—{b—[e—d+et/} =a—{b—c+d—e—f} =a—b+c—d+e4+f. Exercise 6. : Simplify the following by removing the parentheses and collecting like terms: 1. a—b—[a—(b—c)—el]. m —|n—(p —m)]}. 20 —fy + [42—(y + 22)]}. 8a —{2b—[5ce—(8a+4)]}. a—{b+[e—(d—b)+a]— 28}. 32 —[9—(224 7)4+ 32]. 2a —[y—(%—2y)]}. a—[26+(8¢—26)+a]. (a—2 +y)—(b—2—y) + (a+b —2y). 8a —[—4b+(4a —6)—(2a—5d)]. . 4c—[a—(26 —3c)+c]+[a—(26—5e—a)]. . ©+(y—2)—[B2—2y) +2]+[2—y—2)] . a—[2a+(a—2a)+ 2a]—5a—{6a—[(a+2a)—-]}}. Oo pets oy HO! lomo ee — et Oo te St CS 36 SCHOOL ALGEBRA. 14. 2x—(8y+2)—}b—(e—b)+e - [a—(e—B)}}. 15. a—[b+¢—a—(a+6b)—c]4+(2a—b+e). ‘Nors. The sign — which is written in the above problem before the first term 6 under the vinculum is really the sign of the vinculum, —b +c meaning the same as —() + ¢). 16. 10—x#%—j{—x—[2—(*4#—5—2z)}]}. 17. 2e—f2et(y—z)—824[24—(y—z—2y)—82]+4y}. 18. a—[b—{—ce+a—(a—b)—c}]+[2a—(6—a)]. 19. a— jb —[a—(e—b) +e—a—(a—b —c)—a]+ a}. 20. 5a—{—8a—[8a—(2a—a—b)—a]+a}. Exercise 7. v In each of the following expressions enclose the last three terms in a parenthesis preceded by the sign —, remember- ing that the sign of each term enclosed must be changed. 1. 2a—b—38ce—d+38e-—5f. x—a—y—b—z-e. at+tb—ce+4a—6-+1. ax + by + cz + ba — cy + cz. 38a+2b6+2c—5d—8e—4f. x—y+2—5xry—422+ 8 yz. Considering all the factors that precede a, y, and z, respectively as the coefficients of these letters, we may collect in parentheses the coefficients of x, y, and z in the following expression : i ax — by + ay —az—cz + be =(a + b)a+(a—b) y—(a + edz. In like manner, collect the coefficients of x, y, and z in ‘the following expressions : 7. ax+t by +cz+ bu —cy + az. 8. ax+ 2ay+ 4az— bx + By — 3 bz — Qz. ADDITION AND SUBTRACTION. 3h ax — 2by —5cz—4bz%+ 3cy — Taz. . a2+d8ay + 2by — bz —1lex+ 2cy — cz. . 4by —8axc — bez + 2be — Tex — Sey — cx — cy— ez, - 6az—5by + 38cz — 2bz — Bay + bz — ax + by. - 2—by+ 8az—38cy + 2ax— 2mx — 5 bz. . 2+ ay —az—acx + bez -mny —y —z. Exercise 8. EXAMPLES FoR REVIEW. . Add 42°—5a’?— 5az’*+ 6a'x, 60°+382°+4a2?+ 2072, 19 a2?—112?—15a’2, and 102°+-7a@2+5a*®—18 az’. . Add 8ab+38a+66, —ab+2a+4b, Tab —4a—8), and 6a+126—2ad. Note. Similar compound expressions are added in precisely the same way as simple expressions, by finding the sum of their coefh- cients. Thus, 3(¢—y) + 5(a—y)—2(a—y) =6(#—y). 3. 4. Add 4(5—2), 6(5 — x), 8(5— x), and —2(5 — 2). Add (a+ 6)2+(6b+e)¥Y+(ate)2, (6+¢)2 +(a+te)y+(a+b)2, and (a+c)2+(a+6)y7 + (6+ ¢)2*. Add (a+ 6)a+(b+c)y+(e+a)z, (6+e)ze+(e+a)x —(a+b)y, and (a+c)y+(a+6)z2—(b6+¢e)z. From a’?— 2 take a + 2ax-+ 2”. From 8a?7+2azr+ 2’ take a — ax — 2’, From 827?—3azxr+5 take 527 + 2ax+5. From a? + 3 6’c + ab? — abe take ab? — abc + 6’. From (a+ 6)2+(a+ce)y take (a—b)x—(a—e)y. . Simplify 7a—{3a —[4a—(5a—2a)]}. 388 SCHOOL ALGEBRA. 12. Simplify 8a—fa+6—[a+b+c—(a+b+c+d)}]}. 13. Bracket the coefficients, and arrange according to the descending powers of x x? —ax—c'x? — ba + ba? — cx’? + aa? — 2? — cx. 14. Simplify a’—(6’—¢’)—[&—(e—a’)|+[e—(0 —a’)]. 15. If a=], b=38, c=5, and d=7, findothesvameso: a — 2b—$3e—d—[8a—(5b—c—8d)|—28}. 16. From 2d+1lla+106—5e take 2e+5a—806. Find the value of each of these expressions when a, 8, ¢, and d have the values 1, 3, 5, 7, respectively, and show that the difference of these values is equal to the value of their difference. 17. If a=1, b=— 3, ¢-=— 5, d=0, find the’ vaineus ?+26?+ 38+ 4d". If a=3, 6=4, c=9, and 2s=a+6 +e findethe value of 18. s(s—a)(s— b)(s—e).’ 19. s?+(s—a)’+(s—6)?+(s—e). 20. s’—(s—a)(s —b) —(s — 8) (s—e) —(s—e)(s— a). 21. Ift=a+2b—3c, y=b+2c— 8a, and z=c+2a — 386, show that 7+y+2=0. 22. Ift=a—2b+3c, y=b—2c+3a, and z=e—22a +36, show that a+y+z2=2a+26+4 2e. 23. What must be added to 2’?+ 5y?+ 32° in order that the sum may be 27? — 2”? 24. What must be added to 5a3— 7a7b+ 8ab? in order that the sum may be a® — 2a7b — 2ab? + 6? 25. If H=50°+3a7)—205, F=3a' —7a’b — 6B, G=2¢b—a—B, H=avb—2a'— 38°, find the expression for #'—[/’'—(G— A)]. CHAPTER III. MULTIPLICATION. INTEGRAL EXPRESSIONS. 60. The :aws which govern the operation of multiplica- tion are formulated as follows: ab = ba ax (oc) = (ab) X c= abe a(b+c)=ab-+ac | a(b—c)=ab—ac axa =a" . aX (+4) =-+ab aX (— 6) =— ab (—a)xb =—ab Ge) x (-— 6) = +05 61. Multiplication of Monomials. § 47 The commutative law. The associative law. The distributive law. The index law. The law of signs. When the factors are single letters, the product is represented by simply writing the letters without any sign between them. Thus, the prod- uct of a, 6, and ¢ is expressed by adc. 62. The product of 4a, 56, and 3c is 4ax5bx3c=4X5 xX 38abe = 60abe. Norse. We cannot write 453 for 4x 5x3 because another mean- ing has been assigned in Arithmetic to 453, namely, 400 + 50 + 3. Hence, between arithmetical factors the sign must be written. 40 SCHOOL ALGEBRA. 63. The product of ab and a’? is Cb xX &b == eabh = FOV aa 64, To multiply one monomial by another, therefore, Find the product of the coefficients, and to this product annex the letters, ging to each letter in the product an index equal to the sum of its indices in the factors. Notre. The beginner should determine first the sign of the product by the law of signs, and write it down; secondly, after the sign he should write the product of the coefficients; and lastly, each letter with an index equal to the sum of its indices in the factors, 65. We may have an index affecting an expression as well as an index of a single letter. Thus, (adc)? means abe X abe, which equals aabdbce, or a’b’c?,_ In like manner, (abe)*=a"b*c"®. That is, The nth power of the product of several factors is equal to the product of the nth powers of the factors. 66. By the law of signs, we have (= a). X. 2) = Had, and (+ ab) x (—e) = — abe, that is, (—a) x (— 4) x (—¢e) =—abe; and (— abc) x (— d) =+ abcd, that is, (— a) X (— b) X (—e) X (—d) = + abed. It is obvious, therefore, that The product of an even number of negative factors will be positive, and the product of an odd number of negative factors will be negative. 67. Polynomial > MULTIPLICATION. 41 s by Monomials. We have (§ 47), a(b + ¢)=ab + ac. In like manner, a(b—c+d—e)=ab—ac+ad— ae. To multiply a polynomial by a monomial, therefore, Multiply each term of the polynomial by the monomial, and add the partial products. oO fF WO DN & Exercise 9. Find the product of Te and 58. 6. Tab and 8ac. 3x and 8y. 7. —2aand 7a*a’y. 8a’ and 6a’. 8. —38a’b and — 8ab’. 3a and 24°. 9. —5mnp’ and —4m’n*p’. 2mn and 3m’n. 10. — 8a’, —20’, and —3 ab. 11. —2277y, xy’, and —3x’y. 12. —3a*y, —2a7b, and — a’y'a°b". 13. 5a+36 and 2a’. 14. ab—be and 5avbe. 15. ab —ac— be and abc. 16. 6a°b — Ta’b’c and a’b’c. 17. 2+ b—e' and a®bc’. 18. 5a’? — 367+ 2¢ and 4ab'e’. 19. abe— 38a°b2 and — 2ab’e. 20. —xy2?+ a2’y*z and — xyz. 21. —2m’np* —mnp’ and — m’np. os 42, SCHOOL ALGEBRA. 299/799 22. x—y—szand — 32°y'2". 23. —32’ and w+ 27’ —z. 24. 3x—2y—4 and 52’. 68, Polynomials by Polynomials. If we have m+n-+p to be multiplied by a+ 6-+¢, we may substitute I for the multiplicand m+n-+ p (§12). Then (a+b+c)M=aM+bM+ cM. § 28 If now we substitute for JZ its value m-+7-+- p, we shall have a(m+n+p)+b(m+n+p)+e(m+n+p) =am-+an+ap+bm-+ bn+ bp+em-+en+ep. That is, to find the product of two polynomials, Multiply every term of the multiplieand by each term of the multiplier, and add the partial products. 69. In multiplying polynomials, it is a convenient ar- rangement to write the multiplier under the multiplicand, and place like terms of the partial products in columns. ak 5 aes 7606 Dee teb 15a@— 18ab — 20ab + 246? 15a? — 38ab + 240° We multiply 5a, the first term of the multiplicand, by 3a, the first term of the multiplier, and obtain 15a’; then — 66, the second term of the multiplicand, by 3a, and ob- tain —18ab. The first line of partial products is 15a’ —18ab. In multiplying by —40, we obtain for a second line of partial products —20ab+ 240’, which is put one place to the right, so that the lke terms —18ad and MULTIPLICATION. 43 —20ab may stand in the same column. We then add the coefficients of the like terms, and obtain the complete prod- uct in its simplest form. (2) Multiply 424+ 3-4 52°— 62’ by 4—62?—5xz. Arrange both multiplicand and multipher according to the ascending powers of z. 8+ 4a2+ 52°7— 62° 4— 5br— 6x 12+ 162+ 202? — 242° — 1ldx2 — 202? — 252° + 302* — 1827 — 242° — 302* + 362° 12+ #—182?— 732’ + 362° (3) Multiply 1+ 22+ z*— 327 by a —2— 22. Arrange according to the descending powers of z. z*—327+2r+1 | v—32°4+2et+ 2 —22° + 62° —42°—227 — 2x* Gar 4a S22 z'’— 52° + 7a’?+227—62—2 (4) Multiply a?+ 8’+ c?— ab —be—ac by a+b+e. Arrange according to descending powers of a. @—ab—act+ B-~ be+ ¢ a+ 6+ ¢ &—ab—aetabl’?— abetace + a7b —ab*’— abe + 63 — Be + be? + are — abe—ac’ + b6%e—b’?+é a — 8abe + 6° + 44 SCHOOL ALGEBRA. Norr. The student should observe that, with a view to bringing like terms of the partial products in columns, the terms of the multi- plicand and multiplier are arranged in the same order. 70. A term that is the product of three letters is said to be of three dimensions, or of the third degree. In general, a term that is the product of 7 letters is said to be of x dimensions, or of the nth degree. Thus, 5abe is of three dimensions, or of the third degree; 2.a7’c’, that is, 2aabbce, is of six dimensions, or of the sixth degree. 71. The degree of a compound algebraic expression is the degree of that term of the expression which 1s of haghest dimensions. 72; When all the terms of a compound expression are of the same degree, the expression is said to be homogeneous, Thus, 2°+ 32°y + 3ay’+ 7° is a homogeneous expression, every term being of the third degree. 73. The product of two homogeneous expressions 1s homo- geneous. For the different terms of the product are found by multiplying every term of the multiplicand by each term of the multiplier; and the number of dimensions of each partial product is the sum of the number of dimensions of a term of the multiplicand and of a term of the multiplier counted together. Thus, in multiplying v+0?+¢— ab —be—ace by a+6+ ce, Example (4), each term of the mul- tiplicand is of two dimensions, and each term of the multi- plier is of one dimension; we therefore have each term of the product of 2+-1, that is, three dimensions. This fact affords an important test of the accuracy of the work of multiplication with respect to the hteral factors ; for, if any term in the product is of a degree different from the degree of the other terms, there is an error in the work of finding that term. ; MULTIPLICATION. 45 74. Any expression that is not homogeneous can be made so by introducing a letter, the value of which igs unity. ‘Thus, in Example (3), the expressions can be writ- ten z*—3 a’x’ + 2a%x-+- at and 2° —2a’x—2a*. The prod- uct will then be 2’ — 5a7a°+ Tata + 2a°x? — 6a®x — 2a’, which reduces to the product given in the example, by putting 1 for a. 75. It often happens in algebraic investigations that there is one letter in an expression of more importance than the rest, and this is therefore called the leading letter. In such cases the degree of the expression is generally called by the degree of the leading letter. Thus, a?x?+bx-+¢ 1s of the second degree in x. Exercise 10. Find the product of 1. 2+10 and 2+ 6. 12. 2x—-3 and z+ 8. 2. x—2and x — 3. 13. x—T7 and 2x—1. 3. x—38 and 2+ 5. 14. m—nand 2m+1. 4. +3 and x— 3. 15. m—aandm-ta. 5. a—Illand2a—l. 16.:382+7 and 2x —8. 6. —x+2and—2—3, 17. 54a—2y and Ba + Qy. 7. —x—2 and x — 2. 18. 82—4y and 227+ 3y. 8. —x+4andx—4. 19. 2?+ 7’ and 2° — 7’ 9. —x+7 and 2+ 7. | 20. 227+ 3y’ and wv’ + 7’. 10. <—Tandx+7. 21. at+y+zand x—y+z. 11. -~—8 and 22+ 3. 22. «+2y—z and —y+2z, 23. 2 —ay+y’ and 2+ 2y+y7/’. 24. m’—mn+n? and m+n. 25. M+tmn+n and m—n. 46 st Oo Tm FPF DO WO & SCHOOL ALGEBRA. 26. a@— 3ab+6' and a?— 3ab — 8. 27. a —Ta+2 and a&’—2a+3. 28. 2v°—d3ay+4y’ and 82°+ 42y — dy’. 29. v+ay+y and x2 —xrz2—2’. 30. Y+y7Y+2—ay—az—yzande«+y+z. 31. 4a®—10ab + 250°? and 56+ 2a. 32. w+t4y andy’ +4z. 33. 2+ 22y4+8 and y’ + 2ay — 8. 34. V?@+6+1—ab—a—banda+1+0. 35. 38a7— 27+ 52 and 827+ 27 — 32’, 36. v2 +y' + 2xy—2x—2y—1 anda+y-—l. 37. a®+ 2a"! —3a™’— 1 anda-+1. 38. a"—4a"'+ 5a°? +a" anda—l. 39. ait! — 4a**4+ 2a"! — a" and 2a°— a’ +a. 40. 2*—y"" and a®*+y""". Exercise 11. Simplify : (a+6+c)(a+6—c)—(2ab—c’). (m+n) m—[(m—n)’—n(n—m)]. [ae — (a — 6) (6+ e)] — b[b —(a—c)]. (% —1)(a — 2) — 3x2 (x#+8)+2[(e+4+ 2)(@+ 1) — 3]. 4(a— 3b)(a + 36)—2(a— 66) —2 (a? 4 68"). (etyt 22—2yt2—x)—y(@te—y)—2@+y—2). 5 [(a—b) x—cy]—2[a(w— y)— be] —[S8ax—(S5e—2a)y]. CHO ABA Pw VY 10. Ai. 12. 13. 14. 15. 16. 17. 18. 19. 20. a 22. 23. MULTIPLICATION. 47 Exercise 12. EXAMPLES FoR REVIEW. Multiply z’*—x2x—19 by #4 22—83. 1+22+2? by 1—a2'+ 22°— 382. 22°+2+4+32 by 24-3274 22°. 382°+5—42 by 8+ 62?—7z. ate—y by x#—y'+ zy. 38+ 72°— 5x by 82°—62—102°+ 4. 6? + 6ab?—4a°b by 2a7b — ab? — 8a'. x’+ax—b by #+52—4. e—me’+ne+r by 2+cx4d. x’—(a+b)x+ab by x—c. a + xy + xy? + y* by y— x. | 427+ 9y* — bay by 42°+ 97’ + bay. x —32?+5 by 2°+4. a — xy? +y* by at vy? + y'*. 2a°—32°+42?—5 by 2’ —8. am —amy™ +a" by a™-+y"™. a*—a*+a*—1 by a*¥+1. a + B+ P+ ab — ac — be by a+b+e. Simplify (a — 26) (6—2a)—(a—386) (46 —a)+ 2ad. If a=0, 6=1, and c= — 1, find the value of (a—b)(a—c)+e¢(8a—b—c)+ 2ace—(a—e)4+20. [22+ y+ (@—2y)'][Be—2y)P— (2x —8y)]. a’ (b—c)—b’? (a—ec)+¢ (a—b)—(a—b) (a—c) (6—c). (2a—6)?+26(a+b)—38a’?—(a—b)’+ (a+) (a—B). CHAPTER IV. DIVISION. INTEGRAL EXPRESSIONS. 76. The laws for division are expressed in symbols, as follows : +ab+(+a)=+6 —ab+(+a)=—6b L f signs. 44 abst ( Sey aw of signs § —ab+(—a)=+6 b+e_8,¢ | _ Distributive law. § 47, (5) a a a 77. The dividend contains all the factors of the divisor and of the quotient, and therefore the quotient contains the factors of the dividend that are not found in the divisor. Thus, woe Si a rae a Ste Cc a ——- a 78. If we have to divide a® by a’, a® by a‘, a* by a, we write them as follows: a aaaaa pe >= =aaa=a=a>’, a au a’ aaaaaa --6a7b — 9ab® by — 8a. 38. ay? — 2*y* — ay" by 277’. 34. a’b?c — a’b*ce — a’bc’ by abe. 35. 8a°—4a*b — 6al” by — 2a. 36. 5m'n — 10mn? — 15 mn by 5mn. DIVISION. 51 81. To divide one polynomial by another. If the divisor (one factor) = a+6b-+e, and the quotient (other factor)= ntpt+gq, an+bn+en then the dividend (product) =} -+ap+bp+ep ‘tag+bq-+ cq. The first term of the dividend is an; that is, the product ° of a, the first term of the divisor, by , the first term of the quotient. The first term ” of the quotient is therefore found by dividing an, the first term of the dividend, by a, the first term of the divisor. ‘If the partial product formed by multiplying the entire divisor by » be subtracted from the dividend, the first term of the remainder ap is the product of a, the first term of the divisor, by p, the second term of the quotient; that is, the second term of the quotient is obtained by dividing the first term of the remainder by the first term of the divisor. In like manner, the third term of the quotient is obtained by dividing the first term of the new remainder by the first term of the divisor; and so on. Hence we have the fol- lowing rule: Arrange both the dindend and divisor in ascending or descending powers of some common letter. Divide the first term of the dindend by the first term of the dwisor. Write the result as the first term of the quotient. Multiply all the terms of the disor by the first term of the quotrent. Subtract the product from the dividend. If there be a remainder, consider vt as a new dividend and proceed as before. 52, SCHOOL ALGEBRA. 82. It is of fundamental importance to arrange the divi- dend and divisor wm the same order with respect to a com- mon letter, and to keep this order throughout the operation. The beginner should study carefully the processes in the following examples : (1) Divide 2?+ 182+ 77 by «+ 7. e+ 18a"+ Ae e+ Tx zx+1l1 lla+77 llz+77 Nore. The student will notice that by this process we have in effect separated the dividend into two parts, 2? + 7a and lla +77, and divided each part by «+7, and that the complete quotient is the sum of the partial quotients x and 11. Thus, 4+ 1844+ 77 = 02+ 724+ 1le+ 77 =(0? + 7x) + (112+ 77); elisa tT eae Lee ee il ee ea | “+7 e+7 e+7 (2) Divide a—2ab+0 by a—O. a—2ab+B'?la—b a’— ab a—b — ab+ — ab+P (3) Divide 4a*x?— 4a7a* + 2§—a® by 2?— a’. Arrange according to descending powers of z. ti —4ect+ 4a'r*?—a’le?— oe? x*— a'r | xt — 8a’z*+a* — 8a7x'* + 4a‘z?— af — 8a'x*+ 38 a‘2? a‘? — af a‘z*? — a® DIVISION. 53 (4) Divide 22073? + 150* + 3a* — 100°) — 22.a8® by a&’ +30? — 2ab. Arrange according to descending powers of a. 8a*t — 10.a°d + 22070? — 22.00? +1584] a’'—206+386 8a'— 6a%d+ Ya’? peace — 4a°b + 1807)? — 22ab — 40°6+4+ 8a’b? — 12ab’ 5a’b? — 10 ab? + 15 0 5b? — 10 ab? + 150+ (5) Divide 52°79—xv+1—382* by 1+ 82? —2z. Arrange according to ascending powers of 2. l— 24+52'—82'*|1—224+327 1—224+ 382 1+ z#«- x x — 327+ 52° — 82* xa— 224+ 382° — #+22° —32* — #422? — 32" (6) Divide a+ 7° + 2— d3ayz by r+ y+4+2z. Arrange according to descending powers of x. v—d3yuatyteletytez Bt yn? + 2H eB are — ye — 20° — 8y2e + ye +2 ye Yc yee —204+. wa —Qyze+y+2 — 22" — Yeu — 22 yx— yaet+ertyte pac ++ yz — yee+ern—yet2 eee — Yt — Ye" 2a + 2” + 2° oe ye +2 54 SCHOOL ALGEBRA. (7) Divide 4a7*? — 8007+ 19a*!+ 5a*”’+ 9a** by a* ?— Ta®*+ 2a — 8a**, 4a7t1_30a7+19a7"4+ 5a® 49a" oa a 4a7_98 a7 8a*—12a*” 4a'—2a°— 8a — 2a*+lla®* "+17 a*?+9 a*“* — 2a*+14a*""— 4a**4+6a** —' 30°19] o*_ Ga" sas —3 a? 1-21 a? 7-6 a* 9 a" Note. We find the index of a in the first term of the quotient by subtracting the index of a in the first term of the divisor from the index of a in the first term of the dividend. Now (# + 1)—(«#—3) =e2+1—x2+3=4. Hence 4 is the index of a in the first term of the quotient, In the same way the other indices are found. Exercise 14. Divide 1. @+7a+12bya+4. 6. 42?+122+4+9 by 22+3. 2. a@—5a+6bya—38. 7%. 62°—11x+4+4 by 8x4—4. 3. @+2xry+y by x+y. 8. 8x2°—-l0axr—8a? by 42+. 4. «—2xey+ybyx—y. 9. 8a°—4a—4 by 2—a. 5. a —y by x—y. 10. a°— 8a—3 by 3—a. 11. a*+11la’?— 12a—5a®+6 by 8+ a?— 8a. 12. ¥°—9y+7>—l6y—4 by 7+4+4y. 13. 36+ m*—138m by 6+ m+ 5m. 14. 1—s—3s’—s° by 14+ 2s+4+ s*. 15. b°— 26+ 1 by &—26+1. 16. 24 227y°+ 9% by v—2xry+ 3y’. 17. a2 + 6° by at—a’d 4 ab? — ab} + bt. 18. 1+52°—62* by 1—2x+ 382". 19. 20. 21. 22. 23. 24. 25. 26. 27. 28. 29. 30. 31. 32. 33. 34. 35. 36. 37. 38. 39. 40. 41. 42. 43. DIVISION. oO 82’°y? + 9y*+ 1l6x* by 427+ 3y’— 4zy. e+yte+ 3a’y+3uy by e+y+z. J+ 6+ c¢— sabe by a+b+e. + 8y+2—6ayz by +474 2—4x2—2ary—2yz. 227—3y'+ xy —xz—4yz2—2 by 2a+3y+2. v—y—2yz—-2 by x+y+z. a+ ay + y! by 2+ ay + y. xt — 9a? + 122—4 by 2? +32—-2. y — 2y'— 6 +4y? + 18y+6 by ¥+3y4+3y4+1. y—dyz' + 42° by y— y2— 22’. —Ay?— 924 1l2yz by x+2y — 82. 2 —41%—120 by #4 42+ 5. a—3+52+a—42' by 8-—2x—2’. 6 — 2a*+ 102°—1lz’+2 by 4x—3-— 22’. 1—62°+ 52° by 1-244 2”. a+ 81+92? by 8x—2°—9. e—y by 2#+2y+y’. a+y®> by 2+ 7’. v+ax+at by 2?—axr+a’. v7 —20?+ab—8e+ 7Tbe+2ac by 8c+a—b. ab + 2a?—30b’—4be—ac—e by c+ 2a4 36. 15a*+ 10a*x+ 4072? + 6az*?— 32* by 3a?+ Zax — 2’. a®— 86>—1—6ab by a— 26-1. on" — 3 xy" +. 8.2%" — y™ by 2*—y". amtnhn — 4 gmtn-1h% _ 97 qmtn—2h 3m 4. 40 qntn 3h by a”-+ 3a”'b" — 6a™ 6". 56 SCHOOL ALGEBRA. 83, Integral expressions may have fractional coefficients, since an algebraic expression is integral if it has no detter in the denominator. The processes with fractional coefficients are precisely the same as with integral coefficients, as will be seen by the following examples worked out: (1) Add 4a?—4ab+10’, and £a0°+ 2ab — 30’. 4@—4tab+1i0 gv+ 2ab— 3h 4a°+4ab—10 (2) From 4a’?—+4ab+40? take ta’—4ab+ 30’. ta’—tab+ 10 4a0°—tab+ 20° tv? +4ab— 750 (3) Multiply 4a?—4ab+40° by 4a — 20. ta@—tab + 10 da —26 ta’—ta’b+ tab’ —4tvb+ 2ab?—16' ta —1a’b + 28 ab?— 18° (4) Divide 20°+ 4. d?— 448d — 5, d* by 8b $d 25% — 448 3°d + 115d? — 5d’ 3b — 3d 2h3— 20b%d eo — ,0d+1Lbd?— 3d’ — 2 bd+ bd? 8 hd? — 5, d? 3b? — 5d? oA PT RP w DIVISION. Exercise 15. Add 407 + 40?ct+ and — 3,0°b — 10% 4. From 32’+ 3ax— 4a’ take 22°— 8ax2—1a’. From +y—3a— $a+40 take ty +1a—2z. Multiply $¢?>—4ce—4 by 4°—4e+Hh. Multiply 42—42?+412° by da+42°+ 12%. 57 Multiply 0.5m‘ —0.4m'n + 1.2m?n? + 0.8mn'—14nt - by 0.4m? —0.6mn — 0.8 n’. Divide 5% at—{a*b + 42070’? +1ab*® by 3a4+40. Divide —4d°+ d’— $1d*+%d* by —3da’+ 2d. Exercise 16. EXAMPLES FOR REVIEW. . Find the value of 2° + ¥°+ 2— 382yz, if x=1, y= and 2=— 3. 2, Find the value of V2c —a, and of V2c—a, if b=8, ¢=9, and a= 23. Add a’b — ab’+ 6° and a’ — 40° + ab’ — 30°. Multiply a”— a™b™+ 6 by a”™+ b”. Multiply 40°** + 6a™t*+ 9a? by 2a"™*— 3a’. Divide 2° + 87? — 1252+ 30xyz by 7+ 2y — 5z. Simplify -(# — a)? — (a — 6)? —(a— 6) (a+ 6 — 82). Find the coefficient of w in the expression zta—2[2a— b(e—2z)]. Multiply A gimtin-1 _ Tam in? 5 gn tim—2 by 5 gman 58 10. sin 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26. 21 »- SCHOOL ALGEBRA. Divide &d*™— cd *— ci"d* * by eo eee Divide my" — m'tyt™ + m?-*y"—* by m*y" 4, Divide ait” —a'+ a? by a? *t¥. Divide a ha by eee Divide:4? — 4° + oy? by at Divide 22"— 62°"y" + 6a"y™ —2y™ by u™—y”. Divide 2° — 2aa?+ ava — abe —b’x# + a@7b+ ab by 2?—ax-+ bx — ab. Divide 2+ 2"+1 by a™—a"+1. Divide 3 a™t?— 4g™6_— 12 q+ _ 9 gmt by gimt4 an. 38 gms. Divide 62> — 182° + 182" — 1327 =p Dyt2o on ee Divide 12a"? — a”? — 20a"! + 19a"— 10a"! by 4a"— 3a" + 2a’. 7 Arrange according to descending powers of the fol- lowing expression, and enclose the coefficient of each power in a parenthesis with a minus sign before each parenthesis except the first : x — 2 be — a? — ax — av’? — cx — ae — bcx. Divide 1.2a‘v — 5.494.452? + 4.8 072° + 0.9 aa*— 2° by 0.6 a% — 22”. Multiply 4a? —4ab+10° by 4a +210. Multiply ¢a?+. ab +30 by 4a—40. Divide 1a? +7A,a0?+ 70° by 4a+ 40. Subtract £2’ + 42y+147’ from 42?—txy+ 47. Subtract 2+ 1ay—ty from 22?—tay+y’. CHAPTER V. SIMPLE EQUATIONS. 84. Equations. An equation is a statement in symbols that two expressions stand for the same number. Thus, . the equation 82-++2=8 states that 32-+ 2 and 8 stand for _ the same number. 85. That part of the equation which precedes the sign of equality is called the first member, or left side, and that which follows the sign of equality is called the second mem- ber, or right side. 86. The statement of equality between two algebraic expressions, if true for all values of the letters involved, is called an identical equation; but if true only for certain particular values of the letters involved, it is called an equation of condition. Thus, (a+ 6)?=a’?+ 2ab+ 0’, which is true for all values of a and 34, is an zdentical equation ; and 32+2=8, which is true only when z stands for 2, is an equation of condition. For brevity, an identical equation is called an identity, and an equation of condition is called simply an equation. 87. We often employ an equation to discover an unknown number from its relation to known numbers. We usually represent the unknown number by one of the /ast letters of the alphabet, as x, y, 2; and, by way of distinction, we use the first letters, a, }, c, etc., to represent numbers that are supposed to be known, though not expressed in the number- 60 SCHOOL ALGEBRA. symbols of Arithmetic. Thus, in the equation ax+b=c, x is supposed to represent an unknown number, and a, 3, and ¢ are supposed to represent known numbers. 88. Simple Equations. An integral equation which con- tains the first power of the symbol for the unknown number, x, and no higher power, is called a simple equation, or an equation of the first degree. Thus, av-+6=c is a simple equation, or an equation of the first degree a wx. 89. Solution of an Equation. To solve an equation is to find the unknown number; that is, the number which, when substituted for its symbol in the given equation, renders the equation an identity. This*number is said to satesfy the equation, and is called the root of the equation. 90. Axioms. In solving an equation, we make use of the following axioms: Ax. 1. If equal numbers be added to equal numbers, the sums will be equal. Ax. 2. If equal numbers be subtracted from equal num- bers, the remainders will be equal. Ax. 8. If equal numbers be multiplied by equal numbers, the products will be equal. | Ax. 4. If equal numbers be divided by equal numbers, the quotients will be equal. If, therefore, the two sides of an equation be increased by, diminished by, multiplhed by, or divided by equal numbers, the results will be equal. Thus, if 8a = 24, then 84+4= 2444, 8a—4=24—4, 4x 8x4=4~x 24 and 84#+4= 24 +4. 91. Transposition of Terms. It becomes necessary in solv- ing an equation to bring all the terms that contain the SIMPLE EQUATIONS. 61 symbol for the unknown number to one side of the equation, and all the other terms to the other side. This is called transposing the terms. We will illustrate by examples: (1) Find the number for which x stands when 162—11=72+ 70. The first object to be attained is to get all the terms which contain x on the left side of the equation, and all the other terms on the right side. This can be done by first’ subtracting 72 from both sides (Ax. 2), which gives 92—11= 70, and then adding 11 to these equals (Ax. 1), which gives 9xz= 81. If these equals be divided by 9, the coefficient of x, the quotients will be equal (Ax. 4); that is, x= 9. (2) Find the number for which z stands when +-6=a. The equation is z+b=a. Subtract 5 from each side, x+b—b=a—Bb., (Ax. 2) Since +6 and —6 in the left side cancel each other (§ 36), we have zs=a—b. (3) Find the number for which x stands when w~- b=a. The equation is t—b=@. Add +6 to each side, 2—b+b=a+b. (Ax. 1) Since —b and +6 in the left side cancel each other (§ 36), we have x=at+b. (4) What number does w stand for when ax+b=cr+d? This is the general form which every simple equation in x will assume when the like terms on each side have been 62 SCHOOL ALGEBRA. collected. In this equation x represents the unknown num- ber, and a, 6, c, d represent known numbers. If now we subtract (Ax. 2) cx and 6 from each side of the equation, we have ax —cu=d—b; or, bracketing the coefficients of 2, (a—c)x=d—b, Whence, dividing both sides by a —e, the coefficient of z, we get p asina Li = a—c 92, The effect of the operation in the preceding equa- tions, when Axioms (1) and (2) are used, is to take a term from one side and to put it on the other side with its sign changed. We can proceed in a like manner in any other case. Hence the general rule: 93. Any term may be transposed from one side of an equa- tion to the other provided us sign is changed. 94, Any term, therefore, which occurs on both sides with the same sign may be removed from both without affecting the equality. 95. The sign of every term of an equation may be changed, for this is effected by multiplying by — 1, which by Ax. 3 does not destroy the equality. 96. Verification. When the root is substituted for its symbol in the given equation, and the equation reduces to an identity, the root is said to be verified. We will illustrate with examples: (1) What number added to twice itself gives 24? Let x stand for the number ; SIMPLE EQUATIONS. 63 then 22 will stand for twice the number, and the number added to twice itself will be x + 2-2. But the number added to twice itself is 24; + 22— 4. Combining x and 22, - 3a = 24. Divide by 38, the coefficient of z, x= 8 (Ax. 4) The required number is 8. VERIFICATION. 2+ 2a = 24, 8+2x 8= 24, 8+16= 24, 24 = 24. (2) If 4% —5 stands for 19, for what number does x stand ? We have the equation Az —5=19. Transpose — 5 to the right side, 42 =19+ 5. Combine, 44 = 24. Divide by 4, x= 6. (Ax. 4) VERIFICATION. 474—5=19, 4x6—5=19, 24—5=19, 19 = 19. (3) If 8¢—7 stands for the same number as 14 — 4z, what number does x stand for? We have the equation 82—T=14—A4z. Transpose 4x to the left side, and 7 to the right side, 82+427=—14+ 7, Combine, Wher ered Divide by 7, e= 38, 64. SCHOOL ALGEBRA. VERIFICATION. 82—7=14—4z2, 8xX3—T=14—4~x 3, 2 = 2. (4) Solve the equation (a — 3) (— 4) =2(# — 1) — 30. We have (% — 8) (a — 4) =z(#—1)—80. Remove the parentheses, v’— Tx+12=2?—2—30. Since 2 on the left and x? on the right are precisely the same, including the sign, they may be cancelled. Then —72+12=—2—8380. Transpose — x to the left side, and + 12 to the right side, —Tate=— 30-12. Combine, —6x=— 42. Divide by —6, x2 = 7. VERIFICATION. (7 —3)(7—4)=7(7—1)— 30, 4x3=7x6—830, 12 = 42 — 30, 12 = 12 Exercise 17. Find what number x stands for If « —5 stands for 7. If «+8 stands for 12. If 6x—12 stands for 18. . If d¢e¢—2=382+4. If 72— 8 stands for 25. . If 7¢a¢—5=62—1. If 52+8 stands for 43. 10. If 52%—8= 25 — 22. . If2e¢—-—5=7+2. . If 2e%7—4=5—2. i CO Comet | C5 SIMPLE EQUATIONS. 65 11. If 32 and 2+ 8 stand for the same number. 12. If 22 —5 stands for the same number as 32. Solve the equations : 13. 2a—38=8+2. 16. 84—4=12—2. 14. 54+4=20+4 2. 17. 22—5=7—22. 15. 22 -—3=7—~72. 18. 82+14=2—2. Find x 19. If 22—5 and 4x — 11 stand for the same number. 20. If x(x—7) and x?-- 70 stand for the same number. 21. If x(8x—2) and 82(a—1)+2 stand for the same number. 22. If 832—5=42—10. 23. If 2a—4=14—xz2. 24. If 3x—8 and 42 —11 stand for the same number. 25. If 22—5 and 7 —~2 stand for the same number. 26. If 22?— 23 and (2x+1)(x— 3) stand for the same number. 27. If (~+ 38) (x—7)—(#—4) («+ 1)=25. 28. If (22 —1)(#+3)—(a#— 38) (22 — 3) = 72. Solve the equations: 29. 2(*—5)=2’— 380. 30. r(7#+3)=2°+ 18. 31. («—3)(@+1]l)=2—82+1. 32. (cx—1)(+2)+ (+38) (@—-1)=22(x4+4)—(#+1). 33. 2(x+8)—(2#+1)(a#—2)—5(#+3)+3=0. 34. (w—3)(x+3)—(x— 4) (a+ 4)-—2=0. 66 SCHOOL ALGEBRA. 97, Statement and Solution of Problems. The difficulties which the beginner usually meets in stating problems will be quickly overcome if he will observe the following direc- tions: Study the problem until you clearly understand its mean- ing and just what is required to be found. Remember that z must not be put for money, length, time, weight, etc., but for the required nwmber of specified units of money, length, time, weight, etc. Iixpress each statement carefully in algebraic language, and write out in full just what each expression stands for. Do not attempt to form the equation until all the state- ments are made in symbols. We will illustrate by examples: (1) John has three times as many oranges as James, and they together have 32. How many has each? Let x be the number of oranges James has ; then 3a is the number of oranges John has; and x +3. is the number of oranges they together have. But 32 is the number of oranges they together have; “2 pOx = 32% or, 4a = 32, and t= 8, Since xz = 8, 3a = 24. Therefore James has 8 oranges, and John has 24 oranges. Nore. Beginners in stating the preceding problem generally write: Let « = what James had. Now, we know what James had. He had oranges, and we are to discover simply the number of oranges he had. (2) James and John together have $24, and James has $8 more than John. How many dollars has each? SIMPLE EQUATIONS. 67 Let x be the number of dollars John has; then x + 8 is the number of dollars James has; and «# +(# + 8) is the number of dollars they both have. But 24 is the number of dollars they both have; “. @ + (x + 8) = 24: Removing the parenthesis, we have e+xe+8 = 24. Transposing and collecting like terms, we have 2x2 = 16. Dividing by 2, we get x = 8, Since a = 8, x2+8=16, Therefore John has $8, and James has $16. Notr. The beginner must avoid the mistake of writing Let « = John’s money. We are required to find the number of dollars John has, and there- fore x must represent this required number. (3) The sum of two numbers is 18, and three times the greater number exceeds four times the less by 5. Find the numbers. Let « = the greater number. Then, since 18 is the sum, and z is one of the numbers, the other number must be the sum minus x. Hence 18 ~ x = the smaller number. Now, three times the greater number is 3a, and four times the less number is 4(18 — x), We know from the problem that 3a exceeds 4(18 — x) by 5; or, in other words, we know that the excess of 3x over 4(18 — x) equals 5. It only remains to determine what sign the word “ excess” im- plies. If we are in doubt about it, we can apply the phrase to two arithmetical numbers. We shall have no difficulty in seeing that the excess of 50 over 40 is 10; that is, 50 —40, and hence that the sign — is implied by the word “ excess.” 68 SCHOOL ALGEBRA. Hence, 3a —4(18 — x) = the excess. But a 5 = the excess. “ 3a—4(18 —2) =5, or 3e0—72+4a=5. “. 1e = 77, and y= 11, Therefore the numbers are ll and 7. (4) Find a number whose treble exceeds 40 by as much as its double falls short of 35. Let x = the required number ; then 3a = its treble, and 3a — 40 = the excess of its treble over 40; also, 35 — 2” = the number its double lacks of 35. Hence, 32 — 40 = 35 — 2a. Transposing, 3a +22 = 354 40, “. 52 = 75, and % = 15, Therefore the number required is 15. (5) Find a number that exceeds 50 by 10 more than it falls short of 80. Let « = the required number; then x — 50 = its excess over 50, and 80 — a =the number it lacks of 80. Hence, x — 50 — (80 — x) = the excess. But 10 = the excess. “, « — 50 — (80 — x) = 10, or x—50—804+2=10. “. 22 = 140, and x = 70, Therefore the number required is 70. SIMPLE EQUATIONS. 69 Exercise 18. 1. If a number is multiplied by 7, the product is 301. Find the number. 2. The sum of two numbers is 48, and the greater is five times the less. Find the numbers. 3. The sum of two numbers is 25, and seven times the’ less exceeds three times the greater by 35. Find the num- bers. 4. Divide 20 in two parts such that four times the greater exceeds three times the less by 17. 5. Divide 23 into two parts such that the sum of twice the greater part and three times the less part is 57. 6. Divide 19 into two parts such that the greater part ex- ceeds twice the less part by 1 less than twice the less part. 7. A tree 84 feet high was broken so that the part broken off was five times the length of the part left stand- ing. Required the length of each part. 8. Four times the smaller of two numbers is three times the greater, and their sum is 63. Find the numbers. 9. A farmer sold a sheep, a cow, and a horse for $216. He sold the cow for seven times as much as the sheep, and the horse for four times as much as the cow. How much did he get for each? 10. Distribute $15 among Thomas, Richard, and Henry so that Thomas and Richard shall each have twice as much as Henry. 11. Three men, A, B, and C, pay $1000 taxes. B pays four times as much as A, and C pays as much as A and B together. How much does each pay ? 70 SCHOOL ALGEBRA. 12. John’s age is three times the age of James, and their ages together are 16 years. What is the age of each? 13. Twice a certain number increased by 8 is 40. Find the number. 14. Three times a certain number is 46 more than the number itself. Find the number. : 15. One number is four times as large as another. If I take the smaller from 12 and the greater from 21, the remainders are equal. What are the numbers? 16. The joint ages of a father and son are 70 years. If the age of the son were doubled, he would be 4 years younger than his father. What is the age of each? 17. A man has 6 sons, each 4 years older than the next younger. The eldest is three times as old as the youngest. What is the age of each? 18. Add $24 to a certain amount, and the sum will be as much above $80 as the amount is below $80. What is the amount ? 19. Thirty yards of cloth and 40 yards of silk together cost $330; and the silk costs twice as much per yard as the cloth. How much does each cost per yard? 20. Find the number whose double diminished by 24 exceeds 80 by as much as the number itself is less than 100. 21. In a company of 180 persons composed of men, women, and children there are twice as many men as women, and three times as many women as children. How many are there of each? 22. A banker was asked to pay $56 in five-dollar and two-dollar bills in such a manner as to pay the same num- ber of each kind of bills) How many bills of each kind must he pay ? SIMPLE EQUATIONS. 71 23. How can $3.60 be paid in quarters and ten-cent pieces so as to pay twice as many ten-cent pieces as quarters ? 24. I have $1.98 in ten-cent pieces and three-cent pieces, and have four times as many three-cent pieces as ten-cent pieces. How many have I of each? Nore. In problems involving quantities of the same kind ex- pressed in different units, we must be careful to reduce all the quanti- ties to the same unit. ; 25. I have $17 dollars in two-dollar bills and twenty- _five-cent pieces, and have twice as many bills as coins. How many have I of each? 26. I have $6.50 in silver dollars and ten-cent pieces, and I have 20 coins in all. How many have I of each? 27. A bought 9 dozen oranges for $2.00. Fora part he paid 20 cents per dozen; for the remainder he paid 25 cents a dozen. How many dozen of each kind did he buy ? 28. A gentleman gave some children 10 cents apiece, and found that he had just 50 cents left. If he had had another half-dollar, he might have given each of them at first 20 cents instead of 10 cents. How many children were there? 29. A is twice as old as B and 6 years younger than OC. The sum of the ages of A, B, and C is 96 years. What is the age of B? 30. Divide a line 24 inches long into two parts such that the one part shall be 6 inches longer than the other. 31. Two trains travelling, one at 25 and the other at 30 miles an hour, start at the same time from two places 220 miles apart, and move toward each other. In how many hours will the trains meet? 12, SCHOOL ALGEBRA. 32. A man bought twelve yards of velvet, and if he had bought 1 yard less for the same money, each yard would have cost $1 more. What did the velvet cost a yard? 33. A and B have together $8; A andC,$10; Band C, $12. How much has each? 34. Twelve persons subscribed for a new boat, but two being unable to pay, each of the others had to pay $4 more than his share. Find the cost of the boat. 35. A man was hired for 26 days on condition that for every day he worked he was to receive $3, and for every day he was idle he was to pay $1 for his board. At the end of the time he received $58. How many days did he work? 36. A man walking 4 miles an hour starts 2 hours after another person who walks 3 miles an hour. How many miles must the first man walk to overtake the second ? 37. A man swimming in a river which runs 1 mile an hour finds that it takes him three times as long to swim a mile up the river as it does to swim the same distance down. Find his rate of swimming in still water. 38. At an election there were two candidates, and 2644 votes were cast. The successful candidate had a majority of 140. How many votes were cast for each? 39. Two persons start from towns 55 miles apart and walk toward each other. One walks at the rate of 4 miles an hour, but stops 2 hours on the way; the other walks at the rate of 8 miles an hour. How many miles will each have travelled when they meet? . 40. A had twice as much money as B; but if A gives B $10, B will have three times as much as A. How much has each ? 41. If 22 —8 stands for 20, for what number will 4—2 stand ? SIMPLE EQUATIONS. 73 42. A vessel containing 100 gallons was emptied in 10 minutes by two pipes running one at atime. The first pipe discharged 14 gallons a minute, and the second 9 gallons a minute. How many minutes did each pipe run? 43. A man has 8 hours for an excursion. How far can he ride out in a carriage which goes at the rate of 9 miles an hour so as to return in time, walking at the rate of 3 miles an hour? 44. If 32—4a—22x-—a, find the number for which 42 — Ta stands. 45. If 7x—a=9(«#—a), find the number fois Be zta Norr. When we compare the ages of two persons at a given time, and also a number of years after or before the given time, we must remember that both persons will be so many years older or younger. Thus, if a man is now 22 years old and his son z years old, 5 years ago the father was 22—5 and the son x—5, and 5 years hence the father will be 2% + 5 and the son a + 5, years old. 46. A man is now twice as old as his son; 15 years ago he was three times as old as his son. Find the age of each. 47. A man was four times as old as his son 7 years ago, and will be only twice as old as his son 7 years hence. Find the age of each. 48. A, who is 25 years older than B, is 5 years more than twice as old as B. Find the age of each. | 49. A man is 25 years older than his son; 10 years ago he was six times as old as his son. Find the age of each. 50. The difference in the squares of two consecutive numbers is 19. Find the numbers. 51. The difference in the squares of two successive odd numbers is 40. Find the numbers. CHAPTER VL MULTIPLICATION AND DIVISION. SPECIAL RULES. 98. Special Rules of Multiplication. Some results of mul- tiplication are of so great utility in shortening algebraic work that they should be carefully noticed and remem- bered. The following are important : 99. Square of the Sum of Two Numbers. (a +8) =(a+b)(a+0) =a(a+ 6)+b(a+ 5) =@+ab+ab+ 0° =a?+2ab-+ 6’. Since a and 6 stand for any two numbers, we have Rute 1. The square of the sum of two numbers is the _ sum of ther squares plus twice ther product. 100. Square of the Difference of Two Numbers. (a—b)' = (a—b)(a—8) = a(a—b)— b(a—d) =a—ab—ab+ 6 = a? — 2ab+ 8’. ~Hence we have Rue 2. The square of the difference of two numbers 1s the sum of ther squares minus twice ther product. a SPECIAL RULES OF MULTIPLICATION. 0 101. Product of the Sum and Difference of Two Numbers. (a+ 6)(a— 6) =a(a— b)+b(a—6) =@—ab+tab—6 =a — 6. Hence, we have Rue 38. The product of the sum and difference of two numbers rs the difference of their squares. If we put 2x for a and 3 for 6, we have Rule 1, (22+ 38/?=42?+122+4+ 9. Rule 2, (22— 3 =42?—1244 9. Rule 3, (2x +3) (2% — 3) = 427 —9. Exercise 19. Write the product of 1. (oy) 1. (@ty)(e—y), 2. (e—a)’. 8. (4z2—3)(4z+ 3). 3. (7+ 26)’. 9. (807+ 407) (38¢ — 40’). 4. (8% — 2c)’. 10. (83a—c)(8a—¢). 5. (4y—5)’. 11. (c+ 70’) (x+ 70’). 6. (8a? + 427)”. 12. (ax+2by) (ax — 2 by). 102. If we are required to multiply a+6-+e¢ by a+b—e, we may abridge the ordinary process as follows : (atb+0)(a+b—c)=[(a-+b)+e][(a+b)—] By Rule 8, = (a+ b?—¢ By Rule 1, == q® + 2ab + b? — 0’, 76 SCHOOL ALGEBRA. If we are required to multiply a+ 6—c by a—b-+e, we may put the expressions in the following forms, and per- form the operation : (a+b—0)(a—b +e) =[a+ (6-0) [a—(b—o)] By Rule 8, =a —(b—c/y By Rule 2, = o — (6° — 2bc +c’) =a’ — b?+ 2be— e’. Exercise 20. Find the product of 1. 2+y+2 and 7x—y—2. x—y+z2 and «—y—z. ax -+by+1 and ax-+ by—1. l+a—y and 1—z+¥. a+ 26—8e and a—26+4+3e. a —ab+b* and @+ab+6". m+t+mn+n and m’—mn+n’. 2+2+2? and 2—2—2". @V+tat+l1 and @—atl. da+2y—z2 and 8x—2y+z. CM XH xT Pw W a = 108. Square of any Polynomial. If we put ~z for a, and y +2 for 4, in the identity (a+ 6)? =a? + 2ab +B’, we shall have [w+ (y+z)P=2? + 20(y+z)+ (y+2), or (a@+y+zP =2?+2ay+2a2+ 74 2y2+2 =P LY+L + Qry + Qae+ Qyz. SPECIAL RULES OF MULTIPLICATION. it’ It will be seen that the complete product consists of the sum of the squares of the terms of the given expression and twice the products of each term into all the terms that fol- low it. Again, if we put a—6 for a, and e—d for 3, in the same identity, we shall have [(a—b) + (e—d)} = (a—b)’+2(a—6b)(e—d)+(e—dy = (a —2ab+b)+2a(e— d)—2b(e—d)+(e?—2cd+d’) = @—2ab+b2+2ac—2ad—2bce+2bd+ce— 2cd+d? = 7+674+-¢4+ d?—2ab+2ac—2ad—2be+2bd—2cd. Here the same law holds as before, the sign of each double product being ++ or —, according as the factors com- posing it have like or unlike signs. The same is true for any polynomial. Hence we have the following rule: Rue 4. The square of a polynomial is the sum of the squares of the several terms and twice the products obtained by multiplying each term into all the terms that follow tt. Exercise 21. Write the square of 1.22 —3y. 10. 2+ y'+ 2’. 2. atb+e. 11. 2a—y—z. 3. @+y—Zz. 12. a—2b—3e. 4. «—y+z. _ 13. 8a—6+42e. 6. oty+sd. 14. x+2y—3z. 6. ++2y+8. 15. «+y+2-+1. 7. a—b+e. ; 16. 4¢+y+2-—2. 8. 82—2y+4. 1. 2e—y—2—8. 9. 24—8y-+ 4z. 18. «—2y—324+4. 78 SCHOOL ALGEBRA. 104. Product of Two Binomials. The product of two bino- mials which have the form x-+-a, «+4, should be carefully noticed and remembered. (1) (@+5)(e#+3)=2 («+ 38)+5 (4+ 3) =2+382152+15 = 27+ 82+ 15. (2) (e—5)(e—8)=«(e—8)—5 (e—8) = 27 —82—527+15 = z*—824+165. (3) (¢+5)(¢—38)=2(4—3)+5(e¢—8) = 2?7—32+5x—-15 = 2712x2—15. (4) (*#—5)(«+3)=—2(¢+4+3)—5(e+3) =277+382—5x2—15 = 2? —2a2—15. Each of these results has three terms. The first term of each result is the product of the first terms of the binomials. The last term of each result is the product of the second terms of the binomials. The middle term of each result has for a coefficient the algebraic swum of the second terms of the binomials. The intermediate step given above may be omitted, and the products written at once by wspection. Thus, (1) Multiply «+8 by «+7. 8+7=15, 8x 7=56. (@+8)(@+ 7) =2?+ 15a + 56. SPECIAL RULES OF MULTIPLICATION. 79 (2) Multiply «— 8 by x— 7. (-8)+(—1) =-15, (-8)(-7) =+86. . (a — 8) (x — 7) = 2? —152x-+ 56. (3) Multiply «—7y by x+ 6y. ee SS SS SS ao FF WO WO KF OO 6 1. (—Ty) xX 6By =— 427’. J. (@-—- Ty) (a+ by) = 2? — ay— 42y’. (4) Multiply 27+ 6(a+ 6) by 2? —5(a+0). +6—5=1, 6(a+6)x—5(a+6)=—30(a+6)*. ”. [2°+6 (a+0) | [2’—5 (a+6)] = a*++(a+b)2?—380 (a+)'. Exercise 22. Find by inspection the product of - («+8)(%+3). . (c+ 8)(x — 8). . (xc«—7)(x+ 10). (x — 9) (x — 5). » (c—10)(%+ 9). (a —10)(a—5). . (a—12)(a—8). - (a+ 26)(a+40). - (a—38b)(a+76). 7 (a+ 26)(a— 96). . («—8a)(x4—4a). - («+ 42)(4%— 22). . (e+ by) (z— By). . («—38a)(a-+ 2a). - (a+ 26)(a—40). . (x? — 9) (2 + 8). (+ 2y") (a —8y). (at +8y") (2 4y). . (ab —8)(ab+5). . (ab —Txy)(ab+8zy). . (v—8y)(v—8y). . (x+6)(%-+ 6). . (a—8b)(a—385). . («—e)(e— ad). . («+ a)(x—5). . (c—a)(«+ 8). . [((a+6)+2][(a+6)—4]. - [(e@+y)—2][@t+y)+ 4]. - (@ty—T)(e+y+10). - (e—y—T)(«#—y—10). 80 SCHOOL ALGEBRA. 105. In like manner the product of any two binomials may be written. (1) Multiply 2a—6 by 3a+40. ; (2a— b) (3a+4b) =6a?+4+ 8 ab —8ab —42? = 6a’?+ 5ab — 406. (2) Multiply 22+ 3y by 3a—2y. The middle term is (2x2) x (—2y)+3yxX 382=52y; “(22+ 8y)(82— 2y) = 62? + zy — 67%. Exercise 23. Find the product of 38x2—y and 2x%+ y. 10z2—3y and 10x— Ty. 3a?— 26? and 2a’?+ 36’. at? and a—Qb. a—Ty and 2%—5y. 3a’?— 20? and 2a+ 30. llz—2y and 7x+y. 10. a—O anda+8. 4x2—8y and 3x—2y. 5a2—4y and 8a—4y. o FP WO WD £9: Otay 106. Special Rules of Division. Some results in division are so important in abridging algebraic work that they should be carefully noticed and remembered. 107. Difference of Two Squares. From §101, (a+ 6)(a— 6)=a@ — B’. at Wasted Rue 1. The difference of the squares of two numbers 1s divisible by the sum, and by the difference, of the numbers. =a — 6, and Cath, Hence a- Write by inspection the quotient of 1. 12. cave SPECIAL RULES OF DIVISION. ate 5 a—2 ea 3s+2 loo 4+a a — 25. x—5d SO — a 6+ 2 oa — sf’ fn}. 252° — 360° 5a+6b 49 @ — d? it—a 9a?—1 Ba+l1 16—4@ 4—2a Exercise 24. 14. 15. 16. LT. 18. 19. 20. aie 22. 23. 24. 25. abe — x? ab®ct + 28 xta® — bY at — BE et oye a—(b+c¢) a Lire ed be a—(38b—4c) Dy (ed 1+(@—y) (82—y)'— 162 (82—y)+4z (2+3a) — Qa? (7 -+- 3.a)— 32 li (la pe) 1+ (7a—56b) (82+ 2yf—42* (82+ 2y)—2z (ae7 Oo Gn-V/)" (ap) (e—4¥) a (ya) t—(y +2) (1 —2y)y— 25 (c—2y)+5 (22 +y)? — 92 (22+y)—382 81 82, SCHOOL ALGEBRA. 108. Sum and Difference of Two Cubes. By performing the paren, we find that a + 53 Sti es ah b?, Tse =a?—ab+ Hence, == at + ab + Bt Rue 2. Zhe sum of the cubes of two numbers ts divisible by the sum of the numbers, and the quotient is the sum of the squares of the numbers minus ther product. Rute 8. The difference of the cubes of two numbers is divisible by the difference of the numbers, and the quotient as the sum of the squares of the numbers plus ther product. Exercise 25. Write by inspection the quotient of y Lae of OR — 1—2z ab —e Palais : igp eee 1+ 22 ab+e 3. Bhs ei 64-Pe 38a—b 4+y ae 27 at + OF is 348 — 8a 38a+b 7—2a eM ee Nee 49. Oo eee 4x+3y 2a-+ 6° < bara ta xf 729° 42 —3y z+ Oy 7. LS ces 15. at ee 1—8z a — 3b et 1-272 16. 82° — 64y 1432 2a — Ay? SPECIAL RULES OF DIVISION. 83 109. Sum and Difference. of any Two Like Powers. By performing the division, we find that een 7 =a7t@btab’?+ bd’; a— oat — ab + al? — a a’ — 5 x ; =at+@bt+al’+ab?+ bd; Q— ote =at—ad + ab? — ab’ + bt. ar We find by trial that at 0, at+ bt, a®+ 0°, and so on, are not divisible by a+ 6 or bya—b. Hence, If n is any positive integer, (1) a®+0" ts divisible by a+b if n is odd, and by neither a+b nor a—b if nts even. (2) a®—b" is divisible by a— b if n ws odd, and by both a+b and a—b if ns even. Nore. It is important to notice in the above examples that the terms of the quotient are all positive when the divisor is a —6, and alternately positive and negative when the divisor is a + 6; also, that the quotient is homogeneous, the exponent of a decreasing and of b increasing by 1 for each successive term. Exercise 26. Find the quotient of iy, ane hy Aaa foe t—Yy x+1 x+2 2. wy 5. vt — 16 8. Lah x+y e—2 l—m 4 a e Hegre af = 32 9 Th om 6. . x—1 x—2 l+m CHAPTER VIL. FACTORS. 110. Rational Expressions. An expression is rational when none of its terms contain square or other roots. 111. Factors of Rational and Integral Expressions. By fac- tors of a given integral number in Arithmetic we mean integral numbers that will divide the given number with- out remainder. Likewise by factors of a rational and inte- gral expression in Algebra we mean rational and integral expressions that will divide the given expression without remainder. 112. Factors of Monomials. The factors of a monomial may be found by inspection. Thus, the factors of 14a%b are 7,2, a, a, and 0. 118. Factors of Polynomials. The form of a polynomial that can be resolved into factors often suggests the process of finding the factors. CasE I. 114. When all the terms have a common factor. (1) Resolve into factors 227+ 62y. Since 22 is a factor of each term, we have 2% 20 = 25 “. 22+ 62y=22(4+ 3y). Hence, the required factors are 2x and «+ 3y. FACTORS. 85 (2) Resolve into factors 16a*+ 4a?— 8a. Since 4a is a factor of each term, we have | 16a! + 4a ~8a__16a° , 4a’ 8a 4a 4a '4a Aa =4¢7+a—2. », 16a +4a— 8a=4a(4e?+a— 2). Hence the required factors are 4a and 4a?+a—2. Exercise 27. Resolve into two factors: Biuat — 62". 7. 3a°b — 4a’. 2. 2a?—Aa, 8. Say? + Aaty?! 3. 5ab— 5a’d’. 9. 38a*—92?— 62’, 4. 30—a +a. 10. 8a’a? —4a°b 4+ 12077’. 5B. a? + ay — xy’. 11. 8a°b’c? — 4a7b°C’. 6. at—a’b+a’l’. 12. 15a’x — 10a*y + 5a*z. Case II. 115, When the terms can be grouped so as to show a common factor. (1) Resolve into factors ac + ad + be + bd. ae +ad + be + bd =(ac+ad)+ (be + bd) (1) =a(e+d)+b(e+d) (2) = (a+ 6)(e+d). (3) Nore. Since one factor is seen in (2) to be c+ d, dividing byc+d we obtain the other factor, a+ 6. 86 SCHOOL ALGEBRA. (2) Find the factors of ae + ad — be — bd. ac + ad — be — bd = (ac + ad) — (be + bd) =a(e+d)—b(e+d) =(a—b)(e+d). Norse. Here the signs of the last two terms, — be— dd, being put within a parenthesis preceded by the sign —, are changed. (3) Resolve into factors 32° —52*—6x-+ 10. 82° —52?— 62+ 10 = (82° — 52”) — (6x —10) = 27(82—5)—2(8%—5) = (x? — 2)(8x%—5). (4) Resolve into factors 52° —15axz?— 2x -+ 3a. 52'—15az27—x+ 38a= (52° — 15az”) — (x — 8a) = 52? (x —8a)—1(x#-—38a) = (527 —1)(x— 3a). (5) Resolve into factors 6y — 27 xy — 102+ 452". 6y—27 ey—102+ 452° = (452° — 27 2*y) — (102 — 6y) = 92? (52—3y)—2(d4 —3y) = (92? — 2)(5x—38y). Norts. By grouping the terms thus, (6y — 27a?y) — (10% — 452°), we obtain for the factors, (2 — 92?) (3 y — 52). But (2 —92?)(3y —52) = (9a? — 2)(5%—3y), since, by the Law of Signs, the signs of two factors, or of any even number of factors, may be changed without altering the value of the product. Exercise 28. Resolve into factors : 5. 2? + ax — bx—ab. 6. a’ + 2y — an — ay, ax — cy —ay+ex. 7. 2° —ay—62r+ by. 2ab—8ac—2by4+8cy. 8. 22°—32y+4axr—Bay. ax — ba + ay — by. ax — bx — ay + by. PrP OO DW ra FACTORS. 87 9. wb —abz—ac+ ex. 10. wbx + b’ex — acy — be’y. 11. 32° —5y7’?— 62° + 1l0zy’. 12. 8ar—10bu—12a+415d, 18. 227°—32?>—42+-6. 14. 62*+ 82° — 927— 122. 15. azt+ bx? —ax—b. 16. 3c2*— 2dz* — 9cz’* + 6dz. 17. 1+152* —52z— 382’. 18. av’+t+a@et+atse. 19. (a+6)(c+d)—3c(a+d). 20. (c—y)'+ 2y(%@—y). 116. If an expression can be resolved into two equal factors, the expression is called a perfect square, and one of its equal factors is called its square root, Thus, 162°y? = 42°y xX 4a°y. Hence, 16x°y’ is a perfect square, and 42*y is its square root. Notre. The square root of 16 2*y? may be — 4a°y as well as + 425y, for —4a°y x — 4a°y = 16 a*y?; but throughout this chapter the posi- tive square root only will be considered. 117. The rule for extracting the square root of a perfect square, when the square is a monomial, is as follows: Extract the square root of the coefficient, and divide the index of each letter by 2. 118,. In like manner, the rule for extracting the cube root of a perfect cube, when the cube is a monomial, is, Extract the cube root of the coefficient, and divide the index of each letter by 8, 88 SCHOOL ALGEBRA. 119. By §§ 99, 100, a trinomial is a perfect square, if its first and last terms are perfect squares and positive, and its middle term is twice the product of their square roots. Thus, 16 a? — 24ab + 90? is a perfect square. The rule for extracting the square root of a perfect square, when the square is a trinomial, is as follows: Extract the square roots of the first and last terms, and connect these square roots by the sign of the middle term. Thus, if we wish to find the square root of 16a? — 24ab+ 96?, we take the square roots of 16a? and 90’, which are 4a and 30, respectively, and connect these square roots by the sign of the middle term, which is —. The square root is therefore Te eee = In like manner, the square root of 1l6a@?+ 24ab+ 967? is 4a+ 30, CasE III. 120, When a trinomial is a perfect square. (1) Resolve into factors 2’ + 22y +7’. From § 119, the factors of 2? + 22y+ 7’ are (@ry@ty). (2) Resolve into factors a* — 22°y + 7’. From § 119, the factors of #* — 22°y + 7? are (7? —y)@—y). Exercise 29. Resolve into factors : 1. @—6ab+906". . 3. @&—4ab+48*, 2. 407+ 4ab+ 0’. 4. v?+6xy+9y’. FACTORS. 89 5. 427—12axr+ 9a’. 13. 492° — 282xy-+ 47’. 6. a’ —10ab+ 250’. 14. 1— 206+ 10027. 7 4¢7—4a+l1. 15. 8la’?+126ab+ 490". 8. 497? —l4yz+ 2. ._ 16. mn? — 16mnda? + 64 a+. 9. 2? —162-+ 64. 17. 4a?— 20axz + 252”. 10. 92°+ 24zy+ 167’. 18. 12la?+ 198 ay-+ 817’. 11. 16a?+ 8az + 2’. 19. @b*c® — 2ab’c’x’ + x”. 12. 25+ 802+ 642’. 20. 49—140#?+ 100#*. CasE LV. 121. When a binomial is the difference of two squares. (1) Resolve into factors 2? — y’. From § 101, («+y)(#-—y)=xv7—y’. Hence, the difference of two squares is the product of two factors, which may be found as follows : Take the square root of the first term and the square root of the second term. The sum of these roots will form the first factor ; The difference of these roots will form the second factor. Exercise 30. Resolve into factors : 1. w—4, 6. 25 — 16a’. 11. 812’?— 4y’. 2. 1—2’. 7. 16— 257’. 12. 64a‘ — dt. 3. 2? — 97’. 8. ab? —1. 13. mn? — 36. 4. 4a?— 490’. 9. x?— 100. 14. xt — 144. 5. av? — 4y’. 10. 121a’—360’, 15. 2? — 25. 90 SCHOOL ALGEBRA. 16. 49—10077. 23. 49a*—y”. 30. 25 — 647’. 17. 1— 492°. 24. 64a@— 90°. 31, 1621 = aye 18. 4—121y’. 25. 8la‘dt‘—ct. 32. 252%—l6a*e’. 19. 1—169a%. 26. 4a°’c—Q9ec’. 33. 36 a72’?—49a'. 20. v7h?— 4c. 27. 200°b? —5ab. 34. x? —167/’. 21. 92°— a’. 28. 8a0— 120%. 35. 1—4002*. 22, 42%—y”. 29. 9a’?— 816". 36. 4a?c— 9c’. 122. If the squares are compound expressions, the same method may be employed. (1) Resolve into factors (#-+ 3y) — 16a’. The square root of the first term is x + 3y. The square root of the second term is 4a. The sum of these roots is +3y + 4a. The difference of these roots is a + 3y —4a. Therefore (x + 3y)?— 16a?=(a+3y + 4a)(a+3y—4a). (2) Resolve into factors a? — (36 — 5c)’. The square roots of the terms are a and (3b — 5c). The sum of these roots is a + (36 — 5c), ora +36 —5e. The difference of these roots is a — (36 — 5c), ora—3b+5e. Therefore a? — (3b —5c)l?=(a+3b—5c)(a—3b + 5e), 123. If the factors contain like terms, these terms should be collected so as to present the results in the simplest form. (3) Resolve into factors (8a-+ 56)? — (2a — 3b)’. The square roots of the terms are 3a +56 and 2a—3b, The sum of these roots is (3a + 5b) +(2a— 3b), or 8a+5b+ 2a—3b=5a+ 2b. The difference of these roots is (3a + 5b) ~(2a— 86), or 3a+65b—2a+3b=a4+4 86, Therefore (3a + 5b)? ~ (2a —3 bd)? = (5a + 2b) (a + 85). FACTORS. 91 Exerc:se 31. Resolve into factors : 1. (¢+y)— 2. 11. (a— 6b)’ —(e—dy. 2. («—yy—2’. 12. (2a+ by — 25e’. 3. (x—2yy—42. 13. («+ 2y)— (Qa —y). 4. (a+ 36) — 16’. 14. (x +38)—(382—4). 5. wv? — (y—2)’. 15. (a+6—c)—(a—b—c)’. 6. a’ — (86 — 2c)’. 16. (a—38x)?— (8a— 22). 7. D—(2a+ 3c)’. 17. (2a—1)—(8a+1)’. 8. 1—(#+ 50)’. 18. (x—5)?— (a+ y—5)*. 9. 9a?—(x— 3c) 19. (2a+6—c)—(a—2b+e)*. 10. l6a@’—(2y—382z). 20. (a+26—8c)?—(a+5e)’. 124, By properly grouping the terms, compound expres- sions may often be written as the difference of two squares, and the factors readily found. (1) Resolve into factors a? —2ab+ b? — 9c’. a —2ab4+ 0-9? =(v—2ab+ 0’)—9¢ =(a—byP—9¢? =(a—b+3c)(a—b—3c). (2) Resolve into factors 12ab+ 92°—4a’— 90’. Norse. Here 12 ab shows that it is the middle term of the expres- sion which has in its first and last terms a? and b?, and the minus sign before 4a? and 90? shows that these terms must be put in a parenthesis with the minus sign before it, in order that they may be made positive. The arrangement will be 92’ — (4a? —12ab+ 9b’) =92?—(2a—36)’ = (82+ 2a—36)(8%—2a+30). 92 SCHOOL ALGEBRA. (3) Resolve into factors — a?+ 6?—c’+d?+ 2ac-+ 26d. Nort. Here 2ac, 2bd, and —a?, —c?, indicate the arrangement required. —av+b—e?+d?+2ac4+ 2bd = (6? + 2bd+ d’) — (@ —2ac+c’) = (b+ d)' — (a—oF = Gidto jet ds ia Exercise 32. Resolve into factors : me © tO 9. 10. ii: 12. 13. 14. 15. 16. vt+2ab+ ?— x’ —2ry+y'— 9a’. 6? — 27+ 4ar —4a’. 4¢° + 4ab+ 8? — 2”. av—xz—y—22y. 1—a@’— 2ab— 86’. av + 6? + 2ab — 16076’. 4¢7—9a+ 6a—-1. a+b? — ?— d’?—2ab — 2cd. w+ y? — 2xy — 2ab— a’ — B’. 927—6x2+1—a’—4ab — 467. a + 2ab — 2? — bay — 97+ 8’. 2 —2¢4+1—0?+ 2by—y’. 9—62+27— a — 8ab— 1687. 4—42+2°—4a—1—4¢d. a’ — a? —9+6'+ 6a—2a’8’. OOne te Com Ot 125, A trinomial in the form of a‘t+a’7b?+6' can be written as the difference of two squares. Since a trinomial is a perfect square when the middle term is twice the product of the square roots of the first and last terms, it is obvious that we must add ad? to the middle term of a*-+ a’b?+ b* to make it a perfect square. FACTORS. 93 We must also subtract ab? to keep the value of the expression unchanged. We shall then have (1) at+ a7? + Of = at+ 2070? + bt — a7? = (a+ 0) — ab? = (a+ 0? + ab)(a@ + 0? — ab) = (a + ab + 8’) (av — ab + 0’). If in the above expression we put 1 for 6, we shall have (2) @+e@+4+1=(e+2e+1)-e 3 (a? of . 1) ta fo =(¢7+1+a)(7?+1-a) =(v7+a+I1)(?—a+l). (3) Resolve into factors 4a* — 37 a°y’ + 97. Twice the product of the square roots of 4a* and 9y* is 122747. We may separate the term —37.27y? into two terms, —12a?y? and — 25 2x?y?, and write the expression (4a* — 12277? + 9y*) — 25277’ | = (2277 — 37’) — 2527 = (27 —3y+5xy) (20-37 —52y) = (227?+52y—37")(22°—odoay—3y). Exercise 33. Resolve into factors : 1. at+ay+y. 6. 9a‘t+ 26 a7b? + 25 *. 2. ata’+l. 7. 424*— 21277’ + 97. 3. 9at—15a7?+1. 8. 4at— 29 a7%c? + 25c’. 4. 1l6a*—17a@’+1. 9. 4a*+ 16a’? + 25ct*. 5. 4a*—138¢°+1. 10. 25a*-+ 312°’? + 16y*. 94 SCHOOL ALGEBRA. CasE V. 126. When a trinomial has the form x’?+ ax-+-b. . From § 107 it is seen that a trinomial is often the product of two binomials. Conversely, a trinomial may, in certain cases, be resolved into two binomial factors. 127. If a trinomial of the form 2?+ ax+6 is such an expression that it can be resolved into two binomial fac- tors, it is obvious that the first term of each factor will be xz, and that the second terms of the factors will be two numbers whose product is 4, the last term of the trinomial, and whose algebraic sum is a, the coefficient of x in _the middle term of the trinomial. (1) Resolve into factors «? + lla + 380. We are required to find two numbers whose product is 30 and whose sum is 11. Two numbers whose product is 30 are 1 and 30, 2 and 15, 3 and 10, 5 and 6; and the sum of the last two numbers is 11. Hence, x + 1llx+30=(#+5) («+ 6). (2) Resolve into factors 2? — 72+ 12. We are required to find two numbers whose product is 12 and whose algebraic sum is — 7. Since the product is + 12, the two numbers are both positive or both negative; and since their sum is — 7, they must both be negative. Two negative numbers whose product is 12 are — 12 and —1, —6 and —2, —4 and —3; and the sum of the last two numbers is —7,. Hence, 2’ —T*x+12= (¢ —4)(¢#—8). (3) Resolve into factors 2? +4 2% — 24. We are required to find two numbers whose product is — 24 and whose algebraic sum is 2. FACTORS. 95 Since the product is — 24, one of the numbers is positive and the other negative; and since their sum is +2, the larger number is positive. Two numbers whose product is — 24, and the larger number posi- tive, are 24 and —1, 12 and — 2, 8 and —3, 6 and — 4; and the sum of the last two numbers is +2. Hence, x + 2% — 24 = (x + 6) (x — 4). (4) Resolve into factors 2? — 3x — 18. We are required to find two numbers whose product is — 18 and whose algebraic sum is — 3. Since the product is — 18, one of the numbers is positive and the other negative; and since their sum is —3, the larger number is negative. Two numbers whose product is —18, and the larger number nega- tive, are —18 and 1, —9 and 2, —6 and 38; and the sum of the last two numbers is —3. Hence, x —38x—18=(x# — 6)(x+ 8). (5) Resolve into factors 2? —102y + 97’. We are required to find two expressions whose product is 9 y? and whose algebraic sum is —10y. Since the product is + 9y?, and the sum —10y, the last two terms must both be negative. Two negative expressions whose product is 9y?, are —9y and — y. —3y and —3y; and the sum of the first two expressions is —10y. Hence, 2—=10ay+97 = (#—9y)(4#—y). Exercise 34, Resolve into factors : 1. 2?+82+415. 4. «?—8x—10. 2. 2 —8r+ 15. 5. 27+ 5ar-+ 6a’. 3. 2?+ 24-15. 6. 2’?— 5axr-+ 6a’. 96 SCHOOL ALGEBRA. xv? — 2a — 15, zv’+5a2+6. x*— 52+ 6. . v@ta—ob. . v2 —a—Bb. . 2&+6x-+ 5, . 2 —62+5. . @+4xe—5. . @& —44—5., . 2@+9e418. . v2 —9x+18. . 2+82—18. . #&—dsx— 18. . #2 +9r+8. . 2&—9x+8. . v@t+tTe—8. . #@—Tx—8. . &@+7¢+10. . 2 —Tx+10. . 2+3a—10. . 2 — 5ax— 50a’. . vy? —382y —A4. . &@— Bax — 542". . &—ae— 2c’. - ff — 8yz+ 1872’. 32. 33. 34. 35. 36. 37. 38. 39. 40. 41. 42. 43. 44. 45. 46. 47. 48. 49. 50. BL 52. 53. 54. 55. 56. xv’ +ax— 6a’. x’ —ax— 6a’. e+ bay +4y’. x? — day —Ay’. x” — Say + 4y’. x’ + d32y—4y’. 2 —32y —4y/’. a’*®— Tab + 100’. aa? — 8axn — 54. xv’ — Ta — 44. xv’ + 4 — 182. x’ —15x2+50. a? — 23a+ 120. v+17a— 890. c? + 25¢e— 150. c? — 58c-+ 57. a‘ — 11a’? + 308°. 2+ Ozy + 2077. xy? + 19 xyz + 48 2” vb? — 13 abe + 22’. a’ — 16ab — 860". x +17 ay + 807’. x — Tay — 18y’. c’ + ¢ — 20. a’ + 16ab — 26087. FACTORS. 97 57. yu? — Syz— 842’. 58. x? —1l2— 152. 59. 18—32—2?=— (27+ 327 — 18). 60. 8834+ 8x2—27=— (2? — 8x — 83), 61. 78 — Tx — 2x’. | Case VI. 128, When a trinomial has the form ax?-+ bx-+ce, From § 105, (82 — 2)(5x+ 3) = 152°+9x2—102—6 = 152°— 2x—6: (1) (8x2 — 2)(52 — 8) = 152?—92— 102+6=152'—19246. (2) Consider the resulting trinomials : The first term in (1) and (2) is the product 3a x 52. The middle term in (1) is the algebraic sum of the products 32 xX 3 and (— 2) x 5a. The middle term in (2) is the algebraic sum of the products 3a xX (—3) and (—2) x 52. The last term in (1) is the product (— 2) x 3. The last term in (2) is the product (— 2) x (— 3). The trinomials have no monomial factor, since no one of their factors has a monomial factor. Hence, 1. If the third term of a given trinomial is negative, the second terms of its binomial factors will have unlike signs. 2. If the third term is positive, the second terms of its binomial factors will have the same sign, and this sign is the sign of the middle term. 98 SCHOOL ALGEBRA. 3. If a trinomial has no monomial factor, neither of its binomial factors can have a monomial factor. (1) Resolve into factors 62? + 172+ 12. The first terms-of the binomial factors must be either 62 and 2, or 3a and 22. The second terms of the binomial factors must be 12 and 1, or 6 and 2, or 3 and 4. We therefore write I. (624+ )(@+ J; or Il Cat” Que For the second terms of these factors we must reject 1 and 12; for 12 put in the second factor of I. would make the product 6a x 12 too large, and put in the first factor of I., or in either factor of IL., the result would show a monomial factor. We must also reject 6 and 2; for if put in I. or II. the results would show monomial factors; and for the same reason we must reject 3 and 4 for I. The required factors, therefore, are (3a + 4) and (2a + 3). (2) Resolve into factors 142? — ll2— 15. For a first trial we write (Ter Aten e): Since the third term of the given trinomial is —15, the second terms of the binomial factors will have unlike signs, and the two products which together form the middle term will be one +, and the other —. Also, since the middle term is —1la, the negative product will exceed in absolute value the positive product by —1la. The required factors, therefore, are (77 + 5) and (2a — 3). Exercise 35. Resolve into factors : 1. 22?°+52-+3. 4, 242’ —2ay—15y*. 2. 827—x—2. 5. 38627?— 19ay— by’. 3.0” —82-+ 3. 6. 1527+ 192y-+ 6y’. — FACTORS. 99 a dan oe 14. 152° — 26xy+ 87’. 8. 62° —4— 2. 15. 927+ 62y — 8y’. 9. 1527+ 1427-8. 16. 62? —2y — 85y’. 10. 82?—10x+ 3. 17. 102?— 2lzy — 107’. 11. 1827+ 92—2. 18. 142?—55ay +217. 12. 242*°— l4xy — 5y/’. 19. 62? — 23 zy + 207’. 13. 240° 882y+157%. 20. 627+ 35ay—6y’. Casz VII. 129. When a binomial is the sum or difference of two cubes. : 3 From § 108, tae Herter and cag hay Mme Se a—b a +b =(a+b)(a—ab+0’) and a’ — B= (a — b) (a + ab + 0’). In like manner we can resolve into factors any expres- sion which can be written as the sum or the difference of two cubes. (1) Resolve into factors 8a° + 270°. Since by § 118, 8a* = (2a)*, and 276°= (807)*, we can write 8a? + 270° as (2a)? + (807). Since a + 6? = (a+ b)(a — ab + 0B’), we have, by putting 2a for a and 30? for 6, (2a)8+ (30°) = (2a+30*) (4a°—6ab?+ 90%). 100 SCHOOL ALGEBRA. (2) Resolve into factors 1252* — 1. 12527—1=(5z)—-1 = (54 —1) (2527+ 52+1). (3) Resolve into factors 2° + 7’. += (e+) =@ + y)e— vy ty). 130, The same method is applicable when the cubes are compound expressions. (4) Resolve into factors (x — y)? + 2’. Since a@§+0?=(a+6)(?—ab+0’), we have, by putting «—y for a and z for 3, (ex—yfP+2=[(e—-y)+2][@—-yl—-@—yer2] = («—y+z)(a’—2ayt+y—ax2+y2+2). Exercise S36. Resolve into factors : 1. a& + 86%. 5. 27a y® — 1. 9. 216.a° — b*. 2. a&— 27a’. 6. a+ 270°. 10. 64a’— 276°. 3. a+ 64. 7. vy? — 64. 11. 343 — 2°. 4. 125a°+1. 8. 64a°-+ 1256°. 12. a’b* + 348. 13. 8a°— J. 19. 82° —(x#—y)’. 14. 216m* + n°. 20. 8(x+y)+ 2. 15. (a+6)—1. 21. 7297° — 642°. 16. (a—6b)+1. 22. (a+b) —(a— by’. 17. (2a+y)—(x—y). 23. 729a°4+ 2164 18. 1—(a— 6)’. 24. xy? — 5122. FACTORS. 101 181. We will conclude this chapter by calling the stu- dent’s attention to the following statements : 1. When a binomial has the form 2” — y”, but cannot be written as the difference of two perfect squares, or of two ' perfect cubes, it is still possible to resolve it into two fac- tors, one of which isa—y. Thus (§ 109), a — 88 = (a—b) (at + a'b + a°B? + ad? + 0). 2. When a binomial has the form 2”+ y”, but cannot be written as the sum of two perfect cubes, it is still possible to resolve it into two factors, except when 7 1s 2, 4, 8, 16, or some other power of 2. Thus (§ 109), a + 6° = (a+ b)(at—a*b + oh? — ab’ + D*). But a?+ 0?, at +0, a®+ 6°, cannot be resolved into factors. 3. The student must be careful to select the best method of resolving an expression into factors. Thus, a°— 6° can be written as the difference of two squares, or as the dif- ference of two cubes, or be divided by a — 4, or by a+ 8. Of all these methods, the best is to write the expression as the difference of two squares, as follows (a’)? Se (0°)? == (a? + b*) (a? ac b*) = (a+b)(a’— ab+b’)(a—b)(a’+ab+0’). 4, From the last example, it will be seen that an expres- sion can sometimes be resolved into three or more factors. 8 eA 68 — (at + b*) ey es b*) — (a* + b*) ix + b) Ga Ee 6°) = (a + 5‘) (a? + 8’) (x + b) (a — 4). 102 SCHOOL ALGEBRA. 5. When a factor occurs in every term of an expression, this factor should first be removed. ‘Thus, 82? —50¢0+ 42—1l0a= 2(42? — 25a? + 24 — 5a) = 2[ (42? — 25a”) + (22 —5a)] = 2(2%—5a)(2z2+5a+1). 6. Sometimes an expression can be easily resolved if we replace the last term but one by two terms, one of which shall have for a coefficient an exact divisor or a multiple of the last term. ‘Thus, (1) #—52?+112—15=(2'—52°+62x)+(5x—15). =x (2*—52+6)+5(x—3) =(#—3)[#(@—2) +5] = (%—8) (x’?—2x-+5). (2) 2—92?+ 262—24=(2°—92?4+ 1427)+(124—24) = (0!—9e-+14)412("—2) =x (x—7)(«—2)+12(x—2) = (# — 2) (#’?—T2+12) = («—2)(x—8) (a—A4). (3) a — 26% — 5 = (2° — 25x) — (e+ 5) = x(x? — 25) — (x#+ 5) = («+ 5) (a — 52 —1). (4) a+ 3a? —4= (2+ 227)+4+ (2? —4) = x" (x + 2)+ (a? — 4) = (4 + 2) (2? + x — 2) = (e+2)(2+2)(e—1). vo S) FACTORS. 103 Exercise 37. EXAMPLES FoR REVIEW. Resolve into factors : a — 9a. etevtetl. v—2y+ 2x — xy. x2’ —14x%+ 49. 36 27 — 497’. etna (c7—y)’ — 6. x+y’. . o—y, at -- ¥*, . 2 —(a—b)*. pti rae gies eb i aS a ee ce Ce ee ee — | o Ff oO WD FF} © - @—(m+n)’. . 2@—l1l2e+18s. . 2+4a— 45. . 2@+182+ 86. . #& — 184 — 48. . v2 +9x— 86. - 1027+ 2—21. - 62?—x— 12. no wo —|§ §|-& = & = ©Oo2 © © +t xy’? — 4xy* — 32°y’. 82° + 22°—9x—6. . W+2mn+n?— 1. 23. 24. 25. 26. Bhs 28. 29. 30. 31. 32. 33. 34. 35. 36. 37. 38. 39. 40. 41. 42. 43. 44. 122?—xz—1, 1227 — x — 20. 9a¢7+ 124a+4. av — 6 — + 2be. at + aty?+ y4 2 —6z—40. x’? — Tx — 60. a’ —19a-+ 84. + 2ar+ 38b2+6ab. V+ mn’? —n® —2mxz. 4Ax*— x’. ei ty", Dat + 21 x77? + 25 y'. w—4+4y 4+ 22y. 227+ 3xy —2y’. 2¢7—7Ta+6. x —Te+l. 1—a’?— 0° — 2ab. 32* — 62° + 927. v— 52? — 22+ 10. x’ t+ ax — bx — ab. 2a? — 3xy + 4axr— bay. 104 SCHOOL ALGEBRA. 45. az'+ ba’ — ax— b. 63. 6a —a—T7. 46. &+6+a+b. 64. 5c*— 15c? — 90’. 47. &—b+a—Bb. 65. aux—cx+ay—cy. 48. (x—y)—2y(x—y). 66. 162*—8l. 49. 1+ 102y+ 202’°y’. 67. v+a2’+1. 50. a—0?+ 2be—’. 68. 272° — 64a’. 51. a? +4y7—2—4ay. 69. 2° +7. 52. a —40?—97?+12bc. 70. 2—y’. 53. 49°4+9y—2—122y. 71. a°— 206. 54. (a+b) —(e—d)’. 72. 2+ l6a’2’+ 256a%. 55. w+ y’. 73. 1—(«#—yy. 56. 322° —c’. 74. (a@ty)4+ (Qz—y). 57. a®+ 647%. 1b, 2216: BS ahi ee 176. 382° +2—2. bos go ay", 717. 2—8a— 2a’. 60. (a+ 6)* —1. 78. 4—5e—6e’. 61. v@&—B’?+a—b. 79. 2ry—2#?—y +2. 62. v@ta+36—90’. 80. 4at— 9a@?+ 6a—1. 81. w@ —2ab+ 0’ + l2ay —42°— 97’. 82. 2a7—4ry + 2y’+ Zax — ay. 83. (a+ 6)—1—ab(a+b+1). 84. w—a’?+38x2+5. (See § 181, 6.) 85. 62° — 2327+ 16x—38. (See § 1381, 6.) 86. wv +y7+2—2Qxry—222+2yz. (See § 103.) 87. 4a°b?— (a + 0? — ¢’)’. CHAPTER VIII. COMMON FACTORS AND MULTIPLES. 132. Common Factors. A common factor of two or more numbers is an integral number which divides each of them without a remainder. 133. A common factor of two or more expressions is an integral and rational expression which divides each of them without a remainder. Thus, 5a@ is a common fac- tor of 20@ and 25a; 327y? is a common factor of 1227 and 152%’. 134, Two numbers are said to be prime to each other when they have no common factor except 1. 135, Two expressions are said to be prime to each other when they have no common factor except 1. 136. The highest common factor of two or more numbers is the greatest number that will divide each of them without a remainder. 187. The highest common factor of two or more expres- sions is the expression of highest degree that will divide each of them without a remainder. Thus, 3a’ is the highest common factor of 8a7, 6a’, and 12a‘; 527’ is the highest common factor of 102*y? and 15 2’°y’. For brevity, we use H.C. F. to stand for “ highest com- mon factor.” 106 SCHOOL ALGEBRA. To find the highest common factor of two algebraic expressions : Case I. 138. When the factors can be found by inspection. (1) Find the H.C. F. of 42.a°0? and 6076‘. A200 =2x 3X TX aaa X00" 60a7b*= 2x23 x5 xX aa X bbDd. ., the H.C. F.=2 x 8 x aa x 68, or 6a’d*. 2) Find the H.C. F. of 2a?a-+ 2a2z? and 8abay+ 3 b2’y. y y 2e0¢%+2an7 =2ar(a+2); Babry + 3bx’y = 3bry(a+2z). peg nate va Lik Ba thy =2(a+ 2). (3) Find the H.C. F. of 477+ 42 —48, 62?— 482 +4 90. 4Avt 44—48=—4(2?+ 2-12) =4(x—38)(#+4 4); 62? — 482 — 90 = 6 (2? — 8x4 4+ 15) = 6(e—8)(«—5); RDN Rats oad bed 9 Od = 2(a — 3) = 2a — 6. Hence, to find the H.C. F. of two expressions: Resolve each expression into its simplest factors. Find the product of all the common factors, taking each factor the least number of tumes rt occurs in any of the given CXPTESSLONS. COMMON FACTORS AND MULTIPLES. 107 Exercise 88. Find the H.C.F. of 1. 120 and 168. 4. 36a°2? and 28 2°%y. 2. 362° and 272%. 5. 48a7b%c and 60a°c’. 3. 42a72° and 60 a°2’. 6. 8(a+ 6) and 6(a+ 5)’. 7. 12a(@+y) and 46(x4+ y). 8. (7—1)?(x+ 2) and (x — 8) (x + 2)*. 9. 2407)? (a+b) and 42a0°%b(a+ bd)’. 10. 2’?(a—3) and 2’—3xz. 12. 2 —42 and w—62+8. 11. a — 16 and x + 42. 18. 2? —7Tx2+12 and 2?—16. 14. 92°— 47’ and 122’ — xy — 67’. 15. 2 —Ta2#—8and 2’?+ 5274+4. 16. x2’? + 32y — 107 and 2? — 22y — 35 y?. 17. xt — 22° As? and 62°— 62 — 1802". 18. 2° — 3a’y and 2 — 277. 19. 1+ 642’ and 1—42-+4 162”. 20. 2*— 81 and 2*+ 82’ — 9. 21. #+22—3, 2+ T2+12. 22. x?—62+5, 27+ 382 — 40. 23. 8at+ 15a°b — 727d", 6a? — 38007) + 36.07. 24. 62°y —122y?4+ 6y’*, 32° + 9ay’ — 127%. 25. 1—16c*, 1+ c’—12c*. 26. 82° +22%—1, 6a°+ 77442. 27. 627? +a2—2, 122°—2—6. 28. 152° —19277 + 62y’, 102*— a*y — 32°77. 29. 102*y + 927y? —9ay*, 4a? +157°* — 42’y,. 108 SCHOOL ALGEBRA. Case II. 139. When the factors cannot be found by inspection. The method to be employed in this case is similar to that of the corresponding case in Arithmetic. And as in Arithmetic, pairs of continually decreasing numbers are obtained, which contain as a factor the H.C. F. required, so in Algebra, pairs of expressions of continually decreas- ing degrees are obtained, which contain as a factor the H.C. F. required. 140, The method depends upon the following principles: (1) Any factor of an expression is a factor also of any multiple of that expression. Thus, if ¢ is contained 3 times in A, then ¢ is contained 9 times in 8.4, and m times in m A. (2) Any common factor of two expressions is a factor of ther sum, thew difference, and of the sum or difference of any multiples of the expressions. Thus, if ¢ is contained 5 times in A, and 8 times in B, then e¢ is contained 8 times in A+B, and 2 times in A— B. Also,in5 A+2 Ait is contained 5x 5+28, or 81 times, and in 5A —2 28 it is contained 5X 5—28, or 19 times. (3) The H.C. F. of two expressions is not changed vf one of the expressions is diided by a factor that is not a factor of the other expression, or 1f one is multiphed by a factor that is not a factor of the other expression. Thus, the H.C.F. of 4a7bc? and a’c’d is not changed if we remove the factors 4 and 6 from 4a7bc’, and d from ved; or if we multiply 4a%c? by 7, and a’e*d by 11, . COMMON FACTORS AND MULTIPLES. 109 141, We will first find the greatest common factor of two arithmetical numbers, and then show that the same method is used in finding the H.C.F. of two algebraic expressions. Find the greatest common factor of 18 and 48. 18) 48 (2 36 12)18(1 12 6) 12(2 12 Since 6 is a factor of itself and of 12, it 1s, by (2), a fac- tor of 6+ 12, or 18. Since 6 is a factor of 18, it is, by (1), a factor of 2 x 18, or 86; and therefore, by (2), it is a factor of 86+12, or 48. Hence, 6 is a common factor of 18 and 48. Again, every common factor of 18 and 48 is, by (1), a factor of 218, or 36; and, by (2), a factor of 48 — 36, or 12. Every such factor, being now a common factor of 18 and 12, is, by (2), a factor of 18 — 12, or 6. Therefore, the greatest common factor of 18 and 48 is contained in 6, and cannot be greater than 6. Hence 6, which has been shown to be a common factor of 18 and 48, is the greatest common factor of 18 and 48. 142, It will be seen that every remainder in the course of the operation contains the greatest common factor sought ; and that this is the greatest factor common to that remain- der and the preceding divisor. Hence, The greatest common factor of any dwisor and the corre- sponding dividend is the greatest common factor sought, 110 SCHOOL ALGEBRA. 143, Let A and & stand for two algebraic expressions, arranged according to the descending powers of a common letter, the degree of B being not higher than that of A. Let A be divided by JB, and let Q stand for the quo- ent, and # for the remainder. Then B) A(Q BQ te Whence, Rk = A— BQ, and A= BQ+ BR. Any common factor of B and & will, by (2), be a factor of BQ+ R, that is, of A; and any common factor of A and B will, by (2), be a factor of A — BQ, that is, of A. Any common factor, therefore, of A and B is likewise a common factor of B and R&. That is, the common fac- tors of A and B are the same as the common factors of B and #; and therefore the H.C.F. of & and Z& is the H.C.F. of A and B. If, now, we take the next step in the process, and divide B by R, and denote the remainder by S, then the H.C. F. of S and A can in a similar way be shown to be the same as the H.C.F. of B and A, and therefore the H.C. F. of A and B; and so on for each successive step. Hence, The H.C. F. of any dwisor and the corresponding diw- dend 1s the H. C.F. sought. If at any step there is no remainder, the divisor is a fac- tor of the corresponding dividend, and is therefore the H.C.F. of itself and the corresponding dividend. Hence, the last disor is the H.C. F. sought. Notr. From the nature of division, the successive remainders are expressions of lower and lower degrees. Hence, unless at some step the division leaves no remainder, we shall at last have a remainder that does not contain the common letter. In this case the given expressions have no common factor. COMMON FACTORS AND MULTIPLES. PEL Find the H.C.F. of 227+ 2—3 and 42°+82’?—x—-6. 207 +2—3)42°+82?7— x—6(2¢4+8 4° + 227—6x 627+ 52—6 62°+ 382— 9 22+3)22+ #¢—3(¢4—1 227+ 32 —2x4—8 pees. G.b. = 22 + 3. —2x2—838 Each division is continued until the first term of the remainder is of lower degree than that of the divisor. 144, This method is of use only to determine the com- pound factor of the H.C.F. Simple factors of the given expressions must first be separated from them, and the H.C.F. of these must be reserved to be multiplied into the compound factor obtained. Find the H.C.F. of 12a* + 802° — 7227 and 322°+ 842’— 1762. 122+ + 802° — 722? = 62° (227 + 5x — 12). 322° + 842? —176x=—42(82' + 21x — 44). 62? and 4x have 2x common. Qa? + 5x2 —12)8a7+21a—44(4 827+ 20x --48 e+ 4)20°+52—12(2¢ —3 227° + 82x —38x2—12 ». the H.C. F. = 22(x + 4). —3x—12 112 SCHOOL ALGEBRA. 145. Modifications of this method are sometimes needed. (1) Find the H.C. F. of 42?—8xz—5 and 122°—42—65. 4a? —8x—5)122?— 4x—65(8 122? — 24x%—15 202 — 50 The first division ends here, for 20a is of lower degree than 42”. But if 20%—50 is made the divisor, 42? will not contain 20z an integral number of times. The H.C. F. sought 7s contained in the remainder 20% — 50, and is a compound factor. Hence if the simple factor 10 is removed, the H.C. F. must still be contained in 2a — 5, and therefore the process may be continued with 2 — 5 for a divisor. 2% —5)427— 8xex—5(2r+1 427— 102 Qu 5 9x2—5 .. the H.C. F. = 22 —5,., (2) Find the H.C.F. of 212° — 42? — 154 — 2 and 212° — 322? — 542 — 7. 212° — 4a? — lda— 2) 212° — 322? —54e—7(1 QM e— 427°— 15372 — 2827 —39%—5 The difficulty here cannot be obviated by removing a simple factor from the remainder, for — 282?— 39a%—5 has no simple factor. In this case, the expression 21 #3 — 4a?— 15a — 2 must be multiplied by the simple factor 4 to make its first term exactly divisible by — 28 a”. The introduction of such a factor can in no way affect the H.C. F. sought, for 4 is not a factor of the remainder. The signs of all the terms of the remainder may be changed; for if an expression A is divisible by — F, it is divisible by + F. The process then is continued by changing the signs of the re- mainder and multiplying the divisor by 4. COMMON FACTORS AND MULTIPLES. 113 2827+ 3944+ 5)842?— 1l62°?— 60x— 8(8x 842° + 1172+ 1dex —1832°7— Tdx— 8 Multiply by —4, —4 53227 + 3002 +4 32(19 53822? + 7412+ 95 Divide by — 68, — 63)— 4412 — 63 Tatil Tx + 1)282?+4+ 894+ 5(42+5 2827+ 42 85x2+5 reine HH. O.b=/ 2+ 1. 35x2+5 (3) Find the H.C. F. of 827+ 2x7 —8 and 62°+ 52’ — 2. 627+ B5a’?— 2 these alr | 82? + 24 —3)242°+ 202*?— 8 (82+7 2427+ 6a°?— Ya 1427+ 9x— 8 Multiply by 4, 4 5627 + 86a — 382 5627+ 142 — 21 Divide by 11, YY) 22 eee 2%— 1)824+227—3(424+38 8a’—4a 62—3 fe Luertt() b= 22— 1; 62—3 114 SCHOOL ALGEBRA. The following arrangement of the work will be found most convenient : 827+227—8 | 6a + 5a? — 2 827 —4a 4 64—3 2427+ 2027— 8 32x 62 —3 242°+ 627— Ya 1427+ 9xr— 8 4 5627 + 862 — 32 +7 5627+ 142 —21 | 11) 222 —11 Q2— 1 4a +8 146. From the foregoing examples it will be seen that, in the algebraic process of finding the H.C.F., the follow- ing steps, in the order here given, must be carefully observed : I. Simple factors of the given expressions are to be re- moved from them, and the H.C. F. of these is to be reserved as a factor of the H.C. F. sought. II. The resulting compound expressions are to be ar- ranged according to the descending powers of a common letter; and that expression which is of the lower degree is to be taken for the divisor; or, if both are of the same degree, that whose first term has the smaller coefficient. III. Each division is to be continued until the remainder is of lower degree than the divisor. IV. If the final remainder of any division is found to contain a factor that is not a common factor of the given expressions, this factor 2s to be removed; and the resulting expression is to be used as the next divisor. V. A dividend whose first term is not exactly divisible by the first term of the divisor, is to be multypled by such a number as will make it thus divisible. Sp ple LE gs EES 1 he, Por wo wo DO WD S-& S&S BSB SB BS SF SF SF SF BS Cr es Re Ct CO OO - ered Ole Ol ee tie SC OOe se oy 6 ft COMMON FACTORS AND MULTIPLES. 115 Exercise 39. Find by division the H.C. F. of 427+ 382-10, 42°+ 72? — 82-15. 22°— 627+ 524—2, 82° — 2377+ 172 —6. 2023+ 227—182+ 48, 20a*—172?+ 482 — 83. Ag® — 22?—16a—91, 122° — 28 2? — 874 — 42. 120° +42°+172—8, 242°— 522e?+ 14¢—1. 20° + 52? —I9r74+38, 382° 4+ 22?—17274+ 12. 8a* — 62° — 2? +1527 —25, 42° + Tx’? — 32 — 15. 4¢°>—42?—52+38, 102?—192+ 6. 62t—1382°+ 3277+ 22, 62*-—102'+ 42°? —- 624-4. 22° — 82°? + 20? —2Qa—3, 4a*+32°+42—8. - da°—a?— 2e?4+2xe—8, 62°+182?4+ 38x-+ 20. . o@ + 22+ eg, 8at +22 —3827+2r-—1. . 6a2°—9attlla*+6 2’—102, 42°+10a'+10 2°44 24602. . 2e°—I11a’—9, 42°4+11l2*+81. ee + 102 — 1924-9 at +2274 9, . 2e°—382?—16274 24, 4a°4+ 22t— 282° — 162° — 322. . &—e—l4e4+2+1, Rigid eta a oetigs | on . 62°—14az?+ 6a'x—4a', 2*—az*®— aa’ — aa — 2 a. se20 —2a' — 3a’ — 2a, 3a‘ —a®'— 2a'— 16a. . 22° + Tar? +472 — 8a’, 42°4 9ar’— 2e'r — a’. . 22?—9az’+9aa—Ta’, 42° — 20a2? +20 ae —16.a°, . 2a44+ 9e'8 + 142+38, 824+ 142°4+ 927+ 2. 92° — Ta? + 8274+ Qa —4, 6at—T2*°—102?+52742. 116 SCHOOL ALGEBRA. 147. The H.C. F. of three expressions may be obtained by resolving them into their prime factors; or by finding the H.C. F. of two of them, and then of that and the third expression. For, if A, B, and Care three expressions, and D the highest common factor of A and B, and £ the highest common factor of D and C, Then PD contains every factor common to A and B, and # contains every factor common to D and C. .. # contains every factor common to A, B, and C. Exercise 40. Find the H.C. F. of ’+8a+2, 2+427+3, 2’ +62+5. x’—9x2—10, 2 — Tx — 30, 2 —11lxe#+10.. 2—l1, e—227+1, #P-—22+1. 62+a—2, 22°4+7Tx#—4, 2e?—Txr+38. vt 2ab+0?, 7@—B, a+2a7b + 2ab ? + Bb. xv —dax+ 4a, 2 —B8ar+2d, 32?—10ar+ 7a’. eta—, 2—2e—27+2, 2+327—6r—8. e+ T+ 52-1, 2+3e2—82'—1, 32°+ 527+2—-1. v—62"+1la—s6, v°—8274-192—-12, 2°—92°+ 262—24. O~ Or st OF Cte oe 08 NS 148. Common Multiples. A common multiple of two or more numbers is a number which is exactly divisible by each of the numbers. A common multiple of two or more expressions is an expression which is exactly divisible by each of the ex- pressions. Thus, 48 is a common multiple of 4, 6, and 8; 48 (a? —y’) is a common multiple of 3(a—y) and 8(#+ y). COMMON FACTORS AND MULTIPLES. 117 149, The lowest common multiple of two or more numbers is the least number that is exactly divisible by each of the given numbers. The lowest common multiple of two or more expressions is the expression of lowest degree that is exactly divisible by each of the given expressions. Thus, 24(2*?—y’) is the lowest common multiple of 3(z—y) and 8(#+ y). We use L.C.M. to stand for “lowest common multiple.” To find the L.C.M. of two or more algebraic expressions : CasE I. 150. When the factors of the expressions can be found by inspection, (1) Find the L.C.M. of 42.°0? and 60a70*. Siu) KT a Kb: 60a7b*=2X2X3XK5xXa?x Ot. The L.C.M. must evidently contain each factor the greatest num- ber of times that it occurs in either expression. Rea A x OC XK IKK 0 6 OF, = 420 a®d*. (2) Find the L.C.M. of 4a? + 44—12, 627-482-490, 427-102 —6. Agt 4¢—12=4(2?+a2—12) =2xX2(¢—3)(2#+4); 62°—48 2+ 90=6(2?— 82415) =2 x 3(«— 8) (x%—5); 4z’?—10x— 6=2(22°—52—3) = 2(4—8)(24+1]). ~ L.0.M.=2x 2x 3 x (#—8)(#+4)(e—5)(22+1). Hence, to find the L.C. M. of two or more expressions: Resolve each expression into its simplest factors. Find the product of all the different factors, taking each factor the greatest number of times rt occurs mn any of the given expressions. 118 SCHOOL ALGEBRA, Exercise 41. Find the L.C.M. of a I EF FF EF >» SF S we we mia) OU He OO, (Or et ce .00- wets Co) OU hare OO SN 24, 32, and 60. 24 a7x*, 60 a0*2?, and 32472’, xv’? —2Qeyty’ and 2 —y’. v’—4e+4, a?+4a2+4, and 2 —4. sta’ and 2? — a’. Ytarta’, v’—a’, and 2’— a’. x" (%2— 3) and 27— 52-46. e+ Ta+12 and 2 — 92’. v—Ta+10, 2—42—5, 2? —ax—2. 1—32—421—42—527, 1292-1 . 62+ Try —3y’, 32?4+ llay—4y’, 22°+ llezy+12y’. . 8—14a+6a’, 4a+4a?—38a', 40°+2a?— 6a". . 62+72°—8a, 82°+142—5, 62°+392+45. - 6ax+9bx—2ay—3by, 62° + 3a%—2xy — ay. . l2Zax—9ay—82y+ 67’, bax+ 3ay—4ay—2y’. . 272?—a’, 62? +ar—a’, 152°—5ax+8ba—ab. . @—)l, 2e?—ax—-1, 32?—2—-2. Case IT. 151, When the factors of the expressions cannot be found by inspection. : In this case the factors of the given expressions may be found by finding their H.C.F. and dividing each expres- sion by this H.C. F. COMMON FACTORS AND MULTIPLES. 119 Find the L.C.M. of 62° — llay + 27° and 92° — 22 xy? — 8 y’. 62°—11] ay +2 7° 9a°—222y'— 8y’* 3 62°— 82°y—4 27’ 2 — d7y+4 277127 18 2°—44 x7'—16 7? | — 8ay4+4a77427' 18 2°—338 a’y+ 6y’' lly) 33 v’y—44 zy’—22 y° 82° — 4ay— 2y*?|Qr—y .. the H.C. F. = 32? —42y —2y’. Hence, 62°—11lw’y+2y?=(22—y) (82°?—4xy—2y’), and 92° —22 ay’ —8y=(82+4y) (82°—42y—2y’). .. the L.C.M. = (22—y) (84+ 4y) (82 —42y—2y’). Exercise 42. Find the L.C. M. of dh CO IF TP wD - = & wo wm e OS 62° — Tax’? — 20a7x, 32? + ax — 4a’. 32° — 182? + 234 —21, 6a? + 2? — 4424 21. 382° — 8ay+ay—y’, 4e°— ay — 827’. e& —2c-+c, 2c — 2c —2c— 2. v—82r+3, 2°—32°+ 212-8. a —6a74+12 az? — 82’, 2a? — 8axr+ 82". 22+27—12272+9, 22° —72?+122-9. te — 227 — 5, (2? + 1227+ 102+ 5. zt — 1327+ 36, 2t— 2’ —Te’+ +6. 22°1327—Ta—10, 4a°—42?—9x4-+5. . 1222— 2? —802—16, 62° — 22? — 13824 —6. . 684+ 2? —5r—2, 6a® +52°—382—2. ee Or + 267 — 24, a — 1907 4-47 2 — 60, CHAPTER IX. FRACTIONS. 162. An algebraic fraction is the indicated quotient of two expressions, written in the form Z The dividend a is called the numerator, and the divisor 6 is called the denominator. The numerator and denominator are called the terms of the fraction. 153, The introduction of the same factor into the divi- dend and divisor does not alter the value of the quotient, and the rejection of the same factor from the dividend and divisor does not alter the value of the quotient. Thus ce ap SLE 3 Lorene . It follows, therefore, that 4 2x4 7 422 The value of a fraction is not altered uf the numerator and denominator are both multiplied, or both divided, by the same factor. REDUCTION OF FRACTIONS. 164. To reduce a fraction is to change its form without altering its value. Case I. 155, To reduce a fraction to its lowest terms. A fraction is in its lowest terms when the numerator and denominator have no common factor. We haye, therefore, the following rule ; FRACTIONS. 121 Resolve the numerator and denonunator into ther prime factors, and cancel all the common factors; or, divide the numerator and denominator by ther highest common factor. Reduce the following fractions to their lowest terms : (1) 38.a'bict _ 2 K 19a*b®c* _ 26?c” 57abe =—8x19a°be? — Ba (2) fee (a — a) (ai t-ar+a") a +ar+ x ve—x (a—xz)(a+2) ataz (3) a+fa+10_(@+5)(a+2)_a+5 at+d5a+6 (a+3)(a+2) a+3 (4) ee oe 20 8 (22-38) (82-2) sa-+2 827—2x%—15 (2%—38)(47+5) 424+5 e—4e7+42—1 eee (5) v—2e+424—8 We find by the method of division the H.C.F. of the numerator and denominator to be x — 1. The numerator divided by z—1 gives 2 —3x+1. The denominator divided by «—1 gives x’ —2-+8. See 42-1 abet) See Ae 8 at — e+ 3 Exercise 43. Reduce to lowest terms : i, & ab> 4. 42mid. 7 34 any? " Jab " 49 mn? 5lavay' 2 306°C P 30 xyz 35. a°b'c? " 15a°b*e? "18 2%2? ~ Baibic 3 26 ay* _ 21 mint «BB abict Bony 28 mp 87 a'b*¢ a bap 12. 13. 14. 15. 16. 1? fe 18. 28. 29. 30. 31. SCHOOL ALGEBRA. ZO eae 4 (a? + ae + c’) 2+ 22y+y' aa eT ioe a oh xv’+ 2a—15 piroeeh St zers wi i.e) 227— 132+21 a —~ 27 —20 5 92°— Tx—16 ae? —62—4 32° — 8a2+8 Geena wed e+t427—5 8 ee ad 82° —427—a21+2 zt — 182’? + 36 19. 20. 21. 22. 23. 24. 25. 26. 27. 32. 33. 34. 427+ 12axr+ Ia 82° + 27a’ v—y —- 2yz— Zz w+ 2ry+y—2 at 4. aa? yp 20+ 17a+21 3807+ 26a-+ 385 (a- bye (a+tb+ep creel I Y—x (oa) — oe (7+ by—@ (a+6) — (e+ da) (a-+c)?— (6+ ad) (a+ec)—6 4 ag? — (a? oe Reduce by finding the H.C.F. of the terms : 38° + 17a? + 222448 62° +25 27+ 2582+ 6 2° — 3a — 1b ea Zo et T2’+5a—25 22° 2) aie 82°+8e2+2—2 a + Aa) — Bae z'—a2+ 82-8 FRACTIONS. ees Case II. 156. To reduce a fraction to an integral or mixed expression. (1) Reduce = ; to an integral expression. Pata tetl (§ 108) (2) Reduce aa to a mixed expression. 2—1 |“z+1 +e 2t—etl —x—x eo) x+1 —2 x —l1 ; 2 y Sek ei ee Note. By the Law of Signs for division, 2 a ee, 2 2 ——— and = ; z+1 —(# +1) x+l The last form is the form usually written. 157, If the degree of the numerator of a fraction equals or exceeds that of the denominator, the fraction may be changed to a mixed or integral expression by the following rule: Divide the numerator by the denominator. Note. If there is a remainder, this remainder must be written as the numerator of a fraction of which the divisor is the denominator, and this fraction with its proper sign must be annexed to the integral part of the quotient. 124 SCHOOL ALGEBRA. Exercise 44. Reduce to integral or mixed expressions : 4a°+ lat 38 3 29 eee 3 SE EEERASREEP SES ane Ag atod. 9 ae ete en eA 9 Sat Adee le ; 32 a+t4 fy, wets. 10, © eee ety a+2z ie es he iL ae 54 t—Y yp P aes Hh 12 3 iy pa ae as gt ee ” Shae Ee 13, 4@+6ar+9at ety 22—3a 7 v+81 14 one 4 ate Wine in eg @ ae Case III. 158. To reduce a mixed expression to a fraction. The process is precisely the same as in Arithmetic. Hence, Multiply the integral expression by the. denominator, to the product add the numerator, and under the result write the denominator. (1) Reduce to a fraction “— i +26. gr G38 a ee oe z—4 x—4 ae ore ee x—4 De ee Ly x—4 FRACTIONS. (2) Reduce to a fraction a— b — 125 a Te nO enue até _(a@—6)(a+6) — (a — ab — 0) atbéb _@—B—a+ab +P ee ab . a+b a+b Norr. The dividing line between the terms of a fraction has the force of a vinculum affecting the numerator. If, therefore, a minus sign precedes the dividing line, as in Example (2), and this line is removed, the numerator of the given fraction must be enclosed in a parenthesis preceded by the minus sign, or the sign of every term of the numerator must be changed. Exercise 45. Reduce to a fraction : 1: ee 8. 9; ee 9 22 2ab Le b— 10. Ps atb 2 4. Aes 11. . x—2 pe | el 12. a+6 | gpa Ta a T atx iss a 14. . @-—art+2— ial CN eSios co x—3 3 ata Vtarzt2— BIN at Es es eR By 126 SCHOOL ALGEBRA. CasE IV. 159. To reduce fractions to their lowest common denominator. Since the value of a fraction is not altered by multiply- ing its numerator and denominator by the same factor (§ 153), any number of fractions can be reduced to equiva- lent fractions having the same denominator. The process is the same as in Arithmetic. Hence we have the following rule: Find the lowest common multiple of the denominators ; this will be the required denominator. Divide this denomi- nator by the denominator of each fraction. Multiply the first numerator by the first quotient, the sec- ond numerator by the second quotient, and so on. The products will be the respective numerators of the equwalent fractions. Norse. Every fraction should be in its lowest terms before the common denominator is found. ox 2 5 : (1) Reduce hae a and Ba to equivalent fractions having the lowest common denominator. The L.C.M. of 4a?, 3a, and 6a? = 12a’. The respective quotients are 3a, 4a”, and 2. The products are 9aa, 8a?y, and 10. Hence, the required fractions are Jax 8 ary nid Reo 12a? 1243 12a° in 2 3 2) Red —______, ——_—___., —_______ t fee ccs e+ 5e+t6 +4748 3 eo i equivalent fractions having the lowest common denom- inator. , ibe FRACTIONS. ey ee eee Se e+5e+6 @442743 2+22+1 in 1 2 3 ~ (w@+3)(e +2) (@ +3)(@41) (@+1)(@+1) . the lowest common denominator (L.C. D.) is (w + 3) (@ + 2)(@ +1) (a + 1). The respective quotients are (w+ 1)(@ +1), (w+ 2)(@ +1), and (# + 3) (a + 2). The respective products are 1(@+1)(@+1), 2(@+2)(e+1), and 3(@+3)(a + 2). Hence the required fractions are (w +1)(« +1) 2(w + 2)(e+1) (w + 3)(w@ + 2)(a +1)? (w@+3)(@ + 2)(~@4+ 1)? 3 (x + 8) (x + 2) (x + 3)(x + 2) (a + 1)? Exercise 46. Express with lowest common denominator : ae ak — 2 Oat. 4 Ft. 7G 3a Jax MAgew? Sa — ¢ 1 Bo 5 vty’ Layiphi +2 4+8 20%7—4y? Sa+2y a a" : 6 +2 sid t—a 2a Ve ey ee 7 1 1 2 ety a—-y er y¥ 8 1 1 3 T1422 1—42 1-22 9 5 ’ 7 ; 3 yt) Reta Gaeta 1 w—9r+18 2?—102+ 24 128 SCHOOL ALGEBRA. ADDITION AND SUBTRACTION OF FRACTIONS. 160. The algebraic sum of two or more fractions which have the same denominator, is a fraction whose numerator is the algebraic sum of the numerators of the given frac- tions, and whose denominator is the common denominator of the given fractions. This follows from the distributive law of division. If the fractions to be added have not the same denomi- nator, they must first be reduced to equivalent fractions having the same denominator. (§ 159.) Hence, to add fractions, we have the following rule: Reduce the fractions to equivalent fractions having the same denominator; and write the sum of the numerators of these fractions over the common denominator. 161. When the denominators are simple expressions. Sun pe BOAO. Fa— oO eee =a 1) Simplify ——— — ———— +. ——_. The L. C. D. = 12. The multipliers, that is, the quotients obtained by dividing 12 by 4, 3, and 12, are 3, 4, and 1. Hence the sum of the fractions equals 9a—12b_ 8a~4b+44¢,a—4e 12 12 12 _ 9a—12b6—(8a—4b + 4c) +a—4e Ye _ 9a—12b-—8a+4b—4ce+4a—4e 12 . _ 2a—8b—8¢ 12 _~7—4b—4e 6 FRACTIONS. The oaks work may be arranged as follows : The L. C. D. = 12. The multipliers are 3, 4, and 1, respectively. 3(3a—4b) = 9a—12b = lst numerator. —4(2a—b+c)=—8a+ 4b—4c= 2d numerator, l(a — 4c) = a —4c¢c= 3d numerator. * 2a— 8b—8c 129 or 2(a—4b—4c) =the sum of the numerators. .“. sum of fractions = 2(a—4b—4c) a—46—4e 1 6 Exercise 47. Simplify : 3 6 eS Poh aay tz@—5 38242 ,4¢+1 5¢—10 aes 3 4 12 22+38 9 6 12 3 24+3,2+38 182%4+5 w2«-8 2x 4x 82? x oe 29-1 oes Die tates 2 5) ron eer 12 2 2 ae ey ee A Neral ek Oe ee whi 4 at b ab a ab 4 16. h 12 3 pega a al OE 7. e 2 7 5 dia 8 fe eee Lelie) we PE ees 1380 SCHOOL ALGEBRA. Qr—6 8a—4 , 56xr— 48 1 Osa soe Bi 5a 15x as 45a ; 11 Liay Nt Syhaao! L6at— B : ay? ay? ay lone eluate ot 8 ee 2a'y Oy'2. Daz Ave eae 162. When the denominators have compound expressions. a —b atb @—B The L.C. D. is (a — b)(a + 8). ; The multipliers are a + b, a—}b, and 1, respectively. (a+ 6)(2a+b)= 2a*%+3ab + 0? = Ist numerator. —(a—b)(2a—b) = — 2a? + 3ab — 0? = 2d numerator. —1(6ab) = —6ab = 3d numerator. 0 = sum of numerators. *, sum of fractions = 0. : : Pee ie ees 2) Simplify ~ (2) bra ge Oa oi 4 The L.C.D. is (w — 2) (w — 3) (a — 4), («—1)(@—3)(@a—4)= a — 8a%+19x%—12 = Ist numerator. (w — 2)(w@— 2)(a@—4)= a — 82a? + 20x—16 = 2d numerator. (x — 2)(e@—3)(e—3)= a — 8a? + 21% —18 = 3d numerator. 32° — 242? + 60a — 46 = sum of numerators. 32° — 244? + 60a — 46 ~ gum of:aractions = ——— ee (@ — 2)(# — 8)(«— 4) Exercise 48. Simplify : 1 1 il 2, 1. ; eo nes : Rey Oo eB lta 1-2“ 2, l el 4 ty ee Itz l—z x—y (x£—y) Hint. Reduce the first fraction to lowest terms. FRACTIONS. 131 pee ty)" gy Se Sy tay ary («@t+y) a—y “ty w—y’ 1 1 x x Ve ee ee eS 10, : PEGA 2a(a =) eee, 1a, ve+e2 7 od tes Se een ae ling 11 it, Cage Bie ; ; aa l—#4+27 Sa Sgt os Oe bane te ap Rela oc) 12 Doe awe ey mere a — 27¢ OZR Sih a 28S a ise fea ace, e+2y xw-—2y 2—4y7* i a Veet ay Le ae — x ik 1 15. 1 = x—1l aes Fee aume Mmaber wn nirabin ~atb (a+by¥ (a+6) v7, Sy 22 # ey) Tee ged kar YF) Hint. Reduce the last fraction to lowest terms. 18. 3 A Gee eds sDa* z—a (#a—a) (x—a) 19. alae A cl a x—2a 2 —8a' 2#+2ar+4e ?_94¢+8 x— 2 T 20.0. We aes eth Sy Ee etl b AlsGand x+1 8 x—l ve+tae—s TNE BOS Ope Sok raat oe a We Eh Pe AR Oe oe Be Perea xv — 27 29 Te Oe 10) ke 1 ee Geer te LO biont te 132 SCHOOL ALGEBRA. wi—Ssan+6a r—Ta x—S8ar+15e x«—Dd5a iy 3 See oe xa—-2 #—82r4+2 w#—4244+3 23. 24. Hint. Express the denominators of the last two fractions in ' prime factors. 3 I w@-Tatl2 d@—4a+38 av—5a+4 b Gini se SH ae ee l0¢?+a—38 2¢+7Ta—4 Bibs 3 u Q—x2—-62 1l—x—22* 163. Since 2 a, and pan a, it follows that The value of a fraction is not altered of the signs of the numerator and denominator are both changed. It follows, also, by the Law of Signs, that The value of a fraction is not altered if the signs of any even number of factors in the numerator and denominator of a fraction are changed. 164, Since changing the sign before a fraction is equiva- lent to changing the sign before the numerator or the denominator, it follows that The sign before the denonunator may be changed, provided the sign before the fraction is changed. Note. Ii the denominator is a compound expression, the beginner must remember that the sign of the denominator is changed by changing the sign of every term of the denominator. Thus, x x a—2 x~—a These principles enable us to change the signs of frac- tions, if necessary, so that their denominators shall be arranged in the same order. FRACTIONS. 133 3 22—3 soe 2%—1 1—4z Change the signs before the terms of the denominator of the third fraction, and the sign before the fraction we have (1) Simplify 2— xv The L.C. D. = 2(2a% —1)(2a2 + 1). 2(2e@—1)(2%+1)= 827—2 = 1st numerator. —3a(2%+1) =— 6a?—3e= 2d numerator. —«(2%—3) = — 207 +3x2=3d numerator. —2 =sum of numerators. Sean of the fractions =— —_» “> x(2%—1)(22 + 1) (2) Simplify 1 i 1 a(a—b)(a—c) 6b(6—a)(b—c) c(e—a)(e—)) Nort. Change the sign of the factor (6 —a) in the denominator of the second fraction, and change the sign before the fraction. Change the signs of the two factors (e—a) and (¢ — 6) in the de- nominator of the third fraction. We now have soe ae oe MS a(a—b)(a—c) b(a—b)(b—c) e(a—e)(b—e) The L.C.D. = abe (a — b) (a — c) (6 — ¢). be (b —c) = be — be? = lst numerator. ~—ac(a—c)=—a@e+ae? = 2d numerator. ab(a—b) = a*b — ab? = 3d numerator, ab — ae — ab? + ac? + be — be? = sum of numerators. = a? (b — c) — a(b? — c*) + be (b— 0), = (a? —a(b +c) + be] [b — ¢], = (a? — ab — ac + be}[b —c], = ([(a? — ac) — (ab — bc) |[b — ¢], =[a(a —c) — b(a—c)][b —c], = (a — b)(a—c)(b— Cc). (a—b)(a—c)(b—c) _ * gum of the fractions = Eh. abe(a — b)(a—c)(b—c) abe 134 SCHOOL ALGEBRA. Exercise 49. Simplify : 1. 10. 1 Rs 12. 13. 2 x x x -- v?—1 ah. lee — va ae swuien o LD 2a SISOS ALO atone a a—b 2a a+ ab b APRN LT TOT ST 3 5 24 —7 2 1-22 481 psi ty 2 1 bye bias (ae Gives v—a@ is pen ee ey sabe, r—y Y+ayty y—x 1 2 1 (@—2)(e—8)' @—l)@—a)' @-N@—2) be ac ab CENCE CECE b+te ate a+b Seles pee. a eg nes Se re ee es et a ees 8 (a —b)(6—c) te Res ea 1 1 Ne da c@—y@—-) ¥y—NY—D rye a? — be 67 + ae 4 ce? + ab (Gaby (aan (6—a)(b+c)' (e—a)(e+b) FRACTIONS. 135 MULTIPLICATION AND DIvIsIon oF FRACTIONS. 165. Multiplication of Fractions. ae Ee ons i b da | quotient 7 by a, and divide the result by 0. The expression means that we are to multiply the From the nature of division (§ 76) if we multiply the dividend ¢ by a, we multiply the quotient 4 by a, and obtain oe if we multiply the divisor d by 0, we divide the quotient ee by 6, and obtain a Hence, LIBS A ceageet bd bd Therefore, to find the product of two fractions, we have the following rule: Find the product of the numerators for the required numerator, and the product of the denominators for the required denominator. 166. In like manner, and so on for any number of fractions. 2 2 Again, eae In like manner, Xe OF (5) “FF 167. Division of Fractions. If the product of two numbers is equal to 1, each of the numbers is called the reciprocal of the other. 136 SCHOOL ALGEBRA. The reciprocal of 5 is a a b ameba f ON ee eee ab The reciprocal of a fraction, therefore, is the fraction inverted. Qi deere | ince a he od BO va tol oun eid To divide by a fraction ws the same as to multiply by rts reciprocal. - To divide by a fraction, therefore, Invert the disor and multiply. Nore. Every mixed expression should first be reduced to a frac- tion, and every integral expression should be written as a fraction having 1 for the denominator. If a factor is common to a numera- tor and a denominator, it should be cancelled, as the cancelling of a common factor before the multiplication is evidently equivalent to cancelling it after the multiplication. 2a%b \, bed 5.abrc Scd* 5ab “Sec 20% bed. 5ab’e 2x6~x 5a°bied ine 3cd?°5ab ~~ 8a2etd? 3x5x8ar%bedt 223 (1) Find the product of (2) Find the product of re AE is ay" ey x’ —d32y+27 eee * (oe oy" ye Y= 2y" — «ey —3ay+2y? x4 ay ae y)? (7—y)(@+y) x Ya = 2 Dy a(e—y) “(e—y)@— 29) a@+y) ~@-Ne-H) Swine Cae FRACTIONS. 137 Notz. The common factors cancelled are (x — y), (a + y), (w—2y), w, and w— y. (3) Find the quotient of aaa - Pie ~ OB 5. an ye (4= 2) (a + 2) (a—2)? a?—2? (a—2x)(a—z) ab EY x(a + x) b(a— 2) The common factors cancelled are a and a— wz. Exercise 50. Simplify : 1, Lax, aay Ti Sar x 4° fo Spl NS iad e—B dv+ab+e » 2 arb?c® , 20 mn? 11, DEY Be H+ 8Y 4amn 2la’te ey x+y 3 6 abe? Sm?a* 12 san Gi res Tmay” 8c! al(a+b) al(ai—2) 4 San 42° 13 Ga ke 0 oe 8a%y? ~ 15 mn f+ 4ax~ az+4a* 16 ait? . 40° wu, Say er ey. Bl ary? ~ Baty @—y~ wy &: Tay Be 3522 | 16. z+ a’ a+3a 1277 - 3642’ Cm OU Beto 7 aa ge 16 AE ea hc eae Meas ia ety «x+y OLB a&—ab+ Fb 3 32° — 2 2a 17 xz? — 1 re ae a 227 —4a " Ay —-5 ®t 2x8 (ee) 8 2 82 —6 S a 9. - ; 18. (1— ES, peer 5a — 10 * 4x ( a xy 30. 31. SCHOOL ALGEBRA. > eae y au—y ae, aR ATH aay y 42° + 22ry+y’ acerca ary xv’ +? 8a’b .. od 4ab.. bed — cd? GSB bed «A (paabay u@=), @=¥)_, = aa+y) #+ay+y («+y) a a a? + ab —ac ab —b? — be (EO) Ce ene c—a—b a—(b—cy ce—(a+by ac — a? + ab (2 —a)'— 6 at — (6 —a)' | ax + a! — ab (c—b?v—a@ 2#—(a—byY be—ab+O OOO 0 Ce he “v+2ab+0?—c a—b+e ea ee x a(%+1) xu? +- 4x _2ax’ + Zatz v—a v, z+ a (c—ay(e@+a)y 2(2+0) ax a? — 6b? — c?— 2be ae 2c ) a’ — ab—ae at+bte a+ab?t+ bo até 2573 x ore x. a’ — b§ a+b a 30. e+ Tay + lay, n+ cy — 2y7 | Y+5ayt+6y #+32y—-47 FRACTIONS. 139 168. Complex Fractions. A complex fraction is one that has a fraction in the numerator, or in the denominator, or in both. eee 2 (1) Simplify ey 8a 38x e242 7-1 32 4 foe 4c 1-1 AEE Teele Aig a] 4 So Ae ar ee Nors. Generally, the shortest way to simplify a complex fraction is to multiply both terms of the fraction by the L.C.D. of the frac- tions contained in the numerator and denominator. Thus, in (1), if 12a we multiply both terms by 4, we have at once EE H bid Qe be (2) Simplify a—-#L at+az The L.C. D. of the fractions in the numerator and denom- inator is (a—x)(a+2). Multiply by (a—2)(a+2), and the result is (a+ 2)?—(a—2) (a+a)+(a—zy _ (a + 2a + 2”) — (a — 2axr+ 2) (a? + 2ax + 2”) + (a — 2axr-+ 2’) me -- 20% + me —@V+2axr— 2 VC+2%arte2t+e¢—2ar+2’ etter 2a? + 22° _ 200 C+ 2? 140 2 SCHOOL ALGEBRA. (3) Simplify Fane z s he £ ie CAG Eee lpa2+ (l+2)(l—2+2’)+2 x ee ert eo lt+a2+2 as a(1l+2+2°) 1+2+2°—(*x#—2+ 2°) te pt 1+ 2 Nore. In a fraction of this kind, called a continued fraction, we begin at the bottom, and reduce step by step. Thus, in the last example, we take out the fraction me xv 1+2+———— 1l—a#+2? , and multiply the numerator and denominator by 1 — 2 + 2?, getting for the result, 2 2 3 Saat Nocera which simplified is oer ee (l+a)\(1l—v2+4+2%)+¢e l+a+2° Putting this fraction in the given complex fraction for v ye eee ee l—a+2? x 1 #o wv +e l+toat+a2 we have Multiplying both terms by 1+ 2 + 23, we get a(1 + a” + 2%) l+e2+2—a2+a%-—23 ss Ee 1+ 2 Simplify : FRACTIONS. Exercise 51. 10. Ti 12. 13. 14. 2m+n_ 4 mtn ae m+n vty’ oy et —ay ty dito Y 1 a i GD) Se ee Rae yp ITE 1 1 1 ere ab sae ab — 141 142 SCHOOL ALGEBRA. 15. sae 16. —__#+4 SL rT. tT Ye 1 a ve = re get 17 ee ee ee ee ar y (xyz + x -+ 2) Tie Z gat tet 18. 83 .abe ek Ce 2 bc+ac+ab a te 8 ae 19. Gee) es oe Mn 6 0 90 20. ese See arty #+y7* LL — ae 1 1 =a pews, G+ e—a’ e 1 RET STRESS i> Re Lage Lae 1 (1+ 2be ») a b+e Exercise 52. EXAMPLES FOR REVIEW. 1. Find the value of Va? + B® + — (a—6b—e)’, when a=? bS=—2, and ¢c=4 3° + LO mide J ee 82°+ 1382°+17%+6 2. Reduce to lowest terms Simplify : 1 3 co 4. 2-38 * @—3)@—-2) @-N@—-s) @—-Ne—2 Ua FRACTIONS. 143 ; ee eye Yi + re ad A ete ws Ba ; eee 2) Y—2)\(y—2)) eae Hy z l—2\. x , 12) Geet a JE Gace 2 ) Af oan 1 ay 3 e\a—-e 2x£4+2¢ x +ex—2¢e xt — y* x? ° ae a thc Se aa | vty? Car 3 =) Gere a -2) 8 (a'—z—2) 8a xv—x—2 v’ta—2 x’? —4 (et2y “+s » (oer +5) Ga) Meaaite3)(4 = eas toa ty Pa (y—2y wt xy + y’ ae (e+y) 26 0—C As 26—c—a of Pes xOd (@—B(a—)' @—e(b—a)' C—a)(e—B) s! 2 Leah eels ttl («+2)(2+1) 7 aes ead ae Sra Oe cea ay is a +2 (¢#+1)(#+2) reel DE Ay andl pee oe pl ee ed ty Oe a") P—8y y vm ayty xe —y). : a(a—y),.2+22ay+ 47° e+y?* eee co eta} at be Me ads 0 be ac ab CHAPTER X. FRACTIONAL EQUATIONS. 169. To reduce Equations containing Fractions. (1) Solve eR eS 9 5) 11 Multiply by 33, the L.C.M. of the denominators. Then, lle —32 +3 = 332 — 297, lla —3a— 332 = — 297 — 3, — 252 = — 300. “. & = 12. Nore. Since the minus sign precedes the second fraction, in remov- ing the denominator, the + (understood) before a, the first term of the numerator, is changed to —, and the — before 1, the second term of the numerator, is changed to +. Therefore, to clear an equation of fractions, Multiply each term by the L.C.M. of the denominators. If a fraction is preceded by a minus sign, the sign of every term of the numerator must be changed when the denominator 1s removed. xa—4.a4-—5b 2~T ¢—8 ¢— 5 iy 6 6 ee Note. The solution of this and similar problems will be much easier by combining the fractions on the left side and the fractions on the right side than by the rule given above. (2 — 4) (x — 6) ~(w@— 5) _ (w— 7) (e@—9)—(2@— 8? (a — 5) (a — 6) (a — 8)(« — 9) (2) Solve FRACTIONAL EQUATIONS. 145 ¢€ By simplifying the numerators, we have coe bee ed HS SN ete A (7 —5)(w—6) («—8)(a—9) Since the numerators are equal, the denominators are equal. Hence, (x — 5) (a — 6) = (w — 8) (aw — 9). Solving, we have 2 = 7. Exercise 53. Solve: ‘; Pee eel aes 5g 4 3 5 2. 2 in 5 Le 8 6 4 s Seri, ler? 8e-1_ir—l 3 9 2g 6 feo — 2 lle +2 272 4 eee ee oe I Se a 2 14 3 62—4 18 —42z : —2=> A 5 3 3 +2 62+7 ,Tx—18_ 2xr%4+4 eet gee 9 3 27 5 7 too 00-7 36x +115 Al peel a 14 56 56 a pees i Sa 5), 1 2 a 1 4 ae2-—-1., 2a+1 Reb 1 a— 1 9. 11 — =10— Tea ce on a ae Tx—4 ,382%—1 5(a—1)_ 3(8x2—1),2 ee ee ee 9 e 5 6 20 v7 146 oe 13. 14. 15. 16. A Bs 18. 19. 25. 26. 27. 28. SCHOOL ALGEBRA. 6¢ —2i@—1_2%42—1)_9e—5_ Me—2, 99 4 5 4 Q l0z+1l 122-18 4, 7-62 | 6 3 4 3 5 9 1 yi! 3-H 2y—-H 2 Qi ge 57 8 eee Om 21) 8iv 1) ae 10—"2_5#—4 eis a el 6—Tx 5a 2a—1 42?—1 22+4+1 an ped: aii 3) eter oh 21 9—2¢ .24— (toes 8+2x2 13—22 OE ge 2} zx+1 a2’ —] ¢ 2 iE cael pent? 29. 622) Be 2 oi 6 Feces coeees c+Q x—-2 #- oo x w-—bdxr 2 4 ttl 2 23. SSS ee 8 8x—T 8 loch gah nee 1—2 2ea+tl Ta—-1 22?—38x—45 Te ee WO es WaG ae 42? — 4 e7e#+l #@fetl_» x—]l xc+t+l 9r+5 Oi ORL a oe 6(2—1) 18(a?— 1) 9(#+1) x. eS “f 7 ea ees 2: et+2 £43 a#?+52+6 1 1 1 1 FRACTIONAL EQUATIONS. 147 170. If the denominators contain both simple and com- pound expressions, it is best to remove the simple expressions first, and then each compound expression in turn. After each multiplication the result should be reduced to the simplest form. 874+5; Ta#—8 474+6 1) Sol eR gene OO) IS re aa ere aa Multiply both sides by 14. Then, Be pnee Ce) 8s ai, 32+1 Transpose and combine, ee ax 7; v + Divide by 7 and multiply by 32 +1, T7e@—3=32+1. c=. Sat be 2) Solve es (2) 4 4 10 Simplify the complex fractions by multiplying both terms of each fraction by 9. Multiply both sides by 180. 135 — 20x = 45 — 14a + 54, — 62 =-— 36. “a2=6, Exercise 54. Solve: 1 Ae 38 22k 2e—-1 rh Wee 5 9 ES Na eS ee Beals oer wae! 148 SCHOOL ALGEBRA. lO e--l ySloa tes Eb ara 18 132—16 9 Oa bent © salon oie 9 62+3 5 ae aay 154 2 oka ae 62-41. Se — doe 6. —_ = 15 7x—16 4: 7 Lig pon ee 32a 14 28 2(82+7) ~ At Oe ees 42 le ee 9 liz—82 "38.12 36 9. Sica ca here it’s 2274 15 14(¢—1) 21 6 105 10. = 171. Literal Equations. Literal equations are equations in which some or all of the given numbers are represented by letters; the numbers regarded as known numbers are usually represented by the first letters of the alphabet. (1) (a—2) (a+ 2) = 207+ 2axr—2’. Then, a? — a7 = 2a? + 2axr— 27, — 2ar= a’, Exercise 55. Solve: 1. az+2b6=36bx+4a. 3. (a+2)(6+2)=-2(x—- 0). 2. #+6'=(a—-2*. 4. (x—a)(a+b)=(x—b)(x—C). a 7 FRACTIONAL EQUATIONS. 149 eee eed | be. 6b Cc 20a — bx 26. 5a 27. b 2¢ ie az b—« Rea 2,D 17, 22% — 22 ——a\* ataz cetcz bbe GA Sar eh e+d _m—« “ a2—a_(x—by ab+bz an+n2x EBS it OG a tt+2 m+n meen x—-2 m—n 10g eee Se MEN _m—N me b 2+2 2-2 m a+br__c+dz ax—b axtb b d Re aE oT tg teen ra 6a+a_3824—)b a 6b 4z+6 2x—-a b a a a1, ———==%, a az—b_ br+e_ 7, = c a ORS 7 Sn aside. a— 3 b ee re mee Oe Ord a) = a : a—b atb @—l? Lr) RZ a9) Boe 8. GTO 6 Sea Pitice bite ax—e i 94, tta_x—b_ 2(a+) a b—a b+a x—-a «+6 x 9e—azx 6d—cr _46 OC a 2d a(b—x) _ 34 a. 150 | SCHOOL ALGEBRA. 172. Problems involving Fractional Equations. Ex. The sum of the third and fourth parts of a certain number exceeds 3 times the difference of the fifth and sixth parts by 29. Find the number. Let x = the number. Then j “ ; = the sum of its third and fourth parts, : “2 ; = the difference of its fifth and sixth parts, 3(Z = AS 3 times the difference of its fifth and sixth parts, 5 : =i ; —3 & — 4 = the given excess. But 29 = the given excess. L x 2 oi eh en ata Peete re | me ee! & a) Multiply by 60 the L.C.D. of the fractions. 20” + 152% — 36a + 302 = 60 x 29. Combining, 29 2 = 60 x 29. “. & = 60, Exercise 56. 1. The sum of the sixth and seventh parts of a number is 18. Find the number. 2. The sum of the third, fourth, and sixth parts of a number is 18. Find the number. 3. The difference between the third and fifth parts of a number is 4. Find the number. 4. The sum of the third, fourth, and fifth parts of a number exceeds the half of the number by 17. ‘Find the number. 5. There are two consecutive numbers, 2 and x+1, such that one-fourth of the smaller exceeds one-ninth of the larger by 11. Find the numbers. FRACTIONAL EQUATIONS. 151 6. Find three consecutive numbers such that if they are divided by 7, 10, and 17, respectively, the sum of the quo- tients will be 15. 7. Find a number such that the sum of its sixth and ninth parts shall exceed the difference of its ninth and twelfth parts by 9. 8. The sum of two numbers is 91, and if the greater is divided by the less the quotient is 2, and the remainder is 1. Find the numbers. Dividend — Remainder = Quotient. Divisor Q Hint. 9. The difference of two numbers is 40, and if the greater is divided by the less the quotient is 4, and the remainder 4. Find the numbers. 10. Divide the number 240 into two parts such that the smaller part is contained in the larger part 5 times, with a remainder of 6. 11. If a mixture of alcohol and water the alcohol was 24 gallons more than half the mixture, and the water was 4 gallons less than one-fourth the mixture. How many gallons were there of each? 12. The width of a room is three-fourths of its length. If the width was 4 feet more and the length 4 feet less, the room would be square. Find its dimensions. Ex. Hight years ago a boy was one-fourth as old as he will be one,year hence. How old is he now? Let x = the number of years old he is now. Then «— 8 =the number of years old he was eight years ago, and x +1 =the number of years old he will be one year hence. . ¢—-8=}(rc+1). Solving, e=11. 152 SCHOOL ALGEBRA. 13. A is 60 years old, and B is three-fourths as old. How many years since B was just one-half as old as A? 14. A father is 50 years old, and his son is half that age. How many years ago was the father two and one- fourth times as old as his son? ~ 15. A father is 40 years old, and his son is one-third that age. In how many years will the son be half as old as his father ? Ex. A can do a piece of work in 6 days, and B can do it in 7 days. How long will it take both together to do the work ? Let x = the number of days it will take both together. = the part both together can do in one day, 1 x 4 = the part A can do in one day, + =the part B can do in one day, 1 7 and i-++4=the part both together can do in one day. Ilex Lye Oe Te+6a=—42 13 2 = 42 z= 335. Therefore they together can do the work in 3,3; days. 16. Aan do a piece of work in 3 days, B in 4 days, and Cin 6 days. How long will it take them to do it working together ? , | 17. A can do a piece of work in 23 days, B in 3% days, and C in 42 days. How long will it take them to do it working together ? 18. A and B can separately do a piece of work in 12 days and 20 days, and with the help of C they can do it in 6 days. How long would it take C to do the work? FRACTIONAL EQUATIONS. LS 19. A and B together can do a piece of work in 10 days, A and C in 12 days, and A by himself in 18 days. How many days will it take B and C together to do the work? How many days will it take A, B, and C together ? 20. A and B can do a piece of work in 10 days, A and Cin 12 days, B and Cin 15 days. How long will it take them to do the work if they all work together ? 21. A cistern can be filled by three pipes in 15, 20, and 30 minutes respectively. In what time will it be filled if the pipes are all running together ? 22. A cistern can be filled by two pipes in 25 minutes and 30 minutes, respectively, and emptied by a third in 20 minutes. In what time will it be filled if all three pipes are running together ? 23. A tank can be filled by three pipes in 1 hour and 20 minutes, 2 hours and 20 minutes, and 3 hours and 20 min- utes, respectively. In how many minutes can it be filled by all three together ? Ex. A courier who travels 6 miles an hour is followed, after 5 hours, by a second courier who travels 74 miles an hour. In how many hours will the second courier overtake the first ? Let x = the number of hours the first travels. Then x — 5 =the number of hours the second travels, 6a =the number of miles the first travels, and («#—5)74=the number of miles the second travels. They both travel the same distance. in O@i== (#2 —.5).74; or 12¢=152— 5. x = 25. Therefore the second courier will overtake the first in 20 hours, ' 154 SCHOOL ALGEBRA. 24. A sets out and travels at the rate of 9 miles in 2 hours. Seven hours afterwards B sets out from the same place and travels in the same direction at the rate of 5. miles an hour. In how many hours will B overtake A? 25. A man walks to the top of a mountain at the rate of 2% miles an hour, and down the same way at the rate of 4 miles an hour, and is out 5 hours. How far is it to the top of the mountain ? 26. In going from Boston to Portland, a passenger train, at 27 miles an hour, occupies 2 hours less time than a freight train at 18 miles an hour. Find the distance from Boston to Portland. 27. A person has 6 hours at his disposal. How far may he ride at 6 miles an hour so as to return in time, walking back at the rate of 3 miles an hour? 28. A boy starts from Exeter and walks towards Ando- ver at the rate of 3 miles an hour, and 2 hours later another boys starts from Andover and walks towards Exeter at the rate of 24 miles an hour. The distance from Exeter to Andover is 28 miles. How far from Exeter will they meet? Ex. A hare takes 5 leaps while a greyhound takes 8, but 1 of the greyhound’s leaps is equal to 2 of the hare’s. The hare has a start of 50 of her own leaps. How many leaps must the greyhound take to catch her? Let 3x =the number of leaps the greyhound takes. Then 5a=the number of leaps the hare takes in the same time. Also, let a= the number of feet in one leap of the hare. Then 2a=the number of feet in one leap of the hound. Hence 3x2 xX 2a or 6axr =the whole distance, and (5% +50)a or 5ax + 50 =the whole distance. .6ax = Sax + 50a. xe = 00, and 3a = 150. Therefore the greyhound must take 150 leaps. FRACTIONAL EQUATIONS. 155 29. A hare takes 7 leaps while a dog takes 5, and 5 of the dog’s leaps are equal to 8 of the hare’s. The hare has a start of 50 of her own leaps. How many leaps will the hare take before she is caught ? 30. A dog makes 4 leaps while a hare makes 5, but 3 of the dog’s leaps are equal to 4 of the hare’s. The hare has a start of 60 of the dog’s leaps. How many leaps will each take before the hare is caught? | Notr. If the number of units in the breadth and length of a rectangle are represented by x and x +a, respectively, then «x(x + a) will represent the number of units of area in the rectangle. 31. A rectangle whose length is 23 times its breadth would have its area increased by 60 square feet if its length and breadth were each 5 feet more. Find its dimensions. 32. A rectangle has its length 4 feet longer and its width 3 feet shorter than the side of the equivalent square. Find its area. 33. The width of a rectangle is an inch more than half its length, and if a strip 5 inches wide is taken off from the four sides, the area of the strip is 510 square inches. Find the dimensions of the rectangle. Nore. If « pounds of metal lose 1 pound when weighed in water, 1 pound of metal will lose ! ofa pound. a 34. If 1 pound of tin loses ,% of a pound, and 1 pound of lead loses %, of a pound, when weighed in water, how many pounds of tin and of lead in a mass of 60 pounds that loses 7 pounds when weighed in water? 35. If 19 ounces of gold lose 1 ounce, and 10 ounces of silver lose 1 ounce, when weighed in water, how many ounces of gold and of silver in a mass of gold and silver weighing 530 ounces in air and 495 ounces in water? 156 SCHOOL ALGEBRA. Ex. Find the time between 2 and 8 o'clock when the hands of a clock are together. At 2 o'clock the hour-hand is 10 minute-spaces ahead of the minute-hand. Let x = the number of spaces the minute-hand moves over. Then 2—10= the number of spaces the hour-hand moves over. Now, as the minute-hand moves 12 times as fast as the hour-hand, 12(% — 10) =the number of spaces the minute-hand moves over. “. © =12(x%—10), and lla = 120. *, ©= 1019. Therefore the time is 1012 minutes past 2 o’clock. 36. At what time between 2 and 8 o'clock are the hands of a watch at right angles? 37. At what time between 8 and 4 o'clock are the hands of a watch pointing in opposite directions? 38. At what time between 7 and 8 o'clock are the hands of a watch together ? Ex. A merchant adds yearly to his capital one-third of it, but takes from it, at the end of each year, $5000 for expenses. At the end of the third year, after deducting the last $5000, he has twice his original capital. How much had he at first? Let x = number of dollars he had at first. Then ** — 5000, or £2=18000 will stand for the number of dollars at the end of first year, Th ! (= aaa — 6000, or 162 a will stand for the number of dollars at the end of second year, and 4 ie _ eee — 5000, or 64% — 555000 3 9 27 will stand for the number of dollars at the end of third year. ? FRACTIONAL EQUATIONS. 157 But 2x stands for the number of dollars at the end of third year. . 64a — 555000 | OT We is Whence x = 55,500. 20. 39. A trader adds yearly to his capital one-fourth of it, but takes from it, at the end of each year, $800 for ex- penses. At the end of the third year, after deducting the last $800, he has 1% times his original capital. How much had he at first ? 40. A trader adds yearly to his capital one-fifth of it, but takes from it, at the end of each year, $2500 for ex- penses. At the end of the third year, after deducting the last $2500, he has 154 times his original capital. Find his original capital. 41. A’sage now is two-fifths of B’s. Eight years ago A’s age was two-ninths of B’s. Find their ages. 42. A had five times as much money as B. He gave B 5 dollars, and then had only twice as much as B. How much had each at first ? 43. At what time between 12 and 1 o’clock are the hour and minute hands pointing in opposite directions ? 44. Hleven-sixteenths of a certain principal was at in- terest at 5 per cent, and the balance at 4 per cent. The entire income was $1500. Find the principal. 45. A train which travels 36 miles an hour is 3 of an hour in advance of a second train which travels 42 miles an hour. In how long a time will the last train overtake the first ? 46. An express train which travels 40 miles an hour starts from a certain place 50 minutes after a freight train, and overtakes the freight train in 2 hours and 5 minutes. Find the rate per hour of the freight train. 158 SCHOOL ALGEBRA. 47. A messenger starts to carry a despatch, and 5 hours after a second messenger sets out to overtake the first in 8 hours. In order to do this, he is obliged to travel 24 miles an hour more than the first. How many miles an hour does the first travel ? 48. The fore and hind wheels of a carriage are respec- tively 94 feet and 11 feet in circumference. What distance will the carriage have made when one of the fore wheels has made 160 revolutions more than one of the hind wheels ? 49. When a certain brigade of troops is formed in a solid square there is found to be 100 men over; but when formed in column with 5 men more in front and 38 men less in depth than before, the column needs 5 men to complete it. Find the number of troops. 50. An officer can form his men in a hollow square 14 deep. The whole number of men is 3136. Find the num- ber of men in the front of the hollow square. 51. A trader increases his capital each year by one- fourth of it, and at the end of each year takes out $2400 for expenses. At the end of 38 years, after deducting the last $2400, he finds his capital to be $10,000. Find his original capital. 52. A and B together can do a piece of work in 14 days, A and © together in 1$ days, and B and C together in 1% days. How many days will it take each alone to do the work ? 53. A fox pursued by a hound has a start of 100 of her leaps. The fox makes 3 leaps while the hound makes 2; but 3 leaps of the hound are equivalent to 5 of the fox. How many leaps will each take before the hound catches the fox ? FRACTIONAL EQUATIONS. 159 178. Formulas and Rules. When the given numbers of a problem are represented by letters, the result obtained from solving the problem is a general expression which includes all problems of that class. Such an expression is called a formula, and the translation of this formula into words is called a rule, We will illustrate by examples : (1) The sum of two numbers is s, and their difference d; find the numbers. + Let x = the smaller number ; then — a +d =the larger number. Hence x+x+ds=s, or 24 =s—d. Set oe d 2 and peda Eee _std 2 Therefore the numbers are = : aq and = es z As these formulas hold true whatever numbers s and d stand for, we have the general rule for finding two numbers when their sum and difference are given: Add the difference to the sum and take half the result for the greater number. Subtract the difference from the sum and take half the result for the smaller number. (2) If A can do a piece of work in a days, and B can do the same work in 0 days, in how many days can both together do it? 160 SCHOOL ALGEBRA. Let ‘ a = the required number of days. Then, 1 _ the part both together can do in one day. x Now 1 _ the part A can do in one day, a and ; = the part B can do in one day ; therefore : + ; — the part both together can do in one day. a NG ee a45.D Gis ab Whence em PEA The translation of this formula gives the following rule for finding the time required by two agents together to produce a given result, when the time required by each agent separately is known : Duwide the product of the numbers which express the units of tume required by each to do the work by the sum of these numbers ; the quotient is the tume required by both together. 174, Interest Formulas. The elements involved in com- putation of interest are the principal, rate, time, interest, and amount. Let =the principal, r = the interest of $1 for 1 year, at the given rate, ¢ = the time expressed in years, 2 = the interest for the given time and rate, a = the amount (sum of principal and interest). 175. Given the Principal, Rate, and Time; to find the Interest. Since 7 is the interest of $1 for 1 year, pr is the interest of $p for 1 year, and prt is the interest of $p for ¢ years. ‘1 pr. (Formula 1.) Rue. Multiply together the principal, rate, and time. FRACTIONAL EQUATIONS. 161 176. Given the Principal, Rate, and Time; to find the Amount. Since the amount a is the sum of the principal and interest, a==p- prt. (Formula 2.) 177. Given the Amount, Rate, and Time; to find the Principal. From formula 2, ptprt=a, or pitri)=a. Divide by 1+ rt, P=j mE cs (Formula 3.) r 178, Given the Amount, Principal, and Rate; to find the Time. From formula2, p-+pri=a. Transpose p, pri=a—p. Divide by pr, t= sate (Formula 4.) 179. Given the Amount, Principal, and Time; to find the Rate. From formula2, p+prt=a. Transpose p, prt=a—p. Divide by pt, y carn (Formula 5.) Exercise 57. Solve by the preceding formulas: 1. The sum of two numbers is 40, and their difference is 10. Find the numbers. 2. The sum of two angles is 100°, and their difference is 21° 30’. Find the angles. 3. The sum of two angles is 116° 24! 80", and their difference is 56° 21! 44", Find the angles. 162 SCHOOL ALGEBRA. 4. A can do a piece of work in 6 days, and B in 5 days. How long will it take both together to do it? 8. Find the interest of $2750 for 3 years at 4% per cent. 6. Find the interest of $950 for 2 years 6 months at 5 per cent. 7. Find the amount of $2000 for 7 years 4 months at 6 per cent. 8. Find the rate if the interest on $680 for 7 months is $35.70. 9. Find the rate if the amount of $750 for 4 years is $900. 10. Find the rate if a sum of money doubles in 16 years and 8 months. 11. Find the time required for the interest on $2130 to be $456.65 at 6 per cent. 12. Find the time required for the interest on a sum of money to be equal to the principal at 5 per cent. 13. Find the principal that will produce $161.25 interest in 8 years 9 months at 8 per cent. 14. Find the principal that will amount to $1500 in 3 years 4 months at 6 per cent. 15. How much money is required to yield $ 2000 interest annually if the money is invested at 5 per cent? 16. Find the time in which $640 will amount to $1000 at 6 per cent. 17. Find the principal that will produce $100 per month, at 6 per cent. 18. Find the rate if the interest on $700 for 10 months is $25, i — CHAPTER XI. SIMULTANEOUS EQUATIONS OF THE FIRST DEGREE. 180. If we have two unknown numbers and but one rela- tion between them, we can find an unlimited number of pairs of values for which the given relation will hold true. Thus, if z and y are unknown, and we have given only the one relation z+ y=10, we can asswme any value for z, and then from the relation + y= 10 find the correspond- ing value of y. For from 2+y=10 we find y=10—z2. If x stands for 1, y stands for 9; if x stands for 2, y stands for 8; if x stands for — 2, y stands for 12; and so on with- out end. 181, We may, however, have two equations that express different relations between the two unknowns. Such equa- tions are called independent equations. Thus, «+ y¥y=10 and «—y=2 are independent equations, for they ey express different relations between x and y. 182. Independent equations involving the same unknowns are called simultaneous equations, If we have two unknowns, and have given two independ- ent equations involving them, there is but one pair of values which will hold true for both equations. Thus, if in § 181, besides the relation z+ y= 10, we have also the relation x — y = 2, the only pair of values for which both equations will hold true is the pair = 6, y= 4. Observe that in this problem wz stands for the same num- ber in both equations; so also does y. 164 SCHOOL ALGEBRA. 183. Simultaneous equations are solved by combining the equations so as to obtain a single equation with one unknown number; this process is called elimination. There are three methods of elimination in general use: I. By Addition or Subtraction. II. By Substitution. III. By Comparison. 184, Elimination by Addition or Subtraction. (1) Solve: 5a—3y= a} (1) 22+ 5y = 39 (2) Multiply (1) by 5, and (2) by 3, 252 —15y = 100 (3) 62+ 15y=117 Sale: Add (3) and (4), 31a = 217 . o= 7. Substitute the value of « in (2), 14+ 5y =39, “y=, In this solution y is eliminated by addition. (2) Solve: 62+ 385y=177 (1) 82—2ly= 33 (2) Multiply (1) by 4, and (2) by 3, 24a + 140y = 708 (3) 244— 63y= 99 (4) Subtract, 203 y = 609 “ ¥ =3. Substitute the value of y in (2), 8% —63 =33. - * e=12. In this solution 2 is eliminated by swbtraction. SIMULTANEOUS EQUATIONS. 165 185. Hence, to eliminate by addition or subtraction, we have the following rule: Multiply the equations by such numbers as will make the coefficients of one of the unknown numbers equal in the resulting equations. Add the resulting equations, or subtract one from the other, according as these equal coefficients have unlike or like signs. Nors. It is generally best to select the letter to be eliminated which requires the smallest multipliers to make its coefficients equal ; and the smallest multiplher for each equation is found by dividing the L.C.M. of the coefficients of this letter by the given coefficient in that equation. Thus, in example (2), the L.C.M. of 6 and 8 (the co- efficients of x) is 24, and hence the smallest multipliers of the two equations are 4 and 3 respectively. Sometimes the solution is simplified by first adding the given equations, or by subtracting one from the other. (3) 2+49y= 51 (1) 49x+ y= 99 (2) Add (1) and (2), 502 + 50y = 150 (3) Divide (3) by 50, e+y =3. (4) Subtract (4) from (1), 48 y = 48. “y=. Subtract (4) from (2), 48% = 96. “ = 2, Exercise 58. Solve by addition or subtraction : if > ema 4. cee atc at 22— y= 3 32—6y=15 2. aad was? 5. Sa Yi Bo - 2a—4y= 4 3842—2y=17 3. "Sele 6. ete cantet x+by= 28 22 —d3y= 26 166 SCHOOL ALGEBRA. Vs pease Lis te? Qe - Sly = 8 38%—- 8y=90 8. ek See f 12. 4¢%— asi 22+ dy = 43 8x2— 4y=17 ay Sone mua 13. Tx— ae S42 —2y=15 2x —l0y= 39 10. 24+ y= it 14, 82+ Lae Ta+5y=21 22+ 5y=13 186, Elimination by Substitution. (1) Solve: Ha ae 4z+3y=25 5a +4y =32 (1) 4a +3y = 25 (2) Transpose 4 y in (1), 5a=32—4y. (3) Divide by coefficient of 2, Pyles od (4) 5 Substitute the value of a in (2), 128 —16y + 15y =125, —y=-—83. ap eee Substitute the value of y in (2), 42 +9 = 25. Ye Hence, to eliminate by substitution, From one of the equations obtain the value of one of the unknown numbers in terms of the other. Substitute for this unknown number its value in the other equation, and reduce the resulting equation. ° SIMULTANEOUS EQUATIONS. 167 Exercise 59. Solve by substitution : Cy Ey \ 8. 382— ised sa— Sy=11 2z+ dSy=—68 oye = hy = at 9. 2x— eae oz— 2y=—10 52+ 2y= 29 3. 2Ze— 3y= } 10. 6x%— peiaw 82— 2Zy= 29 oz— Oy= 8 Ae et aise Vii Sti test 22+ Ty=88 2a+ S5y=82 Sos a = >} 12. «2+ pte x+ 2y=25 or+ Zy = 46 6. peers” = S| 13. 82— Lelia l3a— 5y=21 4%x— 3y= 9 fe ie! 14. 54+ athe ae if — Ay 7 82+lly= 1 187. Elimination by Comparison. Solve: pate caeat 84+ 2y = 23 2%—5y = 66. (1) 35 4+ 2y = 23. (2) Transpose 5y in (1), and 2y in (2), 2x = 66 +5y, (3) 3a” = 23 —2y. (4) Divide (3) by 2, pie a (5) Divide (4) by 3, Se eh aet, (6) Equate the values of 2, = epee (7) vo 168 SCHOOL ALGEBRA. Reduce (7), 198 + 15y = 46 —4y, 19y =— 152. . y=—8. Substitute the value of y in (1), 2x + 40 = 66. “. = 13, 188. Hence, to eliminate by comparison, From each equation obtain the value of one of the unknown numbers in terms of the other. Form an equation from these equal values and reduce the equation. Exercise 6O. Solve by comparison : by so x + farce 8%2— 2y=25 Tx+ Cae or— 4y= 9ua+ se 4r+ 9y=89 Tx+ as 824—-— y¥= ee ee 38x2--47y = 46 247—- y= ie 5a— 38y=14 llz— Ty= att 92— 5y=10 ee Aan 138%—2y= 57 15. 16. 24— 8y= a 5a+ 2y=126 50z— Iy= t 2 y= BD 2+2ly= ae 2iy-+- 2%= 19 10a + a | 82+ 10y=125 62—138y= i 52—1l2y=. 4 Qa + Yen 10%+ 2y= 60 82— 5y= i Ta+ y=265 12”%+ her 38y—1l9r%= 3 SIMULTANEOUS EQUATIONS 169 189. Each equation must be simplified, if necessary, before the elimination. Solve: $a—L(iyt+ seit Mo+1+iy-1)=9 | e—3(yt1)=1. | (1) 3(@+1)+ey—1)=9. (2) Multiply (1) by 4, and (2) by 12, 34a—2y—2=4, (3) 4%7+4+9y—9= 168. (4) From (3), 3a—2y=6. (5) From (4), 44+ 9y=118. (6) Multiply (5) by 4, and (6) by 3, 12%— 8y= 24 122 4+ 27y = 339 35y = 315 ¥=9. Substitute value of y in (1), x = 8, Exercise 61. Solve Bee. 4.) ae aed ae fie ice = 4 ey 315 3 | 8 1 5 x pen eee gi OL 3 a a28 6. ae J eee Fg |, Pe by gs ee ry) 3 y 9, 5 oy at+y_p¢ g ta 2y_ rt yy | Pegi: 1 | Rec pin | 40 zy 22—Y 1 9 ons Die | [og Sst cal 170 SCHOOL ALGEBRA. 12-422 — SY by _ Ae easy a ed ee e+2y+1l_o 3 a 5 2z—y+1 e210.) 0, ares Dee | 38z—y+1_, 4 3 2 J z—yt+3 9 aA 10> sya ae5h 8 4 AYP th eae yen 3 8 10. tery hs Oa eye 3 4 da—4y+3 _ 2y—4e+421 4 3 11. UR ee ee | 5 4 oy—it , 4@—8 | 5 + —__ 5 = 18— ae 1B ig ae ee OY eee | Pai eC ee Ma ieeee 2 6 2r-+by Je— Ti lyse 2 8 16 14, ee 2 ne at 3 19 SIMULTANEOUS EQUATIONS. 171 15, SEY _? ig, Yo 4 et! | 5a —6 shee ere wa) ESO et eka 10 3 ie! OYE Leen ey Pe OD ) 5 30 | y—l1 zt 10z—3y—20_ 84+2y+3 3 20 30 Norg. -In solving the following problems proceed as in @ 171. ro ou te fe ey $3 ty — 4) 18. 8 4%—2y 12 S2+3 (2—dy _62—1 4 7—2 3 ey ee 20. ¢— =r — 9 93 — a 4 — 3 Gemini fer7, de—4y _lit 8% 1. : 3 2x2+8 6 Pie pao 6y-13. 10a 5384 4 2x—s8y 8 J poem Sle 2y) ise) ci Bale. 2(x — 4) 5 eer eo Tee i ok By+ 5a 4 4 2(2y— 8) 72 SCHOOL ALGEBRA. 190, Literal Simultaneous Equations. Solve: ax + by =e } az + bly =c' Note. The letters a’, b’ are read a prime, b prime. In like man- ner, a/’, a//’ are read a second, a third, and ay, dy, ds are read @ sub one, a sub two, a sub three. It is sometimes convenient to represent different numbers that have a common property by the same letter marked by accents or suffives. Here a and a’ have a common property as coefficients of a. av +by =c. (1) a/c + b/y=c?’. (2) To find the value of y, multiply (1) by a’, and (2) by a, aa’x + a/by =a/e | aa’x + ab/y = ac’ a/by — ab’y = a/c — ac’ To find the value of x, multiply (1) by 0’, and (2) by 5. ab’ + bb’y = b/c a/ba + bb’/y = be’ ab/a — a/bu = b/c — be’ __ b/c — be? H 4; ee ab/—a/b Exercise 62. Solve: i Diem) 5. eae zr—y=d x= dy 2. mx Meee 6. bx bbsas, ma + n'y =r! b'x—aly=1 3. Si are ie pee ale + bly =! 4bx— 8ay= tab 4, «£— eit 8. 22 eae, cu + aby = ms 3a —2y=a+b ee SIMULTANEOUS EQUATIONS. LS bx Ce, ek cing ra high Gath bz +cy=a+b y—n db bz +ay=c b = he o4 15 8a? + samen es ie ott | ax+2by=d BO 2r yl geen ae celle Ws Liste) ab 3 | Oy a eee 1 12. 2 Yorn. | ane pegs) gi 5 a+b a—b a-—6b| 2+y=2a x Yan oe PE fe ‘ a | 13. an, ty _3y 4) ape ee pei ey n'y x—y=a—b n—a n—b 191, Fractional simultaneous equations, of which the de- nominators are simple expressions and contain the unknown numbers, may be solved as follows: (1) Solve: oo =m | YT, aa | — + —-—7N oe RD We have ed aa (1) way and e+ San, (2) To find the value of y. Multiply (1) by «¢, = + = = cm, (3) 174 SCHOOL ALGEBRA. Multiply (2) by a, weds aos an. (4) x y Subtract (4) from (3), bc — ad = cm— an, y Multiply both sides by y, be —- ad =(em — an)y. hs: be — ad °F om — an To find the value of x. ad bd Multiply (1) by d, — +—=dm. (5) wy Multiply (2) by 3, be + bigs bn. (8) J es, Subtract (6) from (5), nacelle bn. v Multiply both sides by a, ad — be = (dm — bn)z. _ ad—be dm — bn 4) 2 2) Solve: ea Backes = hy (2) ate ace Se Rs a 627 10 We have =. + = noite (1) 7 1 ant AS oe a0 6x 10y (?) Multiply (1) by 15, the L.C.M. of 3 and 5, and (2) by 30, oP EN TOG: (3) | 35 3 sei Pb a 4 ney 90. (4) Multiply (4) by 2, and add the result to (3), 2 _ 286, ae ere : 3 Substitute the value of a in (1), and we get 1 ere SIMULTANEOUS EQUATIONS. Exercise 63. Bg se x | et ae. | 7 ee eee leah weersoy | Esthet Dies 2h) ee fe EEA 8 Tela, | y zy ro nm ol 2208 eee | y aly J a) oe mn y 3 | 9. miyr’ | Y aly Cf ee (ty 40 4 3 co aatk ee a7 ale eet ipa y coy a) 5 oun eee an 1) 11. ———= wee aa | Ct bY | ah Se Re Bot BUM, ax a= 04) J Bere | [ome ere arene 3Yy OL Ly 4 Sere pe — — == GQ" 3 is, ia ath 175 176 SCHOOL ALGEBRA. 192. If three simultaneous equations are given, involy- ing three unknown numbers, one of the unknowns must be eliminated between two pavrs of the equations; then a second unknown between the two resulting equations. Likewise, if four or more equations are given, involving four or more unknown numbers, one of the unknowns must be eliminated between three or more pairs of the equations ; then a second between the pairs that can be formed of the resulting equations; and so on. Norr. The pairs chosen to eliminate from must be independent pairs, so that each of the given equations shall be used in the process of the eliminations. Solve: 2%—8y+4z2= 4 (1) Seb 5y—Te=12| (2) S2— y—8z2= 5 (3) Eliminate z between the equations (1) and (8). Multiply (1) by 2, 4x—6y+8z= 8 (4) (3) is 5e— y—8z2= 5 Add, 9x—Ty = 13 (5) Eliminate z between the equations (1) and (2). Multiply (1) by 7, 14”%—21ly + 282 = 28 Multiply (2) by 4, 124+ 20y—28z=48 Add, 26a—- y = 76 (6) We now have two equations (5) and (6) involving two unknowns, «and y. Multiply (6) by 7, 182% -— Ty = 532 (7) (5) is 9a—Ty= 13 Subtract (5) from (7), 173 2 = 519 *. i mak, Substitute the value of x in (6), 78 — y = 76. y=2. Substitute the values of x and y in (1), 6—6442=4. *, =. Oe i SIMULTANEOUS EQUATIONS. 177 Exercise 64, ~ oty— 8=0 y+-2—28=0 a+z2—14=0 44+ 3y+22= 25 34% —2y+5z2= 20 we Tot Sy ever x— a Zt sy a7 0 . loez— y+3z2=42 Ta+t2y+ z2=951 8a+38y— z2= 24 . 5a+2y—82=—160 82+9y+82=115 22—8y—5z= 40 . 62—2y+5z2=53 5a+38y+7z2= 33 e+ yt z= 9d 62+ 2y—Tz= 5 | oe | | | : Se-h2y—Te= | 22 — yt+8z2=45 . 22+ 7y+10z2=25 ety— 424=9 Tx—Ty—11z2=78 | 10. 12. 13. 14. 15. 18.., 82—6y+ 7z=51 4x+-8y— 92= 53 2+2y+10z2= | or oyy reall 82+3y+t Tz= 3884 2e+ yt 2=256 2z2= zxr—d3y—18 aaa 1 zx— yt+38z2=13 l0y+5x2—22=48 2¢-+ 3y— 4z2=1 10z2— -6y+12z2= xtl2y+ 2z=5 82+ by+2z2=3 l2y+ 42—62= 9¢4+18y—4z2=4 5y—4x—42=1 sa+9y+ 2=9 sat2y+t 2=203 20-4 Y + 22 = 26h | a+ y+10z2=55 . 2a+ tai SCHOOL ALGEBRA. 178 ae N t CO co —H CD | | | RIM Alo alow ae, eee aloo AH Dw + + +4 SIN Sie Stix sian O!1a on ee ee mary eR ee | | | N SH OD | | | AID AIH Gs!aR + | + CQ es 8 CO em rt SS 16 RN a ee N m I | | Alr» Olr ©Hla ee eerie AIA HIR OLD ee a a ieeldlice? tes) Te )ES & Y a a a RE sla a8 4/8 | | | COlMm lS wtls | | | HIS Ala Hla R R CHAPTER XIL PROBLEMS INVOLVING TWO UNKNOWN NUMBERS. 193. It is often necessary in the solution of problems to employ two or more letters to represent the numbers to be found. In all cases the conditions must be sufficient to give just as many equations as there are unknown numbers employed. 194, If there are more equations than unknown numbers, some of them are superfluous or inconsistent; if there are less equations than unknown numbers, the problem is inde- terminate. (1) If A gives B $10, B will have three times as much money as A. If B gives A $10, A will have twice as much money as B. How much has each? Let x = number of dollars A has, and y = number of dollars B has. Then, after A gives B $10, x — 10 =the number of dollars A has, y + 10 =the number of dollars B has. “ ¥ +10 =3(e— 10). (1) If B gives A $10, x +10=the number of dollars A has, y — 10 = the number of dollars B has. . ©+10=2(y— 10). (2) From the solution of equations (1) and (2), # = 22, and y = 26. Therefore A has $22, and B has $26. 180 SCHOOL ALGEBRA. (2) If the smaller of two numbers is divided by the greater, the quotient is 0.21, and the remainder 0.0057; but if the greater is divided by the smaller, the quotient is 4 and the remainder 0.742. Find the numbers. Let « = the greater number, and y = the smaller number. Then Ye OOO oe (1) 2 and %— 0.742 _ 4. (2) _ “. y — 0.21 4% = 0.0057, (3) e x—4y =0.742. (4) Multiply (3) by 4, 4 —0.84.¢ = 0.0228 (5) (4) is —4y+ x = 0.742 By adding, 0.16 2 = 0.7648 eel oy Substituting the value of z in (4), —4y = — 4.088, “. y = 1.0095, Exercise 65. 1. If A gives B $100, A will then have half as much money as B; but if B gives A $100, B will have one-third as much as A. How much has each? 2. If the greater of two numbers is divided by the smaller, the quotient is 4 and the remainder 0.37; but if the smaller is divided by the greater, the quotient is 0.23 and the remainder 0.0149. Find the numbers. 3. A certain number of persons paid a bill. If there had been 10 persons more, each would have paid $2 less; but if there had been 5 persons less, each would have paid $2.50 more. Find the number of persons and the amount of the bill. a a oe ae PROBLEMS. 181 4, A train proceeded a certain distance at a uniform rate. Ifthe speed had been 6 miles an hour more, the time occupied would have been 5 hours less; but if the speed had been 6 miles an hour less, the time occupied would have been 74 hours more. Find the distance. Hint. If «=the number of hours the train travels, and y the number of miles per hour, then xy = the distance. 5. A man bought 10 cows and 50 sheep for $750. He sold the cows at a profit of 10 per cent, and the sheep at a profit of 80 per cent, and received in all $875. Find the average cost of a cow and of a sheep. 6. It is 40 miles from Dover to Portland. A sets out from Dover, and B from Portland, at 7 o’clock A.m., to meet each other. A walks at the rate of 34 miles an hour, but stops 1 hour on the way; B walks at the rate of 24 miles an hour. At what time of day and how far from Portland will they meet? 7. The sum of two numbers is 35, and their difference exceeds one-fifth of the smaller number by 2. Find the numbers. 8. If the greater of two numbers is divided by the smaller, the quotient is 7 and the remainder 4; but if three times the greater number is divided by twice the smaller, the quotient is 11 and the remainder 4. Find the numbers. 9. If 3 yards of velvet and 12 yards of silk cost $60, and 4 yards of velvet and 5 yards of silk cost $58, what is the price of a yard of velvet and of a yard of silk? 10. If 5 bushels of wheat, 4 of rye, and 3 of oats are sold for $9; 3 bushels of wheat, 5 of rye, and 6 of oats for $8.75; and 2 bushels of wheat, 3 of rye, and 9 of oats for $7.25; what is the price per bushel of each kind of grain? 182 SCHOOL ALGEBRA. Nore I. A fraction the terms of which are unknown may be repre- sented by ”. y Ex. A certain fraction becomes equal to 4 if 2 is added to its numerator, and equal to 4 if 3 is added to its denomi- nator. Find the fraction. Let : = the required fraction. a 9 Then shililed ie y and tS yts The solution of these equations gives 7 for #, and 18 for y. Therefore the required fraction is 7%. 11. A certain fraction becomes equal to 4 if 3 is added to its numerator and 1 to its denominator, and equal to 4 if 3 is subtracted from its numerator and from its denomi- nator. Find the fraction. 12. A certain fraction becomes equal to 5% if 1 is added to double its numerator, and equal to 4 if 3 is subtracted from its numerator and from its denominator. Find the fraction. 13. Find two fractions with numerators 11 and 5 respec- tively, such that their sum is 14, and if their denominators are interchanged their sum is 24. 14. There are two fractions with denominators 20 and 16 respectively. The fraction formed by taking for a nu- merator the sum of the numerators, and for a denominator the sum of the denominators, of the given fractions, is equal to 4; and the fraction formed by taking for a numerator the difference of the numerators, and for a denominator the difference of the denominators, of the given fractions, is equal to 4. Find the fractions. PROBLEMS. 183 Norse If. A number consisting of two digits which are unknown may be represented by 104+, in which x and y represent the digits of the number. Likewise, a number consisting of three digits which are unknown may be represented by 1002 + 10y +2, in which a, y, and z represent the digits of the number. For example, the expres- sion 364 means 300 + 60+ 4; or, 100 times 3 + 10 times 6 + 4. Ex. The sum of the two digits of a number is 10, and if 18 is added to the number, the digits will be reversed. Find the number. Let x = tens’ digit, and y = units’ digit. Then 10x + y =the number. Hence x+y =10, (1) and 10a+y+18=10y+2. (2) | From (2), 92—-9y=— 18, or x—y=—2. (3) Add (1) and (8), 22 = 8, and therefore ton A, Subtract (3) from (1), 2y = 12, and therefore y= 6. Therefore the number is 46. 15. The sum of the two digits of a number is 9, and if 27 is subtracted from the number, the digits will be reversed. Find the number. 16. The sum of the two digits of a number is 9, and if the number is divided by the sum of the digits, the quotient is 5, Find the number. 17. A certain number is expressed by two digits. The sum of the digits is 11. If the digits are reversed, the new number exceeds the given number by 27. Find the number. 18. A certain number is expressed by three digits. The sum of the digits is 21. The sum of the first and last digits is twice the middle digit. Ifthe hundreds’ and tens’ digits are interchanged, the number is diminished by 90. Find the number. 184 SCHOOL ALGEBRA. 19. A certain number is expressed by three digits, the units’ digit being zero. If the hundreds’ and tens’ digits are interchanged, the number is diminished by 180. If the hundreds’ digit is halved, and the tens’ and units’ digits are interchanged, the number is diminished by 336. Find the number. ° 20. A number is expressed by three digits. If the digits are reversed, the new number exceeds the given number by 99. Ifthe number is divided by nine times the sum of its digits, the quotient is 8. The sum of the hundreds’ and units’ digits exceeds the tens’ digit by 1. Find the number. Nore III. If a boat moves at the rate of x miles an hour in still water, and if it is on a stream that runs at the rate of y miles an hour, then x +y represents its rate down the stream, x —y represents its rate up the stream. 21. A boatman rows 20 miles down a river and back in 8 hours. He finds that he can row 5 miles down the river in the same time that he rows 3 miles up the river. Find the time he was rowing down and up respectively. 22. A boat’s crew which can pull down a river at the rate of 10 miles an hour finds that it takes twice as long to row a mile up the river as to row a mile down. Find the rate of their rowing in still water and the rate of the stream. 23. A boatman rows down a stream, which runs at the rate of 21 miles an hour, for a certain distance in 1 hour and 80 minutes; it takes him 4 hours and 80 minutes to return. Find the distance he pulled down the stream and his rate of rowing in still water. Note LV. It is to be remembered that if a certain work can be done in # units of time (days, hours, etc.), the part of the work dong in one unit of time will be represented by —. ¢ i et be . oe, ee ee PROBLEMS. 185 Ex. A cistern has three pipes, A, B, and C0. A and B will fill the cistern in 1 hour and 10 minutes, A and C in 1 hour and 24 minutes, B and C in 2 hours and 20 minutes. How long will it take each pipe alone to fill it? 1 hour and 10 minutes = 70 minutes. 1 hour and 24 minutes = 84 minutes. 2 hours and 20 minutes = 140 minutes. Let x = number of minutes it takes A to fill it, y = number of minutes it takes B to fill it, and z= number of minutes it takes C to fill it. Then 1 1 es the parts A, B, and C can fill in one minute ma” respectively, and : + ; = the part A and B together can fill in one minute. But = = the part A and B together can fill in one minute. Pe AS .-f-=— 1 geet 70 «) In lke manner, 2 “ By (2) ep 64 Lihat Sediag} d imi age a Es 3 on yee HO ®) Add, and divide by 2, ee (4) By a. BO Lis eats bi Subtract (1) f 4), Se ubtract (1) from (4) Si Subtract (2) from (4), fem oe y 210 Subtract (3) from (4), ins Be ge. 105 Therefore x, y, 2 = 105, 210, 420, respectively. Hence A can fill it in 1 hour and 45 minutes, B in 3 hours and 80 minutes, and C in 7 hours. 24. A and B can do a piece of work together in 8 days, A and C in 4 days, B and C in 44 days, How long will it take each alone to do the work? 186 SCHOOL ALGEBRA, 25. A and B can do a piece of work in 24 days, A and C in 34 days, B and © in 4 days. How long will it take each alone to do the work? 26. A and B can do a piece of work in a days, A and O in 6 days, B and C ine days. How long will it take each alone to do the work ? Note V. If represents the number of linear units in the length, and y in the width, of a rectangle, xy will represent the number of its units of surface; the surface unit having the same name as the linear unit of its side. 27. If the length of a rectangular field were increased by 5 yards and its breadth increased by 10 yards, its area would be increased by 450 square yards; but if its length were increased by 5 yards and its breadth diminished by 10 yards, its area would be diminished by 350 square yards. Find its dimensions. 28. If the floor of a certain hall had been 2 feet longer and 4 feet wider, it would have contained 528 square feet more; but if the length and width were each 2 feet less, it would contain 316 square feet less. Find its dimensions. 29. If the length of a rectangle was 4 feet less and the width 8 feet more, the figure would be a square of the same area as the given rectangle. Find the dimensions of the rectangle, Nore VI. In considering the rate of increase or decrease in quan- tities, it is usual to take 100 as a common standard of reference, so that the increase or decrease is calculated for every 100, and there- fore called per cent. It is to be observed that the representative of the number result- ing after an increase has taken place is 100 + increase per cent; and after a decrease, 100 — decrease per cent. Interest depends upon the time for which the money is lent, as PROBLEMS. 187 well as upon the rate per cent charged; the rate per cent charged being the rate per cent on the principal for one year. Hence, Simple interest = /tncipal x Rate Ber.cenb x Tho, where Time means number of years or fraction of a year. Amount = Principal + Interest. In questions relating to stocks, 100 is taken as the representative of the stock, the price represents its market value, and the per cent represents the interest which the stock bears. Thus, if six per cent stocks are quoted at 108, the meaning is, that the price of $100 of the stock is $108, and that the interest derived from $100 of the stock will be ~$,5 of $100, that is, $6 a year. The rate of interest on the money invested will be 499 of 6 per cent. 30. A man has $10,000 invested. For a part of this sum he receives 5 per cent interest, and for the rest 6 per cent; the income from his 5 per cent investment is $60 more than from his 6 per cent. How much has he in each investment ? 31. A sum of money, at simple interest, amounted in 4 years to $ 29,000, and in 5 years to $30,000. Find the sum and the rate of interest. 32. A sum of money, at simple interest, amounted in 10 months to $2100, and in 18 months to $2180. Find the sum and the rate of interest. 33. A person has a certain capital invested at a certain rate per cent. Another person has $2000 more capital, and his capital invested at one per cent better than the first, and he receives an income of $150 greater. A third person has $3000 more capital, and his capital invested at two per cent better than the first, and he receives an income of $280 greater. Find the capital of each and the rate at which it is invested. 188 SCHOOL ALGEBRA. 34. A sum of money, at simple interest, amounted in m years to c dollars, and in years to d dollars. Find the sum and the rate of interest. 35.. A sum of money, at simple interest, amounted in m months to a dollars, and in m months to 6 dollars. Find the sum and the rate of interest. 36. A person has $18,375 to invest. He can buy 3 per cent bonds at 75, and 5 per cent bonds at 120. How much of his money must he invest in each kind of bonds in order to have the same income from each investment? Hint. Notice that the 3 per cent bonds at 75 pay 4 per cent on the money invested, and 5 per cent bonds at 120 pay 43 per cent. 37. A man makes an investment at 4 per cent, and a second investment at 44 per cent. His income from the two investments is $715. If the first investment had been at 44 per cent and the second at 4 per cent, his income would have been $730. Find the amount of each investment. (1) In a mile race A gives B a start of 20 yards and beats him by 80 seconds. At the second trial A gives Ba start of 32 seconds and beats him by 955 yards. Find the number of yards each runs a second. Let 2 = number of yards A runs a second, and y = number of yards B runs a second. Since there are 1760 yards in a mile, 1760 = number of seconds it takes A to run a mile. Since B has a start of 20 yards, he runs 1740 yards the first trial ; and as he was 30 seconds longer than A, 1760 + 30 = the number of seconds B was running. x But Lidia the number of seconds B was running. ALLEL 0 PROBLEMS. 189 In the second trial B runs 1760 — 9,3, = 175038 yards. _ 17508, _ 1760 ee y a! x From the solution of equations (1) and (2), « = 513, and y= 53. + 82. (2) Therefore A runs 532 yards a second, and B runs 5,3, yards a second. (2) A train, after travelling an hour from A towards B, meets with an accident which detains it half an hour; after which it proceeds at four-fifths of its usual rate, and arrives an hour and a quarter late. If the accident had happened 80 miles farther on, the train would have been only an hour late. Find the usual rate of the train. Since the train was detained 4 an hour and arrived 11 hours late, the running time was ~ of an hour more than usual. Let y = number of miles from A to B, and 5x = number of miles the train travels per hour. Then y—5a2=number of miles the train has to go after the accident. Hence ae = number of hours required usually, © and eae = number of hours actually required. © ‘. Y—S5@_Y—52 _ hogs in hours of running time. 4a 5a But 2 = loss in hours of running time. _¥—5e_y—5a_3 (1) "Ae 5a 4 If the accident had happened 30 miles farther on, the remainder of the journey would have been y—(5 + 30), and the loss in running time would have been 4 an hour. _ y—(6e+4+30)_ y—(Gr+30)_1 - 4% 5x 2 From the solution of equations (1) and (2), « =6, and 5a = 30. (2) Therefore the usual rate of the train is 80 miles an hour. 190 SCHOOL ALGEBRA. 38.. Two men, A and B, run a mile, and A wins by 2 seconds. In the second trial B has a start of 184 yards, and wins by 1 second. Find the number of yards each runs a second, and the number of miles each would run in an hour. 39. Ina mile race A gives B a start of 8 seconds, and is beaten by 124 yards. In the second trial A gives B a start of 10 yards, and the race is a tie. Find the number of yards each runs a second. At this rate, how many miles could each run in an hour? 40. In amile race A gives B a start of 44 yards, and is beaten by 1 second. Ina second trial A gives B a start of 6 seconds, and beats him by 9% yards. Find the number of yards each runs a second. 41. An express train, after travelling an hour from A towards B, meets with an accident which delays it 15 min- utes. It afterwards proceeds at two-thirds its usual rate, and arrives 24 minutes late. If the accident had happened 5 miles farther on, the train would have been only 21 minutes late. Find the usual rate of the train. 42. A train, after running 2 hours from A towards B, meets with an accident which delays it 20 minutes. It afterwards proceeds at four-fifths its usual rate, and arrives 1 hour and 40 minutes late. If the accident had happened 40 miles nearer A, the train would have been 2 hours late. Find the usual rate of the train. 43. A and B can do a piece of work in 24 days, A and C in 34 days, B and C in 33 days. In what time can all three together, and each one separately, do the work? 44, A sum of money, at interest, amounts in 8 months to $1488, and in 15 months to $1530. Find the principal and the rate of interest. PROBLEMS. j 191 45. A number is expressed by two digits, the units’ digit being the larger. If the number is divided by the sum of its digits, the quotient is 4. If the digits are reversed and the resulting number is divided by 2 more than the differ- ence of the digits, the quotient is 14. Find the number. 46. A and B together can dig a well in 10 days. They work 4 days, and B finishes the work in 16 days. How long would it take each alone to dig the well? 47. The denominator of the greater of two fractions is 20. The fraction formed by taking for a numerator the sum of the numerators of the two fractions, and for a denominator the sum of the denominators, is equal to 2. The fraction similarly formed with the difference of the numerators, and of the denominators, is equal to 4. The sum of the numerators is twice the difference of the denomi- nators. Find the fractions. 48. A cistern can be filled in 5 hours by two pipes, A and B, together. Both are left open for 3 hours and 45 minutes, and then A is shut, and B takes 3 hours and 45 minutes longer to fill the cistern. How long would it take each pipe alone to fill the cistern ? 49. A man put at interest $20,000 in three sums, the first at 5 per cent, the second at 44 per cent, and the third at 4 per cent, receiving an income of $905 a year. The sum at 44 per cent is one-third as much as the other two sums together. Find the three sums. 50. An income of $335 a year is obtained from two in- vestments, one in 44 per cent stock and the other in 5 per cent stock. If the 44 per cent stock should be sold at 110, and the 5 per cent at 125, the sum realized from both stocks together would be $8300. How much of each stock is there? 192 SCHOOL ALGEBRA. 51. A boy bought some apples at 3 for 5 cents, and some at 4 for 5 cents, paying for the whole $1. He sold them at 2 cents apiece, and cleared 40 cents. How many of each kind did he buy ? 52. Find the area of a rectangular floor, such that if 3 feet were taken from the length and 3 feet added to the breadth, its area would be increased by 6 square feet, but if 5 feet were taken from the breadth and 3 feet added to the length, its area would be diminished by 90 square feet. 53. A courier was sent from A to B, a distance of 147 miles. After 7 hours, a second courier was sent from A, who overtook the first just as he was entering B. The time required by the first to travel 17 miles added to the time required by the second to travel 76 miles is 9 hours and 40 minutes. How many miles.did each travel per hour? 54. A box contains a mixture of 6 quarts of oats and 9 of corn, and another box contains a mixture of 6 quarts of oats and 2 of corn. How many quarts must be taken from each box in order to have a mixture of 7 quarts, half oats and half corn? 55. A train travelling 30 miles an hour takes 21 minutes longer to go from A to B than a train which travels 36 miles an hour. Find the distance from A to B. _56. A man buys 570 oranges, some at 16 for 25 cents, and the rest at 18 for 25 cents. He sells them all at the rate of 15 for 25 cents, and gains 75 cents. Hew many of each kind does he buy ? 57. A and B run a mile race. In the first heat B receives 12 seconds start, and is beaten by 44 yards. In the second heat B receives 165 yards start, and arrives at the winning post 10 seconds before A. Find the time in which each can run a mile, PROBLEMS. 1938 INDETERMINATE PROBLEMS. 195. If a sengle equation is given which contains éwo unknown numbers, and no other condition is imposed, the number of its solutions is wnlumited; for, if any value be assigned to one of the unknowns, a corresponding value may be found for the other. Such an equation is said to be indeterminate. 186. The values of the unknown numbers in an inde- terminate equation are dependent upon each other ; so that, though they are unlimited in number, they are confined to a particular range. This range may be still further limited by requiring these values to satisfy some given condition; as, for instance, that they shall be positive integers. With such restrictions the equation may admit of a definite number of solutions. Ex. A number is expressed by two digits. If the num- ber is divided by the sum of its digits diminished by 4, the quotient is 6. Find the number. The single statement is e+y—-4 Whence 4a =5y— 24, and v=y + 7 _ 6 ; + eae =y Oa We see from that the values of y which will be integral are 4, 8, 12, 16, or some other multiple of 4, and from the relation s=y—6 44 4 that the least positive integral value of y which will give to x a post- tive integral value, is 8. If we put 8 for y in (1), we find a=4, Hence the number required is 48, 194 SCHOOL ALGEBRA. Exercise 66. 1. A number is expressed by two digits. If the number is divided by the last digit, the quotient is 15. Find the number. 2. A number is expressed by three digits. The sum of the digits is 20. If 16 is subtracted from the number and the remainder divided by 2, the digits will be reversed. Find the number. As Here e+y+z2= 20, and eee = 1002 ede Eliminate y and reduce, and we have 4%=72+ 8. 3. A man spends $114 in buying calves at $5 apiece, and pigs at $3 apiece. How many did he buy of each? 4. In how many ways can a man pay a debt of $87 with five-dollar bills and two-dollar bills? 5. Find the smallest number which when divided by 5 or by 7 gives 4 for a remainder. n—4 Let n = the number, then =, and m—4 a 6. A farmer sells 15 calves, 14 lambs, and 138 pigs for $200. Some days after, at the same price, he sells 7 calves, 11 lambs, and 16 pigs, for which he receives $141. What was the price of each? Discussion OF PROBLEMS. 197, The discussion of a problem consists in making various suppositions as to the relative values of the given numbers, and explaining the results. We will illustrate by an example: PROBLEMS. 195 Two couriers were travelling along the same road, and in the same direction, from C towards D. A travels at the rate of m miles an hour, and B at the rate of m miles an hour. At 12 o’clock B was d miles in advance of A. When will the couriers be together ? Suppose they will be together x hours after 12. Then A has tray- elled mz miles, and B has travelled nz miles, and as A has travelled d miles more than B mz = nx + d, or mz — nx = d. d m—mnN _t=> Discussion oF THE Prosiem. 1. If m is greater than n, the value of x, namely, d , 1s positive, and it is evident that A will over- m—n take B after 12 o'clock. 2. If m is less than n, then d_ will be negative. In this case —n m B travels faster than A, and as he isd miles ahead of A at 12 o’clock, it is evident that A cannot overtake B after 12 o'clock, but that B passed A before 12 o’clock by fore, that the couriers were together after 12 o'clock was incorrect, and the negative value of x points to an error in the supposition. hours. The supposition, there- 3. If m equals n, then the value of a, that is, , assumes the m—n form < Now if the couriers were d miles apart at 12 o'clock, and if they had been travelling at the same rates, and continue to travel at the same rates, it is obvious that they never had poe together, and that they never will be together, so that the symbol £ may be regarded as the symbol of impossibility. 4, If m equals n and d is 0, then becomes >. Now if the m—n couriers were together at 12 o'clock, and if they had been travelling at the same rates, and continue to travel at the same rates, it is: obvious that they had been together all the time, and that ae will continue to be together all the time, so that the symbol 2 5 may be regarded as the symbol of indetermination. 196 SCHOOL ALGEBRA. | Exercise 67. 1. A train travelling 6 miles per hour is m hours in advance of a second train which travels a miles per hour. In how many hours will the second train overtake the first ? , bm Ans. a—b Discuss the result (1) when a>); (2) when a=); (3) when a b and 6 and b>m. CHAPTR Ro ALI INEQUALITIES. 198, An inequality consists of two unequal numbers connected by the sign of inequality. Thus, 12>4 and 4 < 12 are inequalities. 199. Two inequalities are said to be of the same direction if the first members are both greater or both less than the second members; that is, if the signs of inequality point in the same direction. 200. Two inequalities are said to be the reverse of each other if the signs point in opposite directions. 201. If equal numbers are added to, or subtracted from, the members of an inequality, the inequality remains in the same direction. Thus, ifa>0, then at+e>6b-+e, and a—c>b—e. Hence, A term can be transposed from one member of an in- equality to the other without altering the inequality, provided ats sign is changed. 202. If unequals are taken from equals, the result is an inequality which is the reverse of the given inequality. Thus, ifa=y, anda> 3, then x—a 6, then —a<—b. (See § 33.) 198 SCHOOL ALGEBRA. 204, Hence, if the members of an inequality are multi- pled or divided by the same positive number, the inequality remains in the same direction, by the same negative num- ber, the inequality is reversed. (1) Simplify 4a —-3 > sz 2 We have Agee ene 2 5 Multiply by 10, 402-— 30> 152—6. Transpose, 25.2 > 24. Divide by 25, a > 24. Therefore the value of x is greater than 24. (2) Find the limits of w, given x—-4>2— 382, 82—-2<24+3. We have e—4>2-— 32a, : (1) and 38a—-2<2+3., (2) Transpose in (1), 4x >: Pg dae dye Transpose in (2), aay, sed ee eas Therefore, the value of w lies between 14 and 23. (3) Ifa and 6 stand for unequal and positive numbers, then a? + 6 > 2ab. Since (a — b)? is positive, whatever the values of a and b, (a—b)? >0. a?—2ab +b? >0. a2 +b? > 2ad. —— INEQUALITIES. 199 Exercise 68. 1. Simplify (v7+1)?<2#+32—65. 2. Simplify — Fscke Pe. 3. Simplify ++ 26 > 72. 4. Simplify 834—2< mi 7h. Find the limiting values of x, given 5. 44—6< 24+4, 242+4>16— 22. 2 6. = +bn—ab > bx Lay f — ae + ab <=. Find the integral value of x, given 8. 7 4(@4+2)+42<4i(e —4)+3, 4@+2)+32>3@t1) +4. Twice a certain integral number increased by 7 1s not greater than 19; and three times the number dimin- ished by 5 is not less than 13. Find the number. If the letters stand for unequal and positive numbers, show that 9. a +30’? > 2b (a+ 3B). 10. a + BF >a’) + ab’. 1. @W4+@84+e>abtacHt be. 12. @b+a@et+ab?+h'ce+ac’?+ be > babe. 13. f4. 252, | 14, ees CHAPTER XIV. INVOLUTION AND EVOLUTION. 205. The operation of raising an expression to any re- quired power is called Involution. Every case of involution is merely an example of multi- plication, in which the factors are equal. 206. Index Law. If mis a positive integer, by definition mx OCs to m factors. Consequently, if m and m are both positive integers, (a")" = a" X a" X a” + to m factors =(aXa---- ton factors)(a X a +++ to n factors) “ois taken m times =axaxXa-- to mn factors. =a, This is the index law for involution. 202 86, (0) a And (ab)"= ab X ab + to n factors =(aXa-~+ to factors)(b X b+ to factors) 208. If the exponent of the required power is a composite number, the exponent may be resolved into prime factors, the power denoted by one of these factors found, and the result raised to a power denoted by a second factor of the exponent; and soon. Thus, the fourth power may be ob- tained by taking the second power of the second power; atk INVOLUTION AND EVOLUTION. BOD the sixth by taking the second power of the third power ; and so on. 7 209. From the Law of Signs in multiplication it is evi- dent that all even powers of a number are positive; all odd powers of a number have the same sign as the number itself. Hence, no even power of any number can be negative ; and the even powers of two compound expressions which . have the same terms with opposite signs are identical. Thus, (6—a)’={—(a—)b)}?=(a-—)). 210. Binomials. By actual multiplication we obtain, (a+ bP=a@+2ab+ 0’; (a+ be=a+3¢b+38ab’?+ 0; (a+ 6)*== at+ 40° + 6070’+ 406? + dt In these results it will be observed that: I. The number of terms is greater by one than the ex- ponent of the power to which the binomial is raised. II. In the first term, the exponent of a is the same as the exponent of the power to which the binomial is raised ; and it decreases by one in each succeeding term. III. d appears in the second term with 1 for an exponent, and its exponent increases by 1 in each succeeding term. IV. The coefficient of the first term is 1. V. The coefficient of the second term is the same as the exponent of the power to which the binomial is raised. VI. The coefficient of each succeeding term is found from the next preceding term by multiplying the coefficient of that term by the exponent of a, and dividing the product by a number greater by one than the exponent of 0. If } is negative, the terms in which the odd powers of 6 occur are negative. Thus, 202 SCHOOL ALGEBRA. (1) (a — bf =a — 3b + 8ab?— 8’. (2) (a — b)*= at — 40°) + 6070? — 4ab°+ Bt. By the above rules any power of a binomial of the form a -+- 6 may be written at once. Note. The double sign + is read plus or minus; and a+b means a+o or a—o. 211. The same method may be employed when the terms of a binomial have coefficients or exponents. Since (a— bf =a'— 3a°b + 8ab?— B’, putting 52? for a, and 27° for 5, we have (5at— 24), = (52*)!— 8(52°)(2y’) + 8(52°)(2y")— (By), = 125 2° — 150 x*y* + 60 2°y° — 87’. Since (a@—b)*=at— 40° + 6a°b?— 4ab*> + Ot, putting 2? for a, and ty for 6, we have @—4y)' = (2°) 42 )EY) + 8) Gy) 40S Yt yy’, = a — Zaty + gay’ — Fey + Tey’. 212, In hke manner, a polynomial of three or more terms may be raised to any power by enclosing its terms in paren- theses, so as to give the expression the form of a binomial. Thus, ) (1) @@+b+cP=[a+(O+oe)f, =a+3a(b+c¢)+38a(6+cl+(b+e)’, =a+8¢b+8ac+3ab’?+ babe . + 3a?+6?+380?e+3b7 +, INVOLUTION AND EVOLUTION. 203 (2) (# —227+ 32+ 4), =[(0? — 20%) + (80-+4)f, = (a! — 2.2%) + 2 (2? — 22") (8244) + (8044), = 2 —42°+ 424 62t—405—16274 9274 242416, = #—44°+ 102*— 42° —72?4+ 244+ 16. Exercise 69. Raise to the required power : 1. (a*)’. 11. (2? — 2), 2. (a®d*)>, 12. (% + 8)°. 3. ane 13. (22+1)% 3 ab® 14. (2m?— 1)*. 15. (24+ 3y)°. 4. (—5ab’c’)‘. : Bo (— Tay)’. 3 a b3et\5 16. (2% — y)*. at Ke ( Sy) 17. (vy — 2)". 7. (— 2aty')’. 18. (l—2+ 2%). im 27.345 ete - 19. (l— 22+ 32%) 3 7203\4 . > (Greah 20. (l—a-+a’). 10. (~+2)°. 21. (83—44+ 527)’, EVOLUTION. 213. The nth root of a number is one of the m equal factors of that number. The operation of finding any required root of an expres- sion is called Evolution, 204 SCHOOL ALGEBRA. Every case of evolution is merely an example of factor- wg, in which the required factors are all equal. Thus, the square, cube, fourth, ..... roots of an expression are found by taking one of its two, three, four ..... equal factors. The symbol which denotes that a square root is to be extracted is ,/; and for other roots the same symbol is used, but with a figure written above to indicate the root; thus, x/, ~/, etc., signifies the third root, fourth root, ete. 214, Index Law. If m and m are positive integers, we have, (§ 206), (a™)” — ra hacen Consequently, Vinee : Thus, the cube root of a® is a’; the fourth root of 8la” is 8a*; and so on. This is the index law for evolution. » 215, Also, since (ab)*= ad”, conversely, Vath" = ab = 0 ane and Vab = Vax Vo. Hence, to find the root of a simple expression : Divide the exponent of each factor by the index of the root, and take the product of the resulting factors. 216. From the Law of Signs it is evident that I. Any even root of a positive number will have the double sign, +. II. There can be no even root of a negative number. For V—2* is neither +2 nor —2; since the square of +2=-+2", and the square of —x=-+ 2”. The indicated even root of a negative number is called an imaginary number. ‘TI. Any odd root of a number will have the same sign as the number, INVOLUTION AND EVOLUTION. 205 Aa Thus, Apia gee V— 27 min’ = — 8 mn?: 18 yo =9y’ [16 2%? __ ate: 22°y* Sl a ek = 217. If the root of a number expressed in figures is not readily detected, it may be found by resolving the number into its prime factors. Thus, to find the square root of 3,415,104 : 2°| 8415104 426888 11 8,415,104 = 2° x 3? x 7x 112, , V8,415,104 = 22x38 x 7 X¥ 11= 1848. Exercise 70. Simplify : 1. V4a%/. grey 21600. 1s ere 2. Vea, 10. 7292, ie 3. V1625y". 11. VW2438729, «18. eee 4, V— 32a". 12. V—1728a%. pirate e020" 5. V— 27H, 13. V— 348°. ae 6. V25 at. 14. V8la*. 20. Af sete ON 15. V512a"b". ines 8. Vb642", 16. V ain oe 216 a 206 SCHOOL ALGEBRA. SQUARE Roots oF COMPOUND EXPRESSIONS. 218. Since the square of a+0 is a’+2ab+0’, the square root of a?+ 2ab+ 0’ is a+. It is required to find a method of extracting the root a+6 when a’+ 2ab + 2 is given: Ex. The first term, a, of the root is obviously the square root of the first term, a*, in the expression. a+ 2ab+b?la+b If the a? be subtracted from the given a? expression, the remainder is 2ab + b?. 2a+6| 2abd+ 8B? Therefore the second term, 0, of the root 2ab + b? is obtained when the first term of this remainder is divided by 2a, that is, by double the part of the root already found. Also, since 2ab + b?=(2a + b)d, the divisor 2s completed by adding to the trial-divisor the new term of the root. (1) Find the square root of 252?—202°y + 4 ay’. 25 x? — 20 ay + 4aty?|5a — 2a7y 25 x 10a” — 2a°y|— 20a%y + 4aty? — 20 a%y + 4 aty? The expression is arranged according to the ascending powers of «. The square root of the first term is 5a, and 52 is placed at the right of the given expression, for the first term of the root. The second term of the root, —22%y, is obtained by dividing — 20a°y by 10a, and this new term of the root is also annexed to the divisor, 10a, to complete the divisor. 219, The same method will apply to longer expressions, if care be taken to obtain the ¢rial-divisor at each stage of the process, by doubling the part of the root already found, and to obtain the complete divisor by annexing the new term of the root to the trial-divisor. INVOLUTION AND EVOLUTION. 207 Ex. Find the square root of 1+ 1027+ 252'+ 162° — 2425 — 2027°— 42. 16 2° — 242° + 25 at — 202° + 10a? 424 1[408—307+ 22-1 16 «6 8a? — 3a?|— 2405 + 252% — 242°+ at 82? — 6a? + 2x) 16 at — 2023 + 102? 16a*—122°+ 42? 8a°®°—62?+4e2—1]/— 822+ 6227-42741 — 8e3 + 647—4241 The expression is arranged according to the descending powers of z. It will be noticed that each successive trial-divisor may be obtained by taking the preceding complete divisor with its last term doubled. Exercise 71. Find the square root of 1. 2*—82'?+ 182? —82r+1. 2. Jat— 60° + 18? —4a-+ 4. 3. 4a*—12a%y + 29x77? — 30xy* + 257%. 4, 1+ 4274+ 1027+ 122° 4+ 92%. 5. 16—96x2+ 2162? — 2162°+ 81 at. 6. v'— 222? + 9527+ 2862+ 169. 7. 42*—l1la2?+ 25 —122*°+ 302. 8. Jat +49 — 122° — 282+ 462’. 9. 49a*+1262°? + 121 — 732? — 198-2. 10. 162*— 302 — 812’ + 242’ + 25. 11. eee Mache alt 208 SCHOOL ALGEBRA. 12. 4at+40°—te4H. 16 4a? 40? LS ieee aha Sb ee se Lat abe eee eee Ho Gale om a a flee ie eee PAL 4c, B82 41 By, yf 16.0 eh ee aig hie tea 2 17. fan 284 a8 80-45 18. l6a*+182°7+82'+ 47° + 4y+1. 19, 92 82, 482% Te 49 + 2 4 2 as 20. 4a?+~.—8_114 4a, a a Find to three terms the square root of 21. a? +0. 24. 1+a. 27. 497+ 3. 22. vty. 25. 1—2a. 28. 4—Ba4, 23. 1+2a. 26. 47+ 20. 29. 4a?—1. 220. Arithmetical Square Roots. In the general method of extracting the square root of a number expressed by figures, the first step is to mark off the figures in groups. Since 1 = 1?, 100 =10?, 10,000 =100?, and so on, it is evident that the square root of any number between 1 and 100 lies between 1 and 10; the square root of any number between 100 and 10,000 lies between 10 and 100. In INVOLUTION AND EVOLUTION. 209 other words, the square root of any number expressed by one or two figures is a number of one figure; the square root of any number expressed by ¢hree or four figures is a num- ber of éwo figures; and so on. If, therefore, an integral square number is divided into groups of two figures each, from the right to the left, the number of figures in the root will be equal to the number of groups of figures. The last group to the left may have only one figure. Ex. Find the square root of 3249. 32 49 (57 In this case, a in the typical form a? + 2ab + 0? oe represents 5 tens, that is, 50, and 6 represents 7. 107) 749 The 25 subtracted is really 2500, that is, a?, and the 749 complete divisor 2a +6 is 2x 50+ 7 = 107. 221, The same method will apply to numbers of more than two groups of figures by considering a in the typical form to represent at each step the part of the root already found. It must be observed that a represents so many tens with respect to the next figure of the root. Ex. Find the square root of 5,322,249. 5 32 22.49 (2307 4607) 32249 32249 222. If the square root of a number has decimal places, the number itself will have twice as many. Thus, if 0.21 is the square root of some number, this number will be (0.21)? = 0.21 x 0.21 = 0.0441; and if 0.111 be the root, the num- ber will be (0.111)? = 0.111 x 0.111 = 0.012321. 210 SCHOOL ALGEBRA. Therefore, the number of decimal places in every square decimal will be even, and the number of decimal places in the root will be hadf as many as in the given number itself. Hence, if a given number contain a decimal, we divide it into groups of two figures each, by beginning at the decimal point and marking toward the left for the integral number, and toward the right for the decimal. We must be careful to have the last group on the right of the deci- mal point contain two figures, annexing a cipher when necessary. Ex. Find the square roots of 41.2164 and 965.9664. 41.21 64 (6.42 9 65.96 64 (31.08 36 9 124) 521 61) 65 496 61 1282) 2564 6208) 49664 2564 : 49664 223. If a number contain an odd number of decimal places, or if any number give a remainder when as many figures in the root have been obtained as the given number has groups, then its exact square root cannot be found. We may, however, approximate to its exact root as near as we please by annexing ciphers and continuing the operation. The square root of a common fraction whose denominator is not a perfect square can be found approximately by reducing the fraction to a decimal and then extracting the root; or by reducing the fraction to an equivalent fraction whose denominator is a perfect square, and extracting the square root of both terms of the fraction. Thus, Rete e ee VP = 0.625 = 0.79057 ; ue N= EE 001 Wine INVOLUTION AND EVOLUTION. 211 Ex. Find the square roots of 3 and 357.357. dark ied 1 27) 200 189 343) 1100 1029 3462) 7100 6924 3 57.35 70 (18.903..... 1 28) 257 244 369) 3335 3321 37803) 147000 Exercise af ae Find the square root of Letteoo. 6. 2. 1225. (ie 3. 12544, 8. 4. 253009. 9. 5. 529984. 10. 150.0625. 118.1569. 172.3969. 5200.140544. 1303.282201. 113409 ibe 12. 13. 14. 15. 640.343025., 100.240144. 316.021729. 454.585041. 5127.276025. Find the square root to four decimal places of 16. 10. 19,405: thas. 2050 (- 18...5. 21. 0.9. 22. 0.607. 23. 0.521. 24. 0.687. 25. 26. 27. 224, Cube Roots of Compound Expressions. of a+b is a&+38@)+ 8ab?-+ 6°, the cube root of Pi Bone $6 a9. §. 4, 80 tae Since the cube +3076 + 38a?+ 6 is at. It is required to devise a method for extracting the cube root a+6 when a’+3a’b+ 3ab’-+ 6 is given: 212 SCHOOL ALGEBRA. (1) Find the cube root of a?+ 3a’) + 3ab’ + 8°. a+ 3a7b +3ab?+ Bla+b Bae os as +3ab +6? 3.0a7b + 3ab? + 68 3a?+ 3ab + 6? 3a7b + 3ab? + 6 The first term a of the root is obviously the cube root of the first term a’ of the given expression. If a® be subtracted, the remainder is 3 a?) + 3ab? + 6°; therefore, the second term 8 of the root is obtained by dividing the first term of this remainder by three times the square of a. Also, since 307) + 3ab? + B= (3a? + 3ab+0?)b, the complete divisor is obtained by adding 3ab + 0? to the trial-dimsor 3a’. (2) Find the cube root of 82°+ 362°y + 54 ay’ + 2777. 8a? + 36 x7y + S4ay? + 27y?|2e + 3y 122? 8 x3 (62+3y)3y= 182y +94? 36 ay + 54 ay? + 27 y 12a?4+18ay+9y?|_ 36 a%y + 54 ay? + 27° The cube root of the first term is 22, and this is therefore the first term of the root. The second term of the root, 3 y, is obtained by dividing 36 a?y by 3 (2a)? = 1247, which corresponds to 3a? in the typical form, and is completed by annexing to 122? the expression {3(2x)+3y33y = 18ay + 9y?, which corresponds to 3.ab + 6? in the typical form. 225. The same method may be applied to longer expres- sions by considering a in the typical form 3a’?+ 3ab+ 6? to represent at each stage of the process the part of the root already found. Thus, if the part of the root already found is «+ y, then 3a’ of the typical form will be represented by 3(2+ y)?; and if the third term of the root be +z, the 3ab-+ &? will be represented by 3(a+y)z+2%. So that the complete divisor, 3a’?-+ 3ab + 0’, will be represented by d(a+tyyl+3(e#+y)z242. INVOLUTION AND EVOLUTION. 213 Find the cube root of 2°— 32°+ 52°— 3a —1. x2? — 7+. I °° — 32° + 52° — 32-1 3 at x (3a? — x)(—2) = soe + —32° +525 - 3a%*—S8a5+ a71/—325 +3at— 2 —3at+ 623-32 —-1 3 (a? — x)? =3at— 623 + 3.2? (3 2?—3a—1)(—l)= —3a? +3a¢ +1 ed Ll ok + bao a The root is placed above the given expression for convenience of arrangement. The first term of the root, x?, is obtained by taking the cube root of the first term of the given expression; and the first trial-divisor, 3x4, is obtained by taking three times the square of this term. The first complete divisor is found by annexing to the trial-divisor (3a? —a)(—«), which expression corresponds to (3a+0)6 in the typical form. The part of the root already found (a) is now represented by 2?—«; therefore 3a? is represented by 3(a?— 2)? =3at—623 + 327, the second trial-divisor; and (3a + 6)b by (83a?—3a—1)(—1); therefore, in the second complete divisor, 3a? + (3a + b)bd is represented by (3 a*—6 a3 + 3a”) + (—3a?—32—1)x (—1)= 3at— 603 + 3241. Exercise 73. Find the cube root of a+ 8a + 8a27+ 2%. 8+122+62'+ 2. x’ — Baz’ + 5a®z*®— 8a°x — af. 1— 62+ 212?— 442° + 682*— 542° + 27 2x® 1— 382+ 62? — 72°+ 62*— 32°+ 2° 2t+1— 62—62°+ 1527+ b2*— 202%. oO nT Pw YP 214 SCHOOL ALGEBRA. 7. 642°—1442°+ 8— 8642+ 1022?— 171 2°+ 204 2%. 8. 27a°— 27a°— 18a'+ 17a°+ 6¢’— 8a—1. 9. 82°— 362°+ 662‘ — 682° + 8382°—9x+1. 10. 27+ 1082+ 902?— 802? — 60 2* + 482°— 82° . i dietel A Lie ol a pitas een Toe OL AOL ee ae a oe? 226. Arithmetical Cube Roots. In extracting the cube root of a number expressed by figures, the first step is to mark it off into periods. Since 1 =1?, 1000=10*, 1,000,000 = 100%, and so on, it follows that the cube root of any number between 1 and 1000, that is, of any number which has one, two, or three figures, is a number of one figure; and that the cube root of any number between 1000 and 1,000,000, that is, of any number which has fowr, five, or siz figures, is a number of two figures ; and so on. If, therefore, an integral cube number be divided into groups of three figures each, from right to left, the number of figures in the root will be equal to the number of groups. The last group to the left may consist of one, two, or three figures, 227. If the cube root of a number have decimal places, the number itself will have three tumes as many. Thus, if 0.11 be the cube root of a number, the number is 0.11 x 0.11 x 0.11 = 0.001331. Hence, if a given number contain a decimal, we divide the figures of the number into groups of three figures each, by beginning at the decimal point and marking toward the left for the integral number, and INVOLUTION AND EVOLUTION. 215 toward the right for the decimal. We must be careful to have the last group on the right of the decimal point con- tain three figures, annexing ciphers when necessary. 228, Notice that if a denotes the first term, and 6 the second term of the root, the first complete divisor is 3807+ 8ab-+ b’, and the second trial-dwisor is 3(a-+ 6)’, that is, 80+ 6ab +36, which may be obtained by adding to the preceding complete divisor zts second term and twice rts third term. Ex. Extract the cube root of 5 to five places of decimals. 5.000 (1.70997 | 1 3 x 10? = 300 4000 3(10 x 7) = 210 = 49 559 3913 259 87000000 3 x 1700? = 8670000 3(1700x9)= 45900 OF es 81 8715981 45981 3 X 1709? = 8762043 78443829 85561710 78858387 67033230 61334301 After the first two figures of the root are found, the next trial- divisor is obtained by bringing down the sum of the 210 and 49 obtained in completing the preceding divisor; then adding the three lines connected by the brace, and annexing two ciphers to the result. The last two figures of the root are found by division. The rule in such cases is, that two less than the number of figures already obtained may be found without error by division, the divisor being three times the square of the part of the root already found. 216 SCHOOL ALGEBRA. Exercise 74. Find the cube root of 1. 4913. 3. 1404928. 5. 3885828.352. 2. 42875. 4, 127263527. 6. 1838.265625. Find to four decimal places the cube root of (Peeve 9. 3.02. 11. 0.05. 13. 2. 15. 3%. Biol) 10. 2.05. 12. 0.677. 14. 3. 16. +s. 229, Since the fourth power.is the square of the square, and the sixth power the square of the cube, the fourth root is the square root of the square root, and the sixth root is the cube root of the square root. In like manner, the eighth, ninth, twelfth, ...... roots may be found. ~ Exercise 75. Find the fourth root of 1. 8lat+ 1082? + 542*°+ 122-41. 2. 162* — 82a2° + 24077? — 8a*x + at. ~ 1+4e7+4+42'+ 1028+ 162? + 102?+ 192*+ 162". J%) Find the sixth root of 4. 1+6d+d°+6d*°+15d‘*+ 20d*+ 15d’. . 729—14582+12152? —540234 135 2*—184°+ 2° 1—18y+ 1357 — 5407? + 12157‘ — 14587 + 729%. oO oa CHE Pighorn Ve THEORY OF EXPONENTS. 230. If nm is a positive integer, we have defined a” to mean the product obtained by taking a as a factor m times. Thus a* stands for axaxa; 6* stands for bX bx6x 6. 231, From this definition we have obtained the following laws for positive and integral exponents : Dad Xiet=aaet™ IL (a™)"=a™, III. us Bee anid 97 TUT. a LV Van = a™. Wie LOLs sa any™: 232. Since by the definition of a” the exponent n denotes simply repetitions of a as a factor, such expressions as a3 and a~* have no meaning whatever. It is found convenient, however, to extend the meaning of a” so as to include fractional and negative values of n. 233, If we do not define the meaning of a" when 7 is a fraction or negative, but require that the meaning of a” must in all cases be such that the fundamental index law shall always hold true, namely, q™ x q” = Cmte: we shall find that this condition alone will be sufficient to define the meaning of a” for all cases. 218 SCHOOL ALGEBRA. 234. To find the Meaning of a Fractional Exponent. Assuming the index law to hold true for fractional expo- nents, we have 3 3 3 3 iz at x at x xataatth lt gi =e 4 : = he to n terms ms A X Ah wee to n factors = an'2™™” = 4 en x = os < = ton terms ter Gn X QA vee to n factors =an ‘2 == 8 sae That is, a? is one of the two equal factors of a, a? is one of the three equal factors of a, kes a‘ is one of the four equal factors of a’, 1 : an is one of the m equal factors of a, an is one of the n equal factors of a”. Hence at= Va; at = Va; at= Va; a® am (§ 218) 1 1 1 Also, am X ar X an Xe to m factors. 1 i 1 m —~+-—+-—.... to m terms -- =—— (im ey ——i(]ien “aa (Va). The meaning, therefore, of an. where m and are posi- tive integers, is, the nth root of the mth power of a, or the mth power of the th root of a. Hence the numerator of a fractional exponent indicates a power, and the denominator a root; and the result is the same when we first extract the root and raise this root to the required power, as when we first find the power and extract the required root of this power, THEORY OF EXPONENTS. 219 235. To find the Meaning of a’. By the index law, me eer ees OL: *. a’ =1, whatever the value of a is. 236. To find the Meaning of a Negative Exponent. If n stands for a positive integer, or a positive fraction, we have by the index law, OG AY ipa Eee god But ural, .axar=i. That is, a” and a~ are reciprocals of each other (§ 167), so that a” = as andia == as a a 237. Hence, we can change any factor from the numerator of a fraction to the denominator, or from the denominator to the numerator, provided we change the sign of its exponent. Bas La abd? Thus ae may be written al’c*d~™, or 238. We have now assigned definite meanings to frac- tional and negative exponents, meanings obtained by subjecting them to the fundamental index law of positive integral exponents; and we will now show that Law II., namely, (a”)”= a", which has been established for positive integral exponents, holds true for fractional and negative exponents. (1) Ifn is a positive integer, whatever the value of m, We have (a™)n = a™ xX a™ x a™..... to n factors, = gm+m+m eevee to n terms J = amr, 220 SCHOOL ALGEBRA. (2) If nis a positive fraction 2, where p and q are posi- tive integers, we have q (amr = (amy = Vamp 2 234 = Vam (1) exe ae % 234 = q™ xa =a, (3) If is a negative integer, and equal to —p, we have 1 (a™)" = (a™)- P= (amp 2 236 1 ) 8. <6. Va’. 8.9 Vax? + Vato. 222, SCHOOL ALGEBRA. Express with root-signs : 3 os hey Fee fae Pac 9.5.04. 11. a*b8; 13.007. 15. a?— 2% ¢8. 2 1,2 2 8 2 2 2 10. 3, 12. a5bd8, 14. 32%y 4. 16. aS + x55, Express with positive exponents: —1 ,.2 17. a-*. 19. Bay’. 21. 4aty, 23, ee By gh sisya 5, . 20. 4 ye ae Oe 3a-*b?. Write without denominators: 2 2,3 a yy eka a6. eee, 28.4 a eed aren | rae Ais. 2 —1 },-2 ,-3 —4f—-5 ,—6 ST ees. 27. een 29. OVE SY CEUs C7 Open Find the value of 5 2 3 2 5 SiR ema y peyE 34, 36, 36. (—27)5 x 257, 31. 16-2. 33. (—8)-#. 35. (—27)8. 37. 817? x 164. Simplify : 38. 8? x 472, 40. Gis x (ay)? 42. (a $B)" : 39. (se)? x 1674. 41. wa 1bF x atdt. 48. (a 25) If a=4, b=2, c=, find the value of 44. ab 46. a 20%, 48. 8(ab)*. +50. (ad*e)t. 45. ab. 47. ate 8. 49. 2(ab) 4%. 51. (ab%)?, THEORY OF EXPONENTS. 223 240. Compound expressions are multiplied and divided as follows : 1 : 1 Ad 1 aL ae 1 (lye Multiply 2? + 2*7* + 4% by 2? —2x77* +77. 1 Tot a Me ak oy i) 1 1 1 1 24 — pFy* + 74 he HI 1 Us ary) Pet ys = 44/2 Sip Ug + aby? + atyd + y x + why? +y (2) Divide V2?+ Vx—12 by Vz—8. aot 4 at —12|at — 3 ai — 348 at +4 +4a3 — 12 +4a% — 12 Exercise 77. Multiply : 1. at +53 by at — 32. 3. at —b? by a? —}?. 2. at +b? by at + 82. 4. ee by xt — 2m. Bp aty ty? by ot aty ty , ee a tay +y* by at —y. eee po b-* by 1 —b 248-7. ee Ady by at yt +42: 10. «86-24 2at — 383 by 20-4? 4a°# 6a 288. 3 224 SCHOOL ALGEBRA. Divide : 11. a—b by at — BF. 13. a—b by a* — BF, 12. a+6 by at + 8. 14. a+b by ak +55. 15. Qa-8+ 62 y — 162°y— by Qa4+2a%y 4 4ahy 16. ety tz—S8atyte by bt yo. 17. 2-323 +4+82'—1 by a1. 18. statytty by a — at ot + of? 19. 2 —4e5+146a 3 by 22. 20. 92—122? 2444 2427 by 8222-273. Find the square root of 21. x 42341, 23; x — Ags + 4, 22. 4at—4a3bt+ 9°. 24. 4a7+4a7+1, 25. 9a—12a?+10—4a F447, 26. 49a? — 282+ 1823 —423+1. 27. m+ 2m—1—2m "+m. 28. 144y7'—2y-8— 4474 253 — 24 $4 164, Expand : 29. (w@—a). 31. (2at—a-®). 33. (LV 2 —4-V2)'. 30. (Vai—42). 32. (Qa2%+a%). 34. (AVa%+427) CHAPTER XVI. RADICAL EXPRESSIONS. 241, A radical expression is an expression affected with the radical sign; as, Va, V9, Va, Va +6, V32. 242, An indicated root that cannot be exactly obtained is called a surd, or irrational number. An indicated root that can be exactly obtained is said to have the form of a surd. The required root shows the order of a surd; and surds are named quadratic, cubic, biquadratic, according as the second, third, or fourth roots are required. The product of a rational factor and a surd factor is called a mixed gurd; as, 83V2, bVa. The rational factor of a mixed surd is called the coefficient of the radical. When there is no rational factor outside of the radical sign, that is, when the coefficient is 1, the surd is said to be entire; as, V3 2, Va. 243, A surd is in its simplest form when the expression under the radical sign is integral and as small as possible. Surds which, when reduced to the simplest form, have the same surd factor, are said to be similar. Norr. In operations with surds, arithmetical numbers contained in the surds should be expressed in their prime factors. REDUCTION OF RADICALS. 244, To reduce a radical is to change its form without changing its value. 226 SCHOOL ALGEBRA. Case I. 245. When the radical is a perfect power and has for an exponent a factor of the index of the root, (1) Vai=at=at= Va; (2) V36 0%? = V(6ab)? = (6ab)? = (6.ab)? = V6ab; (3) V25 abd = V(b abc! = (5 arbct)§ = (5 arbe)s = V5 abet. We have, therefore, the following rule: Divide the index of the root by the exponent of the power. Exercise 78. Simplify : ia \/ 25. 6. V/ 02d. We Nee 2. VTC. ses Wri ee 6/57 8/474 13:., 4) Loge Shan 27! 8. Va'dbt. (x — 3) 4. \/49, 9. \/ 27 a8b®. ey xe) ee 13. 6} @0° 5. V64. 10. V16a‘d'. 8 xy? CasE II. 246. When the radical is the product of two factors, one of which is a perfect power of the same degree as the radical. Since Va"b = Va" x Vb =aVb (§ 215), we have (Oiney Woe Xa ee ny ae (2) W108 = -V27 x 4 == V27 x V4 =38V4; RADICAL EXPRESSIONS. DOT (3) 4V 72020? = 4-36.00? x 26 = 4-V36 ab? x V26 =4x 6abV2b = 24ab V26- (4) 2V54 ab = 2V27 a? x Zab = 2V 2708 x V2ub = 2x 8aV2ab = 6a V2ab. We have, therefore, the following rule: Resolve the radical into two factors, one of which vs the greatest perfect power of the same degree as the radical. Remove this factor from under the radical sign, extract the required root, and multiply the coefficient by the root obtained. Exercise 79. Simplify : 1. V28. 13. 7V144. os. 3| O425y Oye) F 3,3 Sioa 12. 14. 8Vmn. 27min 3, -V72. 15. 3VBb%. 26. te 4. ~/500. 16. 2Wakc*, a ee ov. 4: 125 2° 5. W482. t2 L1lVa®, : 216y3 6. 192. 18. 7V8a°d. ae BT ap. 24 7. W128. 19. 6V27 mn. 943 8. 243. 20. 4Va"y’. oo tl aut apes — V 1296 9. V176. 21. 1029. Lae 10. 405. 22. /— 2187. 30. yee. Tre —— 11. BV 112. 23. 1250. 3ab_ [B02 12. 3-864. 24, 4r/648. ole OWN Gait 228 SCHOOL ALGEBRA. Case III. 247. When the radical expression is a fraction, the denomi- nator of which is not a perfect power of the same degree as the radical, 5 10 eT VOX <—=~4/--=4/10 ae \ 7 Sr 10. Ge Tee we 9150 tess ee —=—sx = * Ty Ne aie 4x9 * 36 eval; seo aeest et 35x 8x4 : 1 | 2 Se ae a> a= 27 x8 27 x 8 We have, therefore, the following rule: Multiply both terms of the fraction by such a number as will make the denominator a perfect power of the same degree as the radical; and then proceed as in Case LT. Exercise 80. Simplify : ues a Mein # Tan as 10. 2V3%. yh se. 5B. W238. 8. V2. Lea Beet £ 6.08 Pr. SimeNate, 12. ee ate sf cat acy? 1Sii | se 15. 17., Ales 5 NG Ne 14 Abe 16. 412. 1g a® ~ N125¢ 3x’ y2* RADICAL EXPRESSIONS. 929 Case IV. 248. To reduce a mixed surd to an entire surd. Since aVb = Va" x Vb = Va"b, we have (1) 8V5=V3'x5=V9x5=V45; (2) vbVbe = V(b) x be = Vail? x be = Vaibic : (3) 2aVay = V(2a) Xx cy = V8e xX ay = VB ay ; (4) 8yVei=VBy)! xX @ = VB1y 0". We have, therefore, the following rule: Fase the coefficient to a power of the same degree as the radical, multiply this power by the given surd factor, and dicate the required root of the product. Exercise 81. Express as entire surds: 1. 5V5. 5. 2N/8. CD ean aia kame SVE evil 6.81/72. 10..— 8V7. 14. —3-Val. 3. 33. Fen eel l nh / 1004 15... Bach 4, 2V/4, Base 12.0 D4) 2. 16. —3V mi. Case V. 249. To reduce radicals to a common index. (1) Reduce V2 and V3 to a common index. V2 = 2 = = V8 = V8. 43 = 3t 38 YR = V5. Hence, 230 SCHOOL ALGEBRA. Write the radicals with fractional exponents, and change these fractional exponents to equivalent exponents hawng the least common denominator. Raise each radical to the power denoted by the numerator, and indicate the root denoted by the common denominator. Exercise 82. Reduce to surds of the same order : 1. V3 and V5. 7. V2, V8, and V5. 2. V14 and V6. 8. Va?, Vb, and Ve. 3. -/2and V4. 9. Vai, Ve, and V2". 4. Vaand Vb". 10. V2y, Vabe, and W2z. 5. V5 and V7. ll. V2—yand Va+y. 6. 22 23 and 23. 12. Va+b and Va—b. Nore. Surds of different orders may be reduced to surds of the same order and then compared in respect to magnitude. Arrange in order of magnitude : 13. V15 and V6. 15. /80, V9, and V8. 14. V4 and V3. 16. V3, V5, and V7. ADDITION AND SUBTRACTION OF RADICALS. 250. In the addition of surds, each surd must be reduced to its simplest form; and, if the resulting surds are similar, Find the algebraic sum of the coefficients, and to this sum annex the common surd factor. If the resulting surds are not similar, Connect them with their proper signs. RADICAL EXPRESSIONS. ak (1) Simplify V27 + -V48 + V147. V27 = (3? x 3)2=3 x 33 =3V3; V48 = (2! x 3)8 = 2? x 32 = 4 x 38 = 43; V147 = (7? x 8)? =7 x 32 = 7V3. o. V27 4+ V48 + V147 = (8444 7)V3 =14V3. Ans. (2) Simplify 2/320 —8-~/40. 2320 = 2(2 x 5F=2xK Bx 5E=8V5; 3V40 = 3(2 x 5)F=3 x 2x 58 = 6 V5. “. 2320 — 340 =(8 —6)V5=2V5. Ans. (3) Simplify 2V3—8V34 V+. 2V5 = 2V/18 = 2V15 xi =2Vv 15; 3V8 =3V15 =3V15 x ge = 3 V15 z= ./4x 15 \ 4 . Se cs se Vi = 1p 15 Xx ip = 1s V15 1 2V8 —3-V8 4+ VA = (8-8 + 3%) V15 =i VI15. Ans. Exercise 83. Simplify : Po 4 bas y 116/11: MeN) 8 = 54/8 95/8. pny Ae PX/89— ~/108. 1. V8 43/4844 75. . BV2+44V72— V/64. 8. 4V147+ 3775+ -V192. .1AW54+21V5414V40. 9. Vativat va. ; 8VB— 548+ -V248. 10. Va?+ iWVai—8V 27a". Zor SCHOOL ALGEBRA. Lis Val OW Ge Ow 12. V25b+ 2V96b — 8V46. 13. 2/175 — 8-V638 + 5-V28. 14, V2+ 8V382-+ 1/128 — 6V/18. 15. V75 + V48—-V 147 + -V800. 16. 20-V245 — V5 + -V125 — 24-180, 17. 2V20-+ 4VI13 — 2-V87 +. SV — VIB. 18. 725+ 4745 — V9 — 2-804 V20—4 V64, 19. V54+ VE— V250 — 8-v32. 20. 2V8+ V60— V15+ V3+ V4. a1. VOT — V8 + V125e. 22. Vabb — Vb'+ W320. 23. Vata + Vor —-V4a°b'x. 24. V4a%pe t+ Vy't2+ Va'x. 25. Vale —av4e+bvare. 26. V8la’— V16a+ V256a°. 27. W27m*t — V125m + V216 m. 28. V8a—V50a?— 8V18a. 29. 6aV63ab) — 8-V1120°8? + 2abV348 ab. 30. 3V125 min? + nV 20m — -V500 min’. 31. V32a'b® + 6V726 + 8-V128 ab? 32. 2N/a% — 8 0V646 + bavVatb + 2aV125 6. 22. 23. 24. 25. RADICAL EXPRESSIONS. 233 MULTIPLICATION OF RADICALS. 251, Since Va x Vb = Vab, we have fms x oV2—3 X 5 xXV/8.x-V2 = 15V16= 60; (2) 83V2 x 4V3 =38V8 x 4W9 = 12V72. We have, therefore, the following rule: Express the radicals with a common index. Find the product of the coefficients for the required coefficient, and the product of the surd factors for the required surd factor. Reduce the result to its sumplest form. . V8x V97. . VEX -V20. oV2x N18. . VBxXV9. BN \/ 52: ETE aN ES Beaise oh 7. VW4x V8. 8. V27x V9. 9. V2x VI12. LO ry avo, 11. V3xVI18. 12. V6x V8. 13 . Wd4 x V9. 2 2/8x V2: . V8xV-=4, Vix V¥—49. . V8lxV—45. . 2V18 x 8V3. 19. (V18 + 2V72—8V38) x V2. 20. (W732 —1V8644 8V4) x V2. a. (AV 27 — 1/2187 + 4-482) x V3. V5 x V4. V64 x V16. V3 x V7. V16 x 0/250. 26. V2 x Vi. OAD gt GN) 28. V81x V3. 29 GaN ROGAN) Fs 30 31 32 33 VEX VE. . V2a x V2. Vy XN oY. VIX V5. 234 SCHOOL ALGEBRA. 252. Compound radicals are multiplied as follows: — Ex. Multiply 2V34+8V2 by 8V38—4Vz. 2V3 + 3V x 3V3 —4vV 2 18 +9V32 —~8V3e—122 18+. V32—122 Exercise 85. Multiply : 1.V5+V4by V5—V4 4. 84+8-V2 by 2— V2. 2. V9—VI7by V9+VI17. 5. 54+2V3 by 8—5V3. 32 3-2 bby Be o: 6. 8— V6 by 6— 8V6. 7. 2V6 —8V5 by V3 + 2V2. Bie Tey ODN ee 9. V9 — 2/4 by 4734+ V2. 10. 2V30—8V5+ 5V38 by V8+ V3 — V5. 11. 8V5 — 2V38 4+ 4y/7 by 83V7— 4V5 = Dae 12. 4V8+1V12 —1V32 by 8-V32 —4V50 — 2v2. 13. V6 —-~V/3 +4 V/16 by 36+ V9 — V4, 14, 2V2—8V84 8-V§ by 8V3— V12 — V6. 15. 2V3 —4V8 — 7V8 by 8-V3 — 5-V30 — 2-48. 16. 2V124 8V34 6V32 by 2V12 + 8-V3 4 6V1. RADICAL EXPRESSIONS. 200 DIvIsIon oF RADICALS. 258, Since SO = Vb, we have Va Va 4/8 _ O) OV 4/3 40/3? 4/3? x DB 8/75, =. i JN BION iy aa We have, therefore, the following rule: (2) Express the radicals with a common index. Find the quotient of the coefficients for the required coefficient, and the quotient of the surd factors for the required surd factor. feduce the result to rts sumplest form. Exercise 86. Divide: Dey ote by V3. 4. Vi by Vz. 7. V4¢ by V2. 2. V81 by V3. 5. VE by V3. 8. WBF by V5E. pm apuby Va’... 6. Vy, by Ve. 9. V$2 by VEE 10. 83V6+ 45-V2 by 8v3. 11. 42/5 — 30V3 by 2V15. 12. 8415+ 168V6 by 3V21. 13. 3074 — 36-V10+ 30-V90 by 3/20. 14, 50V18+ 18-V20—48V5 by 230. 15. V54 by V36. 17. VI2by V6. 19. V2 by W338. 16. V49 by V7. 18. V& by V6. 20. V2z by Vz. 21. +/0.064 by V/10. 22. Vat—y* by x+y. 236 SCHOOL ALGEBRA. 254. The quotient of one surd by another may be found by rationahzng the divisor; that is, by multiplying the dividend and divisor by a factor which will free the divisor of surds. 255. This method is of great utility when we wish to find the approximate numerical value of the quotient of two simple surds, and is the method required when the divisor is a compound surd. (1) Divide 8V8 by V6. 8V8_6V2_6V2xVv6_ 6V12 Ve Ve Teves r = V12 = 2V3. (2) Divide 8V5—4V2 by 2V548V2. 8V5—4V2_ (3V5—4V2)(2V5—3V2)_54—-17V10 2V54+3V2 (2V543V2)(2V5—-3V2) 20-18 —4=17V10 _ 97 _ gi V0, wt 2 = 5 (3) Given V2 = 1.41421, find the value of Rie BSA RPE A) Scilla V2 V3xv2 2 2 Tacidar Exercise 87. 1. Va+vob by Vab. 7. 3+5V7 by 8—5V7. Bee 195 bys 57 6b: 8. 21V3 by 4V8—8v2. 8. 8 by 11+3V7. 9. 75V14 by 8V2+4 2V7. 4. 8V2—1 by 8V2+1. 10. V5—-V38 by V5 +-V3. 6. 17 by 8V7+2-V8... 115 V8 + V7 by eee 6. 1 by V2+ V3, 12, 7—3V10 by 5+4¥V5. RADICAL EXPRESSIONS. 937 -Given V2=1.41421, V8 =1.73205, V5 =2.23607; find to four places of decimals the value of 13. 14. 15. eLO «16. fab 19. mite 22. {crest wes V2 V500 8/2 5+ 4/5 Pen a) = og, BE VS. V3 V/ 2438 V/125 x/5 73 12 18. 1 = 21. ahs 24. BV2—1 mak V5 23 4/5 3V2+41 bel INVOLUTION AND EVOLUTION OF RADICALS. 256. Any power or root of a radical is easily found by using fractional exponents. i. 2. (1) Find the square of 2Va. (2V a) = (2a8)? = 2a? = 408 = 4V a. (2) Find the cube of 2-Va. (2Va)3 = (2.02) = 2a? = 8a? = 8ava. (3) Find the square root of 4a-V ad’. (4. -V 0363)? = (4. 7a2b2)3 = 42 ett? = 43 atadd = OV aOR, (4) Find the cube root of 4a-Va'd'. (4 @-Va3b3)3 = (4 wa2h?)8 = 43 aha 3ht = 46 aba8h8 — V/16 abe2?. Exercise 88. Perform the operations indicated : ao oe rt Pepanene eaci (Vm). 8. (Wat). 5. VV ye (Vm 4. (Vy, 6. Va 5*. 238 SCHOOL ALGEBRA. Teh Bebo yien 10. Aas 13. \/Ga— 2b)", fb DE br 2 4 ee af. aoe 8. (Vaer—y) 11. VV729. 14. V*/320%. 9. (Wa). 12. \V158. 15. V128-/ 248 a", PROPERTIES OF QUADRATIC SURDS. 257. The product or quotient of two dissimilar quadratic surds will be a quadratic surd. Thus, Vab x Vabe = abvVe; Vabe+ Vab = Ve. For every quadratic surd, when simplified, will have under the radical sign one or more factors raised only to the first power; and two surds which are dissvmilar cannot have adl these factors alike. Hence, their product or quotient will have at least one factor raised only to the first power, and will therefore be a surd. 258. The sum or difference of two dissimilar quadratic surds cannot be a rational number, nor can tt be expressed as a single surd. For if Va+ V6 could equal a rational number c, we should have, by squaring, at2Vab+b=e'; that is, +2\/ db Or Oe. Now, as the right side of this equation is rational, the left side would be rational; but, by § 257, Vab cannot be rational. Therefore, Va+ Vé cannot be rational. In like manner, it may be shown that Va + Vé cannot be expressed as a single surd Ve. RADICAL EXPRESSIONS. 939 259, A quadratic surd cannot equal the sum of a rational number and a surd. For if Va could equal ¢-+ Vb, we should have, by squaring, a=e+2cVWb+, and, by transposing, 9eVb =a—b—e*. That is, a surd equal to a rational number, which is impossible. 260. Tf at+Vb=2+Vy, then a will equal x, and b will equal y. For, by transposing, Vb —Vy=2—a; and if 5 were not equal to y, the difference of two unequal surds would be rational, which by § 258 is impossible. pe a) enor 2. In like manner, if a— Vo=x2— Vy, a will equal z, and 6 will equal y. 261. To extract the square root of a binomial surd, Ex. Extract the square root of a+ Vo. Suppose Va+Vb=Vi+ Vy. (1) By squaring, a+ Vbsa4+2Vay +y. (2) .a=a+y and Vb =2V2y. 3 260 Therefore, a—Vb=x-—2Viy+y, (3) and Va—vVb=vVz— Vy. (4) Multiplying (1) by (4), V@e—b=2-y. But a=x2+y. 240 SCHOOL ALGEBRA. Adding, and dividing by 2, «= 2+ V¢=% vat Subtracting, and dividing by 2, a—vVa?—6b = ; ; : an Vetveas a|a— Va? —8, “ Va+vo POL me 5 From these two values of xz and y, it is evident that this method is practicable only when Bee b is a perfect square. (1) Extract the square root of 7+4-V3. Let Va + Vy =V7+4V3. Then Vi —-Vy =V7—4v3. Multiplying, e—y = V49 — 48. “ £—y= 1. But et+y=7. *. w©=4, and y=3. o Ver Vy =24 v3. 0 V744V8=24 V3. A root may often be obtained by inspection. For this purpose, write the given expression in the form a + 2V, and determine what two numbers have their sum equal to a, and their product equal to 0. (2) Find by inspection the square root of 75 —12-V21. It is necessary that the coefficient of the surd be 2; therefore, 75 —12V21 must be put in the form 75 —2V 756. The two numbers whose sum is 75 and whose product is 756 are 63 and 12. Then 15 — 2756 = 63 + 12 — 2V63 x 12, = (vV63 ~— V12)%. That is, V63 — V12 = square root of 75 —12V21; or, 3V7 — 2V3 = square root of 75—12V21. RADICAL EXPRESSIONS. 941 Exercise 89. Find the square root of ey eae Bee 7. 16-2 5v7. 13. 94-4. 49/5. See ee / 72. Be Yo on/ et 14. 1T — oN /80 3. 7-42-10. IS 8-7/3: 15. 4744/38. 4. 18+ 8V5. LOSS Sy/6'-— 90. 16.45.29 2)- 60/22, 5. 8412/15. rH Gass a Lotaw © ems Wet oe ee EET, Geto 4/14, 12. 61+ 36V2.. 18. 65— 12-21. EQUATIONS CONTAINING RADICALS. 262. An equation containing a single radical may be solved by arranging the terms so as to have the radical alone on one side, and then raising both sides to a power corresponding to the order of the radical. Ex. V27—9+2=9. Vz?—9=9—-c. By squaring, oe? —9 = 81 — 182 + 2. 18% = 90. Shes oe 263. If two radicals are involved, two steps may be necessary. Ex. Vz+15+~V2=15. Va +15 + Va = 15. Squaring and simplifying, we have Va? + 152 = 105 — 2. Squaring, we have a? 4+ 15a = 11025 — 210 + a7, 225 2 = 11025. “= 49, 24 19. 20. oy SCHOOL ALGEBRA. Exercise 90. Solve: . 2V2+5 = V8. 8. V82+7=8. . 8V4e—8=V182—3. 9. 144+V4e2—40=10. . Ve+9=5V2 —38. 10. Vl0y—4=Vi7y+11. Aone SS. 11. 2V2—2=V82e= oy Vb VBy 4. 12. Vip +2=84-Vc. .T+2V8a=5. 13. V32+2=16—-V*r. . V22—8=—8. 14. Ve—VeH5=V5. 15. V2z+20—Vx—1—8=0. 16. Ve+15—T=7—V2—18. 17) city en 18. Vx—7T=V2e+1-2. Ve—8 Ve+1 we 1+(L—2)t _ Ve$3 Ve 8 1-0 23. Va+Ve+Va— Vi= Vo. 24. Var —-1=44+1Var—}l. 25. 8V2 —8Va= Va—Va+ 2Va. Oa V9+22 26. V9+22—V2e= CHAPTER XVII. IMAGINARY EXPRESSIONS. 264, An imaginary expression is any expression which involves the indicated even root of a negative number. It will be shown hereafter that any indicated even root of a negative number may be made to assume a form which involves only an indicated square root of a negative num- ber. In considering imaginary expressions, we accordingly need consider only expressions which involve the indicated square roots of negative numbers. Imaginary expressions are also called imaginary numbers and complex numbers. In distinction from imaginary num- bers, all other numbers are called real numbers. 265. Imaginary Square Roots. Ifa and 6 are both posi- tive, we have ee cy Tl (van. If one of the two numbers a and 34 is positive and the other negative, law I. is asswmed still to apply ; we have, accordingly : Re 1 Nie V1 = By = NAG 5 GaliawWex nol = biv-aL: Rats /a ESS AN ab and so on. It appears, then, that every imaginary square root can be made to assume the form aWV—1, where a is a real number. 244 SCHOOL ALGEBRA. 266. The symbol V—1 is called the imaginary unit, and may be defined as an expression the square of which is —1. Hence, -V2 1x V— t= (V1 = V—ax V—6= Vax V—-1x Vb x V=1 = Vax Vb x (V—1) = Vab x (—1) =— Vab. 267. It will be useful to form the successive powers of the imaginary unit. (VD) eee es ee eeow eh lee ek kc (VE TP EE AO Oe i (V—1)' = (V=1)' V=1 = (-1) V-1=- v1; (Val) = (VET (VAI =e (V—)D§= (V— 1 V-1 = 41) V—-1=4+v-1; and soon. We have, therefore, (Vr al ee (v= Ty" = — @/aDtt = =v (V—1)"#=+1. 268. Every nn expression may be made to assume the form a+6V—1, where a and 8 are real numbers, and may be integers, Sera or surds. If 6=O, the expression consists of only the real part a, and is therefore real. If a=0, the expression consists of only the imaginary part 6-V—1, and is called a pure imaginary. IMAGINARY EXPRESSIONS. 245 269. The form a+b~V—1 is the typical form of imaginary expressions. Reduce to the typical form 6+ V—8. This may be written 6+V8xV—l, or G EON Deena: here a= 6, and b=2v2. 270. Two expressions of the form a+ Ves Diab are called conjugate imaginaries. To find the sum and product of two conjugate imagi- naries, Ge bv 1 ey The sum is 2a Get bee GaN aL a abv — 1 —abV—14 2? The product is a + 0? From the above it appears that the swm and product of two conjugate imaginaries are both real. 271. An imaginary expression cannot be equal to a real number. For, if possible, let atbV—l=e. Then transposing a, 6V—1=c—a, and squaring, — 6? =(e—ay)’. Since 4? and (e—a)* are both positive, we have a nega- tive number equal to a positive number, which is impossible. 946 SCHOOL ALGEBRA. 272. If two vmaginary expressions are equal, the real parts are equal and the imaginary parts are equal. For, let a+bV—l=c+dv-1. Then (b—d)V—1l=c—a; squaring, —(b—d)=(e—ay, which is impossible unless 6 =d and a=c. 278, If x and y are real and x+-yV—1=0, then x=0 and y=0. For, yV—1=—2, a y = 2 a + y = 0, which is true only when x=0 and y=0. 274, Operations with Imaginaries. (1) Add 5+ 7V—1 and 8— 9V—1. The sum is Ba Br ed a ede or RET (2) Multiply 8+ 2V—1 by 5—4V—1. (S+2V—NG6—4v eu = 15 —12V—1+4 10V—1-8(—1) O82 i (3) Divide 14+ 5V—1 by 2—8vV—1. 144+5V—1_ (14+5vV—1)(2+3v-1) 2-3V-1 (2—38V—1)(2+3v—]) _13+52V-1 Ears beet _1384+52V—1 13 =144V-1. IMAGINARY EXPRESSIONS. 247 Exercise 914. Reduce to the form 6V—1: ty V9. 9. V— 625. Lye 2. V—16. 10. V— 36. 18. V= i. SPN 25. Wi = G4. Liab eee 144 tO 20 20g Oe pee 169, 13. */—289. 21. V—(22— By)". 6. V—2. TAN LO PA N/a 7. V—8l. 15. > 23. V—(a?+y*). 8. y/— 256. 16. V—2a°. 24. l= (4), Add: 25. V— 25+ -V— 49 — V— 121. age \/ = 64-- V/= 1 — \/— 86. a7, V/— at | V—4a' + V— 16a+. 98.) -/—a? + V—-81a? — V— a’. 29. a—bV—1+a4+5vV—1. 30. 24+38V—1—24+8V—1. 31. d+ b6V—1+e—dv—1. 32. 8aV—1—(2a—8)V—1. Multiply : 33. V—8 by V—5. 36. V—2? by V—z. 34. —V/—5 by V—5. 87. V—a by V—7. ope — 16 by V— 9. 38. V—8 by V—I6. 248 SCHOOL ALGEBRA. 39. V—25 by V— 64. 41. 8V—8 by 2V—2. 40. V—(a+d) by V—(a—6). 42. —5bV 220 by ox 43, A/S 2 Biby: Ve eee 44, pu la as by Pra ssh ss 2 2 45. aV=0+8-V=) by 4 ee 46. 2V—2-48V—3 by 8V—4 = ae 47. V8+2V—8 by V8 —2V—8. 48. m—38V—b by n+4V—ce. Perform the divisions indicated : Ao ee Sheen 4. URGE Vos 38—V—2 50. ats 5G ee 62. 2+V—=2 V— 6 V2 1—V-1 Bite oie 57, Ma. 63, 2a ae V4 —V—2 a—aV—1 py eee ss. Y—8. 64, Oc Meee v—8l vV—2 V6—V—8 53 melts 59. Vv— 102 65. 2a+3bV—1 V =a V—b V—52 2a 8b 54, = ae 60. SV . 66. ta—4bV—1 Va OG 4a—LbV—1 CHAPTER XVIII. QUADRATIC EQUATIONS. 275. We have already considered equations of the first degree in one or more unknowns. We pass now to the treatment of equations containing one or more unknowns to a degree not exceeding the second. An equation which contains the square of the unknown, but no higher power, is called a quadratic equation. 276. A quadratic equation which involves but one un- known number can contain only: (1) Terms involving the square of the unknown number. (2) Terms involving the first power of the unknown number. (3) Terms which do not involve the unknown number. Collecting similar terms, every quadratic equation can be made to assume the form ax’ + bza+e=0, where a, 6, and c are known numbers, and x the unknown number. If a, 6, ¢ are numbers expressed by figures, the equation is a numerical quadratic. If a, 6, c are numbers represented wholly or in part by letters, the equation is a literal quadratic. 277. In the equation aa*+ bx+e¢=0, a, 6, and ¢ are called the coefficients of the equation. The third term ¢ is called the constant term, 250 SCHOOL ALGEBRA. If the first power of x is wanting, the equation is a pure quadratic; in this case 6 = 0, If the first power of x is present, the equation is an affected or complete quadratic, PURE QUADRATIC EQUATIONS. 278, Examples, (1) Solve the equation 52’— 48 = 22”, \ We have 5? —48 = 222, Collect the terms, 3 a? = 48, Divide by 3, ac} zo 16, Extract the square root, v= +4, It will be observed that there are two roots, and that these are numerically equal, but of opposite signs. There can be only two roots, since any number has only two square roots. It may seem as though we ought to write the sign + before the as well as before the 4. If we do this, we have + x=+4, —z=—4, +e=—4 —2=4+4. From the first and second equations, 7=4; from the third and fourth, e=—4; these values of w are both given by the equation w=+4, Hence it is unnecessary to write the + sign on both sides of the reduced equation. (2) Solve the equation 32?— 15=0. We have 3 27 = 15, or mies 5, 2 Extract the square root, r=tvb, The roots cannot be found exactly, since the square root of 5 can- not be found exactly; it can, however, be determined approximately to any required degree of accuracy; for example, it lies between 2.23606 and 2.23607. (3) Solve the equation 32°+ 15=—0. We have 32? = — 15, or n= — HONE 4 Extract the square root, a=+V—5, QUADRATIC EQUATIONS. ZL There is no square root of a negative number, since the square of any number, positive or negative, is necessarily positive. The square root of —5 differs from the square root of +5 in that the latter can be found as accurately as we please, while the former cannot be found at all. 279. A root which can be found exactly is called an exact or rational root. Such roots are either whole numbers or fractions. A root which is indicated but can be found only approx- imately is called a surd or irrational root. Such roots involve the roots of imperfect powers. Rational and surd roots are together called real roots. A root which is indicated but cannot be found, either exactly or approximately, is called an imaginary root. Such roots involve the even roots of negative numbers. Spe dare ee le Exercise 92. Solve: 827—2=—2°+ 6. 9. Sa F Ete _s, 5a? +10=62'7+1. Ta? — 50 = 4a? 4 25. 10, O@ +8 li-2#_, 62? -—1L=—42'+4 4. : ° : wobihae toa 5, ett=10. Taek OA BE B 5 3 4 327°—8 1 ig cata Spe aly 6. 10 a4. Fs Be rie eat 9 gy ee . 7 @a9_ vt ia tpi aed & meee ee BR zs—-l «+1 4 8. 5 ie ie 9 14. PhO pol als At, 252 15. 16. 17. 18, 19. 20. 21. 22. 23. 24. 25. 29. 30. 280. Since (vx + 0? = 2? + 2bx+ 6’, it is evident that the expression x? + 26x lacks only the third term, 6’, of SCHOOL ALGEBRA. 82°+1ll2a=1074+8+2?+2. (w+ 4)(@+5)=3(@+1)(@+2)—4. 8(a—2)(e+8)=(e@+1)(e+2) 4245. (22 +1)(84—2)+(1—2)(84+ 42) =327 — 15. +9 a'—5 , 82?+10 Tak eeremsnarars = Of DAE Ae pe Oe . i 9 ay ==), LO ete Ta2k12 eon Die er: 18 Lior Deg gee lee en 26; 04 @+1l a“—-1 2 i al av’?+b=c. 27 tte, oa z—a' @2ta 2B 26 5at26 28... — + —_ = — v+2bx+e=b(2x2+1). aE 32 25 (x + a)(x+ 6)+ (w@—a)(@ — d)t} =a? +48. azvtt-b6 = ba? +a. 2}(x— a) (a+ 6)+ («@+a)(x—b)} =9a? + 2ab+ B AFFECTED QUADRATIC EQUATIONS. being a perfect square. This third term is the square of half the coefficient of 2. Every affected quadratic may be made to assume the form a +. 2b2 =c, by dividing the equation through by the co- efficient of x”. QUADRATIC EQUATIONS. = 0 To solve such an equation : The first step is to add to both members the square of half the coefficient of x. This is called completing the square. The second step is to extract the square root of each mem- ber of the resulting equation. . The third step is to reduce the two resulting simple equations. (1) Solve the equation 2? — 8x = 20. We have a? — 8x = 20, Complete the square, #?— 8a + 16 = 36. Extract the square root, x—-4=+6., Reduce, x=4+6=10, or a=4—6=—2., The roots are 10 and — 2. % Verify by putting these numbers for # in the given equation x= 10, e=— 2, 10? — 8 (10) = 20, (—2)'—8(—2) = 20, 100— 80 = 20, 4+ 16 = 20. (2) Solve the equation = Dhl Aic eeee zx—l e2«+9 Free from fractions, (w + 1)(# + 9) =(#—1) (4a —3). Simplify, OPS 1T eG, We can reduce the equation to the form x? — 2b by dividing by 3. Divide by 3, ow — lig = 2, Half the coefficient of x is 4 of — 1,7 = — 4%, and the square of — 4! is 282, Add the square of —1Z to both sides, and we have Lix ee 289 pe hE oud DO) ieee Fae cae Tas (a) + 36 L7\2 361 2 Jee Se or x — 1 =+(7) a Extract the root, Sele + 19 254 SCHOOL ALGEBRA. Reduce, pps es 6 6 i oti tn done O6 ae 6 6 6 17°19 2 1 or 1 Negi Se —— =e = — The roots are 6 and 3 Verify by putting these numbers for « in the original equation : iy == 6. Pee. 6+1_ 24-3 3 SEE Te lo 7 _ 21 ae those l cat _i 9 3 37 2 4 26 Exercise 93. Solve: Lee Dee. 3. 5. 2? + 5e= 14. 9. # +22 =40, 2. 7—624=7. 6. 2 —38x= 28. 10. 82°—42=4. Br A La ee ees ll. 62° -- eae 4, 2°+4x2=5. 8. 52° + 8% =2: 12. 627:->@ ae 18. 1227;—1llz+2=0. 14. 1527—2x—1=0. 15. (7 +1 G42) _ Ga DG~)—s 16, 2% 3)e_ (@+4)@—1)_ 47 4 6 QUADRATIC EQUATIONS. 255 ox+5 , 2a—5 et+2 4-—a“_7 17. = 8, 7 fi tn la a eaten 6-4. : E> 2 el Qa 63 Pee FO _ 1. 93, Tr8_oe+8 a—2 22+1 x—-2 -2+4 19. pret 1+ ae _ 9. 24. LEE tile sao Ait l—x 2 3 2x2—1 x z+1. 18 gt+1,2#2+2 18 20. ——- + —— = —. 25. — oo peela x 6 Soe a 6 Sloe —4¢=— 1. 26. T2?—82=—1. ANOTHER METHOD oF CoMPLETING THE SQUARE. 281. When the coefficient of «? is not unity, we may proceed as in the preceding section, or we may complete the square by another method. Since (aa + 6)? = aa? + 2abx + 6’, it is evident that the expression a7? + 2abzx lacks only the third term, b’, of being a complete square. It will be seen that this third term is the square of the quotient obtained from dividing the second term by twice the square root of the first term. 282. Every affected quadratic may be made to assume the form of aa’? + 2abx =e. To solve such an equation: The first step is to complete the square; that is, to add to each side the square of the quotient obtained from dividing the second term by twice the square root of the first term. The second step is to extract the square root of each side of the resulting equation. The third and last step is to reduce the two resulting simple equations. 9256 SCHOOL ALGEBRA. (1) Solve the equation 162’ + 52—3=72?—2x-+ 46. We have 1627 +5¢%—3=72?—2 +46." Simplify, 92? +62= 48. To complete the square, take the square root of 947. This is 32, and twice 3x is 6a. Divide the second term by 6a, and the quotient is 1. Square 1 and add it to both sides. 9a? +64%+4+1=49. Extract the square root, 3e+1=+7. Reduce, 3¢=—1+7o0r—1—-7, 32 =6 or —8. *, @©= 2 or — 22, Verify by substituting 2 for x in the equation: 162? + 54¢—3=72? —2 + 45, 16 (2)? + 5 (2) — 3 = 7 (2)? — (2) + 45, 64+10—3 = 28 —2+ 45, ih il Verify by substituting — 22 for # in the equation: 1647 ++5e¢—3=72? —2 + 45, 16(—3) + o( =F) se 7(=5)= (=3) 445; 1024 40 448 8 on eee Lee oe Se aed ea 1024 — 120 — 27 = 448 + 24 + 405, 877 — B77. (2) Solve the equation 32?—42= 82. Since the exact root of 3, the coefficient of 2?, cannot be found, it is necessary to multiply or divide each term of the equation by 3 to make the coefficient of «? a square number. Multiply by 3, 9a? — 12% = 96. Complete the square, 9a?-—124+4+4=100. Extract the square root, 3a—2=+ 10. Reduce, 32=2+10 or 2—10; 3% = 12 or —8. *, ©=4 or — 22, - QUADRATIC EQUATIONS. 257 Or, divide by 3, Complete the square, 2? = O53 oer oe: Extract the square root, «— : =+ : 2+10 c= ’ 3 = 4 or — 22. Verify by substituting 4 for x in the original equation : 48 — 16 = 32, 32 = 32. Verify by substituting — 22 for x in the original equation: 214 — (— 102) = 32, 32 = 32. (3) Solve the equation — 327+ 54 = — 2. Since the even root of a negative number is impossible, it is neces- sary to change the sign of each term. The resulting equation is 3.0% — 5x = 2. Multiply by 3, 9a? — 15% =6. Complete the square, Sifted 5 7 Extract the square root, 3a— ae Reduce, 352 ; 7 32=6 or —1 x = 2 or —= Or, divide by 3, a2 — “ = 2 Complete the square, 2 — 5a ve 25 ai 49 258 SCHOOL ALGEBRA. Extract the square root, «— : =+ z pee we 6 = 2or— L 3 284, If the equation 32?— 5x = 2 be multiplied by four tumes the coefficient of x”, fractions will be avoided. We have 36.2? — 60a = 24, Complete the square, 36 a? — 604 + 25 = 49. Extract the square root, 6a —5=+ 7. 62=5+4 7. 62=12 or —2. # ete en 3 It will be observed that the number added to complete the square by this last method is the square of the coefficient of x in the original equation 327 — 52 = 2. Nore. If the coefficient of x is an even number, we may multiply by the coefficient of «?, and add to each member the square of half the coefficient of x in the given equation. (1) Solve the equation 42?— 23% = — 30. Multiply by four times the coefficient of 2, and add to each side the square of the coefficient of a, 64 2? —() + (23)? = 529 — 480 = 49. Extract the square root, 8#—23 = + 7. Reduce, 8a=2347; 8a = 30 or 16. *, e = 32 or 2. Nors. Ifa trinomial is a perfect square, its root is found by taking the roots of the first and third terms and connecting them by the sign of the middle term. It is not necessary, therefore, in completing the square, to write the middle term, but its place may be indicated as in this example. QUADRATIC EQUATIONS. 259 (2) Solve the equation 722? — 30% = — 7, Since 72 = 2° x 3?, if the equation is multiplied by 2, the coefii- cient of x? in the resulting equation, 1442?— 60a =— 14, will be a square number, and the term required to complete the square will be ( syle Ge =>. Hence, if the original equation is multiplied by 4 x 2, the coefficient of x? in the result will be a square number, and fractions will be avoided in the work. We shall then have 576 2? — 2402 = — 56. “, 5762? —() +25 =—81. Extract the root, WMge5=tvV—3l, w= 3 (5+ V— 831). Exercise 94. Solve: omer — 6. 14. Ba? +22 = 26, ooo. — os = 27. 3 2 =a ese a, — 5 157 a =p ae ee 4. 2a? —5a=7. 16, 2 —2—2(e—2) 5, o2'e-iz—6 ip is > x 6. 5a?—Tx = 9A ht ete tora if 827+ 32 = 26. 18. OO to a3 say k's 16 8. 12° +52 =150. 20 19. 32°+ 32 =—- 9. 67+57=14. att 10. 72°24 =$. Coie a Preor + iz= 51. 21. wo? xv aie ons 13677 — 202 = 75. ob . (8+ 210 13. llz?—10x2 = 24. mini ions ae 260 SCHOOL ALGEBRA. 23. («+ 2)(2¢+ 1): (Cos) (Ben eee 24. 3x2(22+5)—(#+ 38)(84%—1)=1. ob. Cet eT) eCe FD 5, 26. 2(52° —8x—6)—4 (2? — 3) =22+1. ites eee, $0, +3 2 x—-l «#-—-2 x«—-—4 Beha e ste eee 31, tt 2 ee x—l 2x4+1 838 x—-4 2-2 7 + 24—3 , 5—38-2 29. ———. + ———_= 9. oak mes be Te OW Om ee ; 4 33. 11— 8a, 2744) _ 1 l—=z 1—22 etl, l—#@ 2 34. == . — 4-208 5 (a — 2) 35. eer A z+] of AE 3 Sled ee ek eee " 9Qe—1) Q2e+l) 4—82 22—1 3 x—2 377. ep eet 5 A 3 x—8 peer 38. 382+2 Ca el 2a—l 2¢4+1 427-1 —5 ,x2-—8 80 1 39) = = xt+3 «2-8 e912 40. Zepl, ser) _ 45 eds 1—2 T+tae 49-2 QUADRATIC EQUATIONS. 261 LITERAL QUADRATICS. 285. Examples. (1) Solve the equation az’+ br+e=0. Transpose ¢, ax* + be=— ce. Multiply the equation by 4a and add the square of 8, 47a? +()+ 0 =0? —4ac. Extract the root, 2ax+b=+ Vb? —4ac. ° eae VE Aue a 2a (2) Solve the equation (a? + 1)2 = az? +a. Transpose aa? and change the sign, ax* — (a? + 1)ex=—a. Multiply by 4a, and complete the square, 4 atx? —()+(a?+1P?=—4a? +at+ 20? +41 = at — 2a? +1. Extract the root, 2a%—(a?+1)= + (a?—1). Reduce, 2aux = (a? + 1) + (a? — 1), = 2a? or 2. : 1 . £=¢@ or —- a (3) Solve the equation adx — aca? = bex — bd. Transpose bca and change the signs, acax* + bex — adx = bd. Express the left member in two terms, acu? + (be — ad) a = bd. Multiply by 4ac, and complete the square, Aatcta? + () + (be— ad)? = bc? + 2abced + a?d?, Extract the root, 2acx + (be — ad) = + (be + ad). Reduce, 2acuw = — (be — ad) + (be + ad) = 2ad or — 2be. here! Ret . == or—-=: c a 262 SCHOOL ALGEBRA. (4) Solve the equation pa? — pet qu + gx pe Express the left member in two terms, (p + 9)2?—(p—g)e=— Multiply by four times the coefficient of 2, 4(p + qa? — 4(p* — q°) «= 4g. Complete the square, 4(p + gPat—() + (p— 9) =p + 2pg + g. Extract the root, 2(p -+.9)2—(p—g)—=(p a gy Reduce, 2(p+qe=(p—gQ+(pt+q), = 2p or — 2g. giant. Se P+ Pt+q Notre. The left-hand member of the equation when simplified must be expressed in two terms, simple or compound, one term con-. taining «?, and the other term containing a. Exercise 95. Solve: 1. a7 -+ 2ar=3a'. 9. 202? + ax —1=0. Ott A = at 10. 120°x?— 5brx4=3. 3. a?+8be = 98%. 11. 22 4 211 a(v—8a). re ete 2 4. 2°+3b2 =100'. #5 Bat eee 5. 27+ 5axr = 140’. pes ie ey ait, 6. 827+ 4er = 4’. 13. yi & 8S a 4@ : — 227? = 2a’. Teor x a Sae* , 20 _ 18 8. 62°?—ar—av=0. i et 3 a 2 2 ON Te 15. epee + SHE Hae, QUADRATIC EQUATIONS. 263 19. 20. —— 21. ——— 37. 38. 39. 40. 29. 30. 31. 32. 33. 34. 35. 36. ot, e+ a= 2a’, 22. 2+ (a—b)x=ab. peg) e _-. os 2G, Oa 8 ataz 3 2a—x at+2uz 8 a(de—a)_ a oA 22—3a , 38a+2a_10 + ae 12 apc 4z—a 7 ee «gg Bong wer mn icra ' 68? Gab at xta Bate 5 26. Ga Be ay bus b—a ta atbtez «+6 a+2b,a+6b_5 1 [ieee Ce | = PW ie =-+—+-. | ay atb4-z2 Pee va 28. Pet a Oat. 9a? — 8(a42b)¢4+ 2ab=0. (2a+1)27?+3a2+a—a’=0. fea 2(1+a’)x+1l—a’=0. (a+ 6)? — (a? — 0°) x= ab. ae? + 2bert a= cx? +2a%r+ 0. (a+ 6)x?—(2a+6)r4+a=0. 2e—36 _ a—26 DO Ly RW z2+2a 2(a—b) x—b (8a? + 0?) (2? —x+1)=(?+380")(¢’+2+1). 2 —(a+ b)x+ac+ be—e?=0. a —p2=(p+q+r)(q+7). (a? + ab) (2? —1)—-(¢+0’)2+(a+b)(a—26)=0. 264 SCHOOL ALGEBRA. SOLUTIONS BY FACTORING. 286. A quadratic which has been reduced to its simplest form, and has all its terms written on one side, may often have that side resolved by znspection into factors. In this case, the roots are seen at once without com- pleting the square. (1) Solve 27+ 7%—60=0. Since a* + 72 — 60 = (a + 12) (a —5), the equation x? + 7a—60=0 may be written (x + 12)(a—5)=0. If either of the factors x +12 or «—85 is 0, the product of the two factors is 0, and the equation is satisfied. Hence, x+12=0, and r—5=0, * e=—12, and «=5., (2) Solve 2a°—2?—6x=0. The equation 223 — a? -—62=0 becomes x (2a? —a«2—6)=0, and is satisfied if «#=0, or if 2227—-x%—6=0, By solving 2x? — x —6=0, the two roots 2 and — 3 are found. Hence the equation has three roots, 0, 2, — . (3) Solve a+ 2?—42 —4=0. The equation oe +a2—42-—4=0 becomes a (a +1)—4(e@ + 1)=0, (a? ~ 4)(a + 1) =0. Hence the roots of the equation are —1, 2, — 2. (4) Solve 2° — 22?—11lz+12=0. . By trial we find that 1 satisfies the equation, and is therefore a root (¢ 89). Divide by «—1; the given equation may be written (a — 1) (x? — x —12)=0, and is satisfied if e —1=0, or if 2?—2—-12=0, The roots are found to be 1, 4, —3. QUADRATIC EQUATIONS. 265 (6) Solve the equation x(a — 9) =a(a@ — 9). If we put a for x, the equation is satisfied; therefore a is a root (¢ 89). Transpose all the terms to the left-hand member and divide by a. ; The given equation may be written (aw — a) (a? + ax + a? — 9) =0, and is satisfied if s—a =0, or if 227+ ax+a?—9=0. The roots are found to be ices V36—3a? —a— V36—3a? 2 2 Exercise 96. Find all the roots of 1. 2#—5xe+4=0. 5. a +2°7—62=0. 2. 627—52—6=0. 6. 2 —8=0. 8. 227?—zx—38=0. eae 8 = 0: 4. 102°?+2—3=0. 8. t 160; 9. («—1)(x—38)(#’+52x+6)=0. 10. (2% —1)(«#— 2) (82°—52—2)=0. 11. (a? +2 — 2)(22?+32—5) =0. 12. 2 +a?—4(7+1)=0. 13. 32°+4 22?— (82+ 2) =0. 14. 2° —27—182+39=0. 15. 2484 3(2?—4)=0. 17. 22°—22°—(2’?—1)=0. 16. x(2?—1)—6(a—1)=0. 18. ae —3x—2=0. 19. 2a°+ 227+ (2? —5x2—6)=0. 20. 2*—42°—2?+162—-12=0. 266 SCHOOL ALGEBRA. SoLUTIONS BY A FORMULA. 287. Every quadratic equation can be made to assume the form az’?+ ba+ce=0. Solving this equation (§ 285, Ex. 1), we obtain for its two roots —b1ivVb—4ac —b= Vea 4ac 2a 2a There are two roots, and but two roots, since there are two, and but two, square roots of the expression 6’— 4ac. By this formula, the values of # in an equation of the form az’-+ bz -+-¢=0 may be written at once. Ex. Find the roots of the equation 8z?—5%+2=0. Here a=8, b=—5, c=2. Putting these values for the letters in the above formulas, we have _5+V25—24 9. 5— V25—24 6 6 = or 4 =1 or 2. Fixercise 97. Solve by the above formulas: 1,22) oa= 14, 7 52—Ta=—2. 2.00 — ox = 12, 8. 42° — 9x = 28. Stag ie 38: 9. 5a? -+ fa male 4. 5a?—2 = 42, 10. Mat —92 = 5. 62°—T7x2=10. 11. 72?+52=88. 6. 3a7—lla=—6. 12.5379 Tan QUADRATIC EQUATIONS. 267 EQUATIONS IN THE QUADRATIC Form. 288, An equation is in the quadratic form if it contains but two powers of the unknown number, and the exponent of one power is exactly twice that of the other power. 289, Equations not of the second degree, but of the quadratic form, may be solved by completing the square. (1) Solve: 82° + 632° = 8. This equation is in the quadratic form if we regard a? as the un- known number. We have 8 x8 + 63 a3 = 8. Multiply by 32 and complete the square, 256 a® + ( ) + (63)? = 4225. Extract the square root, 1623 +63 = +65, Hence, a8 = : or — 8. Extracting the cube root, two values of w are } and —2. There are four other values of « which we do not find at present. (2) Solve: Va 28 4/ a = 40. Using fractional exponents, we have a? — 3a = 40, This equation is in the quadratic form if we regard a as the un- known number. Complete the square, 4a?—12a3 +9 = 169. Extract the root, G4 3 = 13, 2x% = 16 or — 10, re a8 of 6, 0 x = 16 or —5 V5. There are other values of « which we do not find at present. 268 SCHOOL ALGEBRA. (3) Solve completely the equation z*°= 1. We have a —1=0. Factoring, («—1)(#+a+1)=0. Therefore, either z—-1=0 or alee or a oa Ge z—1=0. aw? +a2+1=0, owen ld, eae ey : Solving, #= aw. The three values, 1, alts ge rt haat are the three cube roots of 1. et (4) Solve: (24% — 3) — (2% — 38) = 6. Put y for 2% — 3, and therefore y? for (2% — 3). We have y?—y =6. Solving, y =3 or —2. Putting now 2a—8 for y, 2%—3 = 3, 24—3=— 2, x= 3. v=}. Exercise 98. 1. af —5277+4=0. 6. 10z*—2L=a7 2. 2'— 1827+ 36=0. 7. Vei+8Ve= 18 3. 2 — 212? =100. 8. 8Vx2—2Vx = — 20. 4. 42°— 327° = 27. 9. 54"+32"= 68. 5. Qat+ 5a? = 218, 10. (82+3)*+ (8x48) =380. 11. 2(2?—2+1)—V2?—-2+1=1. 12. 2°—92°+8=0. 14. (e+1)4+ Vat TG: af el 13. «+2 =e 15. 2*— 182? =— 36, 16. 2274+4274+94 8V22?4+ 44+9=40. 122° — 1127+ 102 — 78 1 17. mee ee 82° —Ta+6 ag QUADRATIC EQUATIONS. 26 CO RADICAL EQUATIONS. 290. If an equation involves radical expressions, we first clear of radicals as follows: Solve V2 +44 V22+6=Vizc+l4. Square both sides, e+4+2V(e+4j(22+6)+204+6=7x7 +14. Transpose and combine, 2V(«#+4)(2a¢+6)=4a+4+4. Divide by 2 and square, (7 + 4)(2” + 6) = (24+ 2)% Multiply out and reduce, x? — 3a = 10. Hence, x=5or—2. Of these two values, only 5 will satisfy the original equation. The value — 2 will satisfy the equation Vi+4—V20+6=Vi0+ 14. In fact, squaring both members of the original equation is equiva- lent to transposing \/7z + 14 to the left member, and then multiplying by the rationalizing factor Vz +44+ V22+6+ V7 +14, so that the equation stands (Ve +44+ V204+6—V704+14)(V24+44 V204+6+4+V70+4+14)=0, which reduces to V(a + 4)(2a + 6)—(2a + 2)=0. Transposing and squaring again is equivalent to multiplying by (Va +4~—V20+6 4+ V724+14)(ve+4-— V2046~Vi724 14). Multiplying out and reducing, we have x? —3x%—10=0. Therefore, the equation a? —3«2—10=0 is really obtained from (Va +44+ V204+6-—Vi72+14) x (V2 +44 V204+6+4+ V7a2 414) x (Wa +4-— V2e4+6— Via +14) x (Vot4d— Vie tb + Vin +14) =0. This equation is satisfied by any value that will satisfy any one of the four factors of its left member. The first factor is satisfied by 5, and the last factor by — 2, while no values can be found to satisfy the second or third factor, 270 SCHOOL ALGEBRA. Hence, if a radical equation of this form is proposed for solution, if there is a value of # that will satisfy the particular equation given, that value must be retained, and any value that does not satisfy the equation given must be rejected. (See Wentworth, McLellan and Glashan’s Algebraic Analysis, pp. 278-281.) 291, Some radical equations may be solved as follows: Solve 72? —5274+8V727—57+1=—8. Add 1 to both sides, "et? —5a+148V7e—5e24+1=—7. Put V7a?—5a+1=y; the equation becomes y+ 8y=—7. Hence, y=—lor—7, y? = 1 or 49. We now have 72?—52+4+1=1, or 72?—524+1=49. Solving these, we find for the values of a, Baht at 7 7 These values all satisfy the given equation when we take tle negative value of the square root of the expression 727—5a+1; they are in fact the four roots of the biquadratic obtained by clear- ing the given equation of radicals. 0 Exercise 99. Solve: 1. V92+404+ 2Vet+7= V2. 2. V Gee Va—x2= vb. 82+ V4r—2" _» 8a—WV42—2? 4. Vz—38—WV2—14= V4a— 155. os Vae+4—Va=Va+ 8. QUADRATIC EQUATIONS. 8Ve—4 154+38V2_ QtVe 404+-~Vz V142+9+2V2+14V32+1=0. Voth oN et 1 =), Ve—234+V2+3—V4ze2+1=0. ey ie 1 A/8e 10 ino 8 = 0: oy Yd + Vel + 8/5141 =0. See SV Oe | a + 2. . V2+2—V2—2—V2e=0. 1 i ——_______.. + ——______ = 12, SY 2 ee i eee Mysore piso. V/32+13 See Ve oe —2=0. 824—~vV2?—8 es Co HS Ea ae boa 1 Sa 9S 8: soe + ihe — 229 52 1 = 2 ee 89a 82 oO 7, Bigg e/g ee 8 A gO. . 8a— 424+ V382°—42—6=18. Peat ONO e — 160 21162. 972 SCHOOL ALGEBRA. 292. Problems involving Quadratics. Problems which in- volve quadratic equations apparently have two solutions, since a quadratic equation has two roots. When both roots of the quadratic equation are positive integers, they will give two actual solutions of the problem. Fractional and negative roots will in some problems give admissible solutions; in other problems they will not give admissible solutions. ? No difficulty will be found in selecting the result which belongs to the particular problem we are solving. Some- times, by a change in the statement of the problem, we may form a new problem which corresponds to the result that was inapplicable to the original problem. Imaginary roots indicate that the problem is impossible. Here as in simple equations x stands for an unknown number. (1) The sum of the squares of two consecutive numbers is 481. Find the numbers. Let x = one number, and x +1 =the other. Then x? + (# + 1)? = 481, or 207+ 2%+1=481. The solution of which gives «= 15 or —16. The positive 15 gives for the numbers, 15 and 16. The negative root —16 is inapplicable to the problem, as consecu- tive numbers are understood to be integers which follow one another in the common scale, 1, 2, 3, 4..... (2) A pedler bought a number of knives for $2.40. Had he bought 4 more for the same money, he would have paid 8 cents less for each. How many knives did he buy, and what did he pay for each? Let x = number of knives he bought. Then ae = number of cents he paid for each. QUADRATIC EQUATIONS. 273 But if xz + 4= number of knives he bought, 240 _ number of cents he paid for each, x +4 240 __240 _ the difference in price. we 2+4 But 3 = the difference in price. . 240 240 _. “Np ee ee ea Solving, z= 16 or — 20. He bought 16 knives, therefore, and paid 242, or 15 cents for each. If the problem is changed so as to read: A pedler bought a number of knives for $2.40, and if he had bought 4 dess for the same money, he would have paid 3 cents more for each, the equation will be : Fath Bet e—-4 & Solving, x = 20 or — 16. This second problem is therefore the one which the neg- ative answer of the first problem suggests. (3) What is the price of eggs per dozen when 2 more in a shilling’s worth lowers the price 1 penny per dozen? Let « = number of eggs for a shilling. Then >= cost of 1 egg in shillings, and 12 cost of 1 dozen in shillings, But if ay i = number of eggs for a shilling, 12 — cost of 1 dozen in shillings. x+2 ee ea (1 penny being 7; of a shilling). eo @¢+2 12 The solution of which gives = 16, or — 18. And, if 16 eggs cost a shilling, 1 dozen will cost 9 pence. Therefore, the price of the eggs is 9 pence per dozen. 274 SCHOOL ALGEBRA. If the problem is changed so as to read: What is the price of eggs per dozen when two /ess in a shilling’s worth raises the price 1 penny per dozen? the equation will be ae de eek The solution of which gives «= 18, or —16. Hence, the number 18, which had a negative sign and was inappli- cable in the original problem,-is here the true result. Exercise 100. 1. The sum of the squares of two consecutive integers is 761. Find the numbers. 2. The sum of the squares of two consecutive numbers ex- ceeds the product of the numbers by 18. Find the numbers. 3. The square of the sum of two consecutive even num- bers exceeds the sum of their squares by 336. Find the numbers. 4. Twice the product of two consecutive numbers ex- ceeds the sum of the numbers by 49. Find the numbers. 5. The sum of the squares of three consecutive numbers ‘is 110. Find the numbers. 6. The difference of the cubes of two successive odd numbers is 602. Find the numbers. 7. The length of a rectangular field exceeds its breadth by 2 rods. If the length and breadth of the field were each increased by 4 rods, the area would be 80 square rods. Find the dimensions of the field. 8. The area of a square may be doubled by increasing its length by 10 feet and its breadth by 3 feet. Determine its side. QUADRATIC EQUATIONS. aT 9. A grass plot 12 yards long and 9 yards wide has a path around it. The area of the path is 2 of the area of the plot. Find the width of the path. 10. The perimeter of a rectangular field is 60 rods. Its area is 200 square rods. Find its dimensions, 11. The length of a rectangular plot is 10 rods more than twice its width, and the length of a diagonal of the plot is 25 rods. What are the dimensions of the field? 12. The denominator of a certain fraction exceeds the numerator by 8. If both numerator and denominator be increased by 4, the fraction will be increased by 2. Deter- mine the fraction. 13. The numerator of a fraction exceeds twice the de- nominator by 1. If the numerator be decreased by 3, and the denominator increased by 3, the resulting fraction will be the reciprocal of the given fraction. Find the fraction. 14. A farmer sold a number of sheep for $120. If he had sold 5 less for the same money, he would have received $2 more per sheep. How much did he receive per sheep? State the problem to which the negative solution apples. 15. A merchant sold a certain number of yards of silk for $40.50. If he had sold 9 yards more for the same money, he would have received 75 cents less per yard. How many yards did he sell? 16. A man bought a number of geese for $27. He sold all but 2 for $25, thus gaining 25 cents on each goose sold. How many geese did he buy? 17. A man agrees to do a piece of work for $48. It takes him 4 days longer than he expected, and he finds that he has earned $1 less per day than he expected. In how many days did he expect to do the work? 276 SCHOOL ALGEBRA. 18. Find the price of eggs per dozen when 10 more in one dollar’s worth lowers the price 4 cents a dozen. 19. A man sold a horse for $171, and gained as many per cent on the sale as the horse cost dollars. How much did the horse cost? 20. A drover bought a certain number of sheep for $160. He kept 4, and sold the remainder for $10.60 per head, and made on his investment 3 as many per cent as he paid dollars for each sheep bought. How many sheep did he buy? 21. Two pipes running together can fill a cistern in 52 hours. The larger pipe will fill the cistern in 4 hours less time than the smaller. How long will it take each pipe running alone to fill the cistern? 22. A and Bean do a piece of work together in 18 days, and it takes B 15 days longer to do it alone than it does A. In how many days can each do it alone? 23. A boat’s crew row 4 miles down a river and back again in 1 hour and 30 minutes. Their rate in still water is 2 miles an hour faster than twice the rate of the current. Find the rate of the crew and the rate of the current. 24. A number is formed by two digits. The units’ digit is 2 more than the square of half the tens’ digit, and if 18 be added to the number, the order of the digits will be reversed. Find the number. 25. A circular grass plot is surrounded by a path of a uniform width of 8 feet. The area of the path is % the area of the plot. Find the radius of the plot. 26. Ifa carriage wheel 11 feet round took 4 of a second less to revolve, the rate of the carriage would be 5 miles more per hour. At what rate is the carriage travelling? CHAPTER XIX. SIMULTANEOUS QUADRATIC EQUATIONS. 298, Quadratic equations involving two unknown num- bers require different methods for their solution, according to the form of the equations. Case I. 294, When from one of the equations the value of one of the unknown numbers can be found in terms of the other, and this value substituted in the other equation. Ex. Solve: 82° —Qaey=5 \ (1) z—y=2 (2) Transpose x in (2), y =x — 2. In (1) put « —2 for y, 3a? — 2a(x% — 2) =5, The solution of which gives =1, or «= —5. If 2=1, y=1—-2=-1 and if x= — 5, y=—5-—2=—7, We have therefore the pairs of values, fee : Ele ¥ The original equations are both satisfied by either pair of values. But the values «= 1, y=—7, will not satisfy the equations; nor will the values «=—5, y=—1. The student must be careful to join to each value of x the corresponding value of y. 278 SCHOOL ALGEBRA. Case II. 295. When the left side of each of the two equations is homogeneous and of the second degree. Solve: Rae (1) y?— 2? = 16 (2) Let y = va, and substitute vx for y in both equations. From (1), 2072? — 4 ve? + 327 =17, “3 17 2. 02 = ———____.. 2v?— 4043 From (2), va? — xo = 16. ge v—1 17 16 Equate the values of a?, yh eae ET. 32v? — 640 +48 =17r? —17, 15? — 64v = — 65, 225 v? — 960 = — 975, 225 v? — ( ) + (32)? = 49, 159 = 32 = +7; 5 13 .v== or —: 5 5 If eo) If v8 se 5 y 3 5 Substitute in (2), Substitute in (2), 95 op 2 16 169 x “ds a2 = 16, Ce h perme st 25 ff == 9. a? — 2, e=+3, 9 5 r= Fei y= 1 = +5, 3 sag gis d= 5 3 SIMULTANEOUS QUADRATIC EQUATIONS. 279 Case III. 296. When the two equations are symmetrical with respect to x and y; that is, when x and y are similarly involved, Thus, the expressions 20°4327y?+2y', 2xy—8x—3y4+1, e—8axy—38ay+y', are symmetrical expressions. In this case the general rule is to combine the equations in such a manner as to remove the highest powers of x and y. Solve: at + x! = 387 ; (1) ZY 4 (2) To remove 2‘ and 74, raise (2) to the fourth power, at + 4a3y + 6a7%y? + 4ay?+ yt = 2401 Add (1), af + yt= 337 Dat + dady + Gary? + Lay? + 2yt = 2738 Divide by 2, at + 2a%y + 3a%y? + 2ayi+ y* = 1369. Extract the square root, e+ ay + y? = + 37. (3) Subtract (3) from (2)?, wy = 12 or 86. We now have to solve the two pairs of equations, a py= 7). et+y= 7). Ly = 12 ; xy = 86 From the first, fps inet = 3 y=3 y=4 From the second, PERS. i. 295 7*V — 295 y= 280 SCHOOL ALGEBRA. 297. The preceding cases are general methods for the solution of equations which belong to the kinds referred to; often, however, in the solution of these and other kinds of simultaneous equations involving quadratics, a little inge- nuity will suggest some step by which the roots may be found more easily than by the general method. (1) Solve: a+y=40 \ (1). xy = 800 (2) Square (1), x + 2ay + y* = 1600. (3) Multiply (2) by 4, 4 ay = 1200. (4) Subtract (4) from (3), x — 2ey + y* = 400. (5) Extract root of each side, «—y = + 20. (6) From (1) and (6), 30S eet Ae Psi oey 2) Solve: = | == (1) (2) Solve aa Fe DG Lee oes 1 ay 400 ©) 3 biel Square (1), caress: evs (3) Subtract (2) from (3), | Slr Sle aa 4 Subtract (4) from (2), pet nee ee mt ey ~ ay? -400 Extract the root, aed (5) oY 20 From (1) and (5), ae or te y=5)' y=4 SIMULTANEOUS QUADRATIC EQUATIONS. (3) Solve: Square (1), Subtract (2) from (3), x—y= 4 x+y? = 40 x? —2ey+y?=16. — 2ay = — 24. Subtract (4) from (2), Extract the root, From (1) and (5), (4) Solve: Divide (1) by @), Square (2), Subtract (3) from (4), Divide by —3, Add (5) and (3), Extract the root, From (2) and (6), (5) Solve: Divide (1) by (2), Square (2), Subtract (4) from (3), which gives We now have, Solving, we find, a + 2ey + y? = 64. ety=+8. x“ =='6 . z=—2 . yas y=—6 me chet Dey ==) oT a — xy + y? = 13. a + 2ay + y? = 49. dvy = 36. —ay=—12. a? —2ey+y?=1. e+y =12 ot ay + y? = a. w+ 2ey + y? = 144. xy = 32. ety=12 : xy = 32 aga 281 (1) (2) (3) (4) (5) (1) (2) (3) (4) (5) (8) (1) (2) (3) (4) 10. gia Be 12. 13. 14. 15. 16. 17. SCHOOL ALGEBRA. Exercise 101. ety= } 3 patie 5. ee: zy == 10 ry =—8 v+y = 80 aaa 4 ad iat 6 «ty = a xy = 27 vy = 11 e+ y? = 29 2—-y= = 18. Be xv’ +-y? = 45 3842—y=s3 eos ae 19) V+ Y= os 7) ee Gf = 16 aN Baa 5a —4y = 10 at ae 82°—4y= 8 oy ee ae ant 21. ete=2 | ry = 6 pet ad 22 —By =2 y=6 | 2 2ay=—T es a2, -+>—=11 2% —3y = at oy nae 3x" — 4ay = 32 1,1 _6) x’ af Gets aaa xt+y=9 a 68 eG he coe ch a 22+ 3y=7 xy =16 v—y=9 24. 2 +7? =85 x—y=l 25. 26. 27. 28. 29. 30. 31. 32. 33. 34. 35. 36. SIMULTANEOUS QUADRATIC EQUATIONS. 2838 | z—y= 1 age x+t+y= 5 io. zty=8 1 ae ie 5 alll ae + y* = 126 ees 5) 2 —y> = 56 ee 28t e+ Qey = 24 2ey +47 = oi 47 +5xy=14 Tay+9y? =50 37. 39. 41. 42. 44, 46. 47. “e+ay+y = 39 } 27+ 32y+y? = 63 . ae ay +- 27" = 40 Tee ee sy +y'=14 » &+ay+ 27’? = 44 ; 22° —8ay+2y=16 #+3y7=31 ; 4ay+y' = 33 827+ Try = 82 w+ 5ay+9y? = 279 : dae at x+ty= 5 le a zty= 3 : AU wea z2—y= 1 Ce Bod e+ty= 1 ek are “e—y= 2 d eae Cant a+y =xy9—1 : See ed z—y =2 284 SCHOOL ALGEBRA. 50. BR a 61. . x+y =24 x—Yy= 51. BE Se é. 8 ay + y = 63 62. = 4¥=1) 52. 2+ ayt+y= 61 inte 4 28 A Eas ~ + =e x + ay? + y* = 1281 a he 53. 2 — ay + Y= 4 63. 2? =ax-+ by e+ xy? + y* = 21 oe tty ,«2—-y_10 30 ples 54, —— => 64. 2 =2(¢7+0? byt eee 3 ee a My a? +? = 20 4 bt 65 2 2. a+ “5: fmm ites tae Oy ae t+y «£—y, 9d ry = 8a +4y = 36 66. P+ Y¥+e+y= 66. gaya xy +16 =0 paar Mi?) 57. xz—y—3=0 ; 7~ (a+b) 2 (2? — y’) = 8xy 67. xv? —2y ae 58. atyat sy— y=2ab— 20 1 4 EC ess 68. yee eee ye OG ay = 6 59. x+y! = 272 i 69. "aT sy=v0—40° 60. eh Sa at 70. 2 a z+y =22y-1 x+ty=a+b SIMULTANEOUS QUADRATIC EQUATIONS. 285 Exercise 102. 1. The area of a rectangle is 60 square feet, and its perimeter is 34 feet. Find the length and breadth of the rectangle. 2. The area of a rectangle is 108 square feet. If the length and breadth of the rectangle are each increased by 3 feet, the area will be 180 square feet. Find the length and breadth of the rectangle. 3. If the length and breadth of a rectangular plot are each increased by 10 feet, the area will be increased by 400 square feet. But if the length and breadth are each dimin- ished by 5 feet, the area will be 75 square feet. Find the length and breadth of the plot. 4. The area of a rectangle is 168 square feet, and the length of its diagonal is 25 feet. Find the length and breadth of the rectangle. 5. The diagonal of a rectangle is 25 inches. If the . rectangle were 4 inches shorter and 8 inches wider, the diagonal would still be 25 inches. Find the area of the rectangle. 6. A rectangular field, containing 180 square rods, is surrounded by a road 1 rod wide. The area of the road is 58 square rods. Find the dimensions of the field. 7. Two square gardens have a total surface of 2137 square yards. A rectangular piece of land whose dimen- sions are respectively equal to the sides of the two squares, will have 1093 square yards less than the two gardens united. What are the sides of the two squares? 8. The sum of two numbers is 22, and the difference of their squares is 44. Find the numbers. 286 SCHOOL ALGEBRA. 9. The difference of two numbers is 6, and their product exceeds their sum by 39. Find the numbers. 10. The sum of two numbers is equal to the difference of their squares, and the product of the numbers exceeds twice their sum by 2. Find the numbers. ; 11. The sum of two numbers is 20, and the sum of their cubes is 2060. Find the numbers. 12. The difference of two numbers is 5, and the differ- ence of their cubes exceeds the difference of their squares by 1290. Find the numbers. 13. A number is formed of two digits. The sum of the squares of the digits is 58. If twelve times the units’ digit be subtracted from the number, the order of the . digits will be reversed. Find the number. 14. A number is formed of three digits, the third digit being twice the sum of the other two. The first digit plus the product of the other two digits is 25. If 180 be added * to the number, the order of the first and second digits will be reversed. Find the number. 15. There are two numbers formed of the same two digits in reverse orders. The sum of the numbers is 33 times the difference of the two digits, and the difference of the squares of the numbers is 4752. Find the numbers. 16. The sum of the numerator and denominator of a cer- tain fraction is 5; and if the numerator and denominator © be each increased by 3, the value of the fraction will be increased by 4. Find the fraction. 17. The fore wheel of a carriage turns in a mile 132 times more than the hind wheel; but if the circumferences were each increased by 2 feet, it would turn only 88 times more. Find the circumference of each. CHAPTER XX. PROPERTIES OF QUADRATICS. 298. Every affected quadratic can be reduced to the form ax’ + bx + e¢=0, of which the two roots are _b ,Vbi=4a0 4g 8 vei—4a0 g ogs, 2a 2a 2a - 2a CHARACTER OF THE Roots. 299. As regards the character of the two roots, there are three cases to be distinguished. I. If b?’—4ac is positive and not zero, In this case the roots are real and unequal. The roots are real, since the ° square root of a positive number can be found exactly or approximately. If b*—4ac is a perfect square, the roots are rational; if 6?—4ae is not a perfect square, the roots are surds. The roots are unequal, since Vb? — 4.ac is not zero. ives pb — 426 is zer0. In this case the two roots are ~ real and equal, since they both become on a III. If b’—4ac is negative. In this case the roots are wmaginary, since they both involve the square root of a negative number. The two imaginary roots of a quadratic cannot be equal, since 6?—4ac is not zero. They have, however, the same ~~ 288 SCHOOL ALGEBRA. real part, —2, and the same imaginary parts, but with a opposite signs; such expressions are called conjugate im- aginaries. The expression 6’—4ac is called the discriminant of the expression az? -+ bx-+e. 300. The above cases may also be distinguished as follows: CasE I. 0?—4ac>0, roots real and unequal. Case II. 6?’—4ac=0, roots real and equal. Case III. 0?—4ac <0, roots imaginary. 301. By calculating the value of 6?—4ae we can deter- mine the character of the roots of a given equation without solving the equation. (1) a? —52+6=0. - Here a=1, b=—5, c=6. b?—4ac = 25 — 24=1., The roots are real and unequal, and rational. (2) 827+ 7x2-—1=0. Here a=3, d=7, c=—1. b?~4ac=49+12=61. The roots are real and unequal, and are both surds. (3) 427—122+9=0. Here a=4 b=-12, c=—¥9. b?—4ac= 144-—144=0. The roots are real and equal. (4) 2a? 382+4=0. Here a=2, b=—3, c=4. b?—4ac=9 — 32 = — 23, The roots are both imaginary. \ PROPERTIES OF QUADRATICS. 289 (5) Find the values of m for which the following equa- tion has its two roots equal : 2mx’ + (5m + 2)x-+(4m+1)=0. Here a=2m, b=5m+2, c=4m+1. If the roots are to be equal, we must have b?—4ac=0, or (5m + 2)? —8m(4m+1)=0. This gives m= 2, or — =. For these values of m the equation becomes 40? +1224+9=0, and 42?— 42+1=0, each of which has its roots equal. Exercise 103. Determine without solving the character of the roots of each of the following equations: . Poe = pr -6=0. 6. 627— Tx—3=0. Bee + 24 —15=0. 7, 52°7—52—3=0. S$. ¢+227+38=0. 8. 2e7—27+5=0. 4. 32°+7T2#+2=0. 9. 627 +2—7T7=0. FO Be ts 1280. 10. bat +82 += 0. Determine the values of m for which the two roots of each of the following equations are equal: 11. (m+1)2?+ (m—1)2+m+1=0. 12. (2m—3)2?+ mz+m—1=0. 13. 2m2’?+a°?+4274+ 2mex+2m—4=0. 14. 2m2z?+ 8mx—6=82— 2m — 2’. 15. m2’?+9x—10=38me—227+2m., 290 SCHOOL ALGEBRA. RELATIONS OF Roots AND COEFFICIENTS. 302. Consider the equation z7—10z%-+24=0. Resolve into factors, (« —6)(a—4)=0. The two values of 2 are 6 and 4; their sum is 10, the coefficient of 2 with its sign changed; their product is 24, the third term. 308. In general, representing the roots of the quadratic equation aa’?+ b4+c¢=0 vt 7, and 7,, we have (§ 285), 5 Vb? —4a0¢ 4 a0 Bik ay o— ene ‘ Mie cianee'| Vo? —4a¢ 4ac : 2a 2a : b Adding, TY) + LE areata peek a Rite: Cc multiplying, Tits oe If we divide the equation ax’ -+ bx+c=0 through by ; b ; a, we have the equation 2 +o2+—=0; this may be written z’?+ px-+gq=0 where p =< g =<. It appears, then, that if any quadratic equation be made to assume the form 2’-++ px + ¢=0, the following relations hold between the coefficients and roots of the equation : (1) The sum of the two roots is equal to the coefficient of x with its sign changed. (2) The product of the two roots is equal to the constant term. Thus, the sum of the two roots of the equation v’—Tx+8=0 is 7, and the product of the roots 8. PROPERTIES OF QUADRATICS. 291 304, Resolution into Factors. By § 303, if 7, and 7 are the roots of the equation 2+ px+q=0, the equation may be written 2—(n+r)e+rnr,= 0. The left member is the product of x —7, and x — 7, so that the equation may be also written (2 —1,)(x#— 1.) = 0. It appears, then, that the factors of the quadratic expres- sion x? + px-+q are x—7, and x— 7, where 7, and r, are the roots of the quadratic equation 2° + px+q=0. - he factors are real and different, real and alike, or imaginary, according as 7, and r, are real and unequal, real and equal, or imaginary. If 7,=7;, the equation becomes (x — 17) (2 — 7) = 0, or (c—7,)?=0; if, then, the two roots of a quadratic equa- tion are equal, the left member, when all the terms are transposed to that member, will be a perfect square as regards x. 305. If the equation is in the form az’?+ bx + ¢=0, the left member may be written (v b C a\ x +045), a(e— 7) (% — 72). 306. If the roots of a quadratic equation are given, we can readily form the equation. Ex. Form the equation of which the roots are 3 and — 2. The equation is (x#—3) (« + 5) = 0, or (a — 3) (2% + 5) =0, or 207 —x2—15=0, 992, SCHOOL ALGEBRA. 3807. Any quadratic expression may be resolved into factors by putting the expression equal to zero, and solving the equation thus formed. (1) Resolve into two factors 7? —5x +3. Write the equation v?—5xe4+3=0., Solve this equation, and the roots are found to be 5 + nae Bee 5 vis Therefore, the factors of a? —5a2+3 are ej Dt Vib ng eciD iene 2 2 (2) Resolve into factors 32°—4z-+ 5. Write the equation 3a7—42+5=0. Solve this equation, and the roots are found to be 2b Vet aS ee ed rir ee eee Therefore, the expression 327—4a+5 may be written (¢ 305), 3 (2-24¥%= 3) (ee a t 3 3 Exercise 104. Form the equations of which the roots are 187; 6. opie Sea 9. 84+ V2,3— v2. 2. 5,—8. 6. —1,,—1]%. (10. 1+ Vee 3. 14,—2. 7. 18, —44. 11. a,a—b. 4. 4, 21. 8. - = 12. at+b,a—b. Resolve into factors, real or imaginary : 13. 122? +2—1. 16. v2 —22+8. 14. 827— 142 — 24, 17. 2te+l. 15. 2? —2Q2 —2. 18. 2?—22+9. CHAPTER XXII. RATIO, PROPORTION, AND VARIATION. 308. Ratio of Numbers. The relative magnitude of two numbers is called their ratio when expressed by the indi- cated quotient of the first by the second. Thus the ratio of a to 6 is 5 or a+b, or a:b. The first term of a ratio is called the antecedent, and the second term the consequent. When the antecedent is equal to the consequent, the ratio is called a ratio of equality; when the antecedent is greater than the consequent, the ratio is called a ratio of greater inequality; when less, a ratio of less inequality. When the antecedent and consequent are interchanged, the resulting ratio is called the inverse of the given ratio. Thus, the ratio 3:6 is the znverse of the ratio 6: 8. 809. A ratio will not be altered if both its terms are multiplied by the same number. For the ratio a:6 is represented by 5 the ratio ma: mb is represented by 1 m and since “@=" we have ma:mb=a:b. mb 6b 310. A ratio will be altered if different multiphers of its terms are taken; and will be increased or diminished accord- ing as the multiplier of the antecedent is greater than or less than that of the consequent. Thus, 294 SCHOOL ALGEBRA. If m>n, If mna, then ma< na, d lake Seo alge d ha ae gilt me nb a nb’ by nb “nb but (cn LS but deans NS - nb b ‘ nbd Ma & . bbs ie Lo Pas * RE re: or ma:nb>a:b. or ma:nb or << as ad > or < be. 313. Proportion of Numbers. Four numbers, a, 8, ec, d, are said to be in proportion when the ratio a: 6 is equal to the ratio ce: d. We then write a:b =c:d, and read this, the ratio of a to 6 equals the ratio of ¢ to d, or ais to 8 as ¢ is to d. A proportion is also written a:6::e:d. The four numbers, a, 0, c, d, are called proportionals; a and d are called the extremes, 6 and ¢ the means, RATIO AND PROPORTION. 295 314, When four numbers are in proportion, the product of the extremes is equal to the product of the means. For, if arb=c% d, th ME en F, Multiplying by dd, ad=be. The equation ad=be gives a=" p= so that an extreme may be found by dividing the product of the means by the other extreme; and a mean may be found by dividing the product of the extremes by the other mean. If three terms of a proportion are given, it appears from the above that the fourth term can have one, and but one, - value. 315, If the product of two numbers is equal to the prod- uct of two others, either two may be made the extremes of a proportion and the other two the means. For, if ad = be, fae ad_ be then, dividing by dd, bd bd or pa. std, Res eB ea ee A 316. Transformations of a Proportion. If four numbers, a, b, c, d, are in proportion, they will be in proportion by: I, Inversion; that is, 6 will be to a as dis to ec. For, if Ob == oS.d, then 296 SCHOOL ALGEBRA. and neta iste bad, or ee SE Cac 21D soli ah es II, Composition; that is, a+ will be to 6 as e+d is to d. For, if Tied ak cielo h egos then i a and Fe heros i ee ater -atb:b=c+d:d. III. Division; that is, a—6 will be to 6 as e— dis to d. For, if Tae tow tes 2 then Fee and lees Re GD. saa b d ” a—b:b=c—d:d.. IV. Oomposition and Division; that is, a+ will be to a—base+dis to c—d. For, from II., ate_ctd and from III., oe a+b etd Dividing, 5 a c— *atb:a—b=ce+ RATIO AND PROPORTION. 297 V. Alternation; that is, a will be to c as 6 is to d. For, if a b= e7'd. then ae war. b 6b be Multipl pees rl Ee a ultiplying by : ae a ae ao et 70 eet ees D4. 317. In a series of equal ratios, the sum of the antecedents is to the sum of the consequents as any antecedent is to its consequent. Orsay aime dow y may be put for each of these ratios Then C=yr, ee Cy Lay, a= OT. ONE = FTR g = AP. “ atetetg=(64+d+ft+h)r. Beri ener Fi ae b+d+tfth b “ atetetg:b+d+f+th=a: 6, In like manner it may be shown that ma+ne-+petgqg:mb+nd+pft+qh=a:b. 318, Continued Proportion. Numbers are said to be in continued proportion when the first is to the second as the second is to the third, and so on. Thus, a, 8, ¢, d, are in continued proportion when 298 SCHOOL ALGEBRA. 319. Ifa, 6, ¢ are proportionals, so that a:b =b:c, then 6 is called a mean proportional between a@ and c, and ¢ is called a third proportional to a and 6. If a:b6=b:c, then b= Vac. For, if i SaID Se a_t then ae and Of Sac: b= Vae. 320. The products of the corresponding terms of two or more proportions are in proportion. For, if a: b= ou then Taking the product of the left members, and also of the right members of these equations, . ack: bfl = cgm: dhn. $21, Like powers, or like roots, of the terms of a propor- tion are in proportion. For, if Wit DiS orate then ; == e Raising both sides to the nth power, a” aoe a ai ot Me RATIO AND PROPORTION. 299 322, The laws that have been established for ratios should be remembered when ratios are expressed in frac- tional form. 2 2 Aye ar y Gece: Sette eee V’—x-—-1l #+ne-2 By composition and division, Sree MS Serna B(x +1). — 2(2 —2) This equation is satisfied when a=0. For any other value of z, we may divide by 2’. We then have l ! and therefore (2) If a:b =c:d, show that v+ab:0?—ab=c’+ cd: d’?—cd. a ce : see a+b c+d sales Gm bi eed dk Bas ae and Deng a at+b_¢ etd. i <0) ee Gee that is Stree acne ab @—cd or a+ab:6?—ab=c?+ cd: d*?—cd. 300 SCHOOL ALGEBRA. (3) If a:b =c:d, and a is the greatest term, show that a+d is greater than b+ c. Since as a and. a& > ¢, (1) the denominator b> d. From (1), by division, © $ aes = se (2) Since psa from (2), a—b>c—d, Now, b+d=b+d, Adding, a+d>b+e. 323. Ratio of Quantities. To measure a quantity of any kind is to find out how many times it contains another known quantity of the same kind, called the unit of measure. Thus, if a line contains 5 times the linear unit of measure, one yard, the length of the line is 5 yards. 324, Commensurable Quantities. If two quantities of the same kind are so related that a unit of measure can be found which is contained in each of the quantities an in- tegral number of times, this unit of measure is a common measure of the two quantities, and the two quantities are said to be commensurable. If two commensurable quantities are measured by the same unit, their ratio is simply the ratio of the two numbers by which the quantities are expressed. Thus, + of a foot is a common measure of 24 feet and 32 feet, being contained in the first 15 times and in the second 22 times. The ratio of 24 feet to 32 feet is therefore the ratio of 15: 22. Evidently two quantities different in kind can have no ratio, 4 RATIO AND PROPORTION. 301 325, Incommensurable Quantities. The ratio of two quan- tities of the same kind cannot always be expressed by the ratio of two whole numbers. Thus, the side and diagonal of a square have no common measure; for, if the side is a inches long, the diagonal will be a2 inches long, and no measure can be found which will be contained in each an integral number of times. Again, the diameter and circumference of a circle have no common measure, and are therefore incommen- surable. In this case, as there is no common measure of the two quantities, we cannot find their ratio by the method of § 324. We therefore proceed as follows: Suppose a and 6 to be two incommensurable quantities of the same kind. Divide 6 into any integral number (7) of equal parts, and suppose one of these parts 1s contained in a@ more than ™ times and less than m-+1 times. Then the ratio ; sa —, but < LOnrae that is, the value of : les n between — an m q mel, n The error, therefore, in taking either of these values for @ is less than 4. But by increasing 7 indefinitely, 1 can n n be made to decrease indefinitely, and to become less than any assigned value, however small, though it cannot be made absolutely equal to zero. Hence, the ratio of two incommensurable quantities can- not be expressed exact/y in figures, but it may be expressed approxvmately to any desired degree of accuracy. Thus, if 6 represent the side of a square, and a the diagonal, ie 302 SCHOOL ALGEBRA. Now V2 = 1.41421856.....,a value greater than 1.414218, but less than 1.414214. . If, then, a mzdlionth part of 6 is taken as the unit, the 1414213 .. , 1414214 1000000 1000000" and therefore differs from either of these fractions by less than bani 1000000 By carrying the decimal farther, a fraction may be found that will differ from the true value of the ratio by less than | a billionth, a trillionth, or any other assigned value what- ever. value of the ratio ;, lies between 826. The ratio of two incommensurable quantities is an incommensurable ratio, and is a fixed value toward which its successive approximate values constantly tend as the error 1s made less and less. 827. Proportion of Quantities. In order for four quanti- ties, A, B, C, D, to be in proportion, A and B must be of the same kind, and C' and D of the same kind (but Cand D need not necessarily be of the same kind as A and 8B), and in addition the ratio of A to B must be equal to the ratio of C to D. If this be true, we have the proportion AA Bata gy: When four quantities are in proportion, the numbers by which they are expressed are four abstract numbers in proportion. 828, The laws of § 816, which apply to proportion of _numbers, apply also to proportion of quantities, except that alternation will apply only when the four quantities in proportion are a// of the same kind. RATIO AND PROPORTION. 303 Exercise 105. 1. Find the duplicate of the ratio 3° 4. 2. Find the ratio compounded of the ratios 2: 3, 3: 4, S714: 8. 3. Find a third proportional to 21 and 28. 4. Find a mean proportional between 6 and 24. 5. Find a fourth proportional to 3, 5, and 42. 6. Find #if 5++-2:1l1—27=8:5. 7. Find the number which must be added to both the terms of the ratio 3:5 in order that the resulting ratio may be equal to the ratio 15: 16. If a:b =c: d, show that me od — ¢*: dd’. 10. @—OB:?-aP=a':e. eri eo. == Cr °C", 11. 2a+6:2c+d=0d:d. . da—b:5ce—d=a:e. . a—8b:a+38b=c—8d:c+8d. ~ @+ab+:a—abtPHe+ed+a’?:e—ced+d’. Find # in the proportion 15. 16. 19. 45°68 = 90* 2. 17. x: l4A=—13: 14. pag a: 1, IBA LA. Find two numbers in the ratio 2:3, the sum of whose squares is 325. 20. Find two numbers in the ratio 5:8, the difference of whose squares is 400. 21. Find three numbers which are to each other as 2:3:5, such that half the sum of the greatest and least exceeds the other by 25. 304. SCHOOL ALGEBRA. 22. Find cif 6%#—a:4r7—b=—82+06: 2z-a. 23. Find x and y from the proportionals er:y=ue+y:42; ri y=e— yi. 24. Find x and y from the proportionals 2ety:y=syi2y—az; 2¢2+1:24+6=y:y+2. 25 dt Go OAR Ed a Dee how that =e atb—e—d a—b—cta et De a VARIATION. 329, A quantity which in any particular problem has a fixed value is called a constant quantity, or simply a constant ; a quantity which may change its value is called a variable quantity, or simply a variable, Variable numbers, ike unknown numbers, are generally represented by a, y, z, etc.; constant numbers, like known numbers, by a, 8, e, etc. 830. Two variables may be so related that when a value of one is given, the corresponding value of the other can be found. In this case one variable is said to be a function of the other; that is, one variable depends upon the other for its value. Thus, if the rate at which a man walks is known, the distance he walks can be found when the time is given; the distance is in this case a function of the time. 331, There is an unlimited number of ways in which two variables may be related. We shall consider in this chapter only a few of these ways. 332, When x and y are so related that their ratio is constant, y is said to vary as x; this is abbreviated thus: VARIATION. 305 you. The sign o, called the sign of variation, is read “varies as.’ Thus, the area of a triangle with a given base varies as its altitude; for, if the altitude is changed in any ratio, the area will be changed in the same ratio. In this case, if we represent the constant ratio by m, y Yi tZ=M, or = M; “Y= me. Again, if y’, z' and y", 2" be two sets of corresponding values of y and 2, then yt ah yl al" or ey ae a § 316, V. 1 z is constant, y is said to vary wversely as x; this is written 1 ya -. Thus, the time required to do a certain amount of xe 333, When x and yare so related that the ratio of y to work varies inversely as the number of workmen employed ; for, if the number of workmen be doubled, halved, or changed in any ratio, the time required will be halved, doubled, or changed in the inverse ratio. In this case, y =m; ox: y=—, and xy=m,; that is, the product xy is constant. i 1 As before, rae ap ay hay a ally"! or, ED Bh ae Ga $315 334, If the ratio of y: xz is constant, then y is said to vary jointly as x and z. In this case, y= mzz, and he eels celal 306 SCHOOL ALGEBRA. 835. If the ratio y:~ is constant, then y varies directly z as x and wmversely as z. In this case, ea, and Je 336. Theorems. I. If you, and xz, then y oz. For y=me«z and «—nz, i “. Y Varies as 2. Il. If you, and xz, then (y+z2)oz. For y= mz ond z—nz. YRZ=(MAN)EZ; “. Y +2 varies as x. Ill. If yox when z is constant, and yoz when @ is constant, then y « #z when z and z are both variable. Let 2!, y', z', and x", y", 2" be two sets of corresponding values of the variables. Let # change from 2! to x", z remaining constant, and let the corresponding value of y be Y. Then IPAS Gran AME nF (1) Now let z change from z! to 2", x remaining constant. Then Y: ay! — gle Zl (2) From (1) and (2), WV y"V = a'a! vale", § 320 or Ue aft ag gh lake ; or of ele! ol bat aE § 816, V. . Y e . .“. the ratio pe constant, and y varies as wz. VARIATION. 307 In like manner it may be shown that if y varies as each of any number of quantities 2, z, wu, etc., when the rest are unchanged, then when they all change, y « xzu, etc. Thus, the area of a rectangle varies as the base when the altitude is constant, and as the altitude when the base is constant, but as the product of the base and altitude when both vary. 337, Examples. (1) If y varies inversely as x, and when y= 2 the cor- responding value of x is 36, find the corresponding value of when y= 9. Here y=", or m=ay. wv .m=2X36= 72. If 9 and 72 be substituted for y and m respectively in _m x 9 the result is 9= Lie or 9a = 72. x *a2=8. Ans. (2) The weight of a sphere of given material varies as its volume, and its volume varies as the cube of its diam- eter. Ifa sphere 4 inches in diameter weighs 20 pounds, find the weight of a sphere 5 inches in diameter. Let W represent the weight, V represent the volume, D represent the diameter. Then Wee V and Vee D3, SAG Vy we oe @ 336, I. Put W=mD>; then, since. 20 and 4 are corresponding values of W and D, 20 = m X 64. 20 5 ie W= aan Dd. *. when D=5, W= 5; of 125 = 3974. 308 SCHOOL ALGEBRA. Exercise 106. 1. Ifxecy,and if y=3 when x=5, find # when yis 5. 2. If W varies inversely as P, and W is 4 when Pis 15, find W when P is 12. . 8. If xocyand yz, show that we 0 y?, 4. If rot and oe show that x « z. y zZ 5. If # varies inversely as 7?—1, and is equal to 24 when y= 10, find when y=5. 6. If x varies as ee and is equal to 3 when y=1 Tet S and z= 2, show that zyz= 2(y-+2). . . 1 . 7. If x—y varies inversely as bis and #-+y varies 1 inversely ‘as uae find the relation between a and z if 2=1, y=38, when 25. 8. The area of a circle varies as the square of its radius, and the area of a circle whose radius is 1 foot is 3.1416 square feet. Find the area of a circle whose radius is 20 feet. 9. The volume of a sphere varies as the cube of its radius, and the volume of a sphere whose radius is 1 foot is 4.188 cubic feet. Find the volume of a sphere whose radius is 2 feet. 10. If a sphere of given material 3 inches in diameter weighs 24 lbs., how much will a sphere of the same material weigh if its diameter is 5 inches? VARIATION, | 309 11. The velocity of a falling body varies as the time during which it has fallen from rest. If the velocity of a falling body at the end of 2 seconds is 64 feet, what is its velocity at the end of 8 seconds? 12. The distance a body falls from rest varies as the square of the time it is falling. If a body falls through 144 feet in 8 seconds, how far will it fall in 5 seconds? The volume of a right circular cone varies jointly as its height and the square of the radius of its base. If the volume of a cone 7 feet high with a base whose radius is 8 feet is 66 cubic feet : 13. Compare the volume of two cones, one of which is twice as high as the other, but with one half its diameter. 14. Find the volume of a cone 9 feet high with a base whose radius is 8 feet. 15. Find the volume of a cone 7 feet high with a base whose radius is 4 feet. 16. Find the volume of a cone 9 feet high with a base whose radius is 4 feet. 17. The volume of a sphere varies as the cube of its radius. If the volume is 1792 cubic feet when the radius is 34 feet, find the volume when the radius is 1 foot 6 inches. 18. Find the radius of a sphere whose volume is the sum of the volumes of two spheres with radii 34 feet and 6 feet respectively. 19. The distance of the offing at sea varies as the square root of the height of the eye above the sea-level, and the distance is 8 miles when the height is 6 feet. Find the distance when the height is 24 feet. CHAPTER XXII. PROGRESSIONS. 338, A succession of numbers that proceed according to some fixed law is called a series; the successive numbers are called the terms of the series. A series that ends at some particular term is a finite series; a series that continues without end is an infinite series. 339, The number of different forms of series is unlimited; in this chapter we shall consider only Arithmetical Series, Geometrical Series, and Harmonical Series. ARITHMETICAL PROGRESSION. 340, A series is called an arithmetical series or an arith- metical progression when each succeeding term is obtained by adding to the preceding term a constant difference. The general representative of such a series will be a, at+d,at+2d, a+8d..... in which a is the first term and d the common difference ; the series will be increasing or decreasing according as d is positive or negative. 341. The nth Term. Since each succeeding term of the series 18 obtained by adding d to the preceding term, the coefficient of d will always be one less than the number of the term, so that the nth term is a+(n—1)d. If the nth term be represented by 7, we have l=a+(n—1)d. 1. ARITHMETICAL PROGRESSION. olL 842, Sum of the Series. If Z denotes the nth term, a the first term, 2 the number of terms, d the common difference, and s the sum of n terms, it is evident that s= a +(atd)+(a4+2d)+---+(U—d)+ 7, or s= 1 4(0—-d)t+ (2d) tt td)+ “. 2s=(a+l)+(a+l) +(atd) +e +(atl) +(a4+0) =n(a-+ 0). n s=5(a+ 4). II. 343, F'rom the equations l=a+(m—1)d, i s=5 (a+!) if. any two of the five numbers a, d, /, n, s, may be found when the other three are given. Here | ‘G=2, d=3 From I., [= 2 Substituting in II., $= (2) The first term of an arithmetical series is 3, the last term 31, and the sum of the series 186. Find the series. _ From I., 31=34(n—1)d. (1) From II., 136 = 530 +81). (2) From (2), n= 8, Substituting in (1), d=4., miueeseries. ig oO, 1, 11, 16; 19, 23, 27, 31. SY SCHOOL ALGEBRA. (3) How many terms of the series 5, 9, 18, ....., must be taken in order that their sum may be 275? From L,, 1=5+(n—1)4. v l=4n41. (1) From II., 275 = rAG + 1). (2) Substituting in (2) the value of 7 found in (1), 275 = 5 (tn +6), or 2n? + 3n= 278. This is a quadratic with n for the unknown number. Complete the square, 16 n? + () + 9 = 2209. Extract the root, 4n+3=+ 47. Therefore, n=11, or — 121. We use only the positive result. (4) Find x when d, J, s are given. From L., a=l1—(n—1)d. From IL., a= 28s — In. n 2s—In Therefore, l—(n —1)d==—. n “. In—dn? + dn=2s—In, “. dn? — (214+ d)n=— 2s. This is a quadratic with n for the unknown number. Complete the square, 4d°n? —() + (21+ dl? =(21+ d)— 8ds. Extract the root, 2dn —(21+d)=+ V(21+ dy — 8ds. gl+d+V(21+ d)—8ds 2d Therefore, n= ARITHMETICAL PROGRESSION. 313 (5) Find the series in which the 15th term is — 25 and Alst term — 41. If a is the first term and d the common difference, the 15th term is a + 14d, and the 41st term is a + 40d. Therefore, a+14d=— 25, (1) and a+40d=—41.- (2) Subtracting, —26d= 16. ben ee) eis Substituting in (1), a = — 16,5. Hence, the series is — 1635,, — 17, — 173%, ..... 344, The arithmetical mean between two numbers is the number which stands between them, and makes with them an arithmetical series. If a and 6 represent two numbers, and A their arithmet- ical mean, then a, A, 6 are in arithmetical progression, and by the definition of an arithmetical series, § 340, A—a=d, and 6—A=d. . A-—a=b—-A até AES omen 345, Sometimes it is required to insert several arithmeti- cal means between two numbers. Ex. Insert six arithmetical means between 3 and 17. Here the whole number of terms is eight; 3 is the first term and 17 the eighth. By L,, ee d= 2. The series is 3, [5, 7, 9, 11, 18, 15,] 17, the terms in — brackets being the means required, 314 SCHOOL ALGEBRA. 346, When the sum of a number of terms in arithmetical progression is given, it is convenient to represent the terms as follows: Three terms by x—y, x, x+y; four terms by t+ SY, BY, oY eee and so on. Ex. The sum of three numbers in arithmetical progres- sion is 36, and the square of the mean exceeds the product of the two extremes by 49. Find the numbers. Let «—y, x, x+y represent the numbers. Then, adding, 3.2 = 36. *, @ = 12. Putting for w its value, the numbers are 12—y, 12, 1244, Then (12)?—(12—y)(12 + y) = the excess. But 49 = the excess. Therefore, 144 — 144 + y? = 49. a ee ie The numbers are 5, 12, 19; or 19, 12, 5. Exercise 107. l=a+(n—1)d; s= (a +1) =5(2a+(n—1)d} Find / and s, if 1 a=l, d=4, n=18. -4.. a= 63. ee Bf cei) 2 Onan eens a=%, d=2, oe 3. eee d=6,n=30,"° 6. a@=3n, d= 2 9 ARITHMETICAL PROGRESSION. 315 Find d and sg, if pence ac lot. te 13. Sara — 0) b= 2005 n= 51, Sie lous hs 6, I. SA. FOr Ose 1450) 2 — 21. Insert eight arithmetical means between ite io and 76. 13. 1 and 3. 2 4 ee GAT 14. 47 and 2. Find a and s, if oe et = 149 2 22, 16. d= 21, (= 242, n= 12. Find n and s, if Reet 000, ad=—9. 18.034 7=—10,d=—2. Find d and 7, if fee a—At, (= 54, s=999. -20.a=—2, 7=—87, s=80l. Find d and J, if meee. 2—14,5—1050) 22° a—1, n= 20, s= 800. Find a and d, if free a (,s— 105. 24. /=105,n—16, s— 840. Find a and J, if 25. n=21,d=4,s=1197. 26. n=25, s=—75, d=5 Find and J, if freee ooo. a-—9,d=—8. 28: s— 7198, a= 18; d=6. Find a and 1, if Some o25 a= 5,1=77. -30. s=1008,d=—4; /=88, 316 SCHOOL ALGEBRA. Find the arithmetical series in which 31. The 15th term is 25, and the 29th term 46. How many terms must be taken of 32. The series —16, —15, —14, ..... to make — 100? 33. The series 20, 18%, 174, .....to make 1621? 34. The sum of three numbers in arithmetical progres- sion is 9, and the sum of their squares is 29. Find the numbers. 35. The sum of three numbers in arithmetical progres- sion is 12, and their product is 60. Find the numbers. GEOMETRICAL PROGRESSION. 847. A series is called a geometrical series or a geometrical progression when each succeeding term is obtained by mul- tiplying the preceding term by a constant multipher. The general representative of such a series will be Oh, OF SI0T OR" Cree : in which a is the first term and r the constant multiplier or ratio. The terms increase or decrease in numerical magnitude according as 7 is numerically greater than or numerically less than unity. 848, The nth Term. Since the exponent of 7 increases by one for each succeeding term after the first, the exponent will always be one less than the number of the term, so that the nth term is a7”-}. If the nth term is represented by /, we have pe ae) ik GEOMETRICAL PROGRESSION. 317 349, Sum of the Series. If / represent the nth term, a the first term, n the number of terms, 7 the common ratio, and s the sum of 7 terms, then s=atartarte- art > mh) Multiply by 7, rs =ar + ar? + ar ar" + ar". (2) Subtracting the first equation from the second, T8§ —s = ar" — a, or (r —l)s=aG"— 1). eee =: Atos); a1. r—1 Since /= ar", r/ = av”, and II. may be written rl —a, sa Jane F y—l 350, From the two equations I. and II., or the two equa- tions I. and III., any two of the five numbers a, 7, J, n, s, may be found when the other three are given. (1) The first term of a geometrical series is 3, the last term 192, and the sum of the series 881. Find the number of terms and the ratio. From I., 1923) (1) From IIL, fale esa (2) | eae From (2), 7 = 2, Substituting in (1), aK BA. en =n. The series is 8, 6, 12, 24, 48, 96, 192. 318 SCHOOL ALGEBRA. (2) Find Z when 7, 1, s are given. From I., pica 7r-l rl — = Substituting in IIL., g§= = , Yi — (r—1)s= usar: 2) ere ea (r— 1)"; rr — | 351. The geometrical mean between two numbers is the number which stands between them, and makes with them a geometrical series. If a and 6 denote two numbers, and G their geometrical mean, then a, G, 6 are in geometrical progression, and by the definition of a geometrical series, § 347, | G b ay and ae Wint fered rags . G=Vab. 352, Sometimes it is required to insert several geometri- cal means between two numbers. Ex. Insert three geometrical means between 3 and 48. Here the whole number of terms is five; 3 is the first term and 48 the fifth. By 1, 48 = 374, rt = 16. r=+ 2. The series is one of the following: Boyt eo eeke: 24), 48; 3, [-—6, 12, —24,] 48. The terms in brackets are the means required. GEOMETRICAL PROGRESSION. 319 3538, Infinite Geometrical Series. When 7 is less than 1, the successive terms become numerically smaller and smaller ; by taking 7 large enough we can make the nth term, ar"~’, as small as we please, although we cannot make it absolutely zero. 1 i The sum of n terms, ie ba a asl ; 7 — l—r l- r “may be written “— by the fraction - ; by taking —r 7 enough terms we can make /, and consequently the fraction this sum differs from i ha , as small as we please; the greater the number of —*7 a terms taken the nearer does their sum approach sare Hence is called the sum of an infinite number of —/ terms of the series. (1) Find the sum of the infinite series 1 — ; a i — ‘ Se Here, a =1, een 2 The sum of the series is l or z Ans | ekg ee We find for the sum of n terms ; - : ig iin this sum evidently v 2 : approaches pias me increasen: (2) Find the value of the recurring decimal 0.12135185 ..... Consider first the part that recurs; this may be written 135 = +..., and the sum of this series is 100000” 100000 100000000 qe 1 mae 1000 which reduces to 740" Adding 0.12, the part that does not recur, we obtain for the value of the decimal 12 a 22 or 449 Ans. 100 740’ 3700 320 SCHOOL ALGEBRA. Exercise 108. a(rm—1)_rl—a fae ye ae r—1 r—l Find Z/ and s, if LS yn ey eels 2d raz n=l. Find r and s, if Sh eh as 4 le 4: Qe de sala a0 Os ss 5. Insert 1 geometrical mean between 14 and 686. 6. Insert 38 geometrical means between 31 and 496. %. Find @ and ¢,11.( = 128, 7-2) =e 8. Find sand n, if a=9, 7= 2304, r=2. 9. Find r and n, if a=2, 7= 1458, s = 2186. 10. If the 5th term is 3 and the 7th term 34, find the series. 11. Find three numbers in geometrical progression whose sum is 14, and the sum of whose squares 84. Sum to infinity : bey ay 1 1 ier eed Ae 2 2 new te OEE 1 i D4 7 49 4° 16 1 9 4 13. Dee LBA Dio eens 17.8, —2 tee 3 8, ? 9 ? ? 5’ q 5] ? 3 Find the value of 18. 0.16. 20. 0.86. 22. 0.736. 19. 0.378. 21. 0.54. 23. 0.363. HARMONICAL PROGRESSION. O21 HARMONICAL PROGRESSION. 354, A series is called a harmonical series, or a harmonical progression, when the reciprocals of its terms form an arith- metical series. ‘The general representative of such a series will be Pee oay MUL eres wines 2D ve 4, pera fat ated Oh aly (on Questions relating to harmonical series are best solved by writing the reciprocals of its terms, and thus forming an arithmetical series. 355. If a and 6 denote two numbers, and # their har- monical mean, then, by the definition of a harmonical series, plete ste. Fe hae cba _ 2ab- evs, 356. Sometimes it is required to insert several harmoni- cal means between two numbers. Ex. Insert three harmonical means between 3 and 18. Find the three arithmetical means between 1 and . These are found to be = =f mi therefore, the harmonical means Pe or 348 Bh, 8: 19 14 9 857, Since A aia b and G@= Vab, and H= 2ab 2 a+b wes _ ri CAL That is, the geometrical mean between two numbers is also the geometrical mean between the arithmetical and harmonical means of the numbers. 4 SCHOOL ALGEBRA. Exercise 109. 1. If a, 6,c¢ are in harmonical progression, show that a—b:b-c=a:e. 2. Show that if the terms of a harmonical series are all multiplied by the same number, the products will form a harmonical progression. 3. The second term of a harmonical series is 2, and the fourth term 6. Find the series. 4. Insert the harmonical mean between 2 and 3. 5. Insert 2 harmonical means between 1 and 6. Insert 5 harmonical means between 1 and * 7. The first term of a harmonical progression is 1, and the third term ; Find the 8th term. 8. The first term of a harmonical progression is 1, and the sum of the first three terms is 18. Find the series. 9. Ifa is the arithmetical mean between 6 and e¢, and b the geometrical mean between a and c, show that ¢ is the harmonical mean between a and 6. 10. The arithmetical mean between two numbers exceeds the harmonical mean by 1, and twice the square of the arithmetical mean exceeds the sum of the squares of the harmonical and geometrical means by 11. Find the num- bers. CHAPTER XXIII. PROPERTIES OF SERIES. 358, Convergent and Divergent Series. By performing the indicated division, we obtain from the fraction L the —2£ infinite series 1+ 2+ 2?+2°+...... This series, however, is not equal to the fraction for all values of x. 859. If x is numerically less than 1, the series is equal to the fraction. In this case we can obtain an approximate value for the sum of the series by taking the sum of a number of terms; the greater the number of terms taken, the nearer will this approximate sum approach the value of the fraction. The approximate sum will never be exactly equal to the fraction, however great the number of terms taken ; but by taking enough terms, it can be made to differ from the fraction as little as we please. Thus, if os, the value of the fraction is 2, and the series 1s ‘Bes clea Boies tagtg i au The sum of four terms of this series is 14; the sum of five terms, 118; the sum of six terms, 134; and so on. The successive approximate sums approach, but never reach, the finite value 2. 360, An infinite series is said to be convergent when the sum of the terms, as the number of terms is indefinitely in- creased, approaches some fixed finite value ; this ae value is called the sum of the series. 324 SCHOOL ALGEBRA. 3861, In the series lt+at+a2e7tegt... suppose 2 numerically greater than 1.- In this case, the greater the number of terms taken, the greater will their sum be; by taking enough terms, we can make their sum as large as we pleuse. The fraction, on the other hand, has a definite value. Hence, when 2 is numerically greater than 1, the series is not equal to the fraction. Thus, if #=2, the value of the fraction is —1, and the series 1s ye a a iy, Si a, (a-+a)'= (ear ar(1 42) SUA Pe aren Bake ) oh egy” - 340 SCHOOL ALGEBRA. 380, Examples. (1) Expand (1 + 28, (l+a)@=1+4 o+fG— Do, iG—DG—2) G2) 58 4 1*2:3 2 2°5 = (be eae tS sane hi The above equation is only true for those values of « which make the series convergent. 1 PAVE ean tis oe ee (2) Exp Fj aay i 1 Vlas law 5 1. 5e a] a Pee ae — 4 4 yt ee ee eee (—g)et+ Ort + ae x + pe 1°59 aay ie 2 Bot cage +4 aa Saletan at if x is so taken that the series is convergent. A root may often be extracted by means of an expansion. (3) Extract the cube root of 344 to six decimal places. 1 344 = 343/14 —)=7(14— ( oe (a cau VG ae eee 343 t ( 1 1 4G—)) (ae 7( 8 alee, e183 a ) 7(1 + 0,000971815 — 0000000944), = 7.006796. RA (4) Find the eighth term of (2 -- rae me 4V/x Here a=%, Bie oe = n= 1 ye 4Vu 423 2 : SAMO Shi . . 11) ee eae Sen The term 18 iy Se Ree) MLA) BOSSE We 2 Sie ; 12:3°4:5°6°7 hy or BINOMIAL THEOREM. Exercise 113. Expand to four terms. - + x)? mkt a). . (1+2)7?. Peri 25 . (l—x)- 16: 16. ks 18. 19. 20. 21. 22. 23. 24. 25. 26. 6. (L+2)}. ll. (224 8y)?, 7 (4a) fh, 12. (2x+8y) 4, 1 see ae Ea ee LS ( Vae—ax 9. (1+ 52)”. 3 14, Aha : 10. (1+52)8. V(a—zx) Find the fourth term of (< seh ) IV x Find the fifth term of ——4—_. Via — 22) Find the third term of (4 — 7 x)i. Find the sixth term of (a — 2ax)s Find the fifth term of (1 —22x)~?. Find the fifth term of (1 — x)~*. Find the seventh term of (1 — x)? Find the third term of (1+ 2) an, Find the fourth term of (1+ x)~ é 2 a ‘ Find the sixth term of (2 -;) Find the fifth term of (Qu —8y)~#. Find the fourth term of (1 —52)~ 3 * CHAPTER XXV. LOGARITHMS. 381. If the natural numbers are regarded as powers of ten, the exponents of the powers are the Common or Briggs Logarithms of the numbers. If A and B& denote natural numbers, a and 6 their logarithms, then 10°= A, 10°= B; or, written in logarithmic form, log A =a, log B=0. 382. The logarithm of a product is found by adding the logarithms of its factors. For AX B= 10 * 10 = ig Therefore, log(A x B)=a+b=logA+ log B. 383, The logarithm of a quotient is found by subtracting the logarithm of the divisor from that of the dividend. For ay ee TOs: Therefore log A _ a—b=logA—log B ’ 5 B S ols 384, The logarithm of a power of a number is found by multiplying the logarithm of the number by the exponent of the power. For A" = (10*)" = 10”. : Therefore, log A"*=an=nlog A. LOGARITHMS. 343 885. The logarithm of the root of a number is found by dividing the logarithm of the number by the index of the root. For VA = V108 = 10" Therefore, log VA= a log A n n 386, The logarithms of 1, 10, 100, etc., and of 0.1, 0.01, 0.001, etc., are integral numbers. The logarithms of all other numbers are fractions. mince, 10°= 1; 107° (=7,;) =0.1, 10F== 10, 107° (=;4,) =0.01, 10? = 100, 10-° (= apy) = 0.001, therefore log 1=0, logO.1 =—l1, oo 1 log 0.01 =— 2, log 100 = 2, log 0.001 = — 3. Also, it is evident that the common logarithms of all numbers between land 10willbe O- a fraction, 10 and 100 will be 1-+ a fraction, 100 and 1000 will be 2-+ a fraction, land 0.1 will be —1-a fraction, 0.1 and 0.01 will be —2-+ a fraction, 0.01 and 0.001 will be — 8-+a fraction. 387. If the number is less than 1, the logarithm is nega- tive (§ 386), but is written in such a form that the fractional . part is always positwe. 888, Every logarithm, therefore, consists of two parts: a positive or negative integral number, which is called the characteristic, and a positwe proper fraction, which is called 344 SCHOOL ALGEBRA. the mantissa, Thus, in the logarithm 3.5218, the integral number 8 is the characteristic, and the fraction .5218 the mantissa. In the logarithm 0.7825 — 2, which is sometimes written 2.7825, the integral number — 2 is the character- istic, and the fraction .7825 is the mantissa. 389, If the logarithm has a negative characteristic, it is customary to change its form by adding 10, or a multiple of 10, to the characteristic, and then indicating the sub- traction of the same number from the result. Thus, the logarithm 2.7825 is changed to 8.7825 —10 by adding 10 to the characteristic and writing —10 after the result. The logarithm 18.9278 is changed to 7.9273—20 by adding 20 to the characteristic and writing — 20 after the result. 390. The following rules are derived from § 386: Rute 1. If the number is greater than 1, make the characteristic of the logarithm one wnit less than the num- ber of figures on the left of the decimal point. Rute 2. If the number is less than 1, make the charac- teristic of the logarithm negative, and one unit more than the number of zeros between the decimal point and the first significant figure of the given number. Rute 38. If the characteristic of a given logarithm is positive, make the number of figures in the integral part of the corresponding number one more than the number of units in the characteristic. Rute 4. If the characteristic is negatwe, make the num- ber of zeros between the decimal point and the first sig- nificant figure of the corresponding number one less than the number of units in the characteristic. Thus, the characteristic of log 7849.27 is 3; the character- istic of log 0.037 is —2=8.0000—10. If the characteristic LOGARITHMS. 345 is 4, the corresponding number has five figures in its integral part. If the characteristic is — 3, that is, 7.0000 — 10, the corresponding fraction has two zeros between the decimal point and the first significant figure. 391, The mantissa of the common logarithm of any inte- gral number, or decimal fraction, depends only upon the digits of the number, and is unchanged so long as the sequence of the digits remains the same. For changing the position of the decimal point in a number is equivalent to multiplying or dividing the num- ber by a power of 10. Its common logarithm, therefore, will be increased or diminished by the exponent of that power of 10; and since this exponent is mtegral, the man- tissa, or decimal part of the logarithm, will be unaffected. mame 27196 — 10". 2.7196 = 10°45, 2G 6 1064. 0.27196 = 10%485-10 ee ttre. use ne 0027 Ob. == 1 een, One advantage of using the number ¢en as the base of a system of logarithms consists in the fact that the mantissa depends only on the sequence of digits, and the characteristic on the position of the decimal pount. 392. In simplifying the logarithm of a root the equal positive and negative numbers to be added to the logarithm should be such that the resulting negative number, when divided by the index of the root, gives a quotient of — 10. Thus, if the log 0.002? = 1 of (7.3010 — 10), the expres- sion 4 of (7.3010 — 10) may be put in the form 4 of (27.3010 — 30), which is 9.1003 — 10, since the addition of 20 to the 7, and of — 20 to the — 10, produces no change in the value of the logarithm. 346 SCHOOL ALGEBRA. Exercise 114. Given: log2=0.38010; log8=0.4771; log 5=0.6990 ; log 7 =0,8451. Find the common logarithms of the following numbers by resolving the numbers into factors, and taking the sum of the logarithms of the factors: 1. log 6. 5. log 25. © 9. log 0.02E "iSie icp 2. log 15. 6. log 30. 10. log 0.85. 14. log 16, 3. log 21. 7. log 42. 11. log 0.0035. 15. log 0.056. 4. log 14. 8. log 420. 12. log 0.004. 16. log 0.63. Find the common logarithms of the following : iF 17; 2 <9) 90.58 laa.05t. 26. 77. 29. 5%, 18: 5X 24.) o3; agate Wied. be 30. 27. 19.47! © 22) 5ileke o50dos. 28, 310) ates 3938. The logarithm of the reciprocal of a number is called the cologarithm of the number. If A denote any number, then colog A = log =log1— log A (§ 383) = — log A (§ 386). Hence, the cologarithm of a number is equal to the log- arithm of the number with the minus sign prefixed, which sign affects the entire logarithm, both characteristic and mantissa. In order to avoid a negative mantissa in the cologarithm, it is customary to substitute for — log A its equivalent (10 — log A) — 10. Hence, the cologarithm of a number is found by subtract- ing the logarithm of the number from 10, and then annexing — 10 to the remainder. LOGARITHMS. 347 The best way to perform the subtraction is to begin on the left and subtract each figure of log A from 9 until we reach the last significant figure, which must be subtracted from 10. If log A is greater in absolute value than 10 and less than 20, then in order to avoid a negative mantissa, it is necessary to write —log A in the form (20 — log A) — 20. So that, in this case, colog A is found by subtracting log A from 20, and then annexing — 20 to the remainder. (1) Find the cologarithm of 4007. 10 — 10 Given: log 4007 = 3.6028 Therefore, colog 4007 = 6.3972 —10 (2) Find the cologarithm of 108992000000. 20 en) Given : log 103992000000 = 11.0170 Therefore, colog 103992000000 = 8.9830 — 20 If the characteristic of log A is negative, then the subtra- hend, —10 or — 20, will vanish in finding the value of colog A. (3) Find the cologarithm of 0.004007. 10 —10 Given: log 0.004007 = 7.6028 — 10 Therefore, colog 0.004007 = 2.3972 By using cologarithms the inconvenience of subtracting the — logarithm of a divisor is avoided. For dividing by a num- ber is equivalent to multiplying by its reciprocal. Hence, instead of subtracting the logarithm of a divisor, its colog- arithm may be added. 348 -- - $CHOOL ALGEBRA. 5 4) Find the logarithm of (4) Find the logarithm o 0.002 5 log D008 2 log 5 + colog 0.002. ~ log 5 = 0.6990 colog 0.002 = 2.6990 log quotient = 3.3980 0.07 O3 2 (5) Find the logarithm of log x = log 0.07 + colog 2°. log 0.07 = 8.8451 — 10 colog 2° = (10 — 3 log 2) — 10 = 9.0970 — 10 log quotient = 7.9421 — 10 Exercise 115. Given: log2=0.38010; log 3=0.4771; log 5 = 0.6990; log 7 = 0.8451; log 11 = 1.0414. Find the logarithms of the following quotients: are Py tk 15, 2:08. 22. oy 7 ie 44 gi 3 2. = 9. a 16. a 23. oS : 2 3. 2. 10. = 123 24, ae 7 2 4. x iL 18. = 25. SS 3 5. a. 12. 2 19) 26nd 6. 1s iS; sa 20. = 27. aaa : 5 re z 14. ce 21. o 28. ar LOGARITHMS. 349 394, Tables. A table of fowr-place common logarithms is given at the end of this chapter, which contains the com- mon logarithms of all numbers under 1000, the decimal point and characteristic being omitted. The logarithms of single digits, 1, 8, etc., will be found at 10, 80, ete. Tables containing logarithms of more places can be pro- cured, but this table will serve for many practical uses, and will enable the student to use tables of five-place, seven- place, and ten-place logarithms, in work that requires greater accuracy. In working with a four-place table, the numbers corre- sponding to the logarithms, that is, the antilogarithms, as they are called, may be carried to four significant digits. 395. To Find the Logarithm of a Number in this Table. (1) Suppose it is required to find the logarithm of 65.7. In the column headed ‘‘N” look for the first two significant figures, and at the top of the table for the third significant figure. In the line with 65, and in the column headed 7, is seen 8176. To this number prefix the characteristic and insert the decimal point. Thus, log 65.7 = 1.8176. (2) Suppose it is required to find the logarithm of 2034’. In the line with 20, and in the column headed 83, is seen 3075; also in the line with 20, and in the 4 column, is seen 3096, and the difference between these two is 21. The dif- ference between 20300 and 20400 is 100, and the difference between 20300 and 20347 is 47. Hence, 45 of 21 = 10, nearly, must be added to 3075; that ig, log 20347 = 4.3085. (3) Suppose it is required to find the logarithm of 0.0005076. In the line with 50, and in the 7 column, is seen 7050; in the 8 column, 7059: the difference is 9. The 350 SCHOOL ALGEBRA. difference between 5070 and 5080 is 10, and the difference between 5070 and 5076 is 6. Hence, ;5 of 9=5 must be added to 7050; that is, log 0.0005076 = 6.7055 — 10. 396, To Find a Number when its Logarithm is Given. (1) Suppose it is required to find the number of which the logarithm is 1.9736. Look for 9786 in the table. In the column headed ‘“N,” and in the line with 9736, is seen 94, and at the head of the column in which 9786 stands is seen 1. Therefore, write 941, and insert the decimal point as the characteristic directs; that is, the number required is 94.1. (2) Suppose it is required to find the number of which the logarithm is 3.7936. Look for 7936 in the table. It cannot be found, but the two adjacent mantissas between which it lies are seen to be 7931 and 7988; their difference is 7, and the difference be- tween 7931 and 7936 is 5. Therefore, # of the difference between the numbers corresponding to the mantissas, 7931 and 7938, must be added to the number corresponding to the mantissa 7931. The number corresponding to the mantissa 7938 is 6220. The number corresponding to the mantissa 7931 is 6210. The difference between these numbers is 10, and 6210 + 3 of 10 = 6217. Therefore, the number required is 6217. (3) Suppose it is required to find the number of which the logarithm is 7.3882 — 10. Look for 8882 in the table. It cannot be found, but the two adjacent mantissas between which it les are seen to be 3874 and 8892; the difference between the two mantissas is 18, and the difference between 3874 and the given man- tissa 3882 is 8. LOGARITHMS. ODL The number corresponding to the mantissa 3892 is 2450. The number corresponding to the mantissa 3874 is 2440. The difference between these numbers is 10, and 2440 + 58 of 10 = 2444. Therefore, the number required is 0.002444. Exercise 116. Find, from the table, the common logarithms of : Peo) ~ “4. ‘7808. Te eNOS: 10. 000.5234. 2. 201, 5. 4825. re ES 11. 0.01423. 3. 888. 6. 8109. 9. 0.00789. 12, 0.1987, Find antilogarithms to the following common logarithms: 13. - 4.1482. 15. 2.3177. 17. 9.0380 — 10. 14. 3.5317. 16. 1.3709. 1B eA ate — 5k). 397, Examples. (1) Find the product of 908.4 x 0.05392 x 2.117. log 908.4 = 2.9583 log 0.05392 = 8.7318 — 10 log 2.117 = 0.3257 2.0158 = log 103.7. Ans. When any of the factors are negative, find their logarithms with- out regard to the signs; write — after the logarithm that corresponds to a negative number. If the number of logarithms so marked is odd, the product is negative; if even, the product is positive. — 8.3709 x 834.637 7308.946 log 8.3709=0.9227 — log 834.637 = 2.9215 He colog 7308.946 = 6.1362 — 10 + 9.9804 — 10 = log — 0.9558. Ans. (2) Find the quotient of 352, SCHOOL ALGEBRA. (3) Find the cube of 0.0497. log 0.0497 = 8.6964 — 10 Multiply by 3, 3 6.0892 — 10 = log 0.0001228. Ans. (4) Find the fourth root of 0.00862. log 0.00862 = 7.9355 — 10 Add 30 — 30, 30, — 30 Divide by 4, 4.)37.9355 — 40 9.4839 — 10 = log 0.3047. Ans. 5} 8.1416 x 4771.21 x 2.71832 80.108! x 0.48432 x 69.897 log 3.1416= 0.4971 = 0.4971 log 4771.21 = 3.6786 = 3.6786 dlog 2.7183 = 0.4343+2 =0.2172 4colog 30.103 = 4(8.5214 — 10) = 4.0856 — 10 dcolog. 0.4343 = 0.3622+2 =0.1811 4Acolog 69.897 = 4(8.1555 — 10) = 2.6220 — 10 11.2816 — 20 30 — 30 5 ) 41.2816 — 50 8.2563 — 10 = log 0.01804. Ans. (5) Find the value of 398. An exponential equation, that is, an equation in which the exponent involves the unknown number, is easily solved by Logarithms. Ex. Find the value of x in 81*= 10. 8lz = 10, . log (81*) = log 10, x log 81 = log 10, 10g 10 _ 1.0000 = ——_ = 0,524. Ans. log 81 1.9085 LOGARITHMS. 353 Exercise 117. Find by logarithms the following products : 1. 948.7 x 0.04887. 5. 7564 x (— 0.008764). 2. 8.409 x 0.008763. 6. 3.764 x (— 0.08349). 3. 830.7 x 0.0003769. 7. —5.845 x (— 0.00178). 4. 8.489 x 0.9827. 8. — 8045.7 x 73.84. Find by logarithms: 9, 7065 stiri 45 0.076540) 5401 83.94 x 0.8395 See 652 yo, 212 (—6.12) x (— 2008), — 0.06875 365 x (— 531) x 2.576 13. 0.1768, 17, (44)" 21. 2.563%, 14, 1.211". 18. 906.8%, 22. (83)%*, 15. 11%. 19. (284)%, 23. (581), 16. (73)". 20. (748,)™. 24, (929)>. 25. 1) 0.0075? x 78.84 x 172.44 x 0.00052 42853 x 54.274 x 0.001 x 86.792 og. .1]0.03271? x 3.429 x 0.7752 32.79 x 0.000371! ‘| 7.126 x V0.1827 x 0.05738 Qi. N \/0.4346 x 17.38 x 0.006372 Find x from the equations: gue.) = 10, RU i herta= 4240 A 32. (0.4)-% = 3. 29. 4% = 20. 31. (1.3)*=42, 33: (0.9) 7*=2 004 10 | 0000 ll | 0414 12 | 0792 13 | 1139 14 | 1461 1761 2041 2304 2553 2788 3010 3222 3424 3617 3802 3979 4150 4314 4472 4624 0086 0492 0864 1593 | SCHOOL ALGEBRA. 0128 0531 0899 1229 1523 | 1553 1818 2095 2355 2601 2833 3054 3263 3464 3655 3838 4014 4183 4346 4502 4654 1847 2122 2380 2625 2856 3075 0170 0569 0934 1271 1584 1875 2148 2405 2648 2878 3096 3284 | 3304 3483 | 3502 a 3692 3856 4031 4200 4362 4518 4669 3874 4048 4216 4378 4533 4683 0212 0607 0969 1303 1614 1903 2175 2430 2672 2900 3118 3324 3522 3711 3892 4065 4232 4393 4548 4698 4914 5051 5185 5315 5441 5563 5682 5798 5911 4800 4942 5079 5211 5340 5465 5587 5705 5821 5933 4814 4955 5092 5224 5353 4829 4969 5105 5237 5366 4843 4983 5119 5250 5378 5478 5099 5717 5832 5944 5490 5611 5729 5843 5955 5502 5623 5740 5855 5966 6021 6128 6232 6335 6435 6042 6149 6253 6355 6454 6053 6160 6263 6365 6464 6064 6170 6274 6375 6474 6075 6180 6284 6385 6484 6532 6628 6721 6812 6902 6990 7076 7160 7243 7324 | 7332 7084 7168 eas feeeee i auees Pale elie 6951 6646 6739 6830 6920 7007 7093 7177 7259 7340 AEAee z faa | 6561 6656 6749 6839 6928 7016 7101 7185 7267 7348 6571 6665 6758 6848 6937 7024 7110 7193 7275 7356 6580 6675 6767 6857 6946 7033 7118 7202 7284 7364 LOGARITHMS. 855 CHAPTER XXVI. GENERAL REVIEW EXERCISE. l'a=6 b=5 cu—4 d= 3. nnd eee 1. 2. Vb + ac+ Ve — 2ac. 3. VB? tact Ve —2ac. a — Vb +a0c 4 e+~vV/di te 2a—Vb—ae c+ 2d(d*—¢) Find the value of 5. 10. 5 12. 13. ene abe —+~ when «= i anger when 2 ony, Mb) ales een when PR ideal a b a a os x _a(b—a) oo i ak maar: , when 2 TOK (a+) (6+2)—a(b+¢)+2*, when deeb a(l+6)+o62 a 1 yi a(l1+6)—be a—2ba renin 5 - Add (a —6) 2°+ (b —c) y+ (e—a) 2’, (b—¢) 2+ (e—a)v + (a — b)2, and (ec —a)2*+ (a— b)y¥’+(6—e)2’. Add (a+6)a+(b+c)y—(e+a)z, (6+e)z2+(e+a)x —(a+b)y, and (a+c)y+ (a+ b)z—(6+0e)2. Show that 23+ 7+ 2— 32yz=0, if#+y+z2=0. Show that 2*— 87°— 2727— 18 xyz =0, if = 2y+3z2. GENERAL REVIEW EXERCISE. ooT Simplify by removing parenthesis and collecting terms: 14. 8a—2(6—c)—[2(a—b)—38(e+a)]—[9e—4(ce—a)]. 15. 7(2a+ 6) —§19b — [18(e—a) + 12(6 —c)}}. 16. x—j4y+[8(2—2)—(#4+ 2y)]— (2y+2—22)}. 17. 14+ 2§4+4—38[4+5—4(2+1])}}. 18. 10% —§4[52—38(2—1)]—38[42—3(¢#+1)}}. 19. 32°—{22?—(82—7)—[2 2°—(8 2—2’)|—[5-—(22°—4 2) }}. 20. (% —2)(x — 8) — (w— 7)(x—1) + (@— 1) (x — 2). 21. (2+ y) — 22(32-+2y) —(y—2z)(—2+y). pomee (2a —(3a—b)) + 3a(2d—8a—") Resolve into lowest factors : 23. (x+y)y— 42’. 27. 92? —8aty?+ yy. 24, (a? +") —42°y’. 28. OF eae yy 25. a&—F—e’+2be. 29. 8lat—1. 26. (#—y?— 2)?— 472. 30. a*— 0b”, 31. (a —b’+ ¢ —d’) — (2ac — 26d)’. 32. x’ —192 + 84. 40. 2-7. 33. 42°?+ Qa — 36. 41. 8+ a°2°, 34. 2? — 8x+ 15. 42. 2°— a", 35. 92? — 1502+ 600. 43. 270° — 64. 36. 562°+ 3xy — 20y’. 44, 2 — 327/. 37. 1227+ 38744 21. 45. a? — b°. 38. 33 — 14xz— 402’. 46. 2° + 10247". 39. 627+52—4. 47. a’ —(b+c)*. 48. 82°—62xy(2x4+3y) + 27y’. 358 ’ SCHOOL ALGEBRA. Find the H.C. F. of 49. 6a*—2a°+ 92?+ 92-4, and 92*+ 802? — 9. 50. 32°—52°+2, and 22°— 527+ 3. 51. wv — 98a — 808, and 2*® — 212’?4 13812 — 281. 52. at —2a°+ 42?—62+8, and a*— 22°— 22°+ 62 — 3. 53. a2 —4¢ — 24-2242, and 2-2 ae 54. 32° +102?+ Ta —2, and 32°4 1382°4+172+6. 55. 4a*— 9a?+6x2—1, and 62°—T2’?+1. 56. 2 +1la—12, and 2° 4+ lla’ 54. Find the L.C.M. of 57. 42°+42—8, and 427+ 24-6. 58. 2° — 47’, and 2? + 2y— 6y’. 59. Ta’x(a—x), 2Zlar(a’?— 2’), and 12a27(a+ 2). 60. 92°—x— 2, and 82° — 102?— Ta —4. 61. 2?—52+6, 2—424+3, and 2 — 382-4 2. 62. 2a°-+ 5ary— 5ay+y’, and 2a°— Ta’y-+ day?— y’. Simplify : sn age Sea ee 63. inset a(x—2). “2(¢+4+1). #—2—2 BA) 1 si cg OR Sec CA I l—x l+e 1l-2# 142 65 z—l1 2-2) 4 x+5 (e+2)(a@+5) («+5)@—-1) («—-1]l@+2) Fs MMB Wiese cate ee axc—a ax+t2e 2wv#+ar—2e 67 a x—l x—8s (w+ Be 1a ee eee (2—«2)(1—2) 68. 70. (Qs T2. 73. 80.. GENERAL REVIEW EXERCISE. 4 12 4 12 1 ETE oc Ely ( Petes) Ry) a) eee tt ( x eae ae Yh Wey y feve a-b. T/atd 1 } a’— bt 2(ae—0?) 2\e+s? a+b 1 i, dees | fap Ep es elias ee | es pateee a B ew b 1: 1 75 : i Peon 1. a@ x—1-- haa 1 a ‘a 1 sires l—a l+a ; My 2) 76. —#/1—40 —2)] mee a i—4[1-+0.—@)] Fe SL Na LS Solve: 62+138 92+15 g 2e+1d a. 15 ape a, 5 78 9x +a, Dae be oy SS(7-— a). 2 4F a) Ra x zt+l «2-8 -w«-9 1 re == Ps : a—-2 x-—1 «2-6 2-T can 5a , 0.3 ) — apg! BM Se ee a | Mii y | z., 10 10z , 9 Ba || Tssnoe ais) 309 360 SCHOOL ALGEBRA. Aap aeayt Banee a vee Got See eee a ee me age . | 83. ne 3 4a, 2y 232 a bie ee — Seeahlen Bham eae San eh — — — — sg at C zy z a ») bz ty 225 | 20: aero e Lee) Se 84. 362: s5. 32%; 648; 81; (88); (Byy)*;. (1ah)*. 86. (0.25)?; (0.027)8; 49°; 32%; 81%", 87. 36-2; 27-3; () ?; (0.16) *; (0.0016) * ) Le Pe ee ars | —3\—-4 88. Interpretia (savas ea 2 (ae. Simplify : 1 s Rpg 1 1 Bai y RAL oe 89. a2 a3 CF 08° 998? igh? 5) Ee ee ie y a are red 1,24¢e 90. abc! x a 2b 3c *: abh2e FX ab 2e2d. ) 91. (2ab + 2be+ 2ac—a? —B'— 2) + (a? +? + c?). 92. V12; V8; V50; V16; 4250; Vi; Vi; VS. 93. 5W— 320; Vaid’; Va; Vax tat; 8154 2°. 94, 2V/18 — 8V8+ 2-750; V81 + V24— V192. 95. §V$+ V80—1V20; 8V 35, + 10-V 39 —2V38. Rationalize the divisor, and find the value of pe es Saal nba sue 100, 2a 2—V3 V2+V7 Vi+V7 97. 3 ; 99. V3+v2 101 2V'6 34-6 VE a2 ' 33 ee GENERAL REVIEW EXERCISE. 361 103. zg+38 2xr-—1 104. Petes 2 Gs pee hb ae 105. = es er a, ab goto cated 106. atbz etdz 107. J ba =a+b. : 7 a 109. 110. az 111. —— 112. x’ + 10ay = an r 9) ory —3y' = Va pps Eee a ry —xz'*=8 Form the equations of which the roots are : 116. a—b, a+. 117. a— 26, a+ 30. 118. a+26, 2a-+ 8. Solve: | 122. x*—527+4=—0. 123. eros B= 124. 92t'—132°+4=0. 125. 42*—172?+4=-0. 126. 22° —5x77+2=0; 119. 120. 121. 127. 128. 129. 130. 131. oben / at eae AK See OF Liv 8, V8, 2a° — 19a? 4+ 24=0, z‘—1=0. gee Vie), x? + 8x3 —9 = 0. 16z* —17a°>+1=0, ee 362 SCHOOL ALGEBRA. 132. 2a% —8x2t+1=0. 137. ott be tae. 133. BVz—5Vz+2=0. 138. 42 $324 27=0, 134. 6VWx—3V2—45=0. 139. 2”+382"—4=0, 135. QVet—5V2—74=0. 140. 32° —2ax*—a?=0. 136. 8V2+4V2—20=0. 141. V2x—~V2r2—2=0. 142. 83V92°+4V32—39=0. 143. V3ar+avV3ar —2a°=0. 144, 3VW2bx — 5bV2br — 28? =0. 145. V2+4+V8¢+1=V9e4+4. 146. Vd5¢7+1+4+ 2V42—3 =10V2—2. 147. V2 +2—8V32—5+ Vb24+1=0 148. Vll—2+vV8— 22—V214+ 227=0. ape: 3\ 5 150. (4 (22+-V3 2) (Va2+1+0')* (1+-22—a'—a")". 151. Expand to four terms (1—32) 4, (1— 4a) ?; 1 — gah? (@ 2a 152. Find the eighty-seventh term of (2a — y)”. 153. Resolve into partial fractions 3—2xe 3— 22 . ee 1—82+22?’ (l—az)(1—82)’ 1-2 3 — 2x 154. E d to five t Umeha °F. xpand to five ia a ENO eerie Ft a % 4“ * ak .s t.. a Aa = —_ a ae eel 5 Seetiel UNIVERSITY OF ILLINOIS-URBANA 12.9W48S1891 C001 A SCHOOL ALGEBRA BOSTON yg nad ert By W i ry 6 eacueas gris {ee ween te id LF aA Ve &