• \_ !. Digitized by the Internet Archive in 2008 with funding from Microsoft Corporation http://www.archive.org/details/briefcourseinalgOOmancrich Brief Course in- Algebra BY RAYMOND E. MANCHESTER Professor of Mathematics. State Normal School. Oshkosh, Wis. Author of the Teaching of Mathematics SYRACUSE, N. Y. C. W. BARDEEN, PUBLISHER Copyright, 1915, by C. W. Bardeen ERRATA P. 41 10th line for 3 read 2 19th line for 3 read 2 81 12th line for -x 2 read +x 2 133 last line for — J read =| EDUCATION DEPT. OUC TO /Rang Sane /fcancbester. 54 1264 During the past year or two there has been a grow- ing demand for a twenty weeks' course in Algebra which will not only serve as a preparatory course for more advanced mathematics, but will have an indi- vidual unity as well. It must offer an opportunity to the short term student to get a well established notion of the subject through quadratic equations. This text is arranged to satisfy such a course. All unnecessary theorems, proofs, and processes have been omitted. The problem lists are very short and are to be supplemented by the instructor from the experience of the class. It is, in fact, a course in generalized arithmetic, arranged to face the de- mands made by the ninety per cent of students who expect to leave school either during or at the expira- tion of the high school course. RAYMOND E. MANCHESTER Oshkosh, Wis. June 1, 1915 SECTION I Treating of the relationship existing between Algebra and Arithmetic. SECTION II The Fundamental Operations. SECTION III Fractions. SECTION IV Powers and Roots. SECTION V Quadratic Equations. Brief Course in Algebra SECTION I Treating of the relationship existing between Al- gebra and Arithmetic. A. Letter Symbols B. Negative Numbers C. The use of the equation Lesson 1 Algebra is an enlarged and continued course in arithmetic. As such it has a definite connection with the grade work in arithmetic and such elementary work in generalization as is attempted in the last year of grammar school. The enlargement is (1) an addition to the Arabic symbol system by letters of the alphabet, (to stand for undetermined numbers); (2) an increase in the usefulness of the symbol system by using the plus ( + ) and minus ( — ) signs to qualify numbers (as well as for signs of operation) ; and (3) the use of the equa- tion as a means of solving problems (made possible by the letter symbols). The study of arithmetical operations is continued with such revisement to the rules and laws as is necessary to cover the enlarged symbol system. LETTER SYMBOLS (1) Throughout the work in arithmetic, number ideas have been expressed by either a word (three), (9) 10 BRIE? COURSE IN ALGEBRA a symbol (3); or a picture (111). But the word, the symbol, or the picture has stood for a definite number, so that the student, upon hearing the word, seeing the symbol, or seeing the picture, has had a distinct group brought to the attention. These names and symbols have been memorized with the connections always made clear. (2) In any problem, however, there is a certain number (or there are in some cases certain numbers) not definitely known. There being no definite num- ber idea, no name, Arabic symbol, or picture can stand for it. We know that the number, whatever it may be, exists, and we hope to determine its value by the solution of the problem. We speak of this number in various ways, such as, "the answer", "the unknown part", "the number to be found", or "the result". Example — In a basket there are three times as many pears as apples. The total number of both is 28. To find the number of apples. In this problem the unknown is the number of apples. We know that there is a certain number of apples, but cannot set down a symbol for the number until the problem has been solved. Example — A certain number added to its half and its double gives a sum of 21. Find the number. Here the unknown number is the number itself. We cannot put down a symbol to stand for it until the problem has been solved. (3) In our course in algebra, we do have a symbol for such unknown numbers. It being impossible LETTER SYMBOLS 11 to write down any one of the Arabic symbols until its value has been determined, it is necessary to use other symbols. Because of the familiarity of all students with the letters of the alphabet, it has be- come a custom to use one of these letters as a symbol until enough is known about the number to make it possible to use one of the Arabic symbols. Example — In the first problem mentioned above we might Let # = the number of apples. Then 3x = the number of pears. Adding, 4# = the number of apples and pears. Or, 4x = 28. Or, x = 7, the number of apples. At the outset, we did not know that there were 7 ap- ples, but we did know that there was a certain number of apples. We could not have used the symbol (3) or (5) or (9) or any of the other Arabic symbols with the certainty that the statement would be correct, so we allowed (x) to stand for the number, whatever it might be. So it is that we may use the letters to stand for numbers until the numbers have been so established as to permit us to use the Arabic symbols. (4) It follows that we may speak of (a) houses, (x) horses, (y) trees, or (b) books when there is no definite number idea in our minds. We know that there is a certain number of houses, or horses, or trees, or books, but we do not know what it is. There may be conditions given which will enable us to determine the number, and until such a determina- tion is made the letters may be used. 12 BRIEF COURSE IN ALGEBRA The advantage of having such an addition to our symbol system is at once apparent when the solution of a difficult problem is attempted. Example — A young man wishes to build a bookcase having 3 shelves arranged so that the lower one may contain larger books than the other two. Suppose he wishes to decrease each shelf space by two inches from the lower to the upper and still keep the book- case in proportion. Given the inside measurement as 32 inches, to find the centers for the shelves. Let x = space of the lower shelf. then x — 2= space of the next shelf. and x — 4 = space of the upper shelf. Adding, x+ (x — 2) + (x — 4) = total space. Adding, x+x-2+x-4= = 32. Or, 3x-6 = 32. Or, 3x = 3S. Or, # = 12§ space of lower shelf. Or, x — 2 = 10 J space of middle shelf. Or, x — 4 = 8f space of upper shelf. (5) It is evident, therefore, that the letter stands for a general number while the Arabic symbol stands for the particular number. We may speak of (a) apples as well as (5) apples, providing we understand that the (a) is undetermined. (6) If we are to use the letters to any great ex- tent in number operations, it is necessary for us to review the arithmetical operations and readjust our rules and laws to accommodate the added sym- bols. Taking these operations in order, those of addition, subtraction, multiplication, and division LETTER SYMBOLS 13 will be discussed first. The discussion will be very brief, however, owing to the fact that the operations are to be considered under separate topics. Problems 1 What are the Arabic symbols? 2 What do they stand for? 3 Does the Arabic symbol stand for a particular or a general number? 4 When letters are used as symbols do they stand for particular or general numbers? 5 What does the expression generalized arith- metic mean? 6 Name other points of enlargement and general- ization. 7 Can you state an advantage in having a general number symbol? 8 What are the definite number symbols? 9 Should the algebra be a continuation of the arithmetic ? 10 Do you understand the distinction between a number and the symbol standing for it? Discuss this point. 11 Suppose 2x horses cost $10. What would x horses cost? 12 If (a) boys are to have 12 apples divided among them, what is the share of each boy? Lesson 2 Definitions and Rules To facilitate discussion, the following definitions should be memorized. 1 Multiplication is expressed by using the dot (.), and, in the case of the presence of literal factors, by writing the factors together. Ex. 3a- 2 = da Ex. aXb = ab = ab Ex. 3a-b = 3ab Ex. x-y-z=xyz 2 Any number or combination of numbers by signs of operation is called an algebraic expression. Ex. 2a-\-b-3 a An expression of one term is called a monomial. b An expression of two terms is called a binomial. c An expression of three terms is called a trinomial. d An expression of more than two terms is called a polynomial. 3 An algebraic expression not separated by the plus or minus sign is called a term. Ex. 4ab 2a-\-b — 3 is an expression made up of three terms. 4 Any one of a group of numbers is called a j actor of their product. Ex. 6ax has the factors 2, 3, a, x. 5 Any one of a group of equal factors is called a root of their product. (14) LETTER SYMBOLS 15 One of two equal factors is called the square root of their product. One of three equal factors is called the cube root of their product. One of four equal factors is called the fourth root of their product, etc. 6 A small number written at the upper right hand of a number is called an exponent, and expresses the fact that the number is taken a certain number of times as a factor. Ex. a 2 The a expresses the fact that (a) is used twice as a factor. 7 Any factor or product of two or more factors is called the coefficient of the remaining factors. Ex. 3a (3) is coefficient of (a) (a) is coefficient of (3) ( Note) We usually refer only to the Arabic symbol as the coefficient. Ex. 5x 2 y 5 is coefficient of (x 2 y) 5x 2 is coefficient of (y) 5y is coefficient of (x 2 ) x 2 y is coefficient of (5) x 2 is coefficient of (5y) y is coefficient of (5x 2 ) (Note) Ordinarily only (5) is referred to as the coefficient. 8 The product obtained by using a number two or more times as a factor is called a power. Ex. 2-2-2 = 8 (8) is the third power of (2) Ex. a a a a = a 4 (a 4 ) is the fourth power of (a) 16 BRIEF COURSE IN ALGEBRA 9 Signs of aggregation are those which are used to signify that certain expressions are to be considered as unified wholes. They are the parentheses (( )), the brackets ([ ]), the braces (\ [), and the vinculum Example 2x + (3y -f- z) = 2 3y-\-z is considered as a unified whole. Example 6a -\-(2b — c) — 3c = S 2b — c is considered as a whole. Example (x — y)— 4z x— y is considered as a whole. Example (6m-\-n — p)-\-3d—4:a 6m-\-n — p is thought of as a whole. Example (6a -\-b)—c = 6 6a-\-b is here considered as a whole. 10 If the sign of aggregation is preceded by a (+■) sign, the quantity within is added; if preceded by a ( — ) sign the quantity is subtracted. 11 In case the expression enclosed by the sign of aggregation is to be multiplied by a number, each term of the expression within is multiplied by the number. Example 2 (3x + 4) = 6x + 8 Example a(x—y)=ax—ay Example 2x (c — d) = 2xc — 2xd 12 The signs +, — , X, ••», — , are used as they have been used in arithmetic. 13 It often happens that letter symbols are given particular values in an expression to find the value of the expression. LETTER SYMBOLS 17 Example — If a = 2, 6 = 3, c = 4 To find the value of 3a-\-b — c+4:ab-\-2b Substituting values, 3(2) + (3) -(4) +4(2) (3) + 2(3) Or, 6+3-4+24+6 Or, 35 Problems 1 What are the signs of operation? 2 What are the signs of aggregation? 3 Define exponent, power, root. 4 Define coefficient, literal coefficient, numerical coefficient. 5 Define monomial, binomial, trinomial, poly- nomial. 6 What new symbol for multiplication has been introduced ? 7 What is the sign of equality. 8 If a sign of aggregation includes certain terms of an expression, what does this signify? 9 How is a root indicated? 10 Do you find that words have the same meaning as in arithmetic? 11 Discuss fully the relationship existing between arithmetic and algebra. Lesson 3 Discussion of the letter symbols with regard to the four fundamental operations. Addition: (1) If 4 is added to 3 it is understood that four units are to be grouped with three of the same kind whenever application is made to a problem. It is a matter of common agreement that the result is a group made up of 7 units. All Arabic symbols are names of certain groups. (2) In abstract numbers the symbols are con- sidered apart from particular objects, and the state- ment is made that 4+3 = 7, a law which holds for all cases regardless of the objects considered. (3) In adding letter symbols there is no possi- bility of saying that (a) units added to (b) units equals a group named by an Arabic symbol because of the fact that (a) and (b) stand for undetermined numbers. (4) To add (a) and (b) then, it is only possible to indicate the sum a+0, but in case (2a) units are to be added to (3a) units, it is possible to write the sum as, 2a+3a = 5a. This last is true, owing to the fact that, even though (a) stands for an undetermined number, twice this undetermined number combined with three times this undetermined number equals five times the undetermined number. In the ex- (18) LETTER SYMBOLS 19 pressions (2a) and (3a), (a) is considered the common factor. (5) We enlarge the rule for addition, therefore, to cover addition of numbers having letter factors, (literal factors). We say that numbers having like literal factors may be added by adding the coefficients of the literal factors and writing the sum as the co- efficient of the common factor. Example 5b+4:b = 9b (Note) (b) is common, so the coefficients (5) and (4) are added, giving the sum (9). This number is written as the coefficient of the common factor (b) Example 6a + 4a = 1 Oa Example lab -f 4a& = 6ab (Note) Here the two factors (a) and (b) are common and may be considered as forming the product (a) times (b) or (ab) , which product is common to both numbers. Therefore, twice the product added to four times the product equals six times the product. Example 2a ■ -f 4a 2 = 6a > (Note) In this problem the exponent (2) in- dicates that the number (a) is taken twice as a factor, so that 2(aa)-f- 4 (a -a) =6 (a a) That is, two times (a -a) plus four times (a a) equals six times (a. a), or 6a 2 , — the 2 indicating that (a) is taken twice as a factor. (6) The addition of expressions involving two or more terms is accomplished by arranging the terms so that the terms having like factors may be added together. 20 BRIEF COURSE IN ALGEBRA Example — Add 3a 4-46 4- 6c and 4a 4- 2& 4- 8c Sum is 7a +66 4- 14c ( Note) Here it will be noticed that the numeri- cal coefficients of the like factors are added and the sum written as the coefficient oi the like factors. Example Add 6xy+4w-{-&z 2 and 4w-\-3z 2 -\-4xy This problem must be rearranged so that terms with like factors are in columns. 6xy-\-4\ 4x^4-4^ w+Sz 2 w-\-3z 2 Adding, lG*y+& +Sp 4:cd+2m 7 +3p Subtracting 2cd + 2m 2 + 5p (5) It will be noted that the conditions for sub- traction are the same as for addition. (6) Terms having the same literal factors with the same exponents are called "like terms". (7) Only like terms may be added or subtracted. Problems 1 From 6 10 16 18 160 take 3 5 9 12 40 2 From 4a 16a6 6a 2 b 9x s y Sxy take 2a 3ab 4a 2 b 10x*y Ixy 3 Subtract 4a+6cd+8d> from 8a+ 10cd+4d 2 4 Subtract 3x+4=y'-\-6zw from 6x-{-8y'-\-lQzw 5 Subtract 2m+n+3rs from 12ra-f-4w-|-8rs 6 Subtract 16kl+23x i y from 26kl+40x'y 7 Subtract 10a>+3b'+6cd+9e t from 15a ' + Sb* + llcd-f-14e* Lesson 5 Discussion of the four fundamental operations (continued) Multiplication : (1) To multiply numbers together having literal factors, it is only necessary to multiply the numerical coefficients together and indicate the multiplication of the literal factors by writing them together. Example 2a ■ 4b — Sab Example 5cd-6cx = 30c 7 dx ( Note) The multiplication of these two numbers involves two (c) factors. Write 30 as the coeffi- cient of the product of the literal factors 30cdcx. There being two (c) factors, it is customary to write them as (c 2 ), showing that (c) is taken twice as a factor, or 30c 2 dx Example lab ■ 5a 3 b > = 20a 4 b 3 (Note) Here 5-4 equals 20 for the numerical coefficient, and writing this number as the co- efficient of the indicated product of the literal factors we have 20aba 3 b 7 Here it will be noted that (a) is taken a total of 4 times as a factor, and {b) a total of three times as a factor, so the product is written 20a A b 3 (Note) As stated before, the exponent indicates the number of times a number is used as a factor. (24) LETTER SYMBOLS 25 (2) It will also be noted that to multiply two like literal factors together it is necessary to add the exponents. Example: x * • x 2 = x 6 (3) In case an expression of two or more terms is to be multiplied by a single term, the process is as follows : Example: Multiply 3x 2 y-{-4:xy 3 +3w by Sxy 9x 3 y 2 + 12x 2 y*+ 9xyw (Note) In this case each term of the larger expression is multiplied by 3xy Problems Perform the following operations in multiplication. 1 Multiply 4a by 2 2 Multiply 16b by 3 3 Multiply 14c by +12 4 Multiply +9ab by 7 5 Multiply +lla 2 cby +8 6 Multiply 4:X-\-3y+z by 4x 7 Multiply 3w-\-6z 7 +4y by 5w+2z 8 Multiply 16a 2 +4b+c+2d by lab 9 Multiply +21w+3w 3 -H£ by ll + 2w Lesson 6 Discussion oj the four fundamental operations (continued) Division : 1 Division is denned as the process of finding one of two factors when their product and one factor are given. Example: Given the problem Sa 2 b-v-lab We may determine the quotient by establishing the number which when multiplied by lab will give Sa 7 b. This number is evidently the product of those factors of 8a *b not contained in lab. 8a*b = 1.1.2.a.a.b lab =l.a.b 2,2 . a — factors not contained in l.a.b Therefore 4a is the quotient. Example 10a s b 2 + 5ab Mentally it is possible to determine the quotient. 10-^5 = 2 a 3 -r-a = a* b* + b = b therefore 1 0a * b ■ 4- Sab — la '* b (Note) The fundamental operations are dis- cussed fully in the next section. 2 To divide a polynomial by a monomial, divide each term of the polynomial by the monomial. (26) LETTER SYMBOLS 27 Example — (6xy 2 +4* 2 y+%x s y *) •*• 2xy 2xy\6xy 2 +Ax 7 y+%x 3 y* 3y +2x +4x 2 y z Problems Perform the indicated division. 8jc-j-2x, 16y+8y 14a4-+7a, 18xy++2xy 16a 2 b -r- 2a, 24a 6 & 4 -^8a 4 6 18a 4 & 6 -v-+9a 3 &>, 40x 3 :y 3 -hl0x;y J Perform the indicated divisions. 15xy, 12a 2 b, 12a 2 bc 3x 6ab +4tab +2Sc*d, -r-10m 4 w , + 180r's* ScH +2m'n +90rs THE NEGATIVE NUMBER Lesson 7 We have discussed the enlargement of the number system by the use of letter symbols. We will now discuss another enlargement of the number system not by the addition of new symbols but by the qualifi- cation of those we have. It is possible to qualify number symbols so that the same symbols may have two meanings. This is accomplished by using the signs (+) and ( — ) as signs of quality as well as for signs of operation. The sign (+) and ( — ) may be used other than as signs oj operation. In such cases the sign is con- sidered as a part of the symbol and is written with it. Let us take, for example, the reading of the ther- mometer. We are all accustomed to speak of degrees above or below zero, (zero being a starting point for measurement), so find it very convenient to indicate ten degrees above zero by writing the symbol (10) with the (+) sign preceding it, as follows (+10), and in like manner ten degrees below zero with the minus sign, ( — 10). Used in this way the signs do not indicate operations, but simply that measurement is made in one or the other of the two directions. + 10 degrees means ten degrees above zero. — 10 degrees means ten degrees below zero. Another example would be in measurement of dis- tance along a straight line with symbols for measure- (28) THE NEGATIVE NUMBER 29 ment in one direction used with the (+) sign connected and symbols for measurement in the opposite direc- tion with the ( — ) sign connected. -3 -2 -1 +1 +2 +3 The symbols stand for number groups as they do in arithmetic, but by the use of the (+) and (— ) signs it is possible to see at a glance the particular group referred to. If (+2), we know that the meas- urement is in one certain direction we may agree upon, and if a ( — 2) the measurement is in the op- posite direction. Another example is to consider a man's assets as (+) and his liabilities as ( — ). $20 to be received would be written $+20, while ten dollars to be paid out would be written $ — 10. In this case it is evi- dent that the man has ten dollars more receivable than he has payable. The balance would be $+10. The absolute value remains unchanged when these signs are used denoting quality. In order to avoid confusion, we will consider addition as combination. In the following examples please note that this idea of combination is stressed. Ex. Add 3b and -4b Result -b Ex. Add 5x and —3x Result +2* Ex. Add 15c and 2c Result 17c Ex. Add -8d and +d Result -Id It follows that to express the combination of two or more numbers additively, it is only necessary to connect the numbers by the signs preceding them. 30 BRIEF COURSE IN ALGEBRA Ex. To combine +8* and — 4x +Sx-4x = 4x Here the combination of Sx and — 4x gives a balance in favor of the (+) units of four units. To write +8jc — 4x = 4# is equivalent to subtracting 4# from 8x. Ex. To combine —5a and 2a. ( Note) If no sign precedes the symbol it is understood that the sign is (+). — 5a+2a = — 3a In such a problem as Sx — 4#+ 7x — 5x — Sx the process of combination calls for a combination of the (+) units, a combination of the ( — ) units, and finally a balancing of the two sums. There are (+10*) units and (— 17x) units, giving a balance of ( — 7x) units. Supposing +5a — 6b is to be added to +46 — 3a Arranging 5a — 6b -3a+46 Combining 2a — 2b Here the balance between (5a) and ( — 3a) gives (+2a), and the balance between ( — 6b) and (+46) gives ( — 2b). It should be remembered that negative numbers, (minus numbers) are meaningful only in relation to positive numbers, (plus numbers). Just as (up) has no meaning except in relation to (down) , or (east) except in relation to (west), or (right) except in rela- tion to (left), so this negative number has no meaning THE NEGATIVE NUMBER 31 except in relation to the positive number. Inasmuch as they have these opposite meanings, to combine additively is merely balancing them. ( Note) It is assumed that only numbers having like literal factors may be so combined. All other combinations are simply indicated. Ex. To combine additively (3a) and (—46) we simply write 3a -46 Problems 1 Add Sa-2b+c-3d+5x -6a+3b-6c+5d-8x 2 Add 17cd-3ax+10cy lScd+Aax— Scy 3 Add lxyz+3ab — 6cd —4xyz — 5ab+&cd 4 Add 4a + 36- 6c -3a+ 46- 4c -8a+146-ll<; +3a- 2b+ 6c 5 If a man's wealth is represented by $ — 500 what is he worth after increasing it by $+800? 6 If the temperature reading is —5° what is it after a rise of 7°. 7 A ship is at latitude 10°. She sails south 640 miles. What is her latitude? (60 miles = 1°). 8 What is meant by the temperature, +3°, -6°, +18°, -20° the latitude, +3°, -16°, +32°, -6°, 10° the date, -162, +1262, 1896, -460 Lesson 8 Negative Numbers Subtraction : In arithmetical subtraction we simply diminish a group named by a symbol by some certain number of units. Ex. 8-2 = 6 Here the group of 8 units is diminished by 2 units, giving us a remainder of 6 units. Suppose ( — 2) units is to be taken from (+8) units. The meaning here is that the negative quality of the (2) units is taken from them ,thus restoring them to positive units. So to subtract ( — 2) units from (+8) units the process is one of removal of the nega- tive quality before combination takes place. Thus 8-(-2) = 10 Or 7-(-4) = ll Or 2a-(-3a)=5a Suppose it is desired to subtract (+3) units from (-7) units. (-7)-(+3) = -7-3=-10 Ex. Subtract 2x — fy+c from —3x+6y — &c Arranging terms so that those with like literal factors are in the same columns, -3*+ 6y-8c +2x- 4y+ c -5*+10;y-9c (32) THE NEGATIVE NUMBER 33 Combining, we have ( — 3x) and ( — 2x), giving (-5*); (+6y) and (+4y), giving C+lOy); and (Sc) and (— c), giving ( — 9c) Thus it is that the rule in algebra is to change the signs oj the subtrahend and proceed as in addition. Problems 1 From Sc Sab 16ao lSbd -15rs take —46 — 4a& — Sac — Sbd —18rs 2 From IScd -3Sab -43mn 2Srs -30xy take — 2Scd — 40a6 60mn — 14rs — 43xy 3 Two boys catch 490 fish. One boy catches fifty fewer than the other. How many does each catch? 4 How many years between the dates +1462 and -530? 5 How many years have elapsed since —622? 6 Discuss the absolute value of a number. 7 An elevator goes from the fourth floor to the basement. If distance above the main floor is + and distance below is — , how would its distance from the main floor be represented? 8 What problems can you think of in which the signs + and — might be applied to the solution? THE EQUATION Lesson 9 (1) The equation is not a new thing. It has been used constantly in arithmetic, as for example, 2+4 = 6 3-2 = 6, etc. (2) Its usefulness has been greatly increased, however, by the use of the literal symbol. So effi- cient does the equation become as a method of solution that an eminent mathematician has said that the equation is the vital thing in algebra. (3) The usefulness lies in the fact that by carrying through certain processes the unknown number may be expressed in terms of the known numbers, thus giving a solution. Example: Suppose a problem is stated as fol- lows: twice a number added to itself equals ten. Let x = number 2x = twice the number Then 3* = 10 Or* = 104-3 Therefore x = 3$ (4) The expressions upon the two sides of the equality sign are called the members of the equation. (5) The members of an equation being equal, it follows that (34) THE EQUATION 35 (a) Equal numbers may be added to or sub- tracted from both members of an equation without changing the relationship. (b) The members of an equation may be mul- tiplied or divided by the same number without chang- ing the relationship. Example If 4 = 4 Then 4+2=4+2 Or 4-2=4-2 Example If 4 = 4 Then 4.2=4.2 (multiplying by 2) Or 4^-2=4-^-2 (Dividing by 2) Problems 1 Why can the equation be used to greater ad- vantage in Algebra than in Arithmetic ? 2 Solve for x in the following equations. 2x = 8 4* = 8+4 3x-2 = 8 4x+3 = 6+5 3+2x = 9 3 Solve for the letter symbol in the following equations. 3a = 9 4a+a = 6+4 3a-2 = 8+6 2b-3b = 4:-b 3s+4 = 8 4 A man had 180 apples in two baskets. In one basket he had twice as many apples as in the other. Find the number of apples in each basket. 36 BRIEF COURSE IN ALGEBRA ( Note) Let the number of apples in one basket be x. In the other were 180— x 5 Two boys had together 60 fish. One had 3 times as many as the other. How many had each? 6 The sum of two numbers is 105. One number is twice the other. Find the numbers. 7 A number added to six times itself equals 49. Find the number. Lesson 10 The simple equation (one unknown number) (6) If the unknown number appears in the equa- tion with no higher exponent than (one) , the equation is called a simple equation. Example 2x+4 = 8 ((*) has the exponent (1)) (7) The solution of a simple equation involving one unknown number consists of so applying the laws mentioned in Part 5 as to get an equation having the unknown number upon one side of the equality sign and known numbers upon the other side of the equality sign. Example 6x+8 = 10 Subtracting (8) from each number, 6*4-8-8 = 10-8 Or 6x = 2 Dividing both members by (6) Or, x = J (8) The use of the equation makes possible the simple solution of many problems which otherwise might offer difficulty. Example — (9) Suppose I have three times as many German students as English students, and have a total of 140. To determine the number of each, Let x = number of English students (37) 38 BRIEF COURSE IN ALGEBRA Then 3x = number of German students Or 4x = total Then 4x = 140 And x = 35 And 3x=105 Ans. 35 English students. 105 German students. (Note) This problem may be worked without using the symbol (x), but it is readily appreciated that to use the symbol (x) simplifies the solution. (b) Three boys have a sack of apples to divide. The oldest boy is to have three times as many as the youngest, while the second oldest is to have twice as many as the youngest. How many does each re- ceive if the sack holds 66 apples? Let x m number of apples youngest boy receives Then 2x = number of apples second boy receives, And 3x = number of apples third boy receives. Then 6x = 66. Or x = 1 1 , number of apples youngest boy receives. 2* = 22, number of apples second boy receives. 3# = 33, number of apples third boy receives. (c) Suppose, four girls sell 620 Red Cross stamps. The second sells 60 more than the first, the third sells 10 more than the second, and the fourth sells 30 less than the first. To find how many each sells. Let x = number first girl sells Then x+ 60 = number second girl sells, and a; +70 = number third girl sells, and x — 30 = number fourth girl sells, Or 4x +100 = total number sold. THE EQUATION 39 Then 4*4-100 = 620, Or 4* = 520. # = 130, number first girl sells. #+60 = 190, number second girl sells. #+70 = 200, number third girl sells. # — 30 = 100, number fourth girl sells. (Note) Any solution may be proved by sub- stitution of the value of the unknown symbol in the original equation. Example — Suppose 2#+4 = 8 Then 2# = 4 and # = 2 Substitution of (2) for (#) in the equation 2(2)+4 = 8 4+4 = 8 8 = 8 Problems 1 Solve the following equations for (#). 3#+2a = 4a 13# = 26a 26+2# = 6& 2x-2c+3a = Sc+6a 3a+b+x = 2b 2 Solve the following equations for (a) 3a-{-2c — d Aa-3b = 6b 6a+4c = & 66+4#=-2a y-4d+3a = 2x 3 If I pay 2 times as much for a hat as a shirt and five times as much for a suit of clothes as for the 40 BRIEF COURSE IN ALGEBRA hat and pay $19.50 for all, how much do I pay for each? 4 If a man has a certain amount of money and adds (a) dollars and then has (c) dollars, how much did he have at first? (Solve in terms of (a) and (c).) 5 Two boys raked a yard. One received 40 cents less than the other. Together they received $1.20. How much did each receive? 6 Two men go into business, The first puts in (a) dollars less than the other. Together they put in (d) dollars. How much has the first man put in the business? (Solve in terms of (a) and (d).) 7 A farmer raised twice as many bushels of oats as corn and three times as many bushels of potatoes as corn. In all he raised 1800 bushels. How many bushels of corn did he raise? Lesson 11 The simple equation (tw.o unknown numbers.) (10) An equation containing two unknown num- bers is called an indeterminate equation, owing to the fact than an indefinite number of pairs of values will satisfy the equation. Example 2x — y = 8 Let x = l, then 2— y = 8, and y — — 6. Let x = 2, then 4 — 7 = 8, and y= — 4. Let x = 3, then 6— y = 8, and 3/= — 2. (11) It follows that an equation containing two unknown numbers cannot be solved definitely. (12) It is possible, however, to determine a pair of values providing a second condition is given upon which a second equation may be built. Example Suppose 2x-{-y = 16 Second condition x—y= 2 Adding equations 3x =18 Dividing by (3) # = 6 Substituting (6) in either equation (y) is found to be equal to (4) (13) It follows that a solution may be determined for two equations involving the same two unknown numbers, but not for one equation considered singly. (14) Such a solution is called the simultaneous solution of two indeterminate equations. (41) 42 BRIEF COURSE IN ALGEBRA (15) Examples: (a) To solve j ?+»"* ( 3x— ;y = 4 Adding 4x = 10. Dividing by 4, * = 2J. Substituting in the first equation,^ 2 J -fy — 6, Or y«3*. ,,. ( 3a-26 = 4 W j 2a-26 = 6 Subtracting the equations, a = — 2 Substituting, 3(-2)-2& = 4, or 6 = -5 , 4c-2d = 4 3c+3d = 3 In this pair of equations it is noticed that neither addition nor subtraction of the equations will eliminate either unknown number. By multiplying the first equation through by (3) and the second through by (2), they become, 12c-6d=12 6c+6d= 6 Adding 18c =18 Dividing by (18), c = l Substituting in the second equation, J = (Note) Solutions are proved by substitution of the values of the unknown symbols in the original equations. Example — Suppose x-]~y = 6 3x-y = 2 Adding, 4x = 8 Or x = 2 Substituting. 2+^ = 6, ;y = 4 THE EQUATION 43 To prove, substitute these values in the original equations. First equation, 2 -f 4 = 6, or 6 = 6. Second equation, 6—4 = 2, or 2 = 2. Problems 1 In the equations used heretofore, how many unknown numbers have been involved? What is an indeterminate equation? Determine five sets of values for each of the following equations. 2 3x+2y = 6 3 4x-3v = 8 4 s+3;y = 16 5 -2x = 10+4^ 6 14 = 8x-J-2;y 7 If one set of values satisfies two equations, why is this set of values called the simultaneous solution of the two equations? Solve the following pairs of equations simul- taneously by the addition or subtraction method. 8 3jc+4y = 8 2x+5y = 3 9 3a+2b = 6 4a-& = 8 10 4;y-2x = 6 3;y-f-* = 4 11 2x = 10-:y 4*-:y = 8 44 BRIEF COURSE IN ALGEBRA 12 2w-4z = 6 4+2w = z 13 6*+4z = 4 43+3* = 5 Check results by substitution of the values in the original equations. Lesson 12 Solve the following problems using the method explained in Lesson 1 1 . 1 4x-:y = 10 3x+y = 4 2 5x+4y = 8 x+y = 2 3 3*+;y = 14 x+2y = 6 4 3a+4& = 6 4a = 8+36 5 3w-\-z = 3 z — A-Jriv 6 5a+4c = 8 26 -4a = 6 Check results. 7 Find two numbers whose sum is 20 and whose difference is 4. 8 Two pounds of butter and three pounds of meat cost $.50. Five pounds of butter and four pounds of meat cost $2.30. Find the cost per lb, of butter and meat. 9 Twice the difference of two numbers is 4. Three times their sum is 42. What are the numbers? 10 In a certain section of Wisconsin the total value of the milk product exceeded the total value (45) 46 BRIEF COURSE IN ALGEBRA of the wheat crop by $300,560. The two were valued at $422,081. Find the value of both. 11 A house and lot are worth $4,230. The house is worth three times as much as the lot. What is each worth? 12 The sum of two numbers is 48. Their dif- ference is 13. What are the numbers? Lesson 13 Graphic representation 1 If two straight lines are drawn upon the plane perpendicular to each other they may be used as lines of reference for measurement. For instance, in the following figure we may consider that meas- V X X 1 1 y' "plaTte'i J. 1 1 1 1 1 urement to the right of yy' is positive and meas- urement to the left of yy' is negative. Also, we may (47) 48 BRIEF COURSE IN ALGEBRA 1 TTT ... | i T Y I I i _J ■ 1 ! ! I ' 5 1 1 J- 5 U, i ,! y=4 -jU H n 1 x' c X ft 1 - ? 1 1 Y PI Ai £ n 1 , BRIEF COURSE IN ALGEBRA 49 consider measurement up from xx'< is positive and measurement down as negative. 2 If the plane is divided in squares and x values measured to the right or left with y values measured up or down, points may be located having given an x and y value for each. Ex. To locate the points represented by (1) *«3 (2) *«1 (3) x=-3(4) x=-2 y = 2 y = 4 y = 5 y= — 5 (1) To locate I _ ? 1 measure three units to the right of the intersection on the xx'- line, then two units up. This locates the point. (2) To locate the point I ■ . I measure one square to the right, then four squares up. (3) To locate the point I _ c I measure three squares to the left, then five squares up. (4) To locate the point i _ _ - J meas- ure two squares to the left, then five squares down. Ex. Locate the following points. (1) * = 3 (2) x = 6 (3) *=-2 y = 2 y = 10 y=-4 50 BRIEF COURSE IN ALGEBRA Problems Locate the following points after drawing the lines xx' and yy r . x = 6 x=-2 x=-3 x = 3 / 1 ! ! I •*°~^L- L L L L L L l 1 ' i s£Ks>5sP5p!§[ii i F r " j r " r p " ! i 1 ! 1 i | 1 j || || ||| IPtAfEVn 1 1 .1 1 1 1 1 1 1 1 ll, 1 I 1 1 » ». _ j 18 y*-. 1 " 3 10 """* " J-J__ " "" a Kelt! ;"T^ £.' — — — "• j dj S 5 SSJ*S 11 1 i ' jJJi hil Vl l.;::.." -XL— 60 BRIEF COURSE IN ALGEBRA 2 It is desired to get a mixture of milk and cream which shall contain 18% fat, from cream testing 28% and milk testing 3% In the upper left hand corner of the rectangle place the number representing precentage of fat in the cream. In the lower left hand corner place the number representing the percentage of fat in the milk. In the centre of the rectangle place the number repre- senting the percentage to be found. By cross sub- traction it is found that 15 parts of cream are to be taken and 10 parts of milk are to be taken. 3 A train leaves Chicago toward Rockford run- ning 40 miles per hour. A train leaves Rockford at the same time toward Chicago running 30 .miles per hour. At the end of one hour, which would mean measurement of one unit up, the train from Chicago has gone 40 miles, represented by measurement of four units to the right. Thus a point is located which forms a straight line with the original point. Measuring from the opposite direction one unit up and three to the left, a point is located which forms a straight line with the original point. The inter- section of these lines represents the meeting point of the two trains. The distance up represents time. The distance to the right and left represents distance from the two cities. Note Chicago and Rockford are 90 miles apart. SECTION II The Fundamental Operations ADDITION Lesson I (i) The addition of polynomials depends directly upon the addition of monomials. All that is neces- sary is to arrange terms so that similar terms are in columns. Ex. To add 3a+2b, 6a+46 Arranging, 3a + 2 b 6a+46 Adding, 9a +66 Ex. To add Axy — 6w, 16w — 3xy Arranging, 4xy — 6w — 3xy-\-16w Adding, xy-\~10w Ex. To add 3ab-2c+4td, 3ab+6c-2d and 4c+3d-2ab Arranging, 3ab — 2c+4d 3ab+6c-2d - 2ab+4c+3d Adding, 4ab-\-&c+5d Ex. To add 2^-3^+42, -3x+ 2y-2z, and Sx+2y-3z (61) 62 BRIEF COURSE IN ALGEBRA Arranging, 2x — 3y + 42 -3x+2y-2z 5x+2y-3z Adding, 4x+ y— z Problems 1 Add 6x+4^y -3x-2;y 2 Add 3w+4z 6w — 2z 3 Add 2a+4&- c -3a+2b-2c 4 Add 2a&+5cd 3ab+8cd 5 Add 5x;y2 — 2a& 6x^2 -\-3ab 2xyz — Sab 6 Add 5x 2 y+6w — 3x 2 y — 2w 7 Add 15m — Sn 2m+3n \2m-2n Lesson 2 (2) Polynomials having literal or mixed coefficients may be added by indicating the sum of the coeffi- cients for a new coefficient. Ex. To add, ax+ by, cx—dy, and 4x+3y Arranging, ax-\-by cx—dy 4x+3y Adding, ( a +c+4:)x+(b-d+3)y (Note) In this problem the letters x and y stand for the variables or unknown numbers. (3) Often terms are to be added involving binomial variables. Ex. To add, 3(a+x)-\-6(a+x)-2(a+x) + S(a+x) Arranging, 3(a+x) 6(a+x) -2(a+x) 5(a+x) and Adding, 12(a+x) Ex. To add, 2x+(a+b), 4x-2(a+b) -3x-6(a+b) Arranging, 2x-\- (a+b) 4*-2(a+6) -3x-6(a+b) Adding, 3x-7(a+b) (63) 64 ADDITION Ex. Toadd6w+3w-(*-2;y),3w-2tt+6(*-2:y), and 2m-\-n Arranging, 6m+3n— (x — 2y) 3m-2n+6{x-2y) 2m-\- n Adding, llm+2»+5(# — 2;y) (4) In case expressions have terms connected and held together by signs of aggregation, a removal of the signs of aggregation does not affect the signs of the terms, provided the signs of aggregation are pre- ceded by the plus (+) sign. Ex. 3x+(±y-3c+d) Removing the parentheses, 3x-\-ky — 3:-\-d Ex. 2a-6b+(4x+y-3a) Removing the parentheses, 2a — 6b+4:X-\-y — 3a Problems 1 Add ay+bz 2cy — 6z 2 Add 5x-\-aw bx-\-cw — 2x — 5w 3 Add 6a+ 4(x-y)-6(x+y) 10a- S(x~ y)— 2(*+y) 2a-10(x-y)+3(x+y) 4 Combine 6x-4:(a+b)+3c+5x+6(a+b)-4c+ 8(a+6), -2x 5 Combine 15(x+y) + 20c-10{a-b)+2(x+y)- 6c+4(x+;y)-2c SUBTRACTION Lesson 3 (1) Algebraic subtraction is the process of deter- mining what must be added to the subtrahend to give the minuend. Ex. From lOx take 3x To find what must be added to 3x to give 10* It is evident from arithmetical subtraction that the answer is 7x. Ex. From 14a take 2a Answer, 12a Ex. From 6mn take 3mn Answer, 3mn (2) By reference to the discussion of the negative number in Section I, it is evident that the subtraction of a larger number from a smaller gives a negative number for a difference. Ex. From 7 take 9 The answer then is —2 Ex. From 2a take 6a If —4a is added to 6a the result is 2a. The answer then is —4a. Ex. From 3b take 12& If -9b be added to 126 the result is 3b. -9b is the answer. (65) 66 BRIEF COURSE IN ALGEBRA (3) The working rule for subtraction, when both positive and negative numbers are involved, is to change the sign of the subtrahend and proceed as in addition. Ex. From 14cd take 3cd Changing the sign of the term 3cd, it becomes — 3cd. Adding lAcd and — 3cd, we get lied Therefore lAcd — Zed — 1 led Ex. From 3xy take 2xy Changing the sign of the subtrahend, 2xy be- comes — 2xy Adding 3xy and — 2xy, we get xy Therefore 3xy — 2xy = xy Ex, From 6bc take — 2bc Changing the sign of the subtrahend, — 2bc be- comes +2bc. Adding 6be and -\-2bc } we get Sbc Therefore 6bc - ( - 2bc) = Sbc (Note) From the first general rule it is evident that it is necessary to add -\-Sbc to — 2bc to give 6bc Ex. From — 8x;ytake — 5xy Changing the sign of the subtrahend, — Sxy becomes -\-5xy Adding — Sxy and +Sxy, we get —3xy Therefore, — Sxy — ( — Sxy) = — 3xy ( Note) From the general rule for subtraction it is evident that it is necessary to add — 3xy to — Sxy to get —Sxy SUBTRACTION 67 Ex. From 2(c+d) take -3(c+d) Changing the sign of the subtrahend, — 3(c+d) becomes +3(c+d) Adding 2(c+d) and +3(c+d) the result is 5(c+d) Therefore, 2(c+d) - ( - 3(c+d)) = 5(c+d) £#. From — &pq take 6/>g Changing the sign of the subtrahend, 6pq be- comes — 6pq Adding — Spq and — 6pq we get —14pq Ex. To simplify 3^+4^ — 2^+^ — 6^ Collecting terms, +3* +4y -2* + y -6jc — 5* +5y Or — 5x+5y (Note) In such a problem the process is one of combination of terms with consideration for the signs preceding them. Although the negative sign is involved, the problem is simplified by direct reference to the method of addition. (Addition is defined as combination of terms.) It follows that the presence of the negative sign does not always mean that the process of algebraic subtraction is indicated. In all cases where mere combination is to be made, the process is one of algebraic addition, even though the negative sign is involved. Problems 1 Subtract 6a from 12a 2 Subtract 2x from 6x 68 BRIEF COURSE IN ALGEBRA 3 Subtract 7 c from 5c 4 Subtract 15xy from 8x3; 5 Subtract Amn from Smn 6 Subtract 6a from —10a 7 Subtract — 5x;y from — Sxy 8 Subtract 6(a+&) from 10(a+b) 9 Subtract 12 (a — c) from 7(a — c) 10 Subtract —15c from 2c 11 Combine 3a-f5&-2c+8a-4c-36-18a+ Ib-c 12 Combine 2x-\-y — 6x-\-4:y — 3#+4;y — 7rc+8y -3y Lesson 4 Subtraction of polynomials (4) The process of subtraction of one polynomial from another is performed by application of the pre- ceding rules to each group of similar terms. Ex. From 3x+2y take 2x — 3y Arranging, 3x-\-2y 2x-3y Subtraction of similar terms is accomplished by changing the sign of each sign of the subtrahend, and proceeding as in addition. Changing terms of the subtrahend, 3x+2y -2x+3y Adding, x+5y The process is one of (1) arranging terms so that similar terms are in columns, and (2) applying the rule for subtraction to each group of similar factors. The terms of an expression may be rearranged so that similar terms are in columns without altering the value of the expression. Ex. From 3x-\-2y take 3y-\-6x Arranging, 3x+2y 6x-\-3y Subtracting, —3x — y (69) 70 BRIEF COURSE IN ALGEBRA Ex. Perform the following indicated operation. From, 2a+3b Take, a-2b Changing the signs of the subtrahend, 2a+3b -a+2b Adding, a-\-Sa Ex. Perform the following indicated operation. From, 6x+2y — 3z Take, -4x+3;y+4s Changing the signs of the subtrahend, 6x-\-2y — 3z -\-4x — 3y — 4z Adding, 10x— y — 7z Ex. Perform the indicated operation. From, Sxy —Aw-\-3m 2 n Take, 2xy — 3w — 2m 2 n Changing the signs of the subtrahend. Sxy — Aw -{-3m 2 n — 2xy-\-3w-\-2m 2 n Adding, 6xy— w+5m 2 n Problems 1 Take 6b-4cd+Sm from 3b-5cd-10m 2 Take lScd 2 +19m z n from Wed 2 - 1 1 m z n 3 Take 6a+3b-3c from Sa-Ab+7c 4 Take 5* —4y-\-6z from x — 2y — 82 5 Take 16vw-Sy+6z 2 from Svw-10y-lSz t 6 From \Sa-Ud 2 +c take -10a+8d 2 -3c SUBTRACTION 71 7 From 42xy+3z+Ud take 14xy-13z-lSd 8 From 25m 5 n+4rs 2 take 16m 5 n-Srs 2 9 From 4a&-6cd+14e/ take 7ab+9cd-5ef 10 From 2a - Zed 4 take -5a+7cd A Lesson 5 (5) Expressions involving literal coefficients for the unknown numbers may be subtracted by indicating results. Ex. Suppose x and y are the unknown numbers. From, ax+by Take, cx—dy Changing the signs of the subtrahend, ax+by — cx-\-dy Adding, {a — c)x-\-(b-\-d)y Ex. Suppose m and n are the unknown numbers. From, rm+sn Take, —pm — qn Changing the signs of the subtrahend, rm-\-sn pm-\-qn Adding (r+p)m+(s-\-q)n (6) Often the coefficients of the unknown numbers are mixed numbers. In such cases the process of subtraction follows directly the process of addition after the signs of the subtrahend are changed. Ex. Suppose x and y are the variables. From, 3ax — 2by Take, — 7ax+3by (72) SUBTRACTION 73 Changing the signs of the subtrahend, 3ax — 2by +7ax — 3by Adding , 1 Oax — 5by Ex. Suppose 6 and d are the unknown numbers. From, 3ac+4:bd Take, 2xc+3yd Changing the signs of the subtrahend, 3ac+4bd — 2xc — 3yd Adding, {3a-2x)c+{4b-3y)d (7) Binomial and polynomial terms are treated as monomial terms are treated, the coefficients alone being considered. Ex. From, 2(x-y)+4:(a+b) Take, -3(x-y)+3(a+b) Changing the signs of the subtrahend, 2(x-;y)+4(a+&) 3(*-y)-3(fl+&) Adding, 5(x-y)+ (a+b) Ex. From 3a(m-\-n) — 6b(r—s) Take -2a(m+n)-3b(r-s) Changing the signs of the subtrahend, 3a(m-\-n) — 6b(r — s) 2a(m+n)+3b(r-s) Adding, 5a(m+n)— 3b(r— s) 74 BRIEF COURSE IN ALGEBRA Ex. From, 6ab(p+q+r)-\-3a(m — y) Take, -2cd(p+q+r) + w(m-y ) Changing the signs of the subtrahend, 6ab(p+q+r)-\-3a(m — y) 2cd(p J rq-\-r) — w{m—y) Adding, (6ab+2cd) (p+q+r) + (3a-w) (m-y) Problems 1 From 12a take 5a 2 From lax take ax 3 From ax take bx. (x is the variable) 4 From my take ny (y is the variable) 5 From abx take cix {x is the variable) 6 Subtract S(a+b) from 10(a+6) 7 Subtract 6(x— y) from 4(x— y) 8 Subtract -18(c-d) from 20(c-d) 9 From 6 Va6 take 4 Va& 10 From 8 \/a take 6 \/a 11 From lS(a+b-c) take 9(a+6-6) 12 From6(x-2;y+d) take S(x-2y+d) 13 From 3(x+y), 6 Va&, 18(x+y-z) Take -6(x+y), -7 \/ab, -24(x+;y r -z) 14 From 3x 2 y 2 z*, -15 VS +18 (ro-w) Take -7x 2 ;y 2 2 3 , -28Vafc, -25(w-») Lesson 6 Removing signs of aggregation (8) If signs of aggregation, preceded by the nega- tive sign, are to be removed, all the signs within the signs of aggregation are changed. Ex. 3ab-2c-(2ab+ 3c) To remove the parentheses, all signs of terms en- closed must be changed. 3ab — 2c — 2ab — 3c Collecting terms, 3ab — 2ab — 2c — 3c Or, ab — Sc (Note) It is evident that the presence of the negative sign preceding the sign of aggregation signifies that the terms within are to be subtracted. From the working rule the signs of the subtrahend are changed, and we proceed as in addition. Ex. 3ab-2c-(6x-2ab+3c) Removing the parentheses, and changing all signs within, 3ab — 2c — 6x+ 2ab — 3c Collecting terms, 3ab -f- 2ab — 2c — 3c — 6x Or, Sab — 5c — 6x (9) Often there are signs of aggregation within signs of aggregation. To remove them the same rule applies. (75) 76 BRIEF COURSE IN ALGEBRA Ex. 3a+[3x+4b-(6a-2x+b)-2a] Removing the parentheses, 3a+[3x+4:b-6a+2x-b-2a] Removing the brackets, 3a+3x+4b-6a+2x-b-2a Collecting terms, 3a-6a-2a+3x+2x+4:b-b Or, -5a+5x+3b ( Note) It is customary to remove the innermost sign of aggregation first. Problems 1 Simplify 16*- (40— b) 2 Simplify 8 + 6a - (46 - c) 3 Simplify 15ef-(6a+3b) - (4b -d) 4 Simplify 13p- (6r+4) 5 Simplify 15a+l3b-4d-d+c 6 Simplify I6a-(a-b) + [c-d]- \a+d\ 7 Simplify a-[b+(c+d)] 8 Simplify x+y — (a +& — (w — w) ) 9 Simplify 4x+ 1 —4a+3/—(6c+^— 5c)+3y — (4a+c) } Enclose the following expressions with signs of aggregation preceded by the negative sign. 10 6a+4b-c 11 8m+16» — y — 3r Enclose the last three terms in parentheses pre- ceded by the negative sign. 12 te-3c+5x-3y-±n 13 18r 2 s+5mn-16a-4c 14 2a-4b-16c+lSd+m MULTIPLICATION Lesson 7 General definitions. (1) Multiplication is the process of performing upon a number (called the multiplicand) the same operation that was performed upon unity to produce a second number (called the multiplier) . The result is called the product. Ex. 3-5 = 5+5+5 The number 5 is taken as many times as 1 was taken to get 3. (Note) It is evident that multiplication is shortened addition. (2) Algebraic multiplication follows arithmetical multiplication directly. Owing to the enlargement of the number system by the use of the letter symbols and negative numbers, certain revisements to the arithmetical rules are necessary. a Letter symbols may be multiplied in any order. Ex. ab = ba b Letter symbols may be grouped in any way Ex. xyz = x (yz) = y (xz) = z (xy) c Polynomial expressions may be multiplied by monomials by multiplying each term of the poly- nomial by the monomial. Ex. (x-\-y—z)>a = ax+ay—az (77) 78 BRIEF COURSE IN ALGEBRA d Either a negative multiplicand or a negative multiplier will give a negative product. Ex. -\-a- — b = — ab Ex. —a-\-b——ab e If both multiplicand and multiplier are either positive or negative, the product is positive. Ex. a-b=+ab Ex. —a-—b=-\-ab (3) Exponents from definition indicate the number of times a number is taken as a factor, so the law follows that to multiply like factors, exponents are added. Ex. 2 3 -2 4 = 2-2-2-2. 2-2-2 Or, 2 3 -2 4 = 2 7 Ex. a A a 2 = aaaa a a Ora A a 2 = a G Multiplication of monomials. (4) Monomials are multiplied by multiplying the coefficients together for a new coefficient, and the unknown factors together for a new unknown factor. Ex. 3a2b = 6ab Ex. 3xy ■ 4xz =\2x 2 yz Ex. 6c 2 4d = 24c 2 d Ex. 2a n -3a m = 6a n+m (adding exponents) (5) To multiply a polynomial by a monomial, mul- tiply each term of the polynomial by the monomial. Ex. 3x ■ (2^+3^-4x2) = 6x 2 y+9xy- \2x H Ex. 2a(4b-c+2d 2 )=8ab-2ac+4:ad 2 Ex. 6c z (4d-5cd+3e) = 24c 3 d-30c*d+18c*e MULTIPLICATION 79 Ex. 2{a n -b n ) = 2a n -2b m (Note) The laws of signs are given in working form l as follows: 1 Like signs give a positive product. 2 Unlike signs give a negative product. Problems 1 Multiply 5* by 2 2 Multiply 6ab by 3a 3 Multiply 10x 2 y by 2xy 4 Multiply 12a* by 4a y (Adding exponents by indicating operation) 5 Multiply 2x+y by 3x 6 Multiply 2*+4 by 5x 7 Multiply 4x 2 +3*+4 by 7x 8 Multiply 6a 2 -4a -2 by 6a 9 Multiply -2x+4 by -3x 10 Multiply 16s 2 +4*-5 by -2x Lesson 8 Multiplication of polynomials by polynomials. (6) To multiply a polynomial by a polynomial, multiply each term of the multiplicand by each term of the multiplier. Then arrange and add the several products. Ex. Multiply 3x+y by 3x — y 3x+y 3x—y Multiplying by 3x, 9x 2 + 3xy Multiplying by — v, —3xy—y* Adding, 9x 2 — y 2 Ex. Multiply 3x 2 +2x-6 by 2x+4 3x 2 +2x-6 2x +4 Multiplying by 2x, 6x 3 + 4x 2 — 1 2x Multiplying by 4, + 12x 2 +8*:-24 Adding, 6x 3 +16s 2 -4x-24 Ex. Multiply Scd - 4 by 2c + 2 5cd-4: 2c +2 Multiplying by 2c, 1 Oc H - 8c Multiplying by 2, + 10cd--8 Adding, 10c 2 d-8c-r-10cd-8 (7) In multiplication, it is always advisable to ar- (80) MULTIPLICATION , 81 range the terms of multiplicand and multiplier so that the exponents of the letter symbols under dis- cussion are in ascending or descending order. Ex. Multiply 3x*+2x 2 -3x*+2-x by 3x+ 2x 2 -3x* Arranging terms, 3x*-3x z +2x 2 -x+2 -3x 3 +2x 2 +3x -9x 7 +9x & -6x*+3x i -6x z + 6x 6 -6x 5 +4:X 4 -2x 3 +4x i +9x 5 -9s 4 +6* 3 -3* 2 +6s Adding,-9x' J +15x«-3x & -2x 4 -2x 3 -x 2 +6x (Note) The method of arrangement makes possible the writing of the products in columns, thus simplifying the process. Definition — The degree of a term is determined by adding the exponents of the literal factors. The degree of an expression is determined by the term of highest degree in the expression. Definition — An expression is homogeneous when the terms are all of the same degree. Problems 1 Multiply 2a +1 by 3a 2 Multiply 6x 2 +3x+2 by 2x 3 Multiply 6b + 3c by 4b 4 Multiply 15m 2 n — 2mn by mn 5 Multiply 6ax 2 - 3a 2 x-f-4 by 8x+2 6 Multiply Scd-3d 2 +5 by 2c-3d+2 82 BRIEF COURSE IN ALGEBRA 7 Multiply lOmn— 4m 2 +2n by 15w 8 Multiply 3x 2 y s -\-5x z y 2 by 6x*y 3 9 Multiply 25ab 2 -6a 2 b by Sdb+3 10 Multiply 14xy+ 1x 2 -3y by 2xyAx 2 +y 2 1 1 Expand 5x(2x 2 +4y+3y 2 ) 12 Expand (3a+b) (6a 2 b+10) 13 Expand (5x+^+z) (3x 2 +4;y-3z) 14 Expand (10* 3 +4x 2 -4x+2) (6*»+3x-2) 15 Expand (6ab*+4ab 2 -3ab+6a) (2ab) Lesson 9 Literal exponents 1 The general rule for multiplication holds for expressions having literal exponents. Ex. {3a 2p +2a p +a) (2a p +l) 3a 2p +2a p +a 2a p +l 6a sP +4a 2p +2a p + 1 +3a 2p +2a p +a 6a zp +7a 2p +2a p + 1 +2a p +a It is essential to arrange terms in ascending or descending powers to get the best results. Problems 4x b 4x a+1 +3y 1 2x a +3y 2 -±x b 2 6a m -4b n -3c 2m 3a 2 +4b m 3 8**-4x*+3 3x p -x 4 \5x 2 y a -4z' 3xy—z n (83) DIVISION Lesson 10 (1) If the product and one of two factors are given, it is possible to determine the other factor. This process is called division. The product is the divi- dend, the given factor is the divisor, and the factor found is the quotient. (2) Division is expressed either by using the sign of division (-§•), or by writing the dividend and divisor inthe form of a fraction, the dividend as the numerator the divisor as the denominator. Ex. Divide 3ax by 2y Expressed as 3ax-$-2y 3ax 2y (3) To divide terms having like factors, subtract the exponents. Ex. Divide 12x 3 by 3x 2 12x 3 12-*-x-* = =s 4# 3x 2 3 xx ( Note) The coefficients are divided to determine a new coefficient, and the given factors are divided to determine a new literal factor. Ex. Divide \6x z y 2 by 2x 2 y 16x 3 y 2 (84) DIVISION 85 (4) The law of signs is the same as for multiplica- tion. Like signs give a positive result. Unlike signs give a negative result. Ex. -2a 2 +-a = +2a + 22x 3 ^ + llx 2 =+2:x: 16x 3 y-. — 2x 2 = — Sxy -Sc*d 2 ++4cd 2 =-2c 2 (d*+d*=l) Problems 1 Divide 15a 4 by 5a 2 2 Divide 10* 6 by 2x 3 Divide 20x 4 y 2 by - lOxy * 4 Divide 12m 6 by 4ra 3 5 Divide - 6a 2 b 2 by - 2ab Lesson 11 Polynomials (5) Polynomials are divided by monomials by dividing each term of the polynomial by the mono- mial. Ex. 2x A -{-Ax 3 — 6x 2 +4x is to be divided by 2x 2x )2x 4 +4x 3 -6x 2 +4x x z +2x 2 -3x+2 (6) In case the exponent of the divisor is larger than the exponent of the dividend, the new exponent is expressed by the negative number. Ex. Divide 3a 4 - 2a 3 +4a 2 by 2a 4 2a 4 )3a 4 -2a 3 +4a 2 |-a- 1 +2a- 2 ( Note) The negative exponent will be discussed at length in a following topic. (7) Polynomials may be divided by polynomials as follows. Divide 6x 6 +7* 5 +6x 4 +4s 3 +* 2 by2x 2 +x )3* 4 +2* 3 +2 2 + 1 6* 6 -f-7* 5 + 6x 4 + 4x i + x 2 )2x 2 + x Multiplying (2* 2 + x) by Sx* 6x 6 + 3x S Subtracting 4x 5 + 6x* + 4x 3 + x 2 Multiplying (2s 2 + x) by 2x 3 4x s + 2x* Subtracting 4x A + 4x 3 + x 2 Multiplying (2x 2 + x) by 2x 2 4* 4 +2* 3 Subtracting 2x> + x 2 Multiylying (2x 2 + x) by 1 2x 3 + x 2 Subtracting (86) DIVISION 87 Divide the first term of the dividend by the first term of the divisor. 2x 2 is contained 3x 4 times in 6x 6 3x 4 is the first term of the quotient. Other terms of the quotient are determined in a similar manner. Illustrative examples. Divide 2a 2 +7a+6 by a+2 )2a+ 3 2a 2 + 7a+6)a+2 Multiplying (a+ 2) by (2a), 2a 2 + 4a Subtracting, 3a+6 Multiplying (a+ 2 by (3), 3aj- 6 quotient = 2a+ 3 Divide 3x A — x 3 +2x 2 — 4x by x — 1 3x*—x 3 + 2x 2 —4 xjx—l Multiplying (x-1) by (3x 3 ), 3^ 4 -3a: 3 Subtracting, 2x 3 + 2x 2 —4x Multiplying (x — 1) by {2x 2 ) , 2x 3 —2x 2 Subtracting, 4x 2 —4x Multiplying '.x— 1) by (4x), 4x 2 —Ax quotient = 3% 3 + 2* 2 + 4x Problems 1 Divide 4a 6 +6a 4 -4a 3 +2a 2 by 2a 2 2 Divide 36a 6 - 27a 4 by 9a 2 3 Divide a 2 b +ab 2 - 4a z b 3 by a& 4 Divide a 2 +3a+2 by a+1 5 Divide 6w 2 — 7w — 3 by 2w — 3 88 BRIEF COURSE IN ALGEBRA 6 Divide 2x 4 -6x 3 +3x 2 -5x+2 by x 2 -3*+l 7 Divide 6a 2 -3lab+35b 2 by 2a-7& 8 Divide 2y 8 -9y 2 +lly- 3 by 2;y-3 9 Divide a 2 +6 2 +l + 2a&+2a+26 by a+b+\ 10 Divide # 2 +;y 2 +z 2 — 2xy+2xz — 2yz by x—y+z Special Rules for Multiplication Lesson 12 Owing to the fact that certain products recur time after time, it is of advantage to memorize certain short methods for multiplication. (1) To multiply a binomial by itself. Given the problem (a-\-b) 2 B y multiplication a + b a+b a 2 +ab +ab+b 2 a 2 +2ab+b 2 This product is made up of the squares of the two terms of the binomial plus twice the product of the two terms. It is found that to square any binomial the product is always equal to the squares of the two terms plus twice the product of the two terms. Ex. (a-b) 2 = a 2 -2ab+b 2 Ex. (x+y) 2 = x 2 +2xy+y 2 Ex. (2a-b) 2 = 4a 2 -4cab+b 2 Problems 1 ix-y) 2 7 (a -3b) 2 2 (c+d) 2 8 (a-b 2 ) 2 3 (m-\-n) 2 9 (2x 2 -4y*) 2 4 {2x+y) 2 10 (a+(a-b)) 2 5 (x-2y) 2 11 (l + 2x) 2 6 (2a+3b) 2 12 {x 2 +3a) 2 (89) Lesson 13 (2) To multiply the sum of two terms by their dif- ference. Given the problem (a+b) (a — b) By multiplication a+b a — b a 2 -\-ab -ab-b 2 a 2 -b 2 It is noticed that the product is made up of the difference of the squares of the two terms. To mul- tiply the sum of any two terms by their difference, the product is found always to equal the difference of the squares of the two terms. Ex. (x-\-y) (x — y)=x 2 — y 2 Ex. (x+3y) (x-3y)=x 2 -9y 2 Problems 1 (a-b) (a-b) 7 (l + 2a) (l-2a) 2 (x+y) (x-y) 8 (6+*) (6-x) 3 (m+n) (m-n) 9 (2a 2 +2) (2a 2 -2) 4 (1+x) (1-x) 10 (4c-d)(4<;+d) 5 (2a-b) (2a+b) 11 (2x 2 -12) (2* 2 +12) 6 (2x-4;y) (2x+4y) 12 (1 -3a 2 ) (1-f 3a 2 ) (90) Lesson 14 (3) To multiply two binomials together which have a common term. Given the problem (x+2) (x+3) By multiplication, x+2 x+3 x 2 +2x +3x+6 x 2 +5x+6 The product is made up of the square of the com- mon term, plus the common term with a coefficient equal to the sum of the unlike terms, plus the product of the unlike terms. Ex. (o+4) (a-2)=a 2 +2a-8 Ex. (c + 7) (6 + 2)=c 2 +9a+14 Problems 1 (a+2)(a+3) 7 (*+6) (x-6) 2 (x-4)(x+2) 8 (2a -3) (2a +4) 3 (ro+2)(ro-l) 9 (*-2)(*+10) 4 (* + 3)(x + 5) 10 (3,-6) (y + 5) 5 (y-l)iy-S) 11 (a 2 +2)(a 2 -5) 6 ( r -2)(f+S) 12 (x+ll)(x-8) (91) Lesson 15 (4) Any two binomials may be quickly multiplied together by, (1) multiplying the first terms of the binomials together, (2) multiplying the second terms of the binomials together, and (3) multiplying the terms together in such a manner as will give cross pro- ducts. Ex. (2a+4)(3a-2) Multiplying first terms together, 6a 8 Multiplying second terms together, —8 The cross products are, 12a and —4a Therefore product equals 6a 2 +8a — 8 (Note) The cross products may be easily de- termined by the following process. (2a+4) (3a- 2) Ex. (6c- 2) (2c+3) = 12c 2 +14c 1 2c 2 = product of first terms — 6 = product of second terms 14c = sum of cross products Problems 1 2 3 4 5 6 (a+2) (a+3) (x+y) (w-z) (2a -3) (4a +4) (6-3*) (4+4y) (5-2z)(2+4z) (l+3a)(6+4a) 7 8 9 10 11 12 (6-3:y) (4-4;y) (2-*) (3-;y) (46+2) (3-26) (a-4)(2a-2) (6+lQ*)(*-3) (a+4)(a-26) (92) Special Rules for Division Lesson 16 The special rules for division are derived directly from the special rules for multiplication, and lead directly into factoring. Factoring will be considered as a special application of the short rules for division. (1) If the squaring of a binomial gives a product made up of, (1) the square of the first term, (2) twice the product of the two terms, (3) the square of the second term, then any trinomial, whose terms are two squares and a third term equaling twice the product of the square roots of the squared terms, is divisible by a binomial made up of the square roots connected by the sign of the second term. Ex. If (a+b) 2 = a 2 +2ab+b 2 a 2 +2ab+b 2 Then ; ~- = a +b a+b Ex. If (a-b) 2 = a 2 -2ab+b 2 a 2 -2ab+b 2 Then : =a-b a — b ( Note) The terms (a — b) and (a — b) are factors of a 2 -2ab+b 2 (2) A trinomial is a perfect square if two of the terms are squares and the third is equal to twice the square roots of the squared terms. Ex. a 2 — 2ab-\-b 2 is a perfect square, For, a 2 and b 2 are squares, And — 2ab is twice the product of the square roots. (93) 94 BRIEF COURSE IN ALGEBRA Ex. 4c 2 -\-4cd-\-d 2 is a perfect square, For 4c 2 and d 2 are perfect squares, And 4cd is twice the product of the square roots. Therefore 4c 2 +4cd+d 2 is divisible by 2c+d. Ex. a 2 x 2 -\-2ax+l is a perfect square, For a 2 x 2 and 1 are perfect squares, And lax is twice the product of the square roots. Therefore a 2 * 2 -f2ax+l is divisible by ax+ 1 Determine which of the following are perfect squares 1 x 2 +2xy+y 2 7 4c 2 +8cd+4<2 2 2 a 2 +4ab+4b 2 8 x 4 +x 2 y 2 +y* 3 x 2 -2xy+y 2 9 9a 2 -6ab+b 2 4 x 2 -\-xy+y 2 10 l + 2x-\-x 2 5 25a 2 +10ab+b 2 11 16y 2 +Sy+l 6 y 2 -2\yz+z 2 ' 12 36w 2 -24w+4 Lesson 17 If (a+b) (a-b)=a 2 -b 2 a 2 -b 2 Then — = a-b a+b a 2 -b 2 And — =a+b a — b (3) The difference of two squares is always divisi- ble by either the sum or the difference of the square roots. Ex. c 2 —d 2 is divisible by either c+d or c—d c 2 -d 2 —c—d c+d c*-d 2 ^. — = c+d c—d (Note) (c+d) and (c — d) are factors of c 2 — d 1 Ex. 4a 2 x 2 — 1 6c 2 is divisible by either (lax + 4c) or (lax — be) 4a 2 x 2 -16c 2 _ = lax — 4c 2ax+4c 4a 2 x 2 -16c 2 „ , , = lax+Ac lax — 4c (Note) (lax+4c) and (lax — 4c) are factors of 4a 2 x 2 -16c 2 Problems 1 Divide (x 2 — y 2 ) by (# — ;y) 2 Divide a 4 -6 4 by a 2 +o 2 3 Divide 2 2 — 9 by x — 3 (95) % BRIEF COURSE IN ALGEBRA 4 Divide x 2 -9 by x+3 5 Divided 4 — 16 by 2x 2 — 4 6 Divide 4a 2 +4 by 2a - 2 7 Divide 9-x 2 by 3+x 8 Divide 16a 2 -b 2 by 4a+& 9 Divide a 2 Z> 2 -4 by a&-2 10 Divide 9x*-c 2 by3x 3 +c 11 Divide 16-4a 2 by 4+2a 12 Divide a 2 -b 2 by a+6 13 Divide 1-a 2 by 1-a 14 Divide 1-a 4 by 1— a 2 Lesson 18 It is found by actual division that % S-j-'V 3 = x 2 — xy-\-y 2 x+y x z — y z And = x 2 -\-xy+y 2 x—y It is noticed that the sum of two cubes may be divided by the sum of the first powers of the quanti- ties, and that the difference of two cubes may be divid- ed by the difference of the first powers of the quanti- ties. The quotients are so striking as to be easily re- membered. (4) Dividing the sum of the cubes of two quanti- ties by the sum of the quantities, we get the sum of the squares of the quantities minus the product of the quantities. (5) Dividing the difference of the cubes of two quantities by the difference of the first powers, we get the sum of the squares of the quantities plus the product of the quantities. 8a 3 +6 3 Ex. - r =±a 2 -2ab+b 2 2a+b 8a 3 -b 3 — -=4a 2 +2a&+6 2 2a — b (97) 98 BRIEF COURSE IN ALGEBRA Ex. ^-^=9x 2 -3xy+y 2 3x+y 27x 3 — V 3 — ■ — — = 9x 2 +3xy+y 2 3x — y w z -\-z 3 Ex. =w 2 — wz+z 2 w-\-z w 3 — z Z = w 2 +wz+z 2 w—z The sum or the difference of two cubes is, therefore, factorable, and with a short drill the factors may be determined by inspection. Problems 1 Divide a z — b z by a — b 2 Divide x z +y z by x-\-y 3 Divide 8 -x z by 2- x 4 Divide 27a 3 -6 6 by 3a-b 2 5 Divide x 6 — v 6 by x 2 — y 2 6 Divide 64 — x z by 4 — # 7 Divide 1 — # 3 by 1 — x 8 Divide 1+y 3 by 1+y 9 Sivide 125+x 3 by 5+x 10 Divide v 3 -125 by v-5 11 Divide 27x 3 -8 by 3x- 2 12 Divide a z b z — d z by ab — d Factoring Lesson 19 The direct application of the short methods of division is in the determination of the factors of an expression. Before considering the special cases of factoring it is advisable to memorize the following definitions. (1) A rational expression is one containing no in- dicated root of the letter considered. 2jc 2 -|-^+4 is rational. 2x 2 +y/x-\-4: is irrational. ( Note) The sign V is used to indicate a root to be found. The particular root is indicated by a number written in the \/. or v second root (square root), third root (cube root), fourth root. If no number is written in the \J the root is un- derstood to be the second. (2) An integral expression is one in which the let- ter considered does not appear in a denominator. Ex. 3y-f-6 is integral. 2 -4-4^+8 is not integral. y (3) The factors of an expression are the expres- sions multiplied together to produce it. (99) Ex. These expressions are in Ex. (m+2n) 2 +2(m+2n) (x-y) + (x-y) 2 Factors are, [(m+2n) + (x — y)] and [(w+2n)-f- (x-y)] (Note) In this problem (m+2n) and (x—y) are considered as single terms. a 2 +2ab+b 2 [These ex- (m+2n) 2 +2(m+2n) (x-y)-\-(x-y) 2 \ pressions are in the same form. Ex. (a+b) 2 -(x+y) 2 Factors are [(a+b) + (x+ y)]and (a+b) — (x+y)] a 2 — b 2 ) These expressions are in the (a -\-b) 2 — (x+y) 2 ) same form. Ex. (x+y) 2 +5(x+y)+ 6 Factors are [(#+;y)-r-3)] and [(#+30+2)] a 2 +5a+6 j These expressions are in the (x-\-y) 2 +5(x+y) + 6\ same form. (105) 106 BRIEF COURSE IN ALGEBRA Ex. 2(a-2b)+4x(a-2b)-6{a-2b) Factors are (a — 2b) and (2+4x — 6) (Taking out the common factor.) 2a-\-4:xa — 6a \ These expressions 2(a+26)+4x(a — 2b)— 6(a — 2b) j are in the same form. Problems 1 x 2 +2x(a+b) + (a+b) 2 2 (a-b) 2 -2(a-b) (a+b) + (a+b) 2 3 (x-y) 2 -(x+y) 2 4 {2a-b) 2 -(x-y) 2 5 (46+2) 2 -4 6 (a-b) 2 +2{a-b) + l 7 (a+6) 2 +5(a+6) + 6 8 (x-3/) 2 -(x-y)-6 9 (c+d) 2 +7(c-d) + 10 10 3(a-6)+4(a-6) + 2(a-6) 11 4(a-b) 2 +2(a-b) 12 6(x+;y) 2 +9(x+:y) 4 Lesson 25 (13) Expressions may be grouped in many cases so that they are factorable. Ex. ax+ay+bx+by By grouping the first two terms and the last two terms, (ax + ay) + (bx + by) Factoring each group, a(x+y)+b(x+y) Considering this expression as made up of two terms, (x-\-y) may be taken out as a common factor. (x+y) (a+b) The factors. Ex. 2xy + 3axy — 6xw — 9axw Grouping the first and third terms and the second and fourth terms, (2xy — 6xw) + (3axy — 9axw) Factoring each group 2x(y — 3w)+3ax(y — 3w) Considering this expression as made up of two terms, it is factorable. (y — 3w) (2x+3ax). The factors. Problems 1 ax+bx+ay+by 2 2x-3x-\-2y-3y 3 4a+66+8a+12& (107) 108 BRIEF COURSE IN ALGEBRA 4 6x-12 + 18x 2 -36x 5 3ax-\-2a+9ax 2 -\- 6ax 6 x z — x 2 y-\-xy 2 — y 3 7 3a*-6ya 2 -a+2y 8 mn — np — m-\-p 9 a 2 +36-3a-a& 10 a+2b-4a-Sb Lesson 26 (14) Often an expression may be grouped to give the difference of a trinomial square term and a monomial square term. Ex. a 2 +2ab+b 2 -x 2 Grouping, (a 2 + 2ab -\-b 2 )—x 2 Or (a+b) 2 -x 2 Considering (a +b) as a single term, this is the difference of two squares. Factoring, [(a+b)+x)] [(a+b)-x)] f Or (a+b+x) (a+b-x) Ex. a 2 -2ab+b 2 -\6 Grouping, (a 2 - lab + b 2 ) - 1 6 Or (a-b) 2 -16 Factoring, [(a-b) +4)] [(a-b) -4)] Or (a-6+4) (a-6-4) Problems 1 x 2 +2xy+y 2 -a 2 2 a 2 -2ab+b 2 -c 2 3 4x 2 +4x+l-x 4 4 c 2 -(a+6) 2 5 x 2 -a 2 -2a&-6 2 6 a 2 -2a + l-16ra 4 (109) 110 BRIEF COURSE IN ALGEBRA 7 a 2 +2ab+b 2 -x 2 -2xy-y 2 8 m 2 -2mn+n 2 -r 2 -2rs-s 2 9 4m*n 2 -c 2 -4cd-4d 2 10 (a-b) 2 -x 2 -2xy-y 2 11 \-x 2 -2x-\ 12 x 2 +2xy+y 2 -49 Highest Common Factor Lowest Common Multiple Lesson 27 By inspection it is possible to determine the highest common factor of two or more expressions. Ex. 2ax-\-2xy. Factors are 2x (a -\-y) a 2 +2ay+y 2 . Factors are (a+y) (a+y) The highest common factor is (a-\-y) ( Note) To determine the highest common factor it is necessary first to factor the expressions under consideration. (15) The lowest common multiple of two or more expressions is the lowest expression which will contain the expressions. It is determined first by factoring the expressions and then taking each factor the great- est number of times it appears in an expression. Ex. x 2 —y 2 Factors are (x+y) (x—y) x 2 +2xy+y 2 Factors are (x+y) (x-\-y) 2wx-\-2wy Factors are 2w(x+y) 2w(x+y) (x—y) (x+y) will contain the three ex- pressions. (16) The working method is to take all the factors of the first expression and all the factors of the other expressions not already selected. L. C. M. = 2w(x+y) (x+y) (x-y) (HI) 112 BRIEF COURSE IN ALGEBRA Ex. 2ay-3b) (2x+y) 2ax+4a 2 y — 6ab Factors are 2a(x-\- \x 2 — y 2 Factors are (2x — y) (2x+y) x 2 +7jc+10 Factors are (x+5) (x+2) 4x 2 +4xy+y 2 Factors are (2x+y) x 2 — x — 6 Factors are (x — 3) (x+2) L. C. M. = 2a(x+2ay-3b) (2x-y) (2x+y) * (x+5) (x+2) (x-3) Problems Determine the H.C.F and L.C.M. of the following. [-2* -j 4x 2 y [ 6xy 2 \24ax 2 j 16a 2 x [ 4a 3 x 3 y i a 2 -b 2 j a 2 +2ab+b 2 a 2 +2ab+b 2 a*-b* x 2 +5x-r-6 x 2 +6x+9 2-ft2 ax+ay x 2 +2xy+y 2 \ x 2 y — xy 2 8 x 2 +y 7 4*-2 2x—y (6s 2 +3;v)(3* 2 +;y 2 ) (4x-2)(2x-y) (128) FRACTIONS 129 (2 + **) +(2x-~\r 2y+3x ' 2x y- 6x * y 2y+3x y_ 2y+3x '2xy-6x 2 2xy-6x 2 Problems x 2a Tb' 6 ' a — x^l ~3~ '.2' 2+a^_a+b a—b ' a — b 1 a+b 9+c 9-c 6 a 2 -b 2 ^_a+b a 2 +2ab + b 2 ' a — b Note Cancel whenever possible. 7 * 2 +ll*+30 ^s+6, x 2 + 10 x + 25 * x+5 Q 2x+4 x+2 3a 2 + 6a 4 + 18a 2 3a 2 130 BRIEF COURSE IN ALGEBRA (a-b) 3 ,a 2 -2ab+b 2 10 11 12 4* 12* 18a 3 6a a + b a 2 —b 2 2 ^2a 2 -4a 3 a—x ' (a—x) 2 Complex Fractions Lesson 6 1 A complex fraction is one in which fractional terms are involved in one or both its members. To simplify a complex fraction, simplify each mem- ber separately. Ex. Simplify a 1+ i a a _ b + a _ b _ ab a b + a b b + a b + a' 1 + b T Ex. Simplify b • t i+(i+p i+d.|) a a a b b b a + b ba + b ab+b 2 1+- a a x c Ex. Simplify — • c * d~2y x c xd+yc yd yd xd+yc 2cy — dx xd+yc c x = 2cy—dx = yd 2yd yd d~2y 2dy 2dy _ 2xd + 2yc 2cy—dx 2cy—dx (131) 132 BRIEF COURSE IN ALGEBRA y 2-- x Ex. Simplify - y+- * x y 2x-y x x 2x—y xy+1 2x—y x 2x—y 1 %y + l xxx xy+1 xy+1 y+- x 1 Ex. Simplify 1+ i 1 + x 1 1 - 1 1 1 =1 ^_ 2X + 1 _ „ , * * + !+* 2*4-1 ' x + l " ac-4-1 # + 1 x + 1 j #4-1 flp+l 2* + l~2* + l' Problems Simplify i » 3 2a ~ 4& 5 __ 3. - _ 5. x ' l+o •+1 3 , 1 2+3* 1-* 4-* '1+x 3 , S-x 2+* ' 1 2 1 14- 2*-J_ 4 - _* 6 - x x Fractional Equations Lesson 7 Solution of equations involving fractions. To solve an equation involving fractions, (1) clear the equation of fractions, (2) solve by methods before mentioned. L.C.M. of 3, 2, 6 = 6. Writing all terms as fractions having the common denominator, 4y 15y_3y 48 T e e o" Multiplying the equation by (6), 4;y+ lSy = 3y+48. Collecting terms, 16^ = 48. Or y = 3. „ „ ,3x+4 6x-2 22*. 2x+ — - — « — — - 4 8 L.C.M. of 4, 8 = 8 Writing all terms as fractions having the denomi- nator (8), 16s 2(3*+4) _ (6* -2) 1T + 8 8 * Multiplying the equation by (8), Simplifying, 1 6x + 6x + 8 = 6x — 2 Collecting terms, 16x =—10 Or, *=-f (133) 134 BRIEF COURSE IN ALGEBRA Ex. -L- 4 8 x + l x-l x 2 -l L.C.M. of (x+l), (x-1), and (x 2 -l) = (x+l) (x-l)=x 2 ^-l Writing all terms in equivalent form having the common denominator. 3(*-l) _ 4(*+l) _ 8 x 2 -l " x 2 -l x 2 -l Multiplying the equation by (x 2 — 1), 3(*-l)-4(*+l) = 8 Expanding and simplifying, 3x — 3 — 4x — 4 = 8 Collecting terms, — # = 15 Or *=-15 Ex. Solve for (x) when (a, c, b) are considered as known values. x 2x _c x a cb b a L.C.M. of a, cb, b and a — acb Writing all terms in equivalent form having the denominator acb, cbx lax _ac 2 cbx acb acb acb acb Multiplying by acb, cbx + lax = ac 2 -\- cbx Collecting terms, cbx + lax — cbx — ac 2 Factoring, x(cb -\-la — cb) =ac 2 ac 2 c 2 Dividing by (cb + la — cb),x cb + 2a—cb 2 FRACTIONS 135 Problems Solve, 3,4 , a-2 a+3 x + 2~x =6 7 -r~ ~ir =6 2x_5* 3x = 4 « *+3 *+2 x+5 6 T + ~3~ 3 2x+3=| * „ „ (a— bV a—b x -+2 X = 6 10 1-J 2~ = 3 5* + 3 2*-3 a+1 + 2a-l 11 2a 3a 5 3* +2 6 6*-4 + 4 (a-6) 2 a-6 3 2 x+3 * + l 3a-2 ~3a+3 3a +2 2a 11 11 6 2y + y = 2+ 3 + ^y M 13 3ax+6a = 2*+10a x 2x x Sx 14 -+— = - — — a o a—b a 15 Two men were in business and made a profit of $3,225 in one year. One man received 20% more than the other. What part of the profit did each receive ? 16 A parent offered a boy one dollar if the average of his marks was 80% or over. If the boy gets 90% in mathematics, 85% in history, 78% in rhetoric, and 70% in spelling, what must he get in agriculture to have an average of 80%. SECTION IV POWERS AND ROOTS Lesson 1 General definitions. 1 A power is a product obtained by using a number two or more times as a factor. Ex. The second power of 2 = 2 • 2 = 4 The third power of 3 = 3-3-3 = 27 2 The laws of signs hold as they do in multiplica- tion. Ex. +2 -+2 = +4 -2--2 = +4 3 The exponent is a small number written at one side and above a given number to indicate the number of times the given number is to be used as a factor. Ex. 3 2 , 2 is the exponent. . 3 2 = 9 Ex. 2 3 = 8. 3 is the exponent. Ex. a 2 2 is the exponent. 4 The process of multiplication performed on like factors having equal or unequal exponents is per- formed by adding exponents. Ex. 3 2 -3 4 = 3 6 Ex. a 2 -a z = a h Ex. 4°-4 6 = 4° +b (136) POWERS AND ROOTS 137 5 The above law is explained as follows: a 2 a 3 = a-aaa'a=(a) taken five times as a factor. This is expressed as a 5 b*b A = b 7 = b-b-b-b'bbb = (b) taken seven times as a factor, and is expressed as b 7 . 6 In case the terms have more than one factor, the exponents of like factors are added. Ex. a 3 x 2 a 2 x b = a 5 x 7 b 2 y 3 -by = b 3 y A x 2 y 2 z 4 -x 3 y 5 z* = x 6 y 7 z 10 7 Should there be coefficients to one or both terms, these become factors of the product. Ex. 3a 2 b-a 3 b = 3a 5 b 2 Ex. 4a 3 b 2 '5ab 3 = 20a A b 5 8 The root of a number is one of a certain number of equal factors of a number. Ex. 2 is a root of 4. Ex. 3 is a root of 27 9 The square root is one of two equal factors. Ex. 2 is the square root of 4. The cube root is one of three equal factors. Ex. 4 is the cube root of 64. The fourth root is one of four equal factors. Ex. 2 is the fourth root of 16. 10 The sign V is called the radical sign, and indicates that a root is to be found. 138 BRIEF COURSE IN ALGEBRA 11 A number placed in the V indicates the root. Ex . \/ indicates square root . Ex. <\/ indicates cube root. Ex. \/ indicates fourth root. This number is called the index number. (Note) The square root is usually expressed by writing the sign with no number in the V- Ex. Va', V^, Vl6 Problems 1 Define power, exponent, radical. 2 Multiply 3x 2 y by 2y 3 Multiply 4a 2 b 2 by lab 4 Multiply lOab* by b z 5 Multiply dm 2 n by mn 6 Multiply m 3 n 2 by Smn 2 7 Multiply 6x 2 y by 3xy 2 8 Express the square root of 3a +6 9 Express the cube root of 4# — 2y 10 Express the fourth root of 3xy — 4 11 Multiply 16x by 3xy 12 Multiply 4m 2 by In 2 POWERS Lesson 2 Powers of monomials Integral exponents. 1 If a number having an exponent is to be raised to a power, the exponent of the number is multiplied by the number expressing the power. Ex. (a 2 ) 3 = o 6 Ex. (a 3 ) 4 = a 12 Ex. (a x ) 2 = a 2X Ex. (a x ) y = a xy 2 The explanation of the above law is as follows : a 2 = aa (a -a) 3 = aaaaa-a — a 6 3 If the product of two numbers, each of which has an exponent, is to be raised to a power, the ex- ponent of each factor is multiplied by the number indicating the power. Ex. (a 2 6 3 ) 2 = a 4 & 6 Ex. (x z y 4 )* = x 9 y 12 Ex. (4a 2 fc 3 ) 4 = 4 4 a 8 & 4 c 12 4 The above is evident when one considers that a 2 b 3 is equivalent to aabbb. To take this value twice as a factor is to take (a) four times as a factor and (b) six times as a factor. 5 The process of raising an expression to any power is called involution. (139) 140 BRIEF COURSE IN ALGEBRA 6 Involution has to do with the raising of mono- mial, binomial, and polynomial expressions to powers. 7 The raising of any monomial to a power is accomplished by raising each factor of the monomial to the required power. Ex. a 2 raised to the second power is (a 2 ) 2 — a K Ex. 2a raised to the second power is (2) 2 . (a) 2 = 4a 2 Ex. (3a z b) 2 = 9a«b 2 Ex. (2xy 2 )* = Sx 3 y« Ex. (4a- 2 x n ) 2 =16a- i x 2n ( Note) In case there is no literal factor, the number is not usually factored. Ex. 4 2 = 16 Ex. 5 3 = 125 8 The same method is followed, regardless of the seeming difficulty of the problem. Ex. (x 2 y?) 3 = x 6 y* Ex. (4a°&- 3 ) 2 = 16&- 6 (Note) o°=l 9 To raise a fraction to a power, raise both mem- bers of the fraction to the power. Ex . (§>-* E~offtV4|-| V3/ 9 V3/ 3 3 9 (a\ 3 a 3 b) = b* Ex. Ex. 4s_ 4 9y* 10 Monomials having fractional factors may be expanded by using the above rule. POWERS 141 Ex. Aa 2 fc 3 Y=ja 4 (2 \» 8 -a 6 bc J =^ 18 < 3/- 3 11 In case a monomial is to be raised to a power expressed by a literal exponent, the same rules hold. Ex. (a 2 ) w = a 2n Ex. (aby) x = a x b x y x Ex. (2a 2 b) y = 2 y a 2y l> y Problems 1 Define involution. Expand, 2 (2x 2 ) 2 14 (ax) 2 3 (4xy) 2 15 (2ab) 3 4 (3*?%) 2 16 (2a M ) n 5 (4x 2 ;y) 2 17 (x 2 ;y) 30 6 (2rry 2 ) 3 18 (|a 2 6) 2 7 (5^ 2 2 3 ) 4 19 (4^) 3 8 ($ab 2 c) 3 20 (2a 2 &*c~ 2 ) 2 9 (10c 2 d) 5 21 (4w 2 w*£ -2 ) 3 10 (2a*>c 3 ) 2 22 (a*6 2 c~ 3 ) 2 11 (3a*) 2 23 (*Y«f)* 12 (2**30* 24 (2w 2 w)* 13 (J) 2 25 (4a* 6~ V)* Lesson 3 Fractional exponents. 1 If a* a* a? —a 5 =a 2 Then a 2 is the product of three equal factors, and from the definition of a root, a* is the cube root of a 2 Or ~ 3 In case a root is to be taken of a power. Ex. 78 = 2 Ex. ^4 = 2 Ex. ^/27 = 3 Ex. y/a*=a ( Note) If no index number is given the root to be found is the second. 2 Irrational roots. An indicated root that cannot be exactly found is called an irrational root. Ex. ^16, Vo 7 ; = a -= 1 _ n 4 4a 3 +6a 2 +4a+l Dividing the first term of the remainder by twice the first term of the quotient, Problems 1 2x 2 -2x=12 6 3;y 2 -15;y-42 = 2 3x 2 +10x = 12 7 2x 2 +3x+2=0 3 2x 2 -r-14x=-20 8 3a 2 +4a-3 = 4 7a 2 +7a-42 = 9 2m 2 +3m-7 = 5 2y 2 +18y+40 = 10 4x 2 +3x-2 = 11 A motor boat goes up stream and back in 6 hours. If the boat goes up 10 miles and the stream has a rate of 4 miles per hour, what is the rate of the motor boat in still water. 12 Sum of squares of two numbers is 73. Their difference is 5. Find the numbers. Lesson 4 (5) The formula. The equation ax 2 ± L bx± 1 c = is called the type form of the complete quadratic equation, (a) represents the coefficient of the squared term, (b) represents the coefficient of the first power term, while (c) repre- sents the constant term. Example — Given the particular equation, 3x 2 +2x+4 = Then, a = 3, 6 = 2, c = 4 If, then, we solve the equation, ax 2 +bx+c = 0, we solve the type of all quadratic equations. It is then possible to substitute the particular numbers for (a), (b), and (c) in the result and get a solution for any particular equation. Example — Given the equation ax 2 +bx+c = Q Solving this by completing the square, we get -&±V& 2 — 4ac x = L 2a (Note) This solution is given later. Now suppose the particular equation is x 2 +8x+15 = ami, 6 = 8, c = 15 (183) 184 BRIEF COURSE IN ALGEBRA Substituting these values of a, b, and c in the solu- tion of the type we get _-& = f = V / 6 2 -4ac_-8±:V / 64-(4.1.15) _ * 2a 2 8±V64-60 -8 = hv / 4 -8±2 -10 -6 _ , or — — or — 5 or — 3 2 2 Example — Given the problem 3x 2 — 5^+6 = We know that the solution of the type gives _ -b±\/b 2 — lac X ~ 2a In the given problem, a = 3, b= — 5, c = 6 Substituting, x .-(-5)±V(-5)'-(4.3.6) e- rr : +5±V25-72 Simplifying, x = 6 0r , g .+5^V-47 6 Example — Given ac 2 +6^-f8 = a = l, 6 = 6, c = 8 ,™ , -6zhV5 T -4ac Type form, x — La Substituting, , = -6 ± V(6)'-(4.1.8) THE QUADRATIC EQUATION 185 (Note) The given equation must be arranged so that all terms are upon the left side of the equality sign, that is, arranged like the type form. The solution of the type form follows. Problems Solve by using the formula. 1 x 2 +10x+21=0 7 3w 2 +4w+3=0 2 2s 2 +24x+54 = 8 6y z +2y=-6 3 3x 2 -3x-90 = 9 5m 2 +2m-5=0 4 5a 2 -3a+2=0 10 3a 2 -2+3a = 5 2y 2 +3y-4 = 11 3b 2 -2b = 6 6 5& 2 +36+2 = 12 4c 2 =-5c+2 Lesson 5 Development of the formula (6) Given, ax 2 +bx+c = Subtracting (c), ax 2 +bx = — c Multiplying both members by (a), a 2 x 2 -\-abx = —ac Completing square of left hand member, -•+*+(£)'- -<+(£)' b 2 b 2 Or, a 2 x 2 +abx+—=-ac+— 4 4 Extracting square root of both members, b 2 ax+--=±-y -ac+-- Or, ax+- = ± b l—lac + b'' - . b b —Aac+b 2 Subtracting -, ax= — ~±\ r 2' 2 b Simplifying, ax = — - ±- V— 4ac+b - — 6±V— 4ac + b 2 Or, a* = -n*- -j- u / \ —b±z^/b 2 —4:ac Dividing by (a), x = ~ - Ad ( Note) This solution may be obtained by division of both members by the coefficient of (x 2 ), or in fact by multiplying or dividing by any number making the (x 2 ) term a perfect square. (186) THE QUADRATIC EQUATION 187 (7) If the solution of the type form of the quad- ratic equation gives, — b± X \ r 1 r -A. ■ X , i (Note) All equations of the second degree have the double values to plot. As in simple indeterminate equations, two equa- tions may be graphically solved by plotting them and ■determining the points of their intersection. These points represent the graphical solution. VB 35928 54 1204 UNIVERSITY OF CALIFORNIA LIBRARY