º e e s e s a • * * o ~ * ， ！?? | ? £， | 2§,.ºz ſ 5 +ſ-º {∞∞Ş <！--*№.# # | | Ă |  | | =-| • ** ™=，- *~*=|#:Aº. ;！<！--ſ . =● ·∞ ſºr34 T. & Lou t . or of Ma themat e's sº e º e º 'º e º 'º e < * S Kalamazo Professor of Profess C §e º º ºr & A. se pro s a sºle e'e - a s M. P iſ . Teachers College. º ... a.m. ºn a º ºſ ºſ ºſ, 'ſ', ſºm "Z."， º， ſ. ºſ ºr // r / 74 47, '/'. /： \，-· · · 3. Tº HS ſae sº gº º P Gº º 'º dº -ſtiftſſºſ { , … •••=.*= *s- º-e， w rºw… • • • •* * * * *| º H & sº B A R T O O Western State Teachers Colle E os B O R. N. ■-| &3.*, ， ,, ! - ∞… |- ------| •••••• - ,}ae| |'№！-ſº º} |_№;©} ſa !ſ.==|- • · }-- - - -- |-; · |lae,' |=====F（~~~~）; }∞===）， • zº - Wº º: * . * *** *** *** * * * ~ *-*- tº º Sº tº tº El Tº gº ºr . P & R tº R. O. W. E. R. G sº By ENU LOUIS, MISSOURI. N C. AV ſ * º PUBLISH O T * @ as WEBST E R Co M P ANY 1808 WASHING ST. PRICE PER COPY 48 cts, NET -Postage fºx ſº Mwº FIRST YEAR ALCEBRA * A T E XT - WORK B O O K V : - , , *- : * G R OVER C : BART OO, M.A. Professor of Mathematics Western State Teachers College Kalamazoo, Michigan AND J E S S E O S B O RN, PH. D. Professor of Mathematics Harris Teachers College St. Louis, Missouri 3. - w" - i ;: - - - * * * * W E B S T E R º ºshi Nc comp ANY 1808 WASHINGTON AVENUE ST. LOUIS, MISSOURI ! p xtſ & s º CopyRIGHT, 1937, BY WEBSTER PUBLISHING Co. All rights reserved, including the right to reproduce any part in any form. PREFACE This is a first-year algebra textbook, together with appropriate workbook material to accompany the text. The combination is complete in itself. No other text is needed; no other workbook is needed. The text and workbook are combined that each may supplement and reënforce the other to the fullest extent. The content has been carefully checked with the requirements outlined in the nationally recognized syllabi. To include more would make for a course too difficult for pupils in the first year of high school; to include less would make for a course that fails to meet accepted Standards. We have endeavored to lean neither the one way nor the other, but to organize a course that pupils can understand and enjoy and by means of which they can master those principles and skills that experts have allocated to first-year algebra. A pupil gets most value and most pleasure from a course in which he has to work up to his capacity, but in which no attempt is made to drive him be- yond his capacity. First-Year Algebra: A Teact- Workbook provides for this by arranging for differ- ent levels of pupil ability. The unstarred material makes up the minimum course and is required of all. Better pupils will do the minimum course and the work preceded by a single star. The very best pu- pils will do all the work, including that preceded by two stars. The work is organized into fourteen teaching units, each a little course within itself, complete with teaching material, tests, and reviews. This makes it possible for the pupil to make a fresh start at the beginning of a new unit. He can begin the new unit with fresh courage and a determina- tion to do better than he has ever done before. The teacher should make the most of this in encouraging each pupil to work up to his capacity. The emphasis throughout is upon pupil under- standing and away from the blind manipulation of symbols. Students may be confused or discouraged in the very beginning of algebra by the strangeness and apparent unreality of what they are asked to do. We have made every effort to make this algebra real and meaningful. As a single illustration of this, observe how naturally and how simply Unit 1, “Formulas and Literal Numbers,” grows out of the arithmetic the pupil already knows. The many historical notes broaden the pupil’s understanding and add the human note so necessary to beginning algebra. The principles to be mastered are presented with- out long textual explanations. Such explanations often serve to bewilder or to bore the pupil. We make the essential explanation and give illustrative examples. This enables the pupil to get quickly to the principle or skill to be learned. This arrange- ment also provides an excellent setup for supervised study, where teacher and pupils together read the explanations and work through the examples. Ob- serve, on pages 25–31, the easy, natural introduc- tion to the laws used in solving equations. The first- year algebra pupil learns by doing, not by reading tedious verbal explanations. It is generally agreed that the idea that a change in one variable makes for a change in another is the cord that should bind the diverse topics of algebra into a unified whole. We have kept this principle continually in mind in Selecting and arranging the material of First-Year Algebra: A Teact-Workbook. The pupil experiences various aspects of this func- tion concept in his study of formulas, equations, graphs, proportion and variation, and trigonomet- ric functions. The idea is specifically called to his attention throughout the course, as on pages 16, 19, and 184. We are indebted to Dr. J. P. Everett, Chairman of the Department of Mathematics, Western State Teachers College, Kalamazoo, Michigan, for many helpful ideas. His suggestions for the units on “The Equation” and “Directed Numbers” were espe- cially valuable. THE AUTHORS CONTENTS Page Page UNIT 1. Formulas and Literal Numbers. . . . 1 UNIT 8. Fractions. . . . . . . . . . . . . . . . . . . . . . . . 126 UNIT 2. The Equation. . . . . . . . . . . . . . . . . . . . 19 UNIT 9. Fractional Equations. . . . . . . . . . . . . 143 UNIT 3. Directed Numbers. . . . . . . . . . . . . . . . 40 UNIT 10. Linear Equations in Two Unknowns 155 UNIT 4. Graphs. . . . . . . . . . . . . . . . . . . . . . . . . 56 UNIT 11. Ratio, Proportion, Variation. . . . . . . 177 UNIT 5. The Fundamental Operations. . . . . . 68 UNIT 12. Powers and Roots. . . . . . . . . . . . . . . . 188 UNIT 6. Special Products and Factoring . . . . 89 UNIT 13. Quadratic Equations. . . . . . . . . . . . . . 207 UNIT 7. Linear Equations in One Unknown, UNIT 14. Numerical Trigonometry. . . . . . . . . . 226 and Problems. . . . . . . . . . . . . . . . . . Supplementary and Review Work. . 239 UNIT 1. FORMULAS AND LITERAL NUMBERS GEORGE WASHINGTON GoFTHALs (1858–1928) The building of a canal through the Isthmus of Panama for the passage of ships between the At- lantic and Pacific oceans was for many years an un- fulfilled dream. In the latter part of the nineteenth century, a French corporation attempted the proj- ect. The cost, mounted to millions of dollars and thousands of lives, and the French gave up hope and quit, leaving behind class at West Point. The reward was an assignment to the Engineering Corps of the United States Army. This preferment is given only to those who excel in mathematics and in engineering. The building of the Panama Canal was one of the greatest engineering tasks ever undertaken. Tropical jungles, turbulent rivers, and high moun- tains added to the diffi- them 40,000 graves and great stores of rusted iron and ruined machin- ery. President. Theodore Roosevelt then deter- mined that the United States should build the canal. He chose Colonel (later General) George Washington Goethals to direct the work. As a student at West Point, º culties. Goethals applied to the task a dauntless courage and a compre- hensive understanding of the mathematics of engi- neering. No officer and no laborer worked harder to bring the job to a suc- cessful finish than did he; and under his leadership, the canal, which many said could never be built, was completed a full year and in later life, Goe- ºr. thals' interests were in mathematics and in en- gineering, rather than in what is often considered the work of a soldier. He had ranked second in his SHIP IN GALLIARD CUT, PANAMA CANAL ahead of scheduled time. Algebra and higher mathematics depending upon algebra were as necessary in building the Panama Canal as steam shovels were. MATHEMATICS ADVANCES As CIVILIZATION ADVANCES It is said that the Apache Indian of Arizona keeps a record of his ponies by carrying with him a bag of pebbles—one pebble for each pony. The Apache has made a beginning step toward the system of mathematics, which is the foundation of the modern business and scientific world. More forward races took this beginning step many thou- sands of years ago. Before civilization had advanced far, mankind took other steps. Men invented number names and number symbols, so that they could count and write numbers. These served only so long as people needed to answer the question, “How many?” and the answer was a whole number. As time went on, wholes had to be divided into parts; thus names and symbols were invented for fractions. As civilization moved forward, need for further extension of our number system became apparent. If each child has two hands, then ten children have 2X10 hands. This is arithmetic; the numbers are definite and specific. If each child has two hands, then n children have 2×n hands. This is algebra. The number n is general, that is, may represent any number. When H represents the total num- ber of hands, and m represents the number of children, then the formula H = 2×n holds true. Notice in this formula that n may have any value you wish to assign to it, and that as m takes on different values, the value of H will change. For the most part, algebra is a study of the uses of general numbers, and of the depend- ence of one upon another. THE USE OF LETTERS IN ARITHMETIC Mr. Stewart has an apple orchard. There are eight rows with ten trees to the row. How many trees are in the first row? In the second row? — In the third row? — In the fifth row? In the seventh row? How many trees are in one row? In two rows? In \ \º four rows? — In six rows? In eight rows? Mr. Smith has an apple orchard. There are r rows with t trees to the TOW. How many trees are in the first row? In the second row? In the sixth row? In the eighth row? In the eleventh row? In any row? How many trees are in one row? In two rows? In five rows? In ten rows? In r rows? FORMULAS Total number of trees equals number of rows multiplied by number of trees in one row. \— —V- ~ \-V-7 S- ~~ _* \ Y _* \ —y- _* N -: r X t Sometimes the cross (X) between the r and the t is replaced by a raised dot, in which case the formula becomes N=r. t. Sometimes the cross (X) is omitted, in which case the formula becomes N=rt. The same relationship is expressed in each of these three ways, and each of the formulas is read the same way. IEXERCISES Most of the formulas called for below appear in seventh and eighth grade arithmetic. Either recall them from your arithmetic, or think them out for yourself. Complete each formula. 1. Area of a rectan- 7. Area of a trap- ! b' b h e I gle ezoid h A = : b 4 =– • S 8. Perimeter of a 2. Area of a square Square A = h; P = S 3. Area of a triangle b ! 9. Sum of the an- C A = gles of a tri- 4. Area of a paral- / h / angle lelogram A +B+C = A —AB A *— 10. Volume of a 5. Area of a circle rectangular 4 =– box 6. Distance traveled in t hours at the rate of r W =– miles an hour 11. Circumference - of a circle D = C = Use this formula to find the distance traveled in 12. Cost of n fountai when each costs d six hours at thirty-two miles an hour. ost of n. Iountain pens Wne OSUS dollars C = Use this formula to find the cost of three foun- tain pens at $1.50 each. Answer: — Answer: — 2– THE USE OF LETTERS IN ALGEBRA The letters which appear in arithmetic formulas are symbols standing for numbers. A great difference between arithmetic and algebra is that letters stand for numbers more generally in the algebra. THE RECTANGLE b The area of a rectangle equals the base multiplied by the height. Write this formula in three ways (cross, raised dot, sign omitted). A =bh A = A = A = h P = 2b–H2h h The perimeter of a rectangle equals the sum of its sides. = 26-- \— —V- —’ \-v- S- Y L^ P = b +h-Hb-H h, or P=2b-i-2h b EXAMPLE Using the formulas above, find the area and the perimeter of a rectangle whose base is twelve feet and whose height is eight feet. A = b : h = 12.8 = 96 P = 2b-H2h = 24-H 16 = 40 Area is ninety-six square feet. Perimeter is forty feet. EXERCISES Find the area and the perimeter of each of the rectangles below. Write your answers in the blanks. 1 2 3 4 5 6 7 8 b 12 ft. 9 in. 15 rol. 6 yd. 11 ft. 5 in. 8 ft. 10 in. h 18 ft. 10 in. 4 rol. 1.5 yd. 11 ft. 3 in. 11 ft. 10 in. A. P Do the formulas for the area and the perimeter of a rectangle hold true for one that is higher than it is wide? — For one wider than it is high? For all rectangles? S THE SQUARE A Square is a rectangle whose height and width are equal. Let s represent the side. Write formulas for the area and the perimeter in terms of s. A = s.s =s* 4 =– P = P = 4s Notice the new symbol s”, which is a short way of writing s. s. EXERCISES S Using the formulas you have written, find the area and the perimeter of each of the squares below. Write your answers in the blanks. 1 2 3 4 5 6 8 S 6 in. 9 ft. 6 yd. 12 ft. 4.1 in. # in .6 rol. 6 mi. A. P THE TRIANGLE The area of a triangle equals one half the base multiplied by the height. ^ _* \— S-2–’ >--—’ A. F # b X EXERCISES h i Using the formula above, find the area of each of the triangles. Write your answers in the blanks. *—ºr- 1 2 3 4 5 6 7 8 b 6 in. 9 ft. 6 yd. 12 ft. 4.1 ft. 2.5 mi .5 ft. # in h 8 in. 5 ft. 7 yd. 8 ft. 10 ft. 8 mi. .8 ft. 4 in. A. Supply the missing words. (a) The area of a rectangle depends on its and its (b) The area of a square depends on its (c) The area of a triangle depends on its and its (d) As the side of a square increases, its area (e) If the base of a rectangle remains the same, and its height increases, the area PROBLEMS In solving these problems, (1) decide upon and write out the formula you need to use, (2) substitute the given values in the formula you have written, (3) perform the indicated operations, and (4) leave your result in the simplest form. The first has been done correctly. 1. A field is 12 rods long and 6.5 rods wide. (a) What is its area? Answer: 78 sq. ra. (b) How many rods of fence will it take to enclose it? Answer: 37 rol. A =bh = 12. 6.5 = 78 Area is 78 sq. ra. P = 2b–H2h = 24-H 13 = 37 Perimeter is 37 rods. 6.5 12 T30 65 wº-ºu 78.0 2. A farm is 120 rods long and 80 rods wide. (a) How many square rods does it contain? (b) How many acres does it contain? — — (160 square rods equal one acre.) 3. Find the cost of sodding a lawn 66 feet long and 32 feet wide at five cents a square foot. 4. A garage floor is 12 feet by 14 feet. (a) Find its area. (b) Find the cost of cementing it at 30 cents a Square foot. 5. Find the cost of covering a kitchen floor 12 feet by 8.5 feet with linoleum at $2.75 a square yard. Answer: 6. Find the area of the gable end of a garage which is a triangle with a base of fourteen feet and a height of six feet. Answer: | 7. A city lot is a parallelogram with dimensions i as shown. Find its area. 80 ft. Answer: 8. Find the volume of a rectangular box three inches by four inches by six inches. Answer: 9. How many cubic yards of space in a room fourteen feet long, twelve feet wide, and nine feet high? Answer: 10. Three streets meet as shown. Find the area of the triangular park formed. Answer: ADDING AND SUBTRACTING IN ALGEBRA EXERCISES Find the perimeter of each of the figures at the right. (a) P= + –– + F (b) P- + + —— - (c) P= + –– E (d) P= + + + - (e) P= + + + - (f) P- —— –– + – An algebraic expression, as 89, is called a term. It is composed of two parts, the 8 or numerical part, and the g or literal part. The 8 is called the coefficient of g. When no coefficient is written, the coefficient 1 is understood. Thus a means 1a. Terms having the same literal parts are called like terms. Thus 3g, 5g, and 2g are like terms. Like terms may be added and subtracted, as in arith- 24, 29 24: (a) [24 y (b) {/ º 2a: 2y 22 2p oxy 2m (d) 2n. ſ (e) 7 27, 4m. 4n 4p metic. In arithmetic, 5 feet plus 4 feet equals 9 feet, and 7 bushels minus 3 bushels equals 4 bushels. Sim- ilarly, in algebra, 5a plus 4a equals 9a, and 7a: minus 3a; equals 42. If terms are unlike, addition and subtraction must be indicated. Thus the sum of 2a, and 3y is in- dicated as 22–H3y, and the difference of 5a and 3b is indicated as 5a–3b. EXERCISES 1. Write the numerical coefficient and the literal part of each of the following terms. 1 2 3 4 5 6 7 8 9 10 Term 3a 9b (U 60 9m. Q 82 .5y ab 3ay Numerical coefficient Literal part 2. Add: 3a 14b 67m, 19e g 23k; 16.5i 27f 11.5m. .23p 2a 6b 8m. 13e 17g 14k. 4.3t 17f 3.5m. .41p 3. Subtract: 7a 21b 17C 3.900 15a, 3.5g 9ary 11.52 1977, .97g 20. 6b 11c 2.1M) 11a: 1.5g 4a:y 3.52 12m. .19q THE WHETSTONE OF WITTE The Whetstone of Witte is the fanciful title given lengthe, thus:= bicause noe .2. things can be to the first algebra printed in England (1557). The author was Robert Recorde. Our symbol for equal- ity (=) first appeared in print in this book. In in- troducing the new symbol, the author said: “And to avoide the tedious repitition of these woordes: is equall to: I will sette as I doe often in woorke use, a paire of paralleles or Gemowe [twin] lines of one moare equalle.” Recorde wrote three other well-known books: The Grovnd of Artes (arithmetic), The Pathewaie of Knowledge (geometry), The Castle of Knowledge (astronomy). Because of the great influence his writings had, Recorde has been called the founder of the English school of mathematics. — 6 — ExPONENTS The formula for the area of a square is A = s.s., or A = s”, read “A equals s square.” Similarly, the formula for the volume of a cube is V= e. e. e=e”, read “V equals e cube.” To square a number means to take it twice as a factor; to cube a number means to take it three S times as a factor. The 2 used with s in s” is called an exponent; the The term a” means that a has been taken as a factor four times, and is read “a to the fourth power,” or “a with exponent 4.” The term 2" means that 2 has been taken ° as a factor five times, and is read “2 to the fifth power,” or “2 with ex- ponent 5.” € When no exponent is written, the exponent is understood to be 1. Thus, S € 3 used with e in e” is also an ex- a means a to the first power, or a ponent. with exponent 1. EXERCISES 1. Write the numerical coefficient and the exponent of each of the terms. 1 2 3 4 5 6 7 8 9 10 Term 362 4a:3 #a 1024 67m,7 11774 g #h lºft 8b7 Numerical coefficient Exponent 2. Given the numerical coefficient, the literal part, and the exponent, write the term. 1 2 3 4 5 6 7 8 9 10 Term Numerical - coefficient 3 2.5 # 11 3# 4 9 1 2 1 Literal part 0. b C d € f (C $/ 2 QU) Exponent 3 2 1 4 3 6 2 1 1 5 POLYNOMIALs An algebraic expression having only one term is called a monomial. Examples: 3a, a”, 2xy, 5aºb. An algebraic expression containing two terms is called a binomial. Examples: 22–y, 3a*-i-2y, a”--b”. An algebraic expression containing three terms is called a trinomial. Examples: a-Hb-Hc, ac”–3y–52. More generally, an algebraic expression having two or more terms is called a polynomial. Examples: 3m-6m, 22-Hy–42. The terms of an algebraic ex- pression are always connected by plus or minus signs. Each of these new words ends with nomial. This comes from a Latin word meaning name. The pre- fixes to this ending, mo (one), bi (two), tri (three), and poly (many) come from the Greek and Latin. Thus, monomial means one name (one term). EXERCISES 1. Write five monomials. ; ; y y 2. Write five binomials. ; 3. Write three trinomials. - ; y 4. Write three polynomials, each having a different number of terms. ; y — 7 — TRANSLATING WORD STATEMENTS INTO FORMULAs EXAMPLE The area of a parallelogram equals the product of its base and its height. \- ~ \-v- S- _* ~~ A - Notice that the subject of the sentence is translated as A, the verb as the equality sign, and the object as bh. in / / EXERCISES Complete the formulas. 1. The diameter of a circle equals twice the radius. d = 2. The perimeter of an equilat- eral triangle equals three times One side. S S P = - S 3. The volume of a rect- | angular prism equals the : C product of its length by its width by its height. 22--------> _2^ b V = (l 4. The perimeter of a quadrilateral equals the sum of the four sides. P = 5. The area of a circle equals one half of the circumference (C) times the radius. A = 6. The perimeter of any tri- angle equals the sum of the three C b sides. (l P = - 7. The volume of a cone equals one third the product of its base (b) and its height (h). V = 8. The area of a rhombus equals one half the product of its two diagonals. ZºZ A = - 9. The time (t) it takes to travel a certain dis- tance (d) equals the distance divided by the rate (r). # = 10. Write the formula for the cost (C) of n ar- ticles at d dollars each. C = 11. Write a formula for the distance (d) that can be traversed in t hours at r miles an hour. d = 12. The cost of one apple (C) equals the cost of a box of apples (d) divided by the number (n) of apples in the box. C = *13. A telephone costs three dollars a month with an extra charge of five cents each for suburban calls. Write a formula showing the total cost for k months during which time n suburban calls were made. **14. A taxicab driver charges thirty-five cents for the first mile and twenty-five cents for each ad- ditional mile. Write a formula expressing the total cost for riding S miles. IMPORTANT LAws The sides of a triangle are 3, 4, and 5. The perimeter is 3+4+5. May the perimeter also be written 4+3+5? Or 5+4+3? % || or 4+5+3? – The sides of another triangle are a, b, and c. May the perimeter be 3 written a-Hb-Hcº Or b-Ha-H ch. Or c-Ha-H b? 0. In column addition, should one get the same answer whether he adds up or down?— From considerations of this sort, one arrives at the law: In addition, the numbers added may be taken in any order. Is 3.4=4:3? Is a b = b : aſ Is 5.6.7 = 7.6: 5? Is 6.5-7 = 5. 7.6% Is a , b, c = c. b. a7 Such considerations as these illustrate the law: In multiplication, the factors may be taken in any order. EXERCISES Write each of the following in as many different orders as you can. 1. d-He-Hj = - F - F 2. w—H22+y= - == - F. 3. m.no = - - + - 4. kwh = - – – - When several operations are indicated in the same expression, the order of performing them is given by the law: In a series of operations involving addition, subtraction, multiplication, and division, first the multiplication and division should be performed in the order in which they occur, and then the addition and subtraction in order. EXAMPLES 1. 2+-6-4-2–H 6, 2–8 = 2+3+ 12–8 =9 2. 3+2, 6–10–3–2 = 3–H 12–5 = 10 - EXERCISES 1. 3+4.6 = 4. 4-H 8+4+2.7 = 2. 7–H4+2+1 = 5. 6-3-2-1-6–4.2 = 3. 6-3-2-H4–2= - 6. 8–4–3–2–1 6–2 = EvALUATING ALGEBRAIC ExPRESSIONS One can find the numerical value of an algebraic expression by substituting arithmetical numbers in place of the literal numbers. The process is called evaluation. Example: Evaluate acº—acy, when a = 3 and y = 2. Solution: 3%–3.2=9–6=3. EXERCISE Given a = 2 and y = 3, evaluate each of the expressions. 2a: Q/-Ha: Q: Q Expression ac-H2y 2y–a: 3acy-y acy-Hya: g”—a: tº 9-ta. ——H 9 3a;+4y y—a: $/ 3U Numerical value PARENTHESEs Parentheses ( ) are used in algebra to indicate quantity b plus h.” Suppose b = 5 and h = 3; then that two or more terms are to be considered as a P=2(5+3)=2(8) = 16. single quantity. Thus 2(5+6) means 2(11), or 22; The formula for the cost of riding s miles when (9–1)+2 means 8+2, or 4. We call the terms in the rate is thirty-five cents for the first mile and a parenthesis “the quantity.” Thus 2(5+6) is twenty-five cents for each additional mile is read, “Two times the quantity five plus six.” C =$.35+$.25(s–1). The cost for riding four miles The formula for the perimeter of a rectangle is may be determined by substituting 4 for s. P=2b-H2h. By use of parentheses, this may be C = $.35 + $.25(4–1) = $.35 + $.25(3) = $.35 written P=2(b+h), read “P equals two times the + $.75 = $1.10. EXERCISES 1. Simplify and write the answers in the simplest form. (a) 4(5–2) = (d) 4-H2(8–5) = (g) 4-H2(3+2)(3–2) = (b) (3–2)(8–4) = (e) (15–3) + (9–5) = (h) (15–2)(3+2) = (c) 10+ (12–9) = (f) 15–3(2+1) = (i) (4-H 5) + (6–3) = 2. Using the formula P=2(b+h), find perimeters of the rectangles. Write answers in the simplest form. 1. 2 3 4 5 6 7 º 8 b 6 3.5 12 yd. 7 in. 8.2 ft. 3% in. 6; in. QC h 3 2.5 6 yd. 6% in. 4.1 ft. 2% in. 4; in. $/ P 3. Using the formula C = $.35+$.25(s–1), find C for the values of s indicated. 1 2 3 4 5 6 7 8 S 5 7 8 6 2 1 10 17 C 4. Substitute w = 2, a =3, y =4, and z = 5 in the algebraic expressions and record answers on the blanks. (a) w—Ha:(y-H2) — (c) (w-Ha)(y-H2) – (e) (w-Ha!-y)2 (b) w(z—a)+y — (d) (y-ay(y-Ha) — (f) wry(2–2) e a-Hb-H c e tº a 5. Using the formula s = , find s and s—a for the values of a, b, and c indicated. 1 2 3 4 5 6 7 8 0. 3 4 5 25 17 7 12 3 b 4 3 12 39 10 8 5 4.6 C 5 5 13 46 21 9 13 6.4 S — O. Notice that the numerator is treated as a single quantity just as if the terms were enclosed in parentheses. — 10 — *THERMOMETERS Examine the thermometer in your home or school. It is probably a Fahrenheit thermometer. This is the kind in common use. On the Fahrenheit scale, water freezes at 32° and boils at 212°. The centigrade thermometer is commonly used for scientific work. On the centigrade scale, water freezes at 0° and boils at 100°. Notice that the interval between freezing and boiling is 180 degrees on the Fahrenheit scale, and 100 degrees on the centigrade scale. This means that 180 degrees Fahrenheit are equivalent to 100 degrees centi- grade or that one degree Fahrenheit is equivalent to five-ninths degree centigrade. Notice also that the Fahrenheit thermometer reads 32 de- grees where the centigrade reads 0 degrees. One can change from a Fahrenheit to a centigrade reading by use of the formula: C =#(F–32) EXAMPLE If a thermometer reads 68° F., what would the reading be on the centigrade scale? C =#(F–32) =#(68–32) =#(36) = 20. This checks with the drawing. EXERCISES When possible, check your answers with the drawing. 1. Water freezes at 32°F. Find its freezing point on the centigrade scale. Answer: o C. 2. Water boils at 212° F. Find its boiling point on the centigrade scale. Answer: º' C. 3. The melting point of cast iron is 2,100°F. Change this to a cen- tigrade reading. Answer: o C. 4. The normal body temperature is 98.6°F. Change this to a cen- tigrade reading. Answer: o C. 5. The temperature of the sun is 14,400°F. Find this temperature as a centigrade reading. Answer: o C. r) ^ Boiling Point 100°–4– _ BOIllng Poin +210° of Water +200° 90°-- —-190° +180° 80°-- --170° —I-1609 70°-- 160 —-150° 60°–H +140° --130° 9–1– 50 +120° —-110° 40°–H O - - +100. Normal Body Temperature O —-90° 30°-- +80° O 20°–4– H-4 +79. -Suitable Room Temperature +60° toºl. --50° —-40° 09+ º--------|--|--|->= 3- Freezing Point —-30 of Water —-20° -10°-- O —H·10 -17.78°-- -- 0° C. F. 6. A thermometer reached 104°F. last summer. Change this to an equivalent centigrade reading. Answer: o C. 7. Translate a Fahrenheit reading of 77 degrees into its equivalent centigrade reading. Answer: o C. 8. Translate a Fahrenheit reading of 72 degrees into its equivalent centigrade reading. Answer: o C. — 11 — THE TRAPEZOID A trapezoid has two parallel sides and two sides which are not parallel. b' The parallel sides (b and bº in the figure) are called bases. The height of the trapezoid is indicated as h. The area of a trapezoid is given by the formula: A =#h(b+b') i - EXAMPLE - h Find the area of a trapezoid in which b = 12 ft., b’= 10 ft., and h = 8 ft. Solution: A =#h(b+b’) =# 8(12–H 10) = 4.22 = 88. Area is 88 sq. ft. EXERCISES Find the area of each of the following trapezoids. Write your answers in the blanks. 1. Given: b = 17 yd., b’ = 6 yd., h = 3 yd. Area is Solution: 2. Given: b = 3% ft., b’=2% ft., h = 1% ft. Area is Solution: 3. Given: b = 3.4 in., b’ = 2.8 in., h = 1.6 in. Area is Solution: EXERCISES: REVIEW OF FORMULAs 1. Find the perimeter of a rectangle when b = 4 ft. and h =2# ft. Answer: Formula: Solution: 2. Find the area of a triangle when b = 8.5 yd. and h = 4.3 yd. Answer: Formula: Solution: 3. Find the area of a square whose side is 2% inches. Answer: Formula: Solution: 4. Find the perimeter of an equilateral triangle of side 48 in. Answer: Formula: Solution: 5. Find the area of a parallelogram when b = 16.4 and h = 10.2. Answer: Formula: Solution: 6. Find the perimeter of a rectangle, when b = 3; and h =2#. Answer: Formula: Solution: 7. Find the area of a triangle when b = 16.4 and h = 8.6. Answer: Formula: Solution: 8. Find the perimeter of an equilateral triangle when s = 8.6. Answer: Formula: Solution: WORD STATEMENTS AND ALGEBRAIC ExPRESSIONS Two of the important procedures of algebra are: I. Translating a word statement into an algebraic expression, and II. Translating an algebraic expression into a word statement. EXAMPLES 1. Write as an algebraic expression: Three times a, plus 4. Answer: 32-H4. 2. Express as a word statement: 4y–3. Answer: Four times y, minus 3. EXERCISES 1. Translate each of the word statements into equivalent algebraic expressions. (a) The remainder when 4 is subtracted from 7b (a) (b) The sum of a, 2b, and c (b) (c) The product of b and h (c) (d) The product of s and s (d) (e) m added to 44; (e) (f) The product of 6 and 20. (f) (g) m decreased by 4a: - (g) (h) The square of a increased by twice a t (h) (i) Four times k plus two times y (i) (j) The sum of a and b less their product (j) (k) The product of a; and y added to 5 (k) (1) The sum of the squares of the three numbers a, b, and c (l) (m) a decreased by 6, times the Square of y (m) 2. Translate each of the algebraic expressions into equivalent word statements. (a) a -3g (a) (b) a-H2b-H3C (b) (c) w”—42? (c) (d) mn–5y (d) (e) 6(a+b) (e) (f) 4–(2-Hy) (f) (g) (c-H d)(c-d) (g) (h) a”--b°–2ab (h) (i) w8 (i) (j) aſb-Hc) (j) (k) #h (b+b') (k) (I) 2(b+h) (l) — 13 — USING LETTERS FOR NUMBERS 1. The sum of two numbers is twenty-three. One number is k. What is the other number? 2. The difference of two numbers is eighteen. The smaller number is y. What is the larger number? - 3. What is the cost of y yards of cloth at eighty cents a yard and 2 yards of ribbon at thirty-five cents a yard? 4. Mr. Young is forty-three years old now. How old will he be a years hence? 5. A ball costs a cents, a kite costs y cents, and a knife costs 2 cents. Find the total cost of four balls, seven kites, and three knives. 6. Joan is two years older than Robert. How old will Robert be when Joan is a years old? 7. There are y girls in Jefferson School, and there are 17 fewer boys than girls. How many pupils in the school? 8. George has five marbles, and Harry has a marbles more than George has. How many have they in all? 9. Jean had d dollars and spent a dollars. How much has she left? 10. Mr. Jones is y years old, and he is seven years older than Mrs. Jones. How old is Mrs. Jones? 11. John had a cents. He spent twelve cents more than one half of what he had. How much did he spend? 12. Fred sold four boxes of tomatoes, at a cents a box, seven baskets of grapes at y cents a basket, and three dozen oranges at 2 cents a dozen. What was the total amount of his sales? *13. Mrs. Smith had d dollars and spent eight dollars more than one half of what she had. How much did she have left? *14. Jack had w cents. He spent twenty-five cents and gave one half of what he had left to his mother. How much did he give to his mother? *15. A rectangle is 2 inches wide and four inches longer than it is wide. Find its length and its perimeter. *16. Jenny and Marian together have 22–H5 dollars. Jenny has five dollars more than Marian. How much has each? J ; M **17. A farm is divided so that the oldest son receives a acres; the middle son receives twelve acres less than the oldest Son; and the youngest Son receives one half as much as the oldest son. Fill the blanks below. Oldest son, – – acres; middle son, — acres; youngest son, acres; whole farm, 3,OI’éS. **18. Farmer Brown raised twelve bushels more potatoes than Farmer Smith; Farmer Smith raised twice as many potatoes as Farmer Green. Farmer Green raised a bushels. Fill the blanks below. Farmer Green, bu.; Farmer Smith, bu. bu.; Farmer Brown, bu.; all together, MULTIPLICATION AND DIVISION OF LITERAL NUMBERS The multiplication of literal numbers is indicated by placing them side by side. Thus, the product of a, b, and c is abc, and the product of 2d and 3a is 6aac. EXERCISES Write the following products in simplest form. 3. 7a .9M) = 4. 86.9/= 1. 3a 4b = 2. a .2y = 5. a . b. 3c = 6. 4a. 7uy. 59 = The division of literal numbers may be indicated by writing them as a fraction. Thus, the quotient of a QC - divided by y is written –, and is read “the fraction a divided by y.” $/ EXERCISES Write the following indicated divisions as fractions. 1. 7a;+y= 2. 3a–4–26 = 3. 10d.--5e = 4. 4g-i-6h = 5. 2k+4m = 6. r-i-4p = WOCABULARY OF ALGEBRA Place a check mark before each of the terms below which you understand and can use correctly. — 1. angle —11. — 2. area —12. 3. binomial —13. 4. centigrade —14. 5. circle —15. 6. circumference —16. 7. coefficient —17. 8. Cone —18. 9. dependence —19. 10. diameter —20. difference equilateral evaluate exponent factor Fahrenheit formula like terms literal numbers monomial —21. —22. —23. —24. —25. –26. —27. –28. 29. –30. parallelogram parentheses perimeter polynomial power prism product Quadrilateral quotient radius —31. —32. —33. –34. —35. –36. –37. 38. —39. —40. rectangle rectangular rhombus Square SULIYl symbol term thermometer trapezoid triangle Look in this book or in a dictionary to learn more about any word above, which you do not thoroughly understand. REVIEW OF AREA FoRMULAs Write the formula for and find the area of each of the following. The answers (areas) are given at the right, but not in correct order. Given b h b’ Formula Area Rectangle 3% 2% Triangle 4.5 3.4 Square 16# Parallelogram 3} 4; Trapezoid 4% 6 3% Answers: 141's 13 24 8% 27.2% 7.65 — 15 — RELATIONSHIPS Formulas show how algebraic quantities depend upon each other. The area of a rectangle depends upon its base and its height. The perimeter of a tri- angle depends on the lengths of the three sides. The area of a square depends on the length of a side. The area and the side are related by the formula A = s”. If s increases or decreases, A must at the same time increase or decrease. EXERCISES Write the missing words. 1. If P=2s, P increases as s P = , and when s =4, P= 2. If P = 4s, P 3.S. S When s is doubled, how is P affected? , and P decreases as s When s = 2, , and P &S S —— 3. In the formula A =bh, A depends on A. ; when h increases and b remains the same, A and b remains the same, A 4. Given P=5s, find values of P when S has the values shown. S 1 2 3 4 5 6 P 5. Write the formula which will show the cor- rect relationship between P and s. The table is given. S 1 2 3 4 5 6 P 4 8 12 16 20 24 Formula: P = ; when both b and h increase, A When b increases and h remains the same, ; when h decreases 6. Given A =s*, find values of A when S has the values shown. S 1 2 3 4 5 6 A 7. Write the formula which will show the correct relationship between A and s. The table is given. S 1 2 3 4 5 6 A. 1. 4 9 16 25 36 Formula: A = PLUS AND MINUS SIGNS One value of algebra lies in translating long word statements into concise algebraic symbols. Symbols for addition and subtraction help make this pos- sible. Our present plus sign (+) and minus sign (–) appeared in print first in J. Widman's book on arithmetic, published in 1489. These symbols had been used in manuscripts many years earlier. Some say that our plus sign came from writing the Latin et (and) hurriedly with a pen. The steps may have been something like this: 3 et 4, 3 SP 4, 3-H4. Both the plus sign and the minus sign appear to have been used in mercantile practices before they appeared in books on mathematics. Bales and barrels of goods were weighed in warehouses, and each was marked + or — a certain number, according as it weighed more or less than the standard weight. The oldest mathematical book known was writ- ten about 1750 B.C. by Ahmes, an Egyptian scribe. He used a pair of legs moving forward (A) as the sign for addition, and a pair of legs moving away (A) as the sign for subtraction. Ahmes wrote from right to left. Except that the direction has been changed (left to right) to fit with our way of writing, the following is an algebraic statement as Ahmes wrote it. Notice the plus and minus signs in this equation. T \s-i Ş A. ſ. s." W = Two thirds added, one third subtracted, ten re- mains. — 16 — 1. 2. 3. INVENTORY TEST The second term in the trinomial 3–44 3c is The cost of n pounds of Sugar at six cents a pound is dollars. If the base of a rectangle remains the same and the height is doubled, the area is 4. When d-4 and e=3, then d”—e”= ‘5. different 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. Whetstone of Witte; of English school of mathematics; 25. In algebra, letters often represent , and different letters usually represent In the expression 5m, the 5 is called the In the expression m”, the 2 is called the 6a, means 6 ac; a” means a Ol, Cl. If k books cost five dollars one book will cost The coefficient of 4b* is and the exponent is The coefficient of y is and the exponent is Write the algebraic expression whose literal part is a , whose coefficient is 7, and whose exponent is 3. Give examples of three like terms. y y - e C The expression d IQ68,1].S C d. The algebraic expression for the sum of 2m and 5 is The algebraic expression for 3 less than the product of a and y is The area of a square depends upon its The sum of 3y, 4y, and y is The formula for the area of a rectangle is ; if b = 7 and h = 6.5, the area of the rectangle The formula for the area of a triangle is - ; if b = 3 and h = 4.5, the area of the triangle is In addition may the terms be taken in any order? Combine like terms: (1) 3a–2b-H4b-H2a = ; (2) 42-y–H39-Ha:= In multiplication may the factors be taken in any order? This means that k . m = Check each of the works you associate with Robert Recorde. founder parentheses; plus and minus signs; equality sign; Panama Canal. Perform the indicated operations: (1) 6–8–3–2–H5.3= ; (2) 1+6.2–9–3–3 = – 17– REVIEW Place a check mark (V) before each of the following which you are sure you understand. — 1. — 2. . – 10. — 11. — 12. — 13. — 14. —*15. What is the formula for the area of a rectangle? Of a triangle? Of a parallelogram? What is the formula for the perimeter of a rectangle? Of a square? Of an equilateral triangle? . What is the order of performing the operations of addition, subtraction, multiplication, and division? . Define literal part, coefficient, exponent. . Give illustrations of term, monomial, binomial, trinomial, polynomial. . What is meant by evaluating an algebraic expression? . What purpose do parentheses serve? Give an illustration. . Give an illustration of (a) translating a word statement into an algebraic expression; (b) trans- lating an algebraic expression into a word statement. . Explain the general idea of relationship or dependence as used in algebra; specifically, how the side and perimeter of an equilateral triangle are related. Name three ways to indicate the product of two or more literal numbers. Explain the meaning of “a squared,” “y cubed,” and “2 raised to the fourth power.” Tell something of the introduction of the symbols =, +, and —. Name one important difference between arithmetic and algebra. What is a trapezoid? What is the formula for its area? Explain the difference between the Fahrenheit and centigrade thermometers. What is the formula used in changing a Fahrenheit reading to a centigrade reading? How many checks marks do you have to your credit? THE PANAMA CANAL High waſer ſeveſ + 10" Jea ſeveſ = o' / | | |- - - - -27- - --Rººft'vºv-U- - - - - - - - - Hi yºu.” tº Bottom of Canal + 4O' Tº cº O S §3 & & Sº N. (S S. Ş © AS O CŞ § & & SY N i º A \- wafer ſeve/ - /O' Bottom of Canal –4O DIA GRAM OF THE CAN AL An engineering feat to which mathematics contributed. The Panama Canal has been called a stairway other side. The ship actually climbs over the moun- of water. The largest battleship or ocean liner can tain. enter one end of the canal and by means of dams In thus traversing these fifty miles, hundreds of and locks mount steps of water to the height of a ships each year save journeys of seven thousand ten-story building. It descends similar steps on the miles around, Cape Horn. — 18 — UNIT 2. THE EQUATION MATHEMATICS IN School AND IN LATER LIFE ARCHIMEDES Archimedes (287–212 B.C.) studied the math- ematical theory of levers and pulleys, and then used this theory in setting up systems of levers and pulleys to do real jobs. Perhaps you know the story of how he proved that the King's crown was not pure gold. 3: W The student asks, “What is algebra good for?” In this course he will learn to solve book problems. But he may still ask, “What is algebra really good for?” The story is told of a boy who had had algebra in school, and had left school to take a position. In his work he came upon a problem which he had to solve before he could go on. When he could not solve it, the employer said, “That's just algebra that you had in school. Why don't you use what you have learned?” The boy had not thought of that. To him algebra had been nothing more than solving book problems for the teacher. Archimedes and Steinmetz to a great extent, and thousands of others to lesser extents have used the principles of algebra to solve the real problems that come up in business, in science, and in indus- try. IDENTITIES In arithmetic we used equations of the sort 2+4=6 and 3(2+5)=21; in algebra, we have learned similar equations b-H h--b-H h =2b-H2h and a(b+c) = ab–Hac. This sort of equation is called an identity. Substitute a =2, b = 3, and c=4 on each side of the identity aſb-Hc) = ab–Hac. The value on the left will equal the value on the right. Substitute a =3, b =5, and c=2, and check again. Substitute any values you choose for a, b, and c; you will find CHARLES P. STEINMETZ Charles P. Steinmetz (1865–1923) studied and taught mathematics in the universities of Europe and America. He is known best, however, for having applied his mathematics to the development of the electrical industry. 7, International News Photos in each case that what is on the left of the equality sign equals what is on the right. An identity is de- fined as an equation which is true for all values of the letters involved. CONDITIONAL EQUATIONS An equation of the type a +3= 5 is called a con- ditional equation. The condition that must be satis- fied to make the equation true is that a = 2, for 2+3=5. If a is any other value, the equation is not true. A conditional equation is an equation that is satisfied for some value (or values) of the unknown, and is not satisfied for any other value. Finding a value of the unknown that satisfies an equation is known as solving the equation. The value found is called a root of the equation. Notice that a conditional equation states that two expressions are equal, and asks a question. In the equation 24-H 5 = a +7, the two equal expres- sions are 24-H 5 and a +7, and the question is, What is the value of acº MEMBERS OF AN EQUATION The part of an equation at the left of the equal- ity sign is called the first member or the left mem- ber of the equation; the part at the right of the equality sign is called the second member or the right member of the equation. LAws of CHANGE The old philosopher, Heraclitus (530–470 B.C.), stated the principle that the only thing which is un- changeable is change itself. There is nothing today as it was a hundred years ago, or last year, or even yesterday, except that all things change now as they did then. Many changes are related one to another. A cen- tral idea of algebra is to find and to state such rela- tionships. As the side of a square changes, the area changes; the distance one travels in eight hours de- pends on the rate an hour; the amount of interest one receives changes with the amount of the loan and the rate. In each instance, a change in one thing makes for or depends upon a change in another. — 19 — WORD STATEMENTS AND EQUATIONS In the left column are word statements, and in the right column are equations. The statements are to be matched with the equations. Let n represent the unknown quantity. Before each equation to the right, write the letter of the word statement which corresponds. The first has been completed correctly. a. What number must be subtracted from 72 to leave 54? g 2n+6=20 b. What number must be added to 16 to make 61? 3m — 15 = 21 c. The sum of 16 and 61 is how much? 2m – 6 = 20 d. Fred divided two dozen marbles into three equal piles. 3n+15 = 21 e. Ten years ago, George was nine years old. tºsams-s n-º. f. After subtracting 5 from twice a certain number, the result is 17. m = 16+61 g. Twice a certain number plus 6 equals 20. 3n - 15 = 27 h. Twice a certain number is 6 more than 20. n+16=61 i. Three times a number less 15 equals 21. 72–?) = 54 j. Three times a number is 15 more than 27. 2n −5 = 17 k. Three times a number is 15 less than 21. n—10=9 Before each word statement at the left, write the letter of the equation (in the column at the right) which corresponds. In each case, n represents the unknown quantity. The first has been completed correctly. c A certain number subtracted from 70 leaves 31. a.. n.--6 = 27 Three times Mary's age is fifteen years. b. 4n+7= 19 A certain number added to 11 makes 17. c. 70— n = 31 Three times a certain number less 6 equals 3. d. m-H 11 = 17 Four times a certain number plus 7 equals 19. e. 3m. – 6 = 3 One half a certain number less 6 equals 3. f. m. — 15 = 22 In eight years, Virginia will be twenty-one years old. g. 3m = 15 Add six years to Ruth's age and you have twenty-seven years. h. 2n+3 = 25 Twice a certain number and 3 make 25. i. m. – 5 = 23 Five years ago, John was twenty-three years old. j. n+8 = 21 Fred gave his sister 15 cards and had 22 left. k. #–6–3 Fill the blanks below, supplying either the word statement or the equation, whichever may be lacking. Let n represent the unknown. The first has been completed correctly. 1. Twice a certain number plus 15 is equal to 97. 2n+15 = 97 2. Three times a certain number is 15 more than 36. 3. John is 7 years older than Mary, who is 5. 4. 3m — 6 = 42 70, º –––3 = 41 5 2 + 6. Twice a certain number is 15 more than 97. - — 20 — WoRD STATEMENTS AND EQUATIONS Translate each of the following word statements into an equation, letting a represent the unknown. Do not solve the equations. The first has been done correctly. What number taken from 17 leaves 9? What number when doubled makes 76? Twice a certain number plus 6 equals 40. What number must be added to 14 to make 27? What number must be added to 31 to make 79? Ten years from now, John will be 23 years old. . If 6 is subtracted from twice a number, the remainder is 10. . Grace is 7 years older than Fred, and the sum of their ages is 19. If one half a certain number is subtracted from 17, the remainder is 9. ac—H·14 = 27 Each of the equations above asks a question. To solve the equation means to find the value of the un- known. The value (root) found is the answer to the question. Solve as many of the equations above as you can, and write the answers below. 1. 2. 3. - 4. 5. 6. 7. 8. 9. Translate each of the following equations into an appropriate word statement. The first two have been done correctly. 1. ac-4=8 Eight is 4 less than what number? 2. #2 = 4 Four is one half of what number? 3. #2+4 = 18 4. 22–4 = 7 5. 42+}=8; 6. 5–H4a: = 17 7. 15 = 5a, 8. 20 =3a – 5 9 #-7 10. 3–ac = 1 11. 5-H2a: =9 12. 3a – 1 = 8 FRANCOIs VIETA (1540–1603) Francois Vieta studied mathematics as a recre- ation. His profession was law, and he became a member of the provincial parliament. During a war between France and Spain, he deciphered secret codes for the French military authorities. In mathematics we remember him principally for his work in algebra. He is credited with being the first to use letters in a systematic way to represent numbers. As you learn more about algebra, you will realize how much it means to be able to translate long word statements into concise algebraic equa- tions. It is difficult to solve an algebraic problem, except the very simplest, without first translating it into symbols. Vieta used a, b, c . . . in the same Sense we now use ac, y, 2 . . . . His claim to fame lies in his influence away from the rhetorical algebra of his predecessors, and toward the symbolic algebra of the present time. — 21 — SOLVING FIRST DEGREE EQUATIONS An equation of the type ac-H3=5 is called a first degree equation, or a linear equation. The un- known, a , occurs to the first degree only, and the equation has only one root, 2. We are speaking here, of course, only of conditional equations, and not of identities. Here are some examples of first de- gree equations: 22–3 = 11, 3–a = 1, 2a:–3=4+ æ, 5–2a: = ºr— 1. We are not concerned here with equations of higher degree. However, mention of such equations may help you to a better understanding of what first degree equations are and are not. An equation of the type a 2–7a;+12=0 is called a second degree equation, or a quadratic equation. The unknown, ac, occurs to the Second degree, and the equation has two roots. The roots of this par- ticular equation are 3 and 4, both of which check when substituted in the equation. An equation of the type acº–6a.”-- 11a: —6=0 is called a third degree equation, or a cubic equation. The unknown, a, occurs to the third degree, and the equation has three roots. The roots of this particu- lar equation are 1, 2, and 3, all three of which check when substituted in the equation. We are now ready to proceed to the solving of first degree equations. EXERCISES Supply the missing numbers so that each of the following is an equality. 1. 3+ = 7 8. 20– = 17 15. 3 times = 27 2. +6 = 11 9. 25 = 7–H 16. of = 16 3. 16+ = 36 10. 27 = 40– 17. § of = 20 4. 7.5–H = 10.5 11. of 5 18. 3 times =36 5. 16 – =9 12. of 8 19. 3 times = 33 6. – 5 = 6 13. # of 5 20. 2 times = 7 7. – 6 = 14 14. 3 times = 21 21. of = 2 Write roots on the blanks at the right of the given equations. See if you can think of a way to check the correctness of your answers. 22. n-H2 = 4 -º- 29. e = 4–2 23. a +6 = 10 -*-* 30. m. — 6 = 24. 7-Hy = 11 -ms-s 31. 3d = 6–3 25. 10 = c-H3 *-* 32. #q=3 26. 2m = 6 -sm- 33. 37 = 7 27. d – 8 = 12 -*- 34. 7 – S = 28. §g = 6 - 35. 4 = a – 2 º- 36. 7 = 32 - g- 37. m. — 2 = 5 - - 38. #a = 2.5 -º- tº- 39. 4g = 18 - *- 40. 7) – 1.5 = 3.5 - *-*- 41. 4–3 =g -s- *m- 42. 3.5 = n+1.5 -*- Four numbers follow each equation below. Put a circle around the one number which satisfies the equa- tion. The first exercise has been marked correctly. 43. a +2 = 5 1 2 (3) 4 47, a + 1 = 4 44. 7-i-y = 8 1 2 3 4. 48. 3a;+1 = 7 45. a' — 2 = 5 3 5 7 9 49. a. — 1 = 4 46. 23 — 1 = 5 1 2 3 4 50. 3a; – 1 = 8 1 3 4 6 51. 2a – 1 = 7 1 2 3 4 1 2 3 4. 52. 3a;+2 = 11 4 5 2 3 4 5 53. 7–1–2 r = 15 4 5 1 2 3 4 54. 1+5y = 16 3 4 CHECKING THE SOLUTION OF AN EQUATION To check a result means to substitute the value found for the unknown in the given equation. If this makes a true number statement, the result found is correct. - EXAMPLES 1. ac-H6 = 10 Answer: a = 4 2. 7+y=11 Answer: y =5 Check: 4-H 6 = 10 True r. Check: 7+5 = 11 False - EXERCISES Check these answers. If the answer is correct, place a T in the blank; if the answer is incorrect, place an F in the blank. 1. 3a –2 = 10 Answer: a = 4 - 5. 39–2=19 Answer: y =7 Check: - - =- Check: - 2. 3b = 45 Answer: b = 15 6. 7 = 2m – 2 Answer: m = 4 Check: -ms Check: —— 3. 29–1 =39–9 Answer: y = 6 7. §m =9 Answer: m = 21 Check: - Check: — — . 4. 7 – a = 13–20. Answer: a = 6 8. #m–H1=5 Answer: m = 8 Check: *-*. Check: -º-, Check each of the answers you found for exercises 22 to 42 on the preceding page. ExERCISES IN UNDERSTANDING THE EQUATION Fill the blanks. The first exercise has been completed for you. 1. If ac-H4 =6, then 6 is 4 greater than a, and a = 2. 2. If a -2 =4, then 4 is 2 than 2, and c= 3. If #a: = 3, then 3 is of a., and a = 4. If 72 = 14, then 14 is 7 a', and a =– 5. If y-i-6= 10, then 10 is 6 than y, and y =– 6. If 20=2–5, then 20 is 5 than 2, and z= 7. If 14 = a-H7, then 14 is 7 than a, and a = - 8. If 8-3b, then 8 is of b, and b = 9. If 21 = 7-0, then 21 is 7 w, and w= 10. If 10=4m, then 10 is 4 ºn, and m = *11. If 24+4+16, then 16 is 4 than 20, and q = *12. If 32–4 =17, then 17 is 4 than 32, and a = *13. If 7=2y–3, then 7 is 3 than 2), and y = - *14. If #2-# 1 = 2, then 2 is 1 than #2, and 2 = *15. If y–2=5, then 5 is 2 than #y, and y = *16. If 2m+5=7, then 7 is 5 than 2m, and m = *17. If 29-–5 = 17, then 2, is 5 - than 17, and y = *18. If 32–3 = 18, then 3a is 3 than 18, and a = *19. If 6 =2) —2, then 29 is 2 than 6, and y = *20. If !---1-2, then #2 is 1 than 2, and 2 = . — 23 — USING EQUATIONS TO SOLVE PROBLEMS EXAMPLES 1. What number decreased by 8 equals 15? 2. What number increased by 7 equals 21? S–y-–’ >-—y—’ *— Y- - S-v- _* * º 77, - 8 = 15 70, + 7 = 21 Solution: The equation is n–8= 15. This means Solution: The equation is n+7=21. This means that 21 is 7 more than m, and that n = 14. that 15 is 8 less than n, and that n=23. 7 = 21 Check: 23 *=e 8 = 15 EXERCISES Check. 14 + 1. What number added to 5 gives 26? 5. What number less 16 equals 11? Solution: Solution: Check: Check: * 6. LeRoy gave his sister 17 cards, and had 21 2. What number decreased by 27 gives 17? left for himself. How many did he have before he gave any away? Solution: Solution: Check: Check: . Fi f Willi ill be 26 3. Five years from now, William will be 7. George has 3 times as much money as Henry. Together they have $2.40. How much does Henry years old. How old is he now? º Solution: Let a = William's age now. have': Solution: Check: Check: *8. Twice a number plus 6 is 1 less than three 4. What number added to 5 times itself gives times the number. Find the number. 42? Solution: Solution: Check: Check: — 24 — LAWS OF THE EQUATION More difficult equations and more difficult prob- lems are ahead. In order to solve these, it is neces- sary that we learn certain general principles. Most of the rest of this unit is given to a study of four important principles, or laws. The Subtraction Law The Addition Law The Multiplication Law The Division Law The two members of an equation must be equal. Think of these as the equal weights on the two sides of a balance (scales). The weights may change, but each must change by the same amount to keep the balance. EXERCISES ON THE SUBTRACTION LAW 1. If the scales are balanced and I take two pounds from one side, then I must take 2. If the scales are balanced and I subtract four pounds from one side, then I must subtract 3. Mary asked for five pounds of Sugar. After the groceryman had weighed it, she found that she had money to pay for only three pounds. The groceryman pounds from the other side to keep the balance. pounds from the other side to keep the balance. removed a two-pound weight from one side of the scales. He then had to remove pounds of sugar to restore the balance. 4. If a +2=4, and I subtract 2 from the first member of the equation, I must subtract from the second member to keep the balance. 5. If 6 = a +4, and I subtract 4 from the second member of the equation, I must subtract from the first member to keep the balance. STATEMENT OF THE SUBTRACTION LAW The same number may be subtracted from each member of an equation without destroying the equality. USING THE SUBTRACTION LAW TO SOLVE EQUATIONS EXAMPLES 1. Solve ac-i-10 = 14 for ac, and check the result. Solution: To find ac, it is necessary to get rid of (eliminate) the 10 which occurs with ac in the first member. Subtract 10 from the first member; then, to maintain the balance, subtract 10 from the sec- ond member. This leaves a = 4. Check: 4-H 10 = 14, or 14 = 14 2. Solve ac-H4 = 16 for ac, and check the result. ac-H4 = 16 4 = 4 (subtract) Solution: a = 12 Check: 12+4 = 16, or 16 = 16. EXERCISES Solve each of the following equations for a, and check your results. 1. a, +3 = 13 2. ac-H1 =9 3. ac-H6 = 16 4. a +4 = 17 5. ac-i-3 = 7 6. 5-Ha = 8 — 25 — EXERCISES Solve each of the following, and check your results. 1. a +11 = 21 Q = 2. y +36=47 gy = 3. 17 = m-i-2 7?? - 4. 15 = 2+4 2 = 5. 17 = w = - 11 QU) = 6. b-i-7} = 10 b = 7. 5 = m-H4 777 – 8. a-H17% = 20 3C = 9. 6=s+} S = 10. y +3.2=5.8 g = 11. r–H34 = 46 * = 12. 42 = 2+-S S = 13. 15 = t +7 i = 14. ac-i-67 = 100 QC = 15. p-H # = } p= 16. ac-H6} = 12% QC = 17. r–H 7.6 = 10.4 * = 18. 2–H .5 = 5.1 2 = 19. e-H = } € = 20. d-H 1% = 3} d = FOUR GREAT HINDU MATHEMATICIANS Between 500 A.D. and 1200 A.D., Europe was passing through the darkest part of the Dark Ages. India, on the other hand, was making advances in algebra and in other fields of mathematics. The four most important mathematicians of this period had the strange names of Aryabhata, Brahmagupta, Mahavira, and Bhaskara. Much of the algebra we study today was developed in India centuries before it was known in Europe. With feet and inches substituted for the Hindu units of length, and with most of the flowery, Ori- ental language omitted, one of Mahavira's problems reads about like this: A huge blackSnake 30 feet long enters a hole at the rate of 7% inches in Hºr of a day; in the course of # of a day its tail grows 2} inches. O ornament of arithmeticians, tell me by what time this serpent enters fully into the hole. — 26 — THE ADDITION LAW The scales are balanced with sugar on one side and with four one-pound weights on the other side. If the groceryman adds a two-pound weight on One side, he will need to add pounds of Sugar to the other side to keep the balance. - Statement of the Addition Law: The same number may be added to each member of an equation without destroying the equality. EXAMPLE 3 = a – 2 Answer: a = 5 Check: 3 = 5–2, or 3 =3 2 = 2 (Add) 5 = 2 EXERCISES 1. ac-7 = 16 QC = 9. a. — 14 = 25 3C = 2. a = 16 = 20 (C = 10. ac-1 = 6 QC = 3. 7 = a – 11 Q = 11. 3 = a –9 QC = 4. 9 = a – 7 QC = 12. 5 = a – 10 (C = 5. a. — 6 = 9 Q = 13. ac-7 = 0 Q = 6. a -3.2 = 6 (C = 14. a -.5 = 1 QC = 7. ac-6 = 3.2 Q = 15. a = 1 = .5 Q = 8. a -6 = 6 Q = 16. at −.8 =.8 Q = SoLVING AND CHECKING MENTALLY Usually, when solving easy equations, it is not One of the beauties and joys of mathematics is necessary to write the steps of the solution. You can that more than in any other school subject you can make these steps mentally. You can also check the check your work and can experience the pleasure of results mentally. knowing that you are right. — 27 — USING THE ADDITION AND SUBTRACTION LAWS In Solving the following equations— (a) Decide whether to use the addition law or the subtraction law and what number to add or subtract. (b) Perform the addition or subtraction mentally, and write the value of the unknown. (c) Check mentally by substituting in the given equation the value found for the unknown letter. EXERCISES 1. a, +3 = 7 Q = 13. 4-Ha! = 4 QC = 2. a -3 = 10 Q = 14. 244–5 (C = 3. a+11 = 12 O = 15. ac-H.5 = 6.5 (C = 4. b – 16 = 4 b = 16. m. — 7 = 0 777, E 5. 4 = m.--2 77) = 17. 4} = a +4; QC = 6. 9 = m —2 }/\, = 18. 15=y—10 g = 7. d – 6 = 7 d = 19. n+16=16 70, E 8. 1 = g-4 g = 20. m. – # = 3 7%, F 9. h--7 = 14 º h = *21. # = 2+} 2 = 10. 10 = p—4 p = *22. g-H.3 = 4.1 g = 11. 6 = 2 – 7 2 F *23. 3.2–H d = 6 d = *24. 3} =h-H # h = 12. 2 = w – 7 QU) = It is better to do a few exercises thoughtfully than a great many mechanically and without knowing what you have done. To be sure that you understand the addition law and the subtraction law, indicate after each of the exercises above exactly what you did. For example, in equation 1 you subtracted 3 from each member; show this by writing S3 after the equation. In equation 4, you added 16 to each member; show this by writing A16 after the equation. Do similarly for the other equations. PACIOLI Pacioli, an Italian monk, published at Venice (1494) a mathematics book which had a great in- fluence in Italy and other countries of Europe. The title he gave to this book was Suma de Arithmetica Geometria Proportion? et Proportionalita. It was a summary of the arithmetic, geometry, and algebra known in Italy at that time. In Pacioli's book, the unknown (which we indi- cate by a or y or some other letter) was called cosa meaning “thing.” This is the origin of one of the names by which algebra has been known. In Ger- many one of the earliest algebras was called the Coss; in England, when algebra was first introduced it was called “the cossic art.” Pacioli abbreviated cosa as co. He also used f for the plus sign, and a long dash for equality. When he wrote the equation 2a:--6= 216, it appeared as 2 co. § 6—216; when he wrote the equation ac—7 = 6, it appeared as 1 co. ffi'ſ 6. Can you write ac-H4=9 and 3a–2=7 as Pacioli did? — 28 — THE MULTIPLICATION LAW AND THE DIVISION LAW You have already learned that you can subtract the same number from each side of an equation (subtraction law), or that you can add the same number to each side of an equation (addition law) without destroying the equality. You will now learn two other important laws. The Multiplication Law: Both members of an equation may be multiplied by the same number without destroying the equality. The Division Law: Both members of an equa- tion may be divided by the same number without destroying the equality. EXAMPLES 1. #2 = 7 2. 6a = 12 4= 4 (Multiply by 4) 6= 6 (Divide by 6) a = 28 a = 2 EXERCISES Solve mentally, check, and record the answer. 1. #a = 5 (C = 21. 4a: = 8 QC = 2. #2 =9 3 = 22. 9a = 27 a = 3. To a = 4 Q = 23. 16 = 25 b = 4. #a =# Q = 24. 4m = 10 7?? = 5. #y=5 g = 25. 15 = 2a Q = — 1. == 6. 3 =#y $/ 26. 4.6 = 2.3c C = 7 6 =#. - är r 27. 59 = 2.5 y = 8. .1s =9 S = 28. 8ac = 72 QC = t 9. 6 = - t = 7 29. 67m = 54 777 = 10. *=2 QD = 30. 2r = } r = 8 *31. 2.5 = 2a: O = 2 11. 6 = — 2 = 9 *32. .2b = 1.8 b = 12. a = 0 (C = 13. #w = 0 QU) = *34. #w = 5 QU) = 14. #b = } b = *35. #m = 6 7?? - 15. #a = 2 O! = :k 4n = 104. = 16. # =4m 7?? - 36. 3}q = 10% Q 17. 2%q = 10 q= *37. # = #m 972 = 18. 16c = 4 C = *38. .4p = 16 p = 19. 7ac = 6 (C = *39, 1}ºy =5 g = 20. .62 = 3 2 = *40. 3.1g =93 g = — 29 — MIXED EquaTIONS Indicate on the blank the law and number you would use to solve each equation. Let A, S, M, and D stand for the addition, subtraction, multiplication, and division laws, respectively. If you would multiply by 2, write M2 on the blank; if you would divide by 6, write D6 on the blank. What does A9 mean? What does S4 mean? 1. a+3 = 4 S3 8. n–2} = 4; *m- 15. 7=#t 2. a -2 = 7 9. e-.4 = 6 sm-ms-s 16. 2#d = 100 10. 7 =#b *- - 3. 7=2+y **-- 17, 7–4– 9 4. a = 6 *- 11. 4-9 --- 18. .1M) = .2 5. 6m = 12 ----- - 12. ac-à = 0 *- - 19. #t=} 12" — 6. 2%r=5 ----- 13. #f = } - 20. 16 = 1.6h 7. 6= }s -s-tº----- 14. ac— 2 = 0 *- - 21. n–2 = 4 TIMED PRACTICE TEST Write the answers only. 1. ac-4 = 10 *- 16. ac-2 = 4 -- 31. #r = } 2. y–H6 = 13 *- 17. y–H4=7 *- 32. 2%g = 2; 3. #b = 7 *-m- 18. 4m = 36 e-- 33. 18.6 = 6.2m 4. 9m = 63 a- - 19. #b = 10 -- 34. +-4.5 5. 7 =q+3 º-º-º---- 20. += 35. .252 =3 6. 10 = a – 7 --- 21. 2+p = 6 - 36. a -3.5 = 4.5 7. #q=7 – 22. ac-13 = 21 *-- 37. #-2s 8. 16 = 4n. — 23. 6+y=12; -* 38. 6 = 4a 9. q+} = 1 --- 24. P+3} = 10 *- - 39. 4.5 = }b 10. #r = 5 --- 25. 7 = r — 2 -- 40. m-H7% =7% - 11. y–.5 = 7.5 *- 26. 16 = 8s *-* al. †-41 12. #b = 10 -- 27. 11 =#t --- 42. 27 = .17. 13. 2.5e = 5 *- 28. #-1s —- 43. 2.7 = .1a: 14. 6.5 = h-H.5 *- 29. wi-fi = 4 --- 44. 32 = 4.2 15. a-H+ = 1 *-*- 30. m. —# = 4; e--- 45. a-H3 = 4.2 I solved of the exercises above in – minutes, with correctly done. -------- — 30 — EQUATIONS SOLVED BY USING MORE THAN ONE LAW An equation arising from an actual problem usually requires more than one law for its solution. You will indicate the laws as on the preceding page. In case two or more laws must be used in Solving a single equation, it is usually better to apply the addition law or the subtraction law before applying the multipli- cation law or the division law. EXAMPLES 1. 2a:-H 1 = 5 Check: 2. #m– # = 3; Check 1 = 1 S1 2.2+1 = 5, or # = # A+ # 12– # = 3}, or 5 = 5. 4–3 = #, OT 2a: = 4 #m = 4 3# =3; 2 = 2 D2 3 = 3 M3 a = 2 7m = 12 EXERCISES Solve each of the following, indicating the laws used, and check. 1. 2a:-- 1 = 7 6. 9=#m–3 2. 6/–1 =23 4. 7. #q-# =4; 3. *w-H3 =36 *8. #2–1 =0 Q/ * 4. 4-7-9 *9. 4a:--3} =43 5. 4c-H7 = 19 *10. #y — ; = 4 PROBLEMS AND EQUATIONS Problems are usually stated in words. The first step toward a solution is to translate the word statement into an algebraic equation. Consider the example. The larger of two numbers is 3 more than twice the smaller. Their sum is 21. Find the numbers. Let a, equal the smaller number. Then 2a:--3 equals the larger number. The equation is ac-H2a+3=21. — 31 — COMBINING LIKE TERMS When like terms occur in the same member of an equation, they should be combined before applying the equation laws. EXAMPLES 1. 32-H42 = 21 2. 4y–3#y = 10–4 7a: = 21 (Like terms combined) #y= 6 (Like terms combined) ac=3 (Divide by 7) g = 12 (Multiply by 2) EXERCISES Solve and check. 1. 7a —2a:--a = 24 8. 8a–H4+7a = 34 2. 6y–4y = 6–4 *9. 7 y–4 =2/+3% 3. 6m —2n+4m = 20 *10. 10s-H4% = 6s–H 10 4. 3v-H2w –2 = 3 5. ac-Ha!—ac-H3a = 12-H4 *11. John is two years older than Harry, and - Harry is three years older than Joe. The sum of their ages is twenty-six years. Find Joe's age. 6. 4–2=2t—t **12. Julia lacks two years of being twice as old as Marie. The sum of their ages is sixteen years. Find Marie's age. 7. 32—33 = 18–3 —32— TRANSPOSITION The addition and subtraction laws taken together make up what is commonly called the principle of transposition. This principle states that an equality is not changed when any term is omitted from one member of an equation and written, with its sign changed, in the other member. 1. ac-H 5 = 13 a = 13–5 a = 8 2. a -4 = 12 a = 12-H4 a = 16 EXAMPLES Omit --5 from the left mem- ber of the given equation, and write — 5 in the right member. Omit –4 from the left mem- ber of the given equation, and write +4 in the right member. 3. 15 = 2+3 15–3 = a, 12 = a, 4. 5 = 10–a: a = 10–5 a = 5 Omit +3 from the right mem- ber of the given equation, and write –3 in the left member. Omit +5 from the left mem- ber and write — 5 in the right member. Also omit – a from the right member and write +a; in the left member. This operation of omitting a term from one member of an equation and writing it, with its sign changed, in the other member is called transposition. Observe in example 4 above that more than one term of an equation may be transposed. It is important that you recognize transposition as nothing more than a short method of using the addi- tion and subtraction laws. EXERCISES Use the principle of transposition to solve each of these exercises. Check mentally. 1. ac-H3 = 9 2. ac-5 = 14 3. S – 16 = 2 4. t-H4 = 12 5. 13 = y–4 6. 9 = n+2 7. 3 = k–2 8. 7=w-H4 QC F 9. 1 = 7–a: 3D = Q = 10. 10 = 7-i-a: (C = | s= 11. 3—a: = 2 (C = # = 12. k--3 = 10 k = gy = 13. k—3 = 8 k = 77 – 14, w—H·3 = 12 (UU) = k == 15. 6-15– # = QU) = 16. 23:--3 = a +5 Q = *PROBLEMS 17. 3i — 1 = 2t+1 i = 18. 5v-H2 = 4v-H 5 w = 19. 3–y = 1 gy = 20. y – 1 = 29–4 gy = 21. S-3 = 2s – 12 S = 22. 4-H t = 7 # = 23. 3 = 5–w QU) = 24. 2n + 1 = a +7 3D = Write the appropriate equation for each of the problems below. Solve and check. Let a stand for the un- known. 1. Gene had twenty-five cents and bought four candy bars; Bruce had seventeen cents and bought two candy bars. If each then had the same amount of money remaining, how much did a candy bar cost? 2. Dorothy is twice as old as Maxine, and Dorothy's age minus ten years equals Maxine's age minus two years. How old is Maxine? — 33 — PROBLEMS Solve and check. In checking, substitute your answer in the problem itself, rather than in the equation you have set up. 1. The larger of two numbers is 5 greater than the smaller. Their sum is 31. Find the numbers. Check: 2. The sum of two numbers is 70; the larger is 6 times the Smaller. Find the numbers. Check: Solution: Let a = the Smaller number. Then = the larger number. and =31, the sum of the two num- bers. 2x = (Subtraction Law) QC = (Division Law) a;+5= 3. The sum of two numbers is 15; the greater exceeds the smaller by 6. Find the numbers. Check: 4. Five increased by four times a number is equal to 33. Find the number. Check: 5. The sum of two numbers is 50. The greater is one less than twice the smaller. Find the num- bers. Check: 6. If 3 is subtracted from two times a number, the remainder is 23. Find the number. Check: — 34 — PROBLEMS Solve and check. 1. Find two numbers such that the second is 5 less than 3 times the first, and their sum is 31. Check: 2. Find three consecutive integers (whole num- bers) whose sum is 42. Hint: Let n be the first integer. Then n+1 and n+2 are the next two consecutive integers. Check: 3. Find three consecutive integers such that the sum of the first and the last is 36. Check: 4. If eight is added to four times a certain num- ber, the result is six times the number. Find it. Check: 5. The sum of three numbers is 67. The second is twice the first, and the third is one more than the sum of the first and second. Find the numbers. Check: 6. Find two numbers whose difference is 27, and whose sum is 39. Check: PROBLEMS Solve and check. 1. An algebra class containing 67 pupils was divided into two sections. There were three more in One section than in the other. How many were in each section? Check: 2. The length of a rectangle is three times its width. The perimeter is 56 feet. Find the length and the width. Check: 3. A house and a lot are worth $10,000, the house being worth three times as much as the lot. Find the value of each. Check: 4. A father is eight years more than twice as old as his daughter. The sum of their ages is 71 years. Find the age of each. Check: 5. The perimeter of a triangle is 40. The second side is six greater than the first, and the third side is two less than the second. Find each side. Check: 6. The sum of the three angles of a triangle is 180 degrees. The second angle is twice the first, and the third is three times the first. Find each angle. Check: PROBLEMS Solve and check. *1. A man invested a certain sum at six per cent, and twice that sum at five per cent. The combined income was $800 a year. How much did he invest at six per cent? At five per cent? Check: *2. John is two years older than his sister, and twice as old as his younger brother. Their combined ages are thirty-three years. Find the age of each. Check: *3. Everett has $1.60 more than Russell. To- gether they have $25. How much has each. Check: **4. Barbara is twenty years younger than her mother. Ten years ago she was exactly one-third as old as her mother. How old is each now? Hint: ac-10 =3(a –30). Check: Let a = mother's age now. Then a – 20 = Barbara's age now, and ac-10=mother's age 10 years ago. **5. A father is eight times as old as his son. In four years, the father will be only four times as old as the son. How old is each now? Hint: 8a;+4=4(a +4). Check: **6. Mr. Fish has $1,000 more invested at 4 per cent than he has at 6 per cent. The annual incomes on the two investments are the same. Find the amounts he has invested at 6 per cent and at 4 per Cent. Hint: .04(a +1000) =.06a. Check: — 37 — INVENTORY TEST 1. Translate each of the following into an algebraic equation. Let n represent the unknown. (a) A certain number less 16 is equal to 5. (b) Sixteen is 5 more than a certain number. e-ºs----— (c) One third of a certain number plus 4 is equal to 13. (d) Ten times a certain number less 5 equals 65. (e) Twice a certain number is 5 more than the number. 2. Each of the equations below is followed by a number in parentheses. If the number is a root of the equation, make a plus sign (+) before the equation; if the number is not a root, make a minus sign (–) before the equation. (a) a +4 = 10 (6) (e) #m--1=3 (2) (i) 4 = a – 7 (11) (b) #y=8 (16) (f) #g–2=0 (6) (j) 7= #m–H1 (27) (c) 4 = 2–2 (5) (g) 32–2 = 7 (3) (k) 6=3a;+1 (2) (d) 6w-H1 = 13 (2) (h) 3y—y = 10 (4) (1) #y-H3 = 5 (8) 3. Solve each of the following mentally, and check. (a) a +6 = 10 a. = (e) .1m = 1 7?? - (i) #2 =8 Q = (b) #y = 4 gy = (f) #y-H1 = 6 y = (j) 3p–1 = 0 p = (c) 2m = 7 7?? - (g) 32–2 = 1 a. = (k) 32–1 =7 a = (d) 6 = q—7 q = (h) 3h = h--6 h = (1) #2–H1 =9 a = 4. Indicate below the laws and numbers used in solving exercise 3 above. Parts (a) and (f) have been correctly indicated. (a) S6 (°) — (e) — (g) — (i) — (k) (b) – (d) – (f) S1, M2 (h) — (i) — (1) 5. Combine like terms. (a) 32–22+2 = (c) 7b —b-i-6b = (e) a-Hb-H3a–H2b = (b) 4a–3a–H5a = (d) 32-H42–H3y—y = (f) 6m-H4a:–2m – a = 6. An equation has two separated by the sign. The part on the left side is called the , and the part on the right side is called the 7. Fill the blanks in the statements below. (a) ac-4 = 10 means that 10 is 4 – than ac, and that a = (b) ac-H4 = 10 means that 10 is 4 than ac, and that a = (c) 6–Ha = 10 means that 10 is 6 than ac, and that a = (d) 6–a = 2 means that 6 is 2 than ac, and that a = (e) 9–ac-4 means that 9 is 4 than ac, and that a = — 38 — CUMULATIVE REVIEW 1. Write the term whose exponent is 3, whose coefficient is 7, and whose literal part is m. 2. The expression 5 y means 5 3. Perform the indicated operations and write the answer in its simplest form: 16–4+2+5.3. 4. The base of a rectangle is 3a and its height is 2g. Its perimeter is 5. Is 3 a root of the equation 22+4= a +7? g; the expression mº means , and its area is (Yes or No) 6. Evaluate each of the following when a = 2, y=3, and z=4. (a) 32°–y”= (c) a 4-Hy”—Hz = (b) 3(yz—ay) = (d) += 24/ 7. The base of a rectangle is 32 and its height is 22. Its perimeter is (e) ++y= J. (f) === , and its area is 8. Represent the product of g and h in three different ways. ; ; 9. Represent the quotient of g divided by h in two different ways. ; 10. Write an equation which can be solved by the addition law. law. By the multiplication law. By the addition and multiplication laws. division laws. By the subtraction By the division law. By the addition and division laws. By the subtraction and multiplication laws. By the subtraction and 11. Find the area of a trapezoid whose bases are 12 and 8, and whose height is 5. 12. Find the centigrade reading which corresponds to a Fahrenheit reading of 99°. BAGDAD Do you know of Bagdad, the city of romance and of magic? From Bagdad came Aladdin and his wonderful lamp, Ali Baba and the forty thieves, Sinbad the Sailor, and the Old Man of the Sea. The Arabian Nights’ Tales were written more than a thousand years ago. Harun al-Rashid was Caliph; the tales were written for him. During the period 750 A.D. to 900 A.D., Bagdad was the center of the intellectual world. Harun al- Rashid and his son al-Mamun were great patrons of learning. They invited Hindu scholars from the East to come to Bagdad to study and to translate the mathematics of the Orient into Arabic. They in- vited Egyptians and Greeks to come to Bagdad to study and to translate western mathematics into Arabic. Scholars from everywhere flocked to the Court of the Caliphs. Thus ancient writings, which might otherwise have been permanently lost, were copied and preserved, and have come down to us. The greatest mathematician at the Court of al- Mamun was al-Khowarizmi. The full title of his most famous writing was 'Ilm Al-jabr Wa'l Muqa- blah, which means “the science of reduction and cancellation.” A simpler meaning is “the science of solving equations by use of the addition, subtrac- tion, multiplication, and division laws.” This book was translated into Latin and had a great influence in Europe. But the Arabic title was too hard for Europeans to pronounce. They kept only the second word, al-jabr. From this our word algebra comes. — 39 — UNIT 3. DIRECTED NUMBERS MICHAEL IDvors KY PUPIN (1858–1935) A Serbian shepherd boy, immigrant to America at the age of fifteen, landed in New York with only five cents, worked as a farm hand while earning his way through college, became an honor student in mathematics, charter member of the American Mathematical Society, and professor of electro-mechanics in Columbia Univer- sity. These are some of the stages in the life work of Michael Pupin. Professor Pupin successfully ap- plied mathematics to the solution of a number of practical problems. One of his successes was the invention of an apparatus which makes long distance telephone calls possible. Before this time, so many unwanted sounds be- came mixed with the vibrations set in motion by the voice speaking that the person listening could not distinguish one from the other. The “Hello, dear” became mixed with the roar of trains, the rumble of trucks, and the rattle of everything else under which and over which the wires passed. Telephone engineers were baffled. They could not eliminate the interfering noises without at the same time eliminating the sounds which should be carried through. International News photos MICHAEL PUPIN Years before, in a second-hand bookstore in Paris, Professor Pupin had bought an old mathe- matics book by Joseph Lagrange. This book had in it a seemingly useless problem: useless except to mathematicians who like to play with such prob- lems. It read: “Given a thin string with heavy beads attached at equal inter- vals, discuss the motion of this string when it is caused to vibrate.” The vi- brations of an electric current in a tele- phone wire seem far removed from the vibrations of a string with beads fas- tened to it. But, in the solution of the problem of the loaded string, Professor Pupin found the solution of the prob- lem which baffled the telephone engi- neers. He built the principle he thus discovered into an apparatus known as the Pupin coil. This apparatus is now used when- ever a long distance call is made; without it the human voice would carry but a few miles before being drowned out by other sounds. Professor Pupin said of his discovery, “I suc- ceeded because I did not guess. I was guided by the mathematical solution of the generalized Lagran- gian problem.” NUMBERS USED IN ARITHMETIC “How far can you count?” What child has not been asked that question? One, two, three, . . . to ten or twelve. This is far enough to play jacks, jump rope, and keep track of possessions, for a while. But it is not long till the child needs more and larger numbers. He must use fractions, too, as well as whole numbers. Apples have to be divided into halves and thirds, as do bags of candy and other things. The child's knowledge of the numbers of arithmetic grows up as he grows up. And all these numbers fit together into a systematic and con- tinuous whole. They form an arithmetic scale which can be represented on a foot rule, a yardstick, or a straight line. Each division on the scale shows the distance from the starting point, and the scale can be extended indefinitely to the right. THE ARITHMETIC SCALE . Locate on the scale shown above, as accurately as you can, each of the numbers: 4, 7, 11%, 3, 2%, 10, 7.5, 3.3, 4, 23, 1}. If this scale were extended indefinitely to the right one could locate on it all the whole numbers, fractions, and decimals used in arithmetic. Think where you would locate on the scale (extended) each of the numbers: 15, 50, 213, 1492. — 40 — THE NUMBER SCALE Distance may be measured on the arithmetic scale. If, in addition, we wish to indicate direction, it is necessary that we use the algebraic scale, with signed numbers at the points of division. e-Negative Directione— —"Positive Direction— | ſº ſº ſ † ſ | ſ 1 | ſº ſ ſº u W W T w I s ſº 1– ſ ſ | ſ º u W I I I I I– f ſ ſº W U W I TI u —I I– º u I — 12 – 11 – 10 – 9 – 8 – 7 – 6 – 5 – 4 — 3 — 2 – 1 0 + 1 + 2 + 3 + 4 +5 –1-6 +7 --8 +9 +10+11 +12 The algebraic scale will greatly increase our knowledge of numbers. The part of the scale shown above gives the numbers from — 12, read “negative twelve,” to 12, read “positive twelve.” These numbers increase in value as you go from left to right, and decrease in value as you go from right to left. EXERCISES 1. Beginning at —4, write the next six numbers to the right. y y —y y y The smallest? Which of these numbers is the largest? 2. How does any number on the scale compare with any number to its left? 3. Which is greater, +3 or +4? — 0 or +12 –2 or –3? –9 or +9? –7 or +4? —6 or 0? 4. What number is 2 units to the right of 1? - 2 units to the left of 1? 5. Is the direction from +4 to +2 positive or negative? 6. Is the direction from — 1 to —3 positive or negative? 7. What number is 3 less than 12 4 less than - -27 8. What whole numbers shown on the scale above are ſess than – 10? 9. How many units from —6 to +4? is the direction positive or negative? 10. Rearrange in order of size, beginning with the smallest: 0, 1, -3, 6, -7, 10, 11, -11. . . , y y y y 11. There are 12. There are units from –4 to +5, and the direction is units from +5 to —4, and the direction is 13. The direction from —8 to —6 is , and there are — units. 14. The direction from +8 to +6 is , and there are — units. 15. Is the direction from --6 to — 1 positive or negative? Why? 16. The greater of any two numbers on the scale is the one to the IMPORTANT FACTS TO OBSERVE ABOUT THE NUM BER SCALE 1. The number scale extends indefinitely in both directions. 2. The - sign is used for numbers to the left of 0, and the + sign is used for numbers to the right of 0. 3. A number preceded by a - sign is called a negative number, and a number preceded by a + sign is called a positive number. If the number is preceded by no sign it is understood to be positive. 4. Toward the right, from any point on the number scale, the direction is positive, and toward the left the direction is negative. 5. Because the + and — signs used with numbers may indicate direction, such numbers are sometimes called signed numbers, or directed numbers. * 6. If we begin at any point of the number scale and move to the right, the numbers increase in value; if we begin at any point and move to the left, the numbers decrease in value. — 41 — ADDITION OF DIRECTED NUMBERS h | l | 1 1–1 | ſº | | I—I U n I n i I i J n – 12 – 11 – 10 – 9 – 8 — 7 – 6 – 5 – 4 — 3 – 2 – 1 ſ ſt ſ ſ | ſ ſ ſ t ſ | u n I u I n H n i I I s 0 + 1 + 2 + 3 + 4 +5 +6 +7 −H8 --9 +10+11 +12 EXAMPLES Notice, in each example, that the sign of the number to be added determines the direction to count on the number scale. When the sign of this number is +, one counts to the right; when the sign is —, one counts to the left. 1. Add +2 to +3. Begin at +3 and count 2 units to the right. The result is +5. 2. Add +4 to —3. Begin at –3 and count 4 units to the right. The result is +1. 3. Add – 5 to +4. Begin at +4 and count 5 units to the left. The result is — 1. 4. Add – 3 to —6. Begin at –6 and count 3 units to the left. The result is —9. 5. Add – 6 to +6. Begin at 6 and count 6 units to the left. The result is 0. EXERCISES 1. Add –2 to +2. Begin at — and count units to the — The result is 2 . Add +2 to —2. Begin at . Add –2 to —2. Begin at — and count . Add +2 to +2. Begin at and count . (a) +5 added to +4 = (b) —3 added to +7= (c) – 4 added to —6= (d) — 5 added to +5= and count units to the units to the units to the — The result is — The result is The result is (e) +10 added to — 5 = (f) – 8 added to +3 = (g) – 4 added to 0 = (h) 0 added to +5 = In exercises like 5(a), the words “added to” may be replaced by a + sign, indicating the operation to be performed. Thus +5 added to +4 may be written (+5)--(+4) Observe that the sign immediately before a number indicates direction on the scale, while the plus sign between the two pairs of parentheses indicates the operation of addition. 6. (a) (+4)+(–7) = (b) (–5) + (+5) = (c) (+7)+(–9) = (d) (–7)+(+4) = (e) (+5)+(–5) = (f) (–9)--(+7) = (g) (–8)+(+10) = (h) (0)+(–7) = (i) (–2)+(–10) = 7. The answers to 6(a) and 6(d) are the same. Is the thinking in getting the answers the same? Explain. The value of a directed number when its sign is omitted is called its numerical value. THE EADS BRIDGE Built by James Buchanan Eads (1820–1887) The Eads Bridge at St. Louis was the first to span the Mississippi River. At the time it was fin- ished (1874), it was considered one of the wonders of the western world. The ordinary person glances at the Eads Bridge today and sees only the steel and stone from which it was made. He does not realize the months of com- putation which preceded the first step in the actual construction. Eads and his assistants computed with exactness the dimensions and weight and strength of every brace and beam and granite block which was to be used. Professor Chauvenet, mathe- malician and chancellor of Washington University, checked all the computations and did not find a sin- gle error. All this was done months before a stone was laid or a beam put into place. º ADDING NUMBERs witH LIKE SIGNs * ſº ſ | | | | ſº ſº ſt | ſº † ſ | | ſ l ſ | ſ _ f ſ n W W ſ W I º I I n i lſ I t n I I I I u I l I I I – 12 – 11 – 10 – 9 – 8 – 7 – 6 – 5 – 4 — 3 — 2 – 1 0 + 1 + 2 + 3 + 4 +5 –1-6 +7 --8 +9 +10+11 +12 EXERCISES 1. Use the number scale in adding the following. +2 — 1 +3 — 4 +2 —6 +2 –4 +1 –4 +8 —5 +1 –4 +2 —2 +4 —3 —H7 — 5 +9 –4 +3 — 3 –H3 –3 +4 –4 2. The sum of two or more positive numbers is a number; the sum of two or more negative numbers is a number. 3. The numerical value of the sum of two or more positive numbers is the of the numerical values of the given numbers. 4. The numerical value of the sum of two or more negative numbers is the of the of the given numbers. The facts learned from the exercises above may be stated: To add numbers having like signs, find the sum of their numerical values and prefix their common sign. In the future we shall use this rule, rather than the number scale, in adding numbers with like signs. 5. Add: +6 –4 — 10 + 6 —9 +60 –74 +63 ––275 — 161 +9 —6 – 20 +12 —7 +72 –92 +86 —H 146 — 274 ADDING LITERAL NUMBERs The unit has been 1 on the number scale we have used. It is not necessary that this be true. The unit could be ac, in which EXAMPLES case successive intervals on the scale would be +a, +2a, +3a, +3a. –3cd +52°y +4a: . . . to the right of 0, and would be — ac, -2a, –3a, +2a: —6cd +2a:”y –4a: . . . to the left of 0. Or the unit could be a, or 2, or mm, ––5a: —5cd +72°y or any other letter or combination of letters. The rule stated +10a: – 14cd +14a:”y above applies, whatever the unit is. EXERCISES 1. Add: —6a: +4b –9ary +4a:” – 132y2 –162°y —H 18qc — 16mm. —7a: +3b –6ay +942 –27ayz —47.cºy +87ac —29mm. 2. (a) (+4)+(+8) = (h) (–4m)+(–2m)--(–7m) = (b) (–22)+(–4a)= (i) (+4a?)-1-(+4a?)+(+9a2) = (c) (–16aº)+(–14a3) = (j) (– #2)+(– #2)+(−}a) = (d) (–2.5g)+(–7.5 y) = (k) (+ab)+(+2ab)+(+3ab) = (e) (+6cd)+(+4cd)= (1) (–1)+(–2)+(–7) = (f) (– #2)+(– #a) = (m) (+.2m)--(+2.3m)-1-(+4.1m) = (g) (–6.7m) +(−2.3m) = (n) (–1}a)+(– #ar)+(−}a) = *3. Augustus Caesar was born 63 B.C. and died 14 A.D. Find his age at the time of his death. (Hint: Using the number seale, think how much you would need to add to —63 to make + 14.) *4. Julius Caesar lived from 102 B.C. to 44 B.C. How long did he live? *5. George Washington lived from 1732 A.D. to 1799 A.D. How long did he live? — 43— ADDING NUMBERS WITH UNLIKE SIGN's à ſº ſº | | | _ſ. R ſº | ſº ſº ſ' ſl ſº | ... I ſl ſº | | _l ſº | —w W W W W I W I w T I W W W Ur I n y U w y I W W u – 12 – 11 – 10 – 9 – 8 – 7 – 6 – 5 – 4 — 3 — 2 – 1 0 + 1 + 2 + 3 + 4 +5 +6 +7 −H8 --9 +10+11 +12 EXERCISES 1. Use the number scale in adding the following. +2 +7 + 10 —9 — 3 + 1 +7 —2 — 11 —6 +5 —8 –3 — 5 — 6 +4 + 10 — 6 —5 +9 + 7 +3 —6 +9 2. The numerical value of the sum of two numbers with unlike signs is the of the numerical values of the numbers. 3. The sign of the sum of two numbers with unlike signs is + or – according as the number having the numerical value is + or — . The facts learned from the exercises above may be stated: To add two numbers with unlike signs, find the difference of their numerical values, and prefix the sign of the number having the greater numerical value. Use this rule, in the future, when adding numbers with unlike signs. (a) (b) (c) (d) (e) (f) (g) (h) 4. +6 +4 — 11 — 16 +25 — 16 +29 — 172 —7 —2 + 7 + 16 – 17 + 11 — 13 + 39 5. –– 16a. — 16ac + 1.6a. +7%g –290b —162°y” +6a. – 8ayz —25a. + 9ac –2.4a: — #y +540b +25a.”y” — QC + 16ayz 6 + 4 +7 —6 — 7 —7 +7 +6 —9 – 14 –9 +9 +6 +9 –2 –2 6 7 26 –47 – 17 —61 17 – 17 —41 –27 — 13 51 39 72 — 19 19 52 0 8. 67a;2 –71a:y — 16m. 39cd –67, & 4a: 0 0 — 360° 29ary 14m, — 1960 17y — QC 4 –4 -º- -*s CoLUMN ADDITION In adding more than two signed numbers, one may add the first two, then add the third to their sum, and so on. One may also add all the positive numbers, then add all the negative numbers, and then add these tWO SumS. EXERCISES (1) (2) (3) . (4) (5) (6) (7) (8) +3 +3a. –4a –6ay +4a:” %t .5a. – 6.2y – 5 –2a: –6a +4a:y –6aº – #t — .34: — .8y + 7 +4a: –90 –9ay —92% –3t 1.4a: +3.19 –4 –7a: 190 –8acy +62% +#t — .6a, — .6y *-*. sm- – 44 — Add. 1. –H4 —8 2. –9 +3 3. –8 —3 4. +8 –4 5. ––3 +6 6. ––3 —6 7. -- 4 — 11 8. —2 —7 9. —4 –4 10. —9 –3 11. --9 –3 12. --8 –2 13. --6 14. --5 15. --3 16. – 5 17. —3 18. --4 19. — 3 20. --3 21. — 4 22. —3 23. --3 24. —8 25. 26. 27. 28. 29. 30. 31. 32. 33. 34. 35. 36. EXERCISES +7 –4 * 37. 38. 39. 40. 41. 42. 43. 45. 46. 47. 48. 49. 50. 51. 52. 53. 54. 55. 56. 57. 58. 59. 60. 61. 62. 63. 64. 65. 66. 67. 68. 69. 70. 71. 72. — 45 — SUBTRACTION OF DIRECTED NUMBERS ! ſº l | | º | | R ſº | | ſº ſt l | | ſ l ſ_ ſ ſ W I U w I I I l I l | l —T I { W l I U \ I º W — 12 – 11 — 10 – 9 – 8 – 7 – 6 – 5 – 4 – 3 – 2 – 1 0 + 1 + 2 + 3 + 4 +5 –H6 -H7 ––8 +9 +10+11 +12 Subtraction is the process of finding one of two numbers when their sum and the other number are given. To subtract 3 from 7 is to find the number 7 minuend which added to 3 gives 7. 3 subtrahend 4 remainder EXAMPLES Use the number scale above in thinking these through. 1. Subtract +4 from +9. Begin at +4 and count to +9. This is 5 units in the positive direction. Thus, (+9) — (+4) = +5. 2. Subtract +9 from +4. Begin at +9 and count to +4. This is 5 units in the negative direction. Thus, (+4)–(+9) = —5. 3. Subtract –6 from +8. Begin at –6 and count to +8. This is 14 units in the positive direction. Thus, (+8) — (–6) = +14. 4. Subtract +8 from —6. Begin at +8 and count to —6. This is 14 units in the negative direction. Thus, (–6)–(+8) = –14. EXERCISES 1. Subtract – 3 from +2. Begin at and count to This is units in the direction. Thus, 2. Subtract –2 from —4. Begin at and count to This is units in the direction. Thus, 3. Subtract –4 from –2. Begin at — and count to This is units in the direction. Thus, 4. Subtract +2 from –3. Begin at — and count to This is units in the direction. Thus, 5. Subtract, using the number scale. +9 +5 — 7 — 6 — 11 —8 +7 — 10 minuends +4 — 3 +4 — 10 — 4 +4 –3 + 5 subtrahends *m-m: *s-tº-mº *mº-º-º-º-º: * remainders 6. Check each of the subtractions of exercise 5 by adding the remainder to the subtrahend. If the sum equals the minuend, the subtraction is correct. An easy way of subtracting: Think of the sign of the subtrahend as changed, and proceed as in addition. In subtracting –2 from +6, think the sign of –2 as changed, and add +2 +6 to +6. Do not actually change the sign of –2. –2 +8 7. Subtract each of the following the easy way. Check by adding the remainder to the subtrahend. – 14 +4 +6 + 16 — 16 + 16 — 23 + 19 – 17 — 23 +10 —6 +2 +20 +12 — 12 +26 +29 — 8 + 13 — 4a: 8a; 8 0 0 — 10 7ary 4. — 67m, —3t — 10a: 10a: 0 8 —8 0 —ary 4 — ???, t 8. Apply the short method to exercise 5 above and see whether the results check with those obtained. — 46 — 10. 11. 12. Subtract. . --8 +5 . --9 +4 . —7 –7 gºmmº- ... +7 —6 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28. 29. 30. 31. 32. 33. 34. 35. 36. EXERCISES 37. 38. 39. 40. 41. 42. 43. 45. 46. 47. 48. +1 +8 +7 —2 49. 50. 51. 52. 53. 54. 55. 56. 57. 58. 59. 60. 61. 62. 63. 65. 66. 67. 68. 69. 70. 71. 72, — 47 — SUBTRACTION OF DIRECTED NUMBERS EXERCISES Subtract. Think of the sign of the subtrahend as changed, and proceed as in addition. (a) (b) (c) (d) (e) (f) 1. – 4a: –9a 0 –8 3.4a: 3}a 164: —90. —9 0 —2.6a: 2}a 2. — 16 17 — 16a: QC 14y” 160. 24 – 17 — 16a, —2a: —y” Ol. 3. 16a: 6.4a: —7% 6} 6.2c —3.1a: 27a; – 6.2a: 2} –2% 1.4c 2.4a: 4. 8d. 6a: —8a; Q: —abe – 10y 4a 9a, –9a: —7a: — abo 11y 5. – 5r 60? — 2 O), 9ac —ab —57. —3a? —82 – O, — QC 7ab (g) —67m.” — 37m.” —ab ab 3m. —8m. –7e 13e (h) —5t 5t 3a; —3a; — .5c 1.8c 5st –2st 6. Ten expressions taken as positive are listed below. Fill the blanks with the related negative expres- sions. The first has been done correctly. (a) above below (f) A.D. (g) above sea level (b) east (c) above zero (h) to the right (d) gain (i) up (j) assets (e) north latitude 7. An elevator went up eight floors and down three floors. How far was it then from where it started? 8. A thermometer read six degrees above Zero. During the night the tem- perature fell eight degrees. What was the reading in the morning? 9. An airplane flying at an altitude of 4,500 feet descended 2,600 feet. What was its altitude then? 10. A business concern failed, having assets of $246,000 and liabilities of $392,400. What was its net financial condition? 11. A ship sailed along a meridian from 23° south latitude to 31° South latitude. Through how many degrees and in what direction did it sail? 12. A ship sailed along a meridian from 14° south latitude to 28° north latitude. Through how many degrees and in what direction did it Sail? floors ZerC) feet — 48 — ADDITION AND SUBTRACTION OF DIRECTED NUMBERs RACE TRACK NUMBER 1 - RACE TRACK NUMBER 2 +14 —2 +6 21 16 —9 S - Sº } N, / \p +7 +17 —4 42 —7 20 -O Ö >r Start Start EXERCISES Write the given number (lightly so that it may be erased for the next exercise) in the inner circle. To win a race means to add the given number correctly to every number on the track, or to subtract the given number correctly from every number on the track. It takes ten rights to win, and only one wrong to lose. Exercise 1 is correct as far as the numbers are written. An unsigned number means a positive number. 1. Write 4 in the inner circle of Track Number 1. Add. 21 0 4 - 3 - - *- -*- -*mams 2. Write 4 in the inner circle of Track Number 1. Subtract. 3. Write —7 in the inner circle of Track Number 2. Add. 4. Write 20 in the inner circle of Track Number 2. Subtract. 5. Write –9 in the inner circle of Track Number 1. Subtract. 6. Write — 16 in the inner circle of Track Number 1. Add. 7. Write —4 in the inner circle of Track Number 2. Add. 8. Write 4 in the inner circle of Track Number 2. Subtract. •– – - *- - - - - - - How many races did you win? How long do you take to go around a track? **The sum of all the numbers on Track Number 1 is 6; the sum of all the answers of exercise 1 is 46. This is a check that the answers to exercise 1 are correct, for 6 plus 10 times 4 equals 46. Use this method to check the answers for the other exercises. — 49 — MULTIPLICATION OF DIRECTED NUMBERS Multiplication is a short method of adding or of subtracting. EXAMPLES 1. --4 multiplied by +3. This Add 2. – 4 multiplied by +3. This Add means -H4 added three times. Begin at + 4 means –4 added three times. Begin at — 4 0 on the scale and count 4 units to the + 4 0 on the scale and count 4 units to the — 4 right, three times in succession. The + 4 left, three times in succession. The — 4 result is +12. +12 result is — 12. — 12 From 0 3. --4 multiplied by –3. This means that 4 is to be sub- | From 0 Subtract + 4 tracted three times in succession. This is shown in the frame | Subtract — 4 Tº at the left. + 4 Subtract + 4 Subtract — 4 — 8 4. – 4 multiplied by –3. This means that —4 is to be + 8 Subtract + 4 || Subtracted three times in succession. This is shown in the Subtract — 4 TT2 | frame at the right. - +12 DISCOVERIES MADE IN EXAMPLES ABOVE 1. (+4) X(+3) = +12 3. (–4) X(+3) = –12 2. (+4) X(–3) = –12 4. (–4) X(–3) = +12 Since the reasoning used would apply to any two numbers, we can state as conclusions: The product of two numbers having like signs is positive. The product of two numbers having unlike signs is negative. EXERCISES Multiply. (a) (b) (c) (d) (e) (f) (g) (h) 1. – 3 —H 6 - +9 +7 +9 +7 +8 +3 +4 +7 +4 +9 +6 +8 +4 +9 2. – 4 — 7 —8 — 12 —9 — 13 — 17 — 18 — 3 —9 —6 — 2 — 4 — 6 — 3 — 5 3. ––9 —9 +8 — 13 – 17 + 14 + 17 — 19 — 6 +6 –4 + 9 + 2 — 8 — 4 + 7 4. – 4 –3 6 12 — 15 — 18 14 12 +3 +4 –3 –4 2 3 —5 —6 MULTIPLICATION SIGNS There are three common ways of indicating the product of two numbers. a Xb (cross) a b (raised dot) Sometimes one is more convenient and sometimes another is. The cross was first used for this purpose by William Oughtred in a book which he published in 1631. In 1698, Leibniz wrote, “I do not like X as a symbol for multiplication, as it is easily confused with a .” Leibniz proposed the raised dot. ab (no symbol) — 50 — MULTIPLICATION EXERCISES 1. (+4) X (–7) = 9. 3a - 2 = 17. (–7)(+8) = 2. (–4) × (+7) = 10. 5.3a. = 18. (+8) (–9) = 3. (–3) X(–9) = 11. 14.3 = 19. (–5)(–7) = 4. (+6)×(+7) = 12. 6.20= 20. (+7)(+8) = 5. (–2)×(+9) = 13. 3.2b = 21. (–5)(+9) = 6. (–7)}{(–5) = 14. 7-8c = 22. (+4)(–4) = 7. (+4) × (–5) = 15. 2(–14) = 23. (–5)(–6) = 8. (+8)X (+4) = 16. 4(+12) = 24. (+9) (+9) = PRODUCT OF THREE OR MoRE FACTORS The product of three or more factors is found by multiplying two of them together and then multiplying this product by a third factor, and so on. EXAMPLES (+3)(–4)(+5) = (–12)(+5) = —60 (–2)(+4)(–7) = (–8)(–7) = +56 EXERCISES 1. (+2)(–2)(+5)=' 6. (+4)(+6)(–2)(–3) = 2 (+0–12)(-)- 7. (+2)(–7)(–2)(–1) = 3. (–3)(–6)(+4) = 8. (+4)(+2)(+2)(–3) = 4. (+5)(+2)(+3) = 9. (–5)(–2)(–4)(–3) = 5. (–6)(–3)(–5) = 10. (+4)(+2)(+5)(+3) = THE SIGN OF THE PRODUCT The sign of a product depends upon the number of negative factors it has. If the number of negative factors is odd, the product will be negative; if the number of negative factors is even, the product will be positive. Test each of the products you obtained above by this principle. EXERCISE In each space below write the product of the two numbers, the one at the head of the column and the other at the left of the row in which the space falls. — 16 + 11 —6 +9 — 7 —H 5 –8 –3 — 10 DIVISION of DIRECTED NUMBERS Division is the process of finding one of two factors when the other factor and the product are known. It is the reverse of multiplication. (+3)(+4) = +12. Thus (+12)-4-(+4) = +3 (+3)(–4)=–12. Thus (–12)+(−4) = +3 One can draw two conclusions from the examples above: EXAMPLES The quotient of two numbers with like signs is positive. The quotient of two numbers with unlike signs is negative. 1. (+4) + (+2) = 2. (–16) + (+8) = 3. (+10) + (–2) = 4. (+8) + (+2) = 5. (–12) + (–6)= 6. (+6) + (–3) = 7. (–14) + (+7) = 8. (–18) + (+2) = 9. 9--(–3)= 10. 16+2= Most newspapers print a stock market report similar to that shown at the right. This report gives the high, the low, and the closing price for the day. It also gives the difference (change) between the last prior clos- ing price and the day's closing price. Find the last prior clos- ing price for each of the stocks shown. Write these in the column of blanks. The first one has been done correctly. - 11 12. 13. 14. 15. 16. 17. 18. EXERCISES . 25+ (–5)= (–12) +4= (–20) + (–5) = 18+6= (–30)+6= – 15+ (–3) = +54+ (–9) = (–49)+7= 19, 48+8 = 20 . (–63) + (–9)= 21. 22. 23. 24. 25. 26. 27. 28. 29. 30. THE STOCK MARKET NEW YORK STOCK EXCHANGE Name of Stock Air Red High 78} A Sug Ref 55% Am Tob 98 Anaconda, 46; Armour Ill 5% Balt, & O 23# Bendix A 29; Bon Ami 95 Brown Sh 49% Bulova, W 49 Childs Co 9; Chrysler 128} (–3)(+4) = –12. Thus (–12)+(+4) = −3 (–3)(–4) = +12. Thus (+12)-4-(-4)=–3 56--(–8) = (72) + (+9) = (–36) + (–6) = (–64) +8 = 81+(–9) = 40+8 = 27+3 = (–42) + (–3) = — 164:--2= +27 y--(–3) = Low Close Ch'ge 77 78} +1} 76? 54% 54% – — 97; 98 – º –– 46%. 46% +1 — 5; 5. – º – 23%. 23% § – 28; 29 § – 95 95 – 1 - 49%. 493 – º – 48; 48% – ; – 9} 9% + 3 — 127; 128} +1; – 10. —4 7 —8 —9 —9 —7 —7ab — 11t? ab 3t? 50b 82 – 14 – 14 — 8 8 120. –4bc 23a. —8bc — 13 — 13 — 7 7 9a — 16bc 3 5 . 63--9 = (–63)-;-(–9) = 72bc_ 12 T –962 – 12 T . 4 w—3v-H7 w—H2u – w = 7–H3.4— 12+3 = 8m. — ???, —7a: 17a: REVIEW ExERCISES ADDITION 11 17 4 — 7 —6 16 37pg –32v3 –4pg — 1898 –3pg 5093 SUBTRACTION 14 —8 —8 14 9h? –3p 2.h3 –5p MULTIPLICATION 13 8 7 —9 76# 177: –4 —8 DIVISION 63--(–7) = (–63) +7= –72bc.— –8 T 962 –8T º 15p 73p —8p 33ry –27 — 5 10 –76b — 6b 2b 17+ 29p, —8 63--(–9) = (–63)--9 = –72bc 9 –963: 16 COMBINING T.IRE TERMS –3a;+4y—y--8a–2y = 139 —9 (–63)+(–7) = 63--7 = –9-H9--(–3)+7 (–2) = º –84c? 10C2 * –9az - O — 53 — INVENTORY TEST 1. Tist four things that can be conveniently represented by negative numbers. (a) (b) (c) (d) 2. The quotient of two numbers having signs is positive, and the quotient of two numbers having signs is negative. 3. To add two numbers having unlike signs, find the of their values and prefix the sign of the number having the numerical value. 4. The product of two numbers having like signs is , and the product of two num- bers having unlike signs is 5. To add numbers having like signs, find the of their numerical values and prefix their common 6. To add +5 to +4 on the number scale, begin at and count units to the The sum is 7. To add – 6 to —3 on the number scale, begin at and count units to the The sum is 8. To add +6 to — 5 on the number scale, begin at and count units to the The sum is 9. To add – 7 to +4 on the number scale, begin at and count – units to the The sum is 10. On the number scale, – 7 is units to the of 0, and +6 is units to the Of 0. 11. If we begin at any point of the number scale and go to the right, the numbers in value, and if we go to the left, the numbers in value. 12. Write two numbers with opposite signs and the same numerical value. and 13. The value of a number represents the number of units it contains without reference to its sign. 14. --4 is 1 than +3, and – 4 is 1 than –3. 15. Evaluate: 20–b-H3c, when a = 1, b = 2, c = 3. . . . . 16. Add. 18. Multiply. 33:2 —60b 7y2 67m, 160: — 77° 8bc –5y — 7a;2 10ab -3/2 — ???, –3 4. 4. — 2 ac? 3ab 1 y2 5m. - *- 19. Divide. 17. Subtract. * 10(1 — 27cd — 67m, Opg 14 – 17 == === — ???, — pg 11 4. rº- 2.7a, –4.2h .9 T 2.1 T 54 — CUMULATIVE REVIEW 1. Translate into algebraic expressions: (a) 4 less than p. (b) m divided by 5 (c) 6 more than a (e) 5 times c (d) 6 less than 3m, (f) aa, divided by g 2. The expression 3a means , and at means 3. The formula for the area of a triangle is A = will be If the area is 10 and the base is 5, the height 4. In —4a:” the coefficient is and the exponent is 5. Given two unlike quantities a and b, express their sum. Express their difference. Express their product. Express their quotient. 6. Write a monomial. Write a binomial. Write a trinomial. 7. Given a = 3 and y = 2. ac-i-y= ; ac-y = ; acy = ; a + y = 9*=—; ay-ya, = ; ac”—ary = 8. Solve and check the following equations. The root of an equation may be negative. (a) ac-H5=2 a = * - (d) m—7 = –11 m = (b) #y = —6 y= (c) 42 = —20 z= (e) #g–1 = -2 q= (f) 6r = –1 7 = 9. There are three quantities such that the second is 2 more than the first and the third is 6 less than the first. Write them. First: Second: Third: 10. (a) (12)+(2) = (e) (–12)+(2) = (b) (12) — (–2) = (f) (12) — (2) = (c) (–12)×(2) = (g) (–12)×(–2) = (d) (–12) + (2) = (h) (–12)--(–2) = 11. 5–H0 = 0–H5 = 0–5 = 0–(–5) = o divided by any number = - 12. Combine like terms: o 3a;+4a – 6ac—8a;--a: = (b) 2a+4b–4a–H6b = (i) (–12)+(–2) = (j) (–12) — (–2) = (k) (12) X (–2) = (l) (12) --(–2) = (0)(–5) = ; a number divided by 0 (too difficult for the present). (c) —3-H4–2–H 7–8–13 = (d) 9–12–H5+3–7 = 13. Simplify. Multiplication and division should be done before addition and subtraction. (a) 3+6+2–8+4+1= 14. (a) (+6)(–2)(–4) = (b) (–2)(–4)(+6)= (c) (–2)(–3)(–3)= 15. To subtract one number from another, change the and (b) 2-H4x(–3)+8+2–4 = (d) (–6)(–b)(-)= (e) (–a)(+2y)(–32) = () (–3)(–2)(+)- — 55 — UNIT 4. GRAPHS Based upon average U.S. prices reported by the U. S. Bureau of Labor Statistics DOLLARS DOLLARS The financier, the 6.50 T. $62, I-I-I-I-T-I-T-T—T-T— 6.50 business man, and the $ 5.92 CONTENTS OF THE BASKET tº “, 8.00 3 lb. round steak 3 lb. cabbage 6.00 Ordinary citizen need to N \ $5.73 iºn doz. oranges be able to read and to sº, lº, |. : 5.50 understand graphs. \ º - º º The illustration at ..., º, tº July 2 - 5.00 the right is based on a J- April 15,193) $4.52 - º ºld * * *. *$4.68 | |#469 alſº graph in the Chicago 4 so 52AFIL (ºS $437 Ry 4.50 Tribune, entitled “The s | Sº ſº § | | º Cost of the Market Bas- 400| Rºšº. 22 * $3.9 4.00 ket.” § Nºm A s/ſ 3.50 N $3.69 3.50 3.00 t $3.26 3.00 1929 1930 1931 | 932 1933 tº 4 1935 BAR GRAPH John's wages vary from week to week. His record for ten weeks is T 80 given in the table below, and is pictured in the graph at the right. HT —ſ i. +50 Week 1 || 2 || 3 || 4 || 5 || 6 || 7 || 8 || 9 || 10 i. --20 Wages 55 65 70 75 80 70 65 75 70 80 cents -j-10 BROKEN LINE GRAPH The table below gives the price of sugar for the years 1915–1921 /\ 20 ct. inclusive. These same data are pictured in the graph at the right. / \ 15 ct. Year 1915 1916 || 1917 | 1918 1919 || 1920 | 1921 Hº Nº. 5 ct. Cents a pound 6.6 8.0 9.3 9.7 11.3 | 19.4 8.0 tº $2 - ? Sº gº si CIRCLE GRAPH The percentage of annual income derived from various Sources, for the state of Okla- homa, is about as follows: agriculture, 46.4%; manufacturing, 14.5%; trade and trans- portation, 16.0%; mining, 5.8%; other occupations, 17.3%. These data are pictured in the graph at the right. Russell budgets his wages, setting aside forty cents out of each dollar for room and board, twenty cents for clothing, twenty cents for transportation, ten cents for savings, and ten cents for all other expenses. RECTANGULAR GRAPH These facts are pictured in the graph at the right. STRAIGHT LINE GRAPH Gene is two years older than Edmund. If a represents Gene's age, and y represents Edmund's age, then y = a –2. Corresponding values of a; and y are given in the table below and in the graph at the right. (U 2 3 4 7 Agriculture $/ 5 2 BAR GRAPHS Weather records show the monthly averages for the maxi- mum daily temperatures in New York City (1933) to have been: - 80 January 47° May 71° September 76° — 70 February 43° June 79° October 64° { March 44° July 82° November 49° – 60 April 58° August 80° December 41° 50 - b Such data as these can be easily shown on a bar graph. January and February have been done correctly on the graph - 40 at the right. Fill in the other months on this graph. - 30 What month has the highest average temperature? – 20 The lowest? Can you answer - 10 such questions as these better by use of the graph, or by use of a table of temperatures such as is shown above? J F M A M J J A S () N D AUTOMOBILE ACCIDENTS IN ST. LOUIS Automobile accidents in St. Louis, as in other large cities, have become a problem of grave 275 concern. Neither drivers nor pe— 25 O destrians are as careful as they ought to be. In order to call the seriousness of this matter to the attention of everyone, forcibly and in a way all could under- stand, the city authorities pre- pared a graph like that shown at the right. This shows the number of automobile accidents for the various hours of the day and night for a recent year. The dark strips indicate night hours and the light strips indicate day hours. // A. Af. AVooay A // /2 / 2 3 4 5 6 7 8 9 /o // /2 / 2 3 4 5 6 7 8 9 /O // A/ooAy AVA. AAGA CAAA A/Af$5. AVAAAG& ©4Ya/G//7. [T] Courtesy Dept. of Streets and Sewers, City of St. Louis GRAPH OF TRAFFIC ACCIDENTS AT STREET INTERSECTIONS DURING HALF-HOUR PERIODS IN ST. LOUIS, 1934 At what period were accidents least frequent? Can you give a reason for this? At what period were accidents most frequent? Can you give a reason for this? Finish the table below by writing the number of accidents which occurred during each of the half hour periods indicated. 9 A.M. 10 A.M. 12 NOON, 2 P.M. 4 P.M. 6 P.M. 9 P.M. 10 P.M. 9:30 A.M. | 10:30 A.M. | 12:30 P.M. 2:30 P.M. 4:30 P.M. 6:30 P.M. 9:30 P.M. | 10:30 P.M. — 57 — COST OF SENDING PACKAGES BY PARCEL POST Parcel post rates in 1936 for the second, third, and fourth zones were: Second zone (50 to 150 miles)–8 cents for the first pound or fraction of a pound, and 1.1 cent for each additional pound or fraction of a pound. Third Zone (150 to 300 miles)–9 cents for the first pound or fraction of a pound, and 2 cents for each additional pound or fraction of a pound. Fourth Zone (300 to 600 miles)–10 cents for the first pound or fraction of a pound, and 3.5 cents for each additional pound or fraction of a pound. A fraction of a cent in the total amount of postage is to be counted as a whole cent. The finished graph for the second Zone is shown at the bottom of the page. Read costs from this graph and fill the blanks in the table. Weight 4 3 # 2 # 5 # 6 1 # 2 pounds # Cost - cents Make similar graphs for the third zone and the fourth zone. Read costs from the graphs and fill the blanks in the table. Distance 180 78 350 573 257 480 92 515 miles Weight # 6 3} 2% 5} 1} 2 5 pounds Cost cents Examine the graphs for the sixth and seventh zones. Notice that the scale used in these graphs differs from that used in the others. What is the difference? Read the data from the graphs and fill the blanks below. (* Sixth Zone (1000 to 1400 miles): cents for the first pound or fraction of a pound, and CentS for each additional pound or fraction. Seventh zone (1400 to 1800 miles): cents for the first pound or fraction of a pound, and CentS for each additional pound or fraction. Second Zone Third Zone Fourth Zone Sixth Zone Seventh Zone § +26 § -H 52 # +24 -24 — T46 | | 1.48 C —-22 +22 -H 44 —H·44 +20 —H2O -20 - -H 40 ->º —H·40 --18 --18 +-18 —-36 –4–36 + 16 —H·16 l, gº +32 sº tºº —|-32 H --14 —-14 —-14 —H28 -j-28 HT +12 —H·12 +12 ſº -H 24 ſº —F24 -- 10 —-10 H +-10 sº-ºº: +20 +-20 _ſ la T —F. 8 +8 -H16 – -H 16 +6 —H6 -F 6 - +12 -H 12 –4–4 —- 4 -H 4 + 8 +8 —H2 -H2 —H·2 —-4 —F4 0 1 2 3 4 5 6 0 1 2 3 4 5 6 0 1 2 3 4 5 6 0 1 2 3 4 5 6 0 1 2 3 4 5 6 Pounds Pounds Pounds Pound 8 Pounds — 58 — BROKEN LINE GRAPHS The monthly rainfall of New York City to the nearest inch is given below, either in a table or a graph. When the graph is given, the student is to finish the table; when the table is given, the student is to make the graph. January Year 1924 1925 1926 1927 1928 1929 1930 1931 1932 1933 Rainfall (inches) March Year 1924 1925 1926 1927 1928 1929 1930 1931 1932 1933 Rainfall (inches) April Year 1924 1925 1926 1927 1928 1929 1930 1931 1932 1933 Rainfall (inches) June Year 1924 1925 1926 1927 1928 1929 1930 1931 1932 1933 Rainfall (inches) November Year 1924 1925 1926 1927 1928 1929 1930 1931 1932 1933 Rainfall (inches) 5 December Year 1924 1925 1926 1927 1928 1929 1930 1931 1932 1933 Rainfall (inches) § § § § § § # § § # . — 59 — CIRCLE GRAPHS A sum of money is to be divided among three boys, Andrew, Henry, and John. Six methods of sharing are shown in the circle graphs below. 1 2 3 4 5 6 \uly // H | J A different distribution is shown in each graph. Estimate the part (percentage) each boy receives accord- ing to each distribution. Record your estimates in the blanks below. 1 2 3 4 5 6 Andrew % % z % % % % Henry ". % % % % % John % % % % % % Suppose $360 is the amount to be divided. Write in the blanks below the amount each boy would receive according to the distributions above. 1. 2 3 4 5 6 Andrew $ $ $ $ $ $ Henry $ $ $ $ $ $ John $ $ $ $ $ $ The whole angle about a point is divided into 360 equal parts called degrees. Degrees are measured by use of an instrument called a protractor. A small circle written above and to the right of a number indicates degrees. Thus, 18° means 18 degrees. The first graph above shows Andrew's share as covering one half of the whole circle, or 180°; Henry’s part covers one fourth of the circle, or 90°; and John's part covers one fourth of the circle, or 90°. Referring to graphs 2, 3, 4, 5, and 6, write the number of degrees covered by each boy's share. 90° 2 3 4. 5 6 - $ 60° Andrew O * O O O O Henry O O O O O John 18O. EXERCISES - Mark off the circles below into circle graphs to picture the following distributions. 1 2 3 4 5 6 Andrew $ 60 $ 90 $50 $40 $25 $ 45 Henry $120 $135 $40 $20 $35 $160 John $180 $135 $30 $30 $30 $155 1 2 3 4 5 () — 60 — ROASTING MEATS Mrs. Eversull places a beef roast in a covered pan, and puts the pan in the oven. The Oven is kept at a temperature of 525°F. for thirty minutes; the temperature is then reduced to 400°, and the roast is allowed to cook an additional fifteen minutes for each pound that it weighs. This means that it takes forty-five minutes for a one-pound roast, sixty minutes for a two-pound roast, seventy-five minutes for a three- pound roast, and so on. Notice that the time needed in cooking any particular roast depends on its weight. 90 This dependence can be clearly shown in a graph. Making the Graph. Locate the first point above 1 on the pound line (horizontal) and to the right of 45 on the minute line (vertical). Proceed by locating the second point above 2 on the pound line and to the right 30 of 60 on the minute line, the third point above 3 on the pound line and to the right of 75 on the minute line, and so on. These three points are indicated on the graph. In a similar manner locate and indicate on the graph four other points. These seven points lie on the same straight line. Draw this line. - Reading the Graph. To determine the time it takes to cook a six-pound roast, find (on the straight line you have just drawn) point P-directly above 6 on the pound line. Then find the point on the minutes line to the left of P. The path your eye would follow in doing this is shown by the arrows on the graph. - Read the time from the graph and write in the table below the lengths of time it takes to cook beef roasts of the weights indicated. 150 120 60 0 1 2 3 4 5 6 : 7 Weight of roast 2 3 3 # 4 # 5 6 # 7 Time Mrs. Goodloe has a different way of cooking a beef roast. She places it in any sort of pan (uncovered), and puts the pan in the oven. She sets the oven at 325° F. and keeps it there, allowing twenty-five minutes for each pound the roast weighs. This means that a One- 175 pound roast needs twenty-five minutes, a two-pound roast needs fifty minutes, and so on. The time it takes to cook a roast depends on the weight of the roast; this dependence can be clearly shown in a graph. 100 Make Such a graph at the right. 75 Read the time from the graph and write in the table 50 below the lengths of time it takes Mrs. Goodloe to Cook beef roasts of the weights indicated. 150 125 25 Weight k of roast 2 3 4. J 6 Time EXERCISES In cooking roast pork, Mrs. Eversull browns the roast in the oven (525°F.) for thirty minutes and allows twenty minutes at a lower temperature (400°F.) for each pound. Mrs. Goodloe keeps the oven at a temperature of 350° F. and allows thirty minutes for each pound. In each case the time needed depends upon the weight of the roast. Show this on a graph. Use a dotted line for the first (Mrs. Eversull) and a full line for the Second (Mrs. Goodloe). — 61 — CooBDINATES A horizontal and a vertical line are given. The horizontal line may be called the a-axis, and the vertical line may be called the y-axis. Distances measured from the y-axis to the right are positive, while dis- tances measured to the left are negative. Distances measured upward from the ac-axis are positive, and distances measured downward are negative. This makes it possible to match every pair of numbers with a point on the graph, and to match every point on the graph with a pair of numbers. Point P (4,2) is 4 units to the right of the y-axis and 2 units above the ac-axis. The two numbers (4, 2) are known as the coordinates of point P. The first number, 4, is called the ac coordinate, or abscissa, and is measured parallel to the a-axis. The second number, 2, is called the y coordinate, or ordinate, and is measured parallel to the y-axis. Points Q (-2, 3), R (–2, —4), and S (3, -2) are plotted on Graph I. Be able to give the abscissa and ordinate of each. Six points, A, B, C, D, E, and F, are shown on Graph II. Write in the blanks below the coordinates of these points. A ( ) C ( ) E ( ) B (–, ) D ( ) I' ( ) Locate on Graph III and identify with the proper letters the points whose coordinates are * G (3, 2) I (–2, 3) R (–1, 3) MI ( 2, —2) H (2, 3) J ( 3, − 1) L ( 2, 0) N (–2, —2) j ) y j j PLOTTING FORMULAs The method is to assign different values to s in the formula, P=3s, and to find the corresponding values of P. We obtain a series of pairs of values. S () 1 2 — 1 || –2 | # 1} | – # P o –3 || – 6 # 4; —# In plotting these pairs of points, we shall call the horizontal axis the S-axis, and the vertical axis the P-axis. In Graph IV three pairs of values shown in the table above have been plotted and a straight line has been drawn through them. EXERCISE Given the formula, P=s+2, make a table giving eight pairs of corresponding values of P and s. Plot these values on Graph IV. Can a straight line be drawn through these eight points? S P RENE DESCARTEs (1596–1650) The method of graphing above was invented by Descartes. After a part of his last name, it is called the Cartesian Coordinate System. Mathematics in Europe had been almost at a standstill for a thousand years at the time Descartes was born. His method of repre- senting points and formulas graphically is considered a beginning of modern progress in mathematics. It furnishes a basis for analytic geometry, calculus, and more advanced mathematics. Graph I Q/' Graph II {/ J’ Graph III {/ (ſ’ Graph IV P — 62 — STRAIGHT LINE GRAPHS A formula is a statement of relationship between quantities; a graph is a picture of this relationship. Make a table of pairs of values and a graph for each of the formulas. 1. N = n+1 f 4. W = 2D – 3 *7. K = 2–4 n – 3 | – 2 −1|0 1 2 D P|- 0 || 4 s 12 w|| || N R 2. N = 1 — m a -3-2-1 2 N 6. T = — 2 — i. *9. C=2(n–1) REVIEW: READING GRAPHS 1. The annual production of anthracite coal in the United States, in millions of tons, is pictured in Graph I. Use this graph for facts to complete the table. In writing, omit the last six 0's. Year | 1920 | 1921 | 1922 1923 || 1924 | 1925 | 1926 1927 | 1928 1929 1930 1931 | 1932 1933 Tons of coal 2. Graph II pictures the number of deaths due to pneumonia per 100,000 population. Use this graph for facts to complete the table. In writing the numbers, omit the last five 0's. Year 1920 | 1921 | 1922 || 1923 || 1924 || 1925 | 1926 1927 | 1928 1929 || 1930 1931 || 1932 Deaths 3. The standard weight in pounds of 19-year-old cadets at West Point is given by the formula: W =4h–135, where h is the height in inches. This relationship is pictured in Graph III. Use Graph III for facts to complete the table. Height 64 || 65 | 66 67 | 68 || 69 || 70 || 71 || 72 73 Weight 140 ! 2 0 100 80 60 40 § § § § § § § § à # § 64 65 66 67 68 69 70 71 72 73 74 inches § § — 64 — REVIEW: MAKING GRAPHS 1. Automobile production in the United States from 1919 through 1933 is given in the table below. The unit is 100,000 cars. Make a bar graph of these figures. The last five 0's are omitted in the numbers. Year 1919 1920 1921 12|10x 192,192. 1926 1927 1928 1929 1930 1931 1932.1% Number of C2,I'S 17 19 15 24 38 33 39 39 31 40 48 29 20 12 16 2. Imports of crude rubber into the United States from 1919 through 1933 are given in the table below. The unit is $10,000,000. Make a line graph of these figures. The last seven 0's are omitted in the numbers. Year 1919 1920 1921 1922 1923 1924 1925 1926 1927 1928 1929 1930 1931 1932 1933 Imports 22 24 7 10 19 17 43 51 34 24 24 a 3 5 3. Mrs. Eversull uses this formula in determining the time it takes to cook a beef roast, T=30+ 15W, where T is the time in minutes and W is the weight in pounds. Make a table of values and a graph of this formula. W 1 5 6 7 10 11 12 T 1. EAR GRAPH 1919 1920 1921 1922 2. E ROKEN LIN E GRAPH 1923 1924 1925 1926 1927 1928 1929 1930 1931 1932 3. STRA GHT LIN E GRAPH 1933 9 10 11 12 INVENTORY TEST 1. The unit is 100,000 cars, and a statistical table gives 17 as the production for a certain year. This means that cars were produced that year. 2. The unit is $10,000,000, and a statistical table gives 83 as the value of the imports of crude rubber for a certain year. This means that dollars’ worth of rubber was imported that year. 3. We have studied five types of graphs in this unit, namely, j y y , and 4. A point two units to the left of the y-axis and three units above the ac-axis has the coordinates ( —y ). The first of these numbers is called the of the point, and the second of these numbers is called the of the point. 5. The point whose coordinates are (4, -1) is units to the of the y-axis and – units the ac-axis. 6. How many values may s have in the formula P=4s? For a given value of s, how many values may P have? For every units increases, P increases units. 7. Name or give illustrations of all the ways you know of expressing a relationship between two quan- tities. 8. Below each of the graphs write the letter which identifies the table from which the graph was made. OTTT2 T3TTT5 * 5 5 (a) 4 || 3 || 5 || 3 || 2 || 4 | 'N 4 3 3 3 0 | 1 || 2 || 3 || 4 || 5 (b) 2 2 2 (d) 4 4 0 || 1 || 2 || 3 || 4 || 5 3 3 0 || 1 || 2 || 3 || 4 || 5 (e) 2 2 to "| || |*|*|" 0 1 2 3 4 5 4 || 5 || 3 || 2 || 3 || 4 tºm-m-m-m-mºº-º-º-º-º-º-º-º-º-º: Kºmºmsºmºmºmºmºmºmº. gººmsºmºsºmºsºmºsºm-º-º-º-º-mm. 9. A sum of money is divided so that one person receives thirty-five percent, two others receive twenty percent each, w—gs—w-w-ºr-w-r- - - -w- – w a fourth receives fifteen percent, and a fifth receives ten percent. Show this on the rectangular graph at the right. 10. A has $360, B has $180, C has $90, D has $60, and E has *——ºs –A–º-–a–a–º-—º-º $30. Show this on the circle graph at the right. CUMULATIVE REVIEW 1. Perform the indicated operations and leave the results in the simplest form. (a) 2–H (–3) = (e) 10+ (–2) = (i) 3– (–2) –4 = (b) 2–(–3) = (f) 10+2= (j) 7–H (–3)–2= (c) 5X (–2) = (g) 10– (–2) = (k) 3X (–2) X5 = (d) 5×2= (h) 10+ (–2) = (1) 10+5+ (–2) = 2. Wherever a question mark (?) appears, a word or a number has been omitted. Write the omitted words on the blanks provided. (a) A trinomial has 7 terms. (C (b) – means a ? by y. $/ (c) m-H m means the 7 of m and n. (d) a -y means the 2 of a and y. (e) cq means the 2 of c and d. (f) – 4 is 2 than 2. 3. Solve the equations and write the roots on the blanks provided. (a) a -i-4 = 7 -*-º-º-º- (e) 32-H1=7 *mº-º-º-º-mm. (i) ºr = 4 &ºmº-mºmºmºmºm, (b) ac-3 = 2 mº-mºmº-mº-mºm-º. (f) 32–1=8 *—ºs.--—se (j) #a = 6 * *mºmºmºmºm. (c) 23:=6 *mº-º-º-º-º-º-mº (g) 1–a: =5 ------ (k) #~-H1 = 5 m-mam-m-m-m-mº-º-º-º- (d) 2a: = —8 *m-rººm-mm-mm, (h) 1–2a: =5 *º-º-º-º-mº (1) 1–4a: = 6 *-*-*-* 4. Set up equations and Solve. (a) A father is twice as old as his son, and the sum of their ages is sixty-three years. Find the age of each. (b) Janet is 1% times as old as Ruth. The sum of their ages is twenty-five years. Find the age of each. (c) Five times a certain number increased by 15 equals 9 times the same number decreased by 13. Find the number. *(d) Thirty men and boys work in a factory. Each man receives forty dollars a week, and each boy receives thirty dollars a week. The payroll for a certain week was $1,110. How many men and how many boys were employed? *(e) Fred has three times as many pennies as nickels. In all he has eighty-eight cents. How many pennies and how many nickels does he have? *(f) Separate 40 into two parts such that of the smaller part equals of the larger part. — 67 — UNIT 5. THE FUNDAMENTAL OPERATIONS CALCULATING MACHINES The basic principle upon Charles Babbage, profes- which modern calculating - sor of mathematics at Cam- machines work was applied bridge University in Eng- first in 1642 by Blaise Pascal, land, saw big possibilities in who was a mathematician Pascal's little machine. Aided and the son of a mathema- by a grant from the British tician. The meter that shows Parliament, Babbage sought how far an automobile has to build a machine that could been driven is a simple exam- carry through the compli- ple of Pascal's principle at cated computations needed work. As soon as ten ones are in astronomy and other sci- recorded, they vanish as ones ences. He worked on this and appear as one ten; as from 1820 to 1856, but failed soon as ten tens are recorded, in carrying through his plans. º they vanish as tens and ap- couras, Burrow. Addin, Machinºco. Babbage's machine was never pear as one hundred; and as A COMPUTING MACHINE finished. soon as ten hundreds are re- Through the efforts of corded, they vanish as hundreds and appear as one scores of mathematicians and inventors, calculating thousand. Pascal built this simple principle into a machines have now reached a high degree of perfec- machine that could add and carry when it needed tion. They have become standard items of office to do so. equipment. ADDITION A few days before Thanksgiving, the principal of Lincoln School asked the children on the upper floor to bring in canned goods to help fill baskets for families whose fathers had no work. In finding how many cans had been given by all the rooms, the principal made a table like this: Corn Beans Peas Pumpkin Peaches Cherries Room 1 14 11 3 7 4 2 Room 2 10 11 4 8 3 5 Room 3 6 12 10 12 5 3 Room 4 12 8 8 14 6 4 Totals 42 42 25 41 18 14 Last Easter the lower-grade children colored eggs. When the tally was made, it was found that the kinder- garten had colored 10 eggs red, 14 blue, 22 green, and 9 yellow; the first grade had colored 18 eggs blue, 14 red, 20 green, and 16 yellow; the second grade had colored 12 eggs yellow, 24 green, and 16 blue. Make a table, similar to the one the principal made for the canned goods, and find how many eggs in all there were of each color. Red Blue Green Yellow Kindergarten -- - - - First Grade - - - - Second Grade - - - - Totals - - - - In all exercises of this kind, things of the same sort must be placed in the same column. There was one column for corn, one for beans, one for peas, and so on; there was one column for red, another for blue, another for green, and so on. – 68 — ADDITION EXERCISES 1. 5 yd. 1 ft. 3 in. 2. 5 wk. 1 da. 3 hr. 3. 5a-Hy-H.32 2 yd. 4 in. 2 wº. 4 hr. 2a ––42 7 yd. 1 ft. 2 in. 7 wº. 1 da. 2 hr. 7a;+y+22 ADDITION OF POLYNOMIALs . —3r +7a: –2a: You learned how to add denominate numbers in arithmetic; in Unit 3, you learned how to add columns of signed numbers. You will combine these two ideas in adding polynomials. This means that like terms must be placed in the same column, as with denominate numbers, and that these columns will be added as signed numbers. These principles are illustrated in the addition exercises above. EXERCISES 1. 32–29-H4a 9. 6d — 3b-H c 17. – 42°-H 6:ry–7y” –62-H4y–3a 20 – 4b-H4c 11a:2+13+y 4a –39-H6a 5a–H 7b — 26 –72°–14 ry-H9y” 2. 3a – 4b-H 6c 10. 47m —3rº-H r 18. 3ré–H2r?-- r –6a +4c 7m —9n–H 67. — 2r.” — r –8a–H 10b — 3C 2m – 2n — 2r j-3 +3r 3. – 6m – 11n +7r 11. 3r—2S-- i. 19. 5/2+11c” — 7m-H 2n — 2r–H3s—3t —yz – 7c.” — 4m-H 7n — 7 r r — S-H-2t 3y2+ 4cº 4. d— e-H f 12. 8p–3q+4r 20. —7w-H42–H 6y–42 –2d-H4e–H7f 3p– g – r –2w – a – 2y–22 d–3e–8f —p-H20+2r w-H2a-H y–H 2 5. –g-H 19h — 7. 13. – 3ry–H4arz–6aw 21. 7 rº–3rs—H2s” g–11h — 87 7ary — 4.rw r”— 7's — 3s” g— 4h – 107 –8.ry–4 rz — Sr.” + s” 6. 9q+13r-16s 14. —3r”–4y2–3 *22. 2.2 r + 1.77) —q-- r –6a 3–4y2+2 .33 – 2.37. 3r—27s –82%–4y?--1 . 1ſt — .4v 7. 2d-H4b-i-8c 15. 23.2–4xy-H.89% *23. 3.2a+4.8b+ 1.8c (l, + c –6aº-H6ay— y” — 5.6a — .2b — . .2c b-i- c 4a2+8:ry--4y” .4a–H .4b + .4c 8. 47m. — 6a: 16. a?--2ab-H b% *24. Alt-Häy–4}z –3m —89-H7a: a?–2ab-i- b” – c +6/-H 4 2 6m-H8y — a —20% –2b” —#r-H+y+ #2 CHECKING ADDITION BY SUBSTITUTION The sum of two or more polynomials may be checked by substituting arithmetic numbers for the letters in the given polynomials and in the sum. |EXAMPLE Check the addition below by substituting 2 for a and 3 for y. 2a:” – a y–H y” becomes 8— 6–H 9 = | + 11 The sum of +11 and — 10 is +1. The —ac”-H2:ry–2y” becomes —4+12–18 = | – 10 sum of 4, 6, and –9 is also +1. This wº-H acy– y” becomes | 4+ 6– 9= | + 1 checks. EXERCISES Copy in the spaces provided (placing like terms in the same column) and add. Check by letting a = 2, y = 3, and 2 = 4. 1. 3a;+4y; 2a:-H6y; — ac-H5y 5. 42+32–59; –42–H 6y–7a:; 22–8) 2. a 4-Hay-Hy”; a 2–acy--y”; a 2–y” 6. 4a – 6y-H82; +3y–42–2a:; —42+3y–2a: 3. –3a;+4y–62; – a –y–32; —2a-H4y-H62 7. 32–42–2y; —3a;+42; –62+4a: 4. –acy-Hy”; 3ay-H2y”; y”–2a:” 8. — 42+62–29; 4y-H92–a:; 22–6, 1 5. 2 6. 3. 7. 4. 8. — 70.- REMOVING PARENTHESEs Parentheses are used in algebra to show that two or more terms are to be treated as a single quantity. Braces { } and brackets [ ] serve the same purpose. EXAMPLES (3a;+4y)+(–2a:-H6y) = ? (4a –3y) — (2x+y) = ? This is an exercise in addition. This is an exercise in subtrac- The addition may be performed 3a;+ 4y tion. The subtraction may be per- 4a – 3y as shown at the right. Another formed as shown at the right. An- –2a:-H 6y 2a –H y method is to remove the paren- IIon other method is to remove the 2-4, theses and collect like terms as ac-H 10y parentheses and collect like terms a — 49 shown below. as shown below. 3a;+4y–2a:-H6y = a +10y 4a –3y–2a – y = 2a:–4y If no sign appears before a parenthesis, the + sign is understood. When the parentheses in example 1 were removed, how were the signs of the terms inside the parentheses affected? What signs preceded these parentheses? Answer the same questions for example 2. TWO IMPORTANT RULES A parenthesis preceded by a + sign, or by no sign, may be removed without changing the sign of any term within the parenthesis. A parenthesis preceded by a - sign may be removed by changing the sign of every term within the parenthesis. EXERCISES Remove the parentheses and collect like terms. The first and second have been done correctly. 1. (a+b)+(a+3b)+(2a–b) = 6. (a —y) — (2a –3y) — (y–4a) = a-Hb-Ha-H3b-H2a–b = 4a–H3b 2. (a +y) — (a)-H3y)– (22–y) = 7. (3a–H2b-i-c)+(a+b)—(b–c) = a;+y—a:-3y–2a:-Hy- -2a:-y 3. (3a–H2b)+(4a–3b)+(2a+5b) = 8. (5:c–4y–32) + (x+2y–32) = 4. (32-H2y+2) — (2a–Hy–2)= 9. — (6a — 4y) — (32–a)+(2y–32) = 5. (a+b+c)+(3a–H2b-H4c) = 10. – (62–4y-Hz)–(–2a+7y–72) = — 71 — ENCLOSING TERMS WITHIN PARENTHESES Terms may be enclosed in parentheses preceded by a + sign without changing the sign of any term. Terms may be enclosed in parentheses preceded by a - sign by changing the sign of every term. EXERCISES For each exercise below, enclose the first two terms in a parenthesis preceded by a + sign, and the last two terms in a parenthesis preceded by a - sign. The first and second have been done correctly. 1. 3a;+4y-Ha-2b = 7. —7r-4S-H-2t — u = (32-H4y)–(–a–H2b) 2. – c.--3d—w-H2v = 8. —e—f–g—h = (—c-H3d) — (w—2v) 3. a+b-c-H d = 9. --2a-H39-H42–H5w = 4. –3a;+4y–6a+7b = 10. —2g+3h–2i +37 = 5. 3a–4b-H 6c.— 7d = 11. – a –2b – c – 2d = 6. – 6m-H4–37) –2 = 12. 5a–b – c.—H·56 = *PARENTHESES WITHIN PARENTHESES EXAMPLE Remove the parentheses and combine like terms. 4b–I6a–(a+2b)+7b)=4b–[6a–a–2b-H7b)=4b —[5a–H5b)=4b–5a–5b = —50—b Inner Terms in Brackets Terms Parentheses Brackets Removed Collected Removed Collected EXERCISES Remove the parentheses and combine like terms. 1. a+[3b-H (4a–H2b)]= 2. 6c--6d—[c-H (c.—2d)]—d= 3. e-Hſ—ſe— (f-H6e)–6f]= — 72 — * * * * * * * * * * * * * * * * * * * * * * MAGIC SQUARES Magic squares came from China. Legend says that the first magic square was found by Emperor Yu (2200 B.C.) on the back of a turtle which came out of the Yellow River. The simplest magic square (shown at the left) is made up of three rows and three columns, with the numbers 1 to 9 written in the small squares; the Chinese way of writing this is shown at the right. The magic lies in the fact that the sum of each row and each column and each di- agonal is the same number, 15. In olden times, magic squares formed a part of the number mysti- cism. They were part of the fortune teller's stock in trade, and when carved on amulets were believed to charm away evil spirits. Today we look upon such beliefs as childish. Magic squares, however, con- tinue as interesting playthings. EXERCISES ( &–. > Q <> The test of a magic square is that the sum of each row, each column, and each diagonal be the same. Apply this test to the squares below, placing the sum of each row and each diagonal to the right on the blanks provided, and the sum of each column in the space below the column. 1. 3. 6a-8y | 11ac – 3y º = = * * * * * *s sº & 2a+4b 9a–H 11b 4a–H 6b |---------- 5a —9y | 73 – 7 y | 9a – 5 y ---------- 7a-H.9b 5a–H 7b | 3a–H 5b |---------. 102 – 49 || 3a – 11y | 8a; – 6y ---------. 6a–H.8b a-H 3b | Sa-H 10b ---------. 2. 6a: 1 1. +5y 4a – 2y --------- 2ab 7a3+9ab | 20°–H4ab |---------. 5a – y 7a;+ y | 9a–H3y |---------- 5a”--7ab | 3a*-H 5ab a*-i-3ab ---------. 10a;+4y 3a – 3y | 8a;+2y |---------- 4a2+-6ab — a”-Hab | 6a–H8ab | --------. SUBTRACTION EXAMPLE Subtract 32-H2y–1 from –62-Hôy-H4, and check by letting a =4 and y=3. Subtraction Checking Minuend –6a +6/-H4 becomes —24-H 18+ 4 = — 2 Subtrahend 3a;+2y – 1 becomes 12+ 6 — 1 = 17 Remainder – 9a;+4y–H5 becomes –36–1–12–H5 = — 19 This checks. Another method for checking subtraction is to add the remainder to the subtrahend; their sum should equal the minuend. EXERCISES Subtract, and check by letting a = 2, y =3, and z=4. 1. 4a – 6y-H2 1. a 2+2a:y-Hy” 7. 52--4y–H32 –2a-i-60ſ –2 2xy-Hy”--2” a — y–32 2. a 2+2a:y-Hy” 5. a.”—y” 8. – a – 29–32 a 2–2acy-Hy” a 2–1-y” 2a – 6y-H42 3. 3a*—4y” 6. acº—acy 9. 62–4y-H 132 3y2–22% 3ay–6 2y – 172 Subtract, and check by adding the remainder to the subtrahend. 10. – 6m-H4m –2p 12. 2d-H b-i-3C 14. 3r–H 7s — 16t 7m – 6m 6a+b — 4c 3r—7s–H 16t 11. 5a–2b–3C 13. —3S-H-2t+2r 15. a-Hy 3a–H b-H c S — i — r 2a –y–3 Rewrite, with like terms in the same column, and subtract. 16. 5a–6b+c from — a +b+3c 17. – witH 3v-v from 2w-Hw-Hv 18. 32-H2y+z from ac-H2y–32 16. 17. 18. — 74 — MAGIC SQUARES Finish the magic squares below and check. Remember, the test of a magic Square is that each row, each column, and each diagonal have the same sum. Hint for finishing the first square: Find the sum of the first column. From this sum subtract the sum of blocks 1 and 5, and write the remainder in block 9; from this same sum of the first column subtract the sum of blocks 4 and 5, and write the remainder in block 6. Proceed in a like manner in filling the other blocks. Space for Computation 10a;+8y | 6a–H4y | | --------- 2 * * * * * * * * * 16a-H2y | 6a–3y 82–2; --------- 2a – 5y | | |--------- 90 — b ---------- 10a || ||-------- tºº 11a–H b 7a –3b ---------- — '5 — ADDITION AND SUBTRACTION OF POLYNOMIALs EXERCISES Add each of the following. From the sum, subtract the first addend; from the resulting remainder, sub- tract the second addend. Continue till you have subtracted each addend. The first exercise has been done correctly, - 1. 52–5g–H52 4. 2d —4b —7c 7. 8w—20—5u, 82+3y—62 a-H b-H c 3u +3M) —a;+4y–22 — a +3b-H2c —4v.–90 122-H2y–32 (sum) 52–5 y–H52 - 7a;+7y–82 (rem.) 8a;+3y—62 —a;+4y–22 (rem.) —a:-H4y–22 0–H0 +0 (rem.) 2. 12bc—3ca-H2ab 5. 5a-H6y—42 8. 2m+3n+4p 3bc–3ca–2ab –3y-H22 — mº, –4p –9bc-H7ca–H5ab 3a; — 2 —m-H2n+ p 3. 5r—3s —8t 6. 2p–3r-H7t 9, 2pg-i-3rs–5up —3r–H S-H-2t 3p–H3r–5t - –pg –2uv —4r-H4S-H-3t —p-H2r–4t rs-H4viv — 76 — REVIEW OF ADDITION AND SUBTRACTION 1. Add. Check by letting a = 2, y=3, and z=4. (a) 22–H3y-H22 (b) 4a2+ y?–6 3a;+ y – 2 . 4y?--4 a — y–H 2 622–5 y?–2 2. Subtract. Check by adding the remainder to the subtrahend. (a) 6m-H4n. (b) —7m-H4m –6r (c) —2a:-H5y–72 7m — 4n+6 6m – 7m —6 r 52-H2y 3. Copy in the spaces provided, and perform the operations indicated. (a) From the sum of –2a+4b-i-c and 4b-7a-H6c, subtract -5c+6a-2b. (b) From the sum of a 2+2xy+y” and –2y”--4a.”–6ay, subtract 2a:*-H2y”. (c) From the sum of 92+4y and –62-H42, subtract –69-102. (a) . (b) (c) 4. Remove parentheses and combine like terms. (a) m—(m+a;)+(32–4m) = (c) — (a +y)+(42-H2y)+2 = (b) —6a–ſ(a-b)+4]–2a = (d) 6-i-[d–(e–H3)+4]= 5. The rule for subtracting is: Think of the sign of the 6. Parentheses preceded by a + sign may be removed Parentheses preceded by a - sign may be removed by 7. Enclose the a's and b's in a parenthesis preceded by a + sign, and the ar's and y's in a parenthesis preceded by a - sign. Rearrange terms when it is necessary. (a) ac-2a–3y-Hb (b) 3y—a –a–b (c) a-b-i-a –y 8. Add the column of signed numbers shown at the right. State the rule you used. -: - — ac 5a: — 77 — MULTIPLICATION You have learned that s” means s's, and that e” means e. e. e. You have also learned that a a a may be written more briefly as ac". EXAMPLES Multiply ac” by a 3. Multiply a” by a′. (r: a) (a a ac) = a a a a . a = azº (a a a a)(a a a) = a- a. a. a. a. a. a = a” In example 1, a is called the base; the base, in example 2, is a. When two or more factors have the same base, the product has the same base, and the exponent of the product is the sum of the exponents of the factors. Thus, æ", a' = a "+". EXERCISES 1. a”. aš = 4. b3. b3 = 7. ac2. ac{= 10. y”, y' = 2. a 4. a6 = 5. b3. b2 = 8. acq. a.4 = 11. y”. y”= 3. a 4. a” = 6. b3. b3 = 9. acă. ac{= 12. y” y = When two or more factors have different bases, we cannot do more than indicate their product. Thus, a” times b% is written a”b3. In the expression, 32°, the 3 is called the , and the 4 is called the MULTIPLICATION OF MONOMIALs To multipſy monomials, multiply the numerical coefficients, observing the law of signs, and add the exponents of like letters. EXERCISES 1. (–4)(6a) = 5. (–2a)(4a) = 9. (–c)(– ac) = 2. (2x)(–4a) = 6. (– ar)(a’a) = 10. (–9a2b”)(–abe) = 3. 4.3a = 7. (–3arº)(–ac”) = 11. (–;km3)(#m’) = 4. (–5)(–3b) = 8. (–4ab)(–a’at) = 12. (3a.”)(–2a") = MULTIPLICATION OF A POLYNOMIAL BY A MONOMIAL To multiply a polynomial by a monomial, multiply each term of the polynomial by the monomial. The first exercise below has been done correctly. EXERCISES 1. 3a ––4a 3. m.--9 5. r–8 –2a: –3a 8r — 6ac”—804: 2. a -3 4. – a –H2 6. m.-H.3m. — QC 20. 2n MULTIPLICATION OF BINOMIALs Example: Multiply a -2b by a-Hb, and check by letting a = 2 and b = 3. Multiplicand a —2b becomes 2– 6 = — 4 Multiplier a + b becomes 2–H 3 = 5 a”—20b – 20 ab–2b? a” — ab–2b” becomes 4–6–18 = −20 To multiply binomials, multiply each term in the multiplicand by each term in the multiplier and add the partial products. We begin the multiplication at the left. EXERCISES Multiply. Give a convenient value to each letter and check as in the example above. 1. ac-H3y 4. a + a 7. a +6 a;+4y ac-H b a – 7 2. ac-39 5. a. – 2 8. a. – 2b ac-H4y ac-H2 ac—H2b 3. ac-H3y 6. a. – 5y 9. m. — 6m. a — 49 ac-H4y m — 2n MAKING A MAGIC SQUARE In each block of the small square, substitute ac-y for a and ac--y for b. Combine like terms and write the results in the corresponding blocks of the larger square. When completed, the larger square should be a magic square. Check to find whether the sum is the same for each row, each column, and each diagonal. 13a–H3b | 6a–4b 11a–Hb 8a–2b 10a |12a-H2b 9a– b 14a–H4b 7a-3b | | | | | | \ ||--------------- - * * * * * * * * * * * * * * Multiply. 1. 2a:-H.3/ 2a:-H2y . 22–39 3a;+2y . 2a:--3y 3a;+2y . 3a;+4y a – 29 . 2m+ n, m–H2n, . 6a–5b a-H2b a – 5 y 52–H y . 3m.--6m. 2m – n. MULTIPLICATION OF POLYNOMIALS 10. 11. 12. 13. 14. 15. 16. EXERCISES . 30 — 7b 2a – b a;+2y+2 —tº 3a –y-H.32 a — 2 5ac—y–2 2y+2 2a2–1 a – 1 2a2+y a +y 2a2+ g” 22°–39? 17. a”-- a--1 a – 1 18. y”—y-H1 ºt'. 19. 2a – 1 20. a.”--ay-Hy” a — y 21. a 9–acy-Hy” a +y *22. 4b2–2bc-H cº 2b–3C *23. a -acy-Hy acy-Hy *24. a”--ab–H b” a?—-ab-H b% — 80 — DIVISION BY A MonoMIAL EXERCISES 1. ––4+ –2 = 2. – 6 + +2= 3. –H8+ +4 = 4. – 9-– –3 = DIVISION OF A MONOMIAL BY A MONOMIAL EXAMPLES –6a Fis a a" & & & 3a*T ºn The coefficient of the quotient of two monomials is obtained by dividing the coefficient of the dividend by the coefficient of the divisor. The exponent of any letter in the quotient of two monomials equals its ex- ponent in the dividend minus its exponent in the divisor. - –2. a . a = –2a? EXERCISES — 15cº 1. 32°– —32 = 7. – 210.4b*-i-3ab? = 13. = 5c 2. 10d4+ 5a? = 8. 27abc”--9C = 14 18a" 30.2 3. —8m2 + — 2m2 = 9. #a’acº-:-}ar = 15 21b% 4. – 4acº ––ac” = 10. – 27aëm --9aºm = e T35T 1203 — ‘D 5 5. 237 -- . 10.5 = 11. *= 16. 25ct_ 2a 5C2 14b.6 17. 27a 7-i-9a 3 = 6. 102°y-:- —2ary = 12. 2. T 18. — 30c^+-6c4 = DIVISION OF A POLYNOMIAL BY A MONOMIAL EXAMPLES (r3+3+2+2)--a = a 2–H3++1 3a*—6a?--a2 =3a7–6 To divide a polynomial by a monomial, divide each term of the polynomial by the monomial. EXERCISES 1 40?--8a. - 7 –6a”(c—y) º 20. – (a –y) - 25cº — 15c 14b5— 7b3 2. — = 8. — - 5c 7b4 8ab–4a? a?b? — a 4 3. — = 9. — = 20. a? 4 204–H4a3+80% 10 a 2–1–23°y-i-43%)? º 2a - • ar? - 3604b*c? 12a3b*(ºr 5. — = 11. 120°b'Gr-Hy). 6abc} — 60% –6a”(a —y) a 6––ac4 6. — = 12. - — 3a. ac2 — 81 — | DIVISION BY A BINOMIAL IEXAMPLE (a) a +y)a?--3a.”y-H2:ry” The first term of the quotient will be ac”, which is found by dividing the first term of the dividend by the first term of the divisor. 3:2 Write this first quotient term above the horizontal line as shown in (b). (b) ac-i-y)a:3+33%)+2:ry? Multiply the divisor, a H-y, by wº. Write this product under like terms of the dividend. a;2 Subtract. The remainder is 2a:”y. Bring down the next term of the dividend, (c) 2-Hy)a”-H3a*y-H2:ry” as shown in (c). a 3–1– arºy The second quotient term is 2a:y, found by dividing the first term of the new 22°y-H2:ry” dividend by the first term of the divisor. Write this second quotient term above the horizontal line, as shown in (d). (d) a. lº +2a:y? Multiply the divisor, a Hu, by 2:cy, and write the product under like * arº-H º 3.0/ terms of the new dividend. e 16 Tº +2ary? Subtract. There is no remainder. 5)80 2aºy--2ary? Check by substituting a = 2, and y =3, in divisor, dividend, and quotient, as shown at the right. The division checks. EXERCISES - Check here. Check here. 1. a, +3)ac” + 52 + 6 Let a = 2 4. a +7)a? -- 15a, + 56 Let a = 3 Check here. Check here. 2. a+2b)a? -- 5ab + 6b? Let a = 2 5. 2d-H3b)6a” + 13ab -- 6b? Let a = 3 b = 3 - b = 4 Check here. Check here. 3. 22–y)8a.” + 2xy — 3/? Let a = 2 6. a -y)3a.” – aſy — 29% Let a = 2 y=3 g = 3 — 82 — DIVISION AND MULTIPLICATION OF POLYNOMIALS Find the quotient in each exercise below. When you have divided, check by multiplying the quotient by the divisor. EXERCISES 1. y–H5)2/*-H 99 – 5 5. a--2y+4)a.” + 2xy + 7a: + 6y + 12 2. a 4–2/)a? -- 7ary + 10y” 6. a 2–H1)2a:” – a 4 + 2a – 1 3. 4-H.3a)24 + 14a – 3a.” 7. y?--y-H1)g” + 0y” + 0y – 1 4. a -i-y--3)2x2 + æy -- 6 - y” – 3g 8. 3a;+2y)3r” – a y – 29% — 83 — DIVISION OF POLYNOMIALS The example at the left is not arranged properly to divide. Can you tell what is wrong? The example at the left is arranged properly to divide. In what 2a2+2 − 1)2.c5+a+–a4+63:?--32–3 way does it differ from the one above? ar–H2a2–1)3a;+63:?—a 3-H2:rº–3––a4 In copying a polynomial to be divided, both the dividend and the divisor should be arranged in de- scending (or ascending) powers of a common letter. EXERCISES Copy, rearranging terms where necessary, and divide. 1. a 4-H 66–17a: –5a.” by a -6 4. 62%--38wy”–27a.”y–20y” by 2a –5y 2. a "+1 by a +1 (See Ex. 7, page 83.) 5. a 4–1898–15ay?--4a”y by a 2–2ay–3y” 3. 12+13++622–1–28 by 3--a: 6. 1+2+2}+a;% by a 2+1 Space for solutions 1. 4. 2 5, 3. 6 — 84 — DIVISION WITH REMAINDERS Find the quotient and the remainder in each exercise below. Check by multiplying the quotient by the divisor and adding the remainder. The first one has been done correctly. EXERCISES Check a +4 ac” — ac 1. a 2–ac)a?--3a.”–2a: +4 a +4 acº — ac” acł — ac” 4a2–2a: 4a2–4a: 42% — 4a: acº—H·3ac”— 4a: 2a –H4 2a ––4 a:4–H3ac”— 2a:--4 2. 23%-H.32)8a.” + 8a.” – 6a, +3 3. ac-H3)2a:” -- 10a: + 12 4. 3a –2)6ac” – 13a; + 10 5. 5a-H2)5acº — 83% + æ + 4 MULTIPLICATION AND DIVISION OF POLYNOMIALs Divide to find the quotient and the remainder. Check by multiplying the quotient by the divisor and adding the remainder. EXERCISES 1. a, -39)a.”–8ay-H 15y” *4. 33:2-H2a – 1)12.c4–7a:8–52%-H 102–5 2. 5:c-i-6).25a:8–52%–87a: — 57 *5. 22-5/)6.cº–27a:”y--38wy?–20y” 3. 32–2)21a:8–17a:”—H·82–6 INVENTORY TEST 1. Copy in the spaces provided (placing like terms in the same column), add, and check by letting a = 2, g =3, and z=4. (a) 32–294-2, a —y–2, 22–H3y-H42, —a –y-H22 (b) a 2–Hay-Hy”, a 2–y”, a 2–2°, -52°–Hay (a) Check - (b) Check 2. Copy in the spaces provided; subtract the second polynomial from the first; check by adding the remainder and the subtrahend. (a) –2a:-Hy–2, a -Hy-Hz (b) 32–7 y, 22-y-H42 3. Perform the indicated multiplications and divisions. (a) 42%. 2a: = (e) –638.22% = (i) —2a : —30%= (b) 629-–32 = - (f) —32%--6a = (i) — 12a"--3a*= (c) 3a;+2y (g) a 9-Hay-Hy” (k) acº—acy-Hy” 32–29 - a — y a;+y (d) a-2b)a?=ab–25. (h) 2d-Hb)8a?--b% (l) a-i-b)a?–4ab–5b” 4. Finish the magic square. Space for computation 4a –H3y —a —2/ 3a;+2y 2a:-Hy CUMULATIVE REVIEW 1. Given the formula A = }h(b+b'), find the value of A when h, b, and bº have the values shown in the table below. (a) (b) (c) (d) (e) (f) (g) (h) h 4 2% 3 4 ... 5 .25 18 3# b 7 8 5 7 2.3 1.28 67 5% b/ 5 4 3 4 1.7 . 64 32 3 A. 2. Write the formula defined by each of the tables. 1 3 5 6 QC 1 2 4 5 6 (a) (b) 3 7 11 || 13 $/ 2 5 11 || 14 || 17 !/ -: 3. A and B form a partnership, agreeing that they will share the first $100 they earn equally, and that A will receive .6 and B will receive .4 of all earnings in excess of $100. Write formulas for A's share and B’s share, when they earn a dollars. (Assume that a is greater than 100.) A’s share = 4. The sum of three consecutive odd numbers is 99. Set up the proper equations and solve. 5. Ethel's mother is twenty years older than Ethel. In eight years, the mother will be twice as old as the daughter. How old is each now? B’s share = *6. Let P represent the principal, I the interest, r the rate, and n the time (number of years). Write a formula for I in terms of P, r, and n. Use this for- mula to find I, when P=$240, r=.06 and n=2. **7. The digit in the tens place of a given two- digit number is twice as great as the digit in the units place. The number formed by reversing the order of these digits is 18 less than the given num- ber. Find the given number. — 88 — UNIT 6. SPECIAL PRODUCTS AND FACTORING SIMON NEwcomb (1835–1909) Simon Newcomb became a great astronomer by making the most of every opportunity. His father was a wandering teacher of country schools. Simon wrote of his own early education, “I began to study arithmetic when I was five years old, and when six, I am told, I was fond of doing sums. At twelve I was studying algebra, and about that time I began to teach. I remember that I was thirteen when I first took up Euclid. There was a copy of it among my father's books.” At the age of nineteen, Simon found employment in the vicinity of Wash- ington, D. C. He spent every hour he could at the Smithsonian Institution. A biographer says of these hours of study, “Books of all kinds, especially those on mathematics, were eagerly sought and quickly mastered.” New- comb mentions his finding Laplace's Mecanique Celeste (Celestial Mechan- ics), and calls this book “the greatest treasure that my imagination had ever pictured.” Friends he made at the Smithsonian Institution helped him to get a better position. At the age of twenty-two, Newcomb became computer for the Nautical Almanac, with headquarters at Cam- bridge, Massachusetts. Harvard University was near. He enrolled in advanced courses in mathe- matics and studied under the inspiring teacher, Professor Benjamin Peirce. Newcomb described his work as a computer. “From these data [facts astronomical observers had recorded], he is to weigh the [heavenly] bodies, pre- dict their motion in all future time, compute their William Thompson SIMON NEW COMB orbits, determine what changes of form and position these orbits will undergo through thousands of ages, and make maps showing exactly over what cities and towns on the surface of the earth an eclipse of the sun will pass fifty years hence, and over what regions it did pass thousands of years ago.” In 1861, Newcomb obtained a better position. Abraham Lincoln commissioned him as professor of mathematics associated with naval work; he was assigned to the Naval Observatory in Washing- ton, D. C. Simon Newcomb devoted his whole life to the advancement of mathemat- ical astronomy, and he became one of the foremost scientists of his time. You may ask, “What is the value of astron- omy?” It guides ships at sea; it en- ables us to locate points on the earth's surface; it makes accurate time-keeping possible. A long list of such practical applications could be cited. But most important of all, it has helped mankind toward a clearer under- standing of the inner workings of the universe. In olden times an eclipse of the sun or the appearance of a comet was a fearful event. It was looked upon as a supernatural display which foreboded disaster. Now such a phenomenon is looked upon as just another aspect of nature's scheme of things; not so frequent as the sun's rise and setting, but just as much a part of nature's plan. This enlightened change has come to pass through the work and writings of mathematical astronomers; among the greatest of these is Simon Newcomb. MULTIPLICATION, DIVISION, AND FACTORING Multiplication is the process of finding the product of two or more factors. Division is the process of finding one of two factors when the other factor and the product are given. Factoring is the process of finding the factors when only the product is given. EXERCISES 1. Multiplication: Given the factors (r-2y) and (x+2y), the product is 2. Division: Given the product a?–4y” and one factor a –2y, the other factor is 3. Factoring: Given the product a?–4y”, the factors are and — 89 — 1. 2. 6a .32 = 18qa: 4c. —2a: = 7y .2b = THE PRODUCT OF Two MONOMIALS EXAMPLES —3b . —20 = 6ab ac. —32 = —33:2 EXERCISES 3. 56 - 2ab = 5. –32. — 42 = 4. –30, 2a:y= 6. acy. –2 = THE SQUARE OF A MONOMIAL To square a monomial, square the numerical coefficient and double each exponent. The sign of the result is always positive, for positive times positive or negative times negative gives positive. 1. 2. 1. 2. (5a)?=5a. 5a = 25a” (—a)?= (–32%)?= 6a+3 = 2a: 12b-i-6= 6qa;+2q = EXAMPLES (–32)?= —3a. —32 = 9.2% (223)2=223.2×8–42% EXERCISES 3. (–2a:y)*= 5. (52%)?= 4. (3aa)*= 6. (–acy?)?= THE QUOTIENT OF Two MONOMIALs EXAMPLES 5acy--a = 5 y * 12bacº -i- 6a: = 2ba; EXERCISES 3. 10ay”--2acy = 5. – 14s?-- – 7s = 4. – 9t?--3 = 6. 12abc-:- —3bc= THE SQUARE ROOT OF A MONOMIAL The square root of a monomial is one of its two equal factors. The square root is indicated by the sign VT. v'9a2=3a –3c. —6g= . (7ab)?= 3. (6ay5)?= 5 V493:2y2= . V}+6= —80m-:- —2d = EXAMPLES —V4a:”y4 = —2ay? b266 =bc} EXERCISES 3. – V. 93-6 = 5. V36x2y2 = 4. V 250?b% = 6. – V 64t2= MIXED EXERCISES 7. (4r2/3)?= 13. (–929)?= 8. —6r-2t = 14. V25aclo = 9. (–5ab)?= 15. 170.bc.--abo = 10. – V16cºyº- 16. (–ac)?= 11. V 1628 = 17. – ac-i- — a = 12. 16ay2%+2a:yz= 18. — ac' – a = —90— MULTIPLICATION OF A PolyNOMIAL BY A MONOMIAL EXERCISES Perform the indicated multiplications. The first has been done correctly. 1. a,(a+b) = az-Hba: 6. –2rs(1—rs-H2rs”) = 2. 2(32–4y) = 7. 3cd(–1–4cd—cºd”) = 3. a.”(2ab–5bc) = 8. 2m (a-b-c) = 4. 5m (–6a+4b) = 9. —4ab(1—a:--ac’) = 5. #ab(6–ab) = 10. 7a:”y°(acy—ac”--y”) = REMOVING A CoMMON MonoMIAL FACTOR FROM THE TERMs OF A POLYNOMIAL EXERCISES Divide each term of the polynomial by the greatest common monomial factor. The first has been done correctly. 1. aa –ay = a(2–y) 23. 9aºacº—3a*ac?--3aa: = 2. 2a:–2y = 24. 9a2b–6ab — 15b = 3. 5a-i-59–5b = 25. #am-H"mb – #cm = 4. –-aē-Ha”--a = 26. ab–bc= 5. 4a:y–8a.”y” = 27. 4cº-H6c”—2c = 6. – a 4–ac = 28. a78–7a: = 7. 7c6–7ary = 29. a”c-Ha”y—a”= 8. 27c8–96 = 30. #303—t?0.2 = 9. #acy — war-Ha!”= 31. #ary”—#acºy = 10. P-H PRT = 32. a Al-H4a”= 11. 2k+2\v = 33. — .5m —2m2 = 12. #hb-i-;hb' = 34. .3a – .62° = 13. cºa — ca.”--ca: = 35. 4ab–H 6aºbº — 10abc = 14. .2a:*— .2a:*-H.1a: = 36. 8b —64b*= 15. —4?n”a:”—H·16ma — 4a: = 37. —bº — bº—b” = 16. 7a3a;8–6aa:3+112a: = 38. — .2a:8–.2a2+.6 = 17. cº–H cº-c = 39. aa"—a'a;+aa: = 18. a 5–28-|-7a:” = 40. 2rfº-2wr?–2000: = 19. 487-2–96) — 48 = 41. 933:8–1822–1–9 = 20. 4a2+8ay-H4y”= 42. –6aºy”2°–12a:yz = 21. #an-Hºnd= 43. #24–4aa’ = 22. 52%–25ac—5 = 44. .8a”—.1a8+.16a = — 91 — THE SQUARE OF A BINoMIAL Perform the indicated multiplications. 1. a, +y 2. 2C-H.3d 3. ac-y 4. 2C-3d ac-Hy 2C-H.3d ac—y 2C-30 Thus (c-Hy)*= Thus (2C+3d)?= Thus (a —y)*= Thus (26–3d)?= As you worked the examples above, perhaps you discovered the laws: The square of the sum of two terms equals the square of the first, plus twice the product of the first and the second, plus the square of the second. The square of the difference of two terms equals the square of the first, minus twice the product of the first and the second, plus the square of the second. As formulas, these laws can be stated: (a+b)?=a^+2ab-i-b” (a-b)?= a”–2ab-i-b” EXERCISES The formula for the square of the sum of two terms is (a+b)*= a”--2ab-H b”. 1. (2-Hy)*= 11. (d+10)” = 21. (4a–H6b)*= 2. (a +a)?= 12. (8+d)?= 22. (4a2+y})*= 3. (a+3)*= 13. (3a–H1)*= *23. (2x2+3y2)?= 4. (y-H4)*= 14. (4a–Hb)*= *24. (#a2++b%)?= 5. (m+2)*= 15. (3a;+y)*= *25. (5m2+})*= 6. (r-H5)?= 16. (2n+m)*= *26. (#e3+%fº)?= 7. (c-i-6)?= 17. (3e-H4)*= *27. (5r-i-4g)?= 8. (4+a;)?= 18. (2+3a)?= *28. (10x2y+4)?= 9. (3+y)*= 19. (2a:--39)*= *29. (7h-H4a?)?= 10. (7+b)?= 20. (2C+5d)*= *30. (9-1-s})*= The formula for the square of the difference of two terms is (a-b)*= a”-2ab-Hb°. 1. (a —y)*= 9. (20–5b)?= *17. (2x2–1)?= 2. (a –3)*= 10. (8–7c)?= *18. (3a?—b%)?= 3. (y–4)*= 11. (a —2cy)*= *19. (9–c4)?= 4. (b —2c)*= 12. (3e–4f)*= *20. (2m – #)*= 5. (4—a)*= 13. (a –3e)*= *21. (3a*—#)?= 6. (10–a)* = 14. (3e—a)*= *22. (#a–3b)*= 7. (32–2)*= 15. (9–4m)?= *23. (6m2–?)?= 8. (2a–3y)*= 16. (3g–1)*= *24. (32°–1)*= —92– TRINOMIAL SQUARES The product obtained by multiplying a binomial by itself is called a trinomial square. EXERCISES Write the trinomial squares obtained by performing the multiplications indicated. 1. (a+b)”= 3. (a-b)*= 2. (a:--2y)*= 4. (2–2)*= The following facts should be true of the trinomial squares you obtained. There are two positive terms which are squares. Underscore these terms in each of the four exercises above. The remaining term equals twice the product of the square roots of the two terms mentioned above. This term may be either positive or negative. Double underscore this term in each of the four exercises above. EXERCISES Place a check mark before each of the following which is a trinomial square. 1. [ ] a 2-H2:ry-Hy” 5. [ ] 4a2–6ay-H9y” 9. [ ] a 2–122–36 2. [ ] a 2–H92- 9 6. [ ] 8b°–4b-H1 10. [ ] 64a2+y–20ay 3. [ ] a 2–12a:--36 7. [ ] 362°–12ary-Hy” 11. [ ] 4a2+16 y?–12a:y 4. [ ] 49b%+1+14b 8. [ ] 4a2+4acy-Ha” 12. [ ] 1–14ab-i-49a2b” Write each missing term below so that the results are trinomial squares. 13. 4a:*— -H 1 15. a”-- +4b* 17. a 2–H6+y+ 14. — 12ab-i-4b* 16. y”— + 16 18. – 202–H 25.2% A trinomial square may be written as the product of two equal binomial factors. To find these factors, take the square root of each of the two square terms in the trinomial and connect them by the sign of the remaining term. EXAMPLES a2+6a+9 = (a+3)? 42°–4a:--1 = (2a – 1)* The square terms in the trinomial are ac” and 9. The square terms in the trinomial square are The sign of the remaining term, 6a, is +. Thus, the 4a:” and 1. The sign of the remaining term, -4.x, binomial factor is a +3. is — . Thus, the binomial factor is 2a – 1. EXERCISES Factor each of the following trinomial squares into their two equal binomial factors, 1. a”--2ab–H b% = 10. dº — 160+64 = 2. a 2–2a:y-Hy”= 11. 49a2+14ab–H b% = 3. a 2–H62–H 9= 12. y?–20xy--100.rº = 4. 40%–4a–H 1 = 13. 25–80a–H 64a? = 5. 1+4b-i-4b*= 14. 9m?--42am-H 49a” = 6. e?--16e-H64 = *15. 9m?--1–6m = 7. 16–H 8a–H a” = *16. 25c”—50cal--25d3 = 8. 9b%–6h-i- 1 = *17. 169g?–26g-H1 = 9. 16A*–8A-H1 = *18. 121e?–154ef-i-49ſ = — 93 — TIMED PRACTICE TEST Factor the following trinomial squares. 1. a”–200–H 100 = 11. 1–20g-H 100g?= 2. 9–1–62-Ha!?= 12. a”--18q-H81 = 3. 4b2+4b-i- 1 = 13. 9a2%–42a:--49 = 4. y?–10by--25b% = 14. 25c”—60cd+36d?= 5. cºd?–4cd+4 = 15. a.”y?--64-H 16ay = 6. ac”— 12a:--36 = 16. b%+16–8b = 7. 36––12bc-i-b°C2 = 17. a 6–1–4a:8–H4 = 8. 92?–6a:-- 1 = 18. yº–2y2+1 = 9. m?–4mm-H4m” = 19. a.4+2a:”y?--yº = 10. 9–12a-H 40” = 20. —8c3+c++16= My time was FACTORING MORE DIFFICULT TRINOMIALS The product of 3m times (a+1)* is 3m (a”--2a+1), or 3ma”--6ma–H3m. The reverse process is to take a polynomial which is the product of a monomial and a trinomial square and break it up into its factors. Given such a polynomial, 2da.”--8aacy-H8ay”, one finds the factors as follows: (1) 2a(x2–4a:y-H4y”) First, remove any monomial factors. (2) 2a(a —2y)*=2a(a –2y)(2–2y) Factor the trinomial square. MIXED EXERCISES The following polynomials contain monomial factors, or trinomial squares, or both. Factor completely. 1. aa – bar = | 12. 4a:y?–8acy-H4a: = 2. a 2–H 18q-H81 = 13. abe-Hbcd—cale = .2 'FTT = 3. 3m.a.”—H·3ma;+6m 14. aºy”22–2ay22-H2°= 4. 2a2+4a:--4 = 15. 6aº–H12a3+6a = 5. 4ac”—8ac = 16. 32a2a2–800a:--50 = 6. – 7a:”—21a: = *17. ała”--2a2b%ry-Hbºy”= 7. 48b+6b% = *18. 1604–8a?--4 = 8. ab–bc—b = *19. 32d2a2+48a?bºa;+180°b% = 9. 4db”—H·8ab–H4a = 19. 32a2a2+48a?bºa;+18a. 10. 303+180°-H27a = *20. m2–3 m--# = 11. a”c?–20°C+a”= **21. 15m.3–6m2m – 217, m2 = —94 — THE PRODUCT OF THE SUM AND DIFFERENCE OF Two TERMS EXERCISES Perform the indicated multiplications. The first three have been done correctly. 1. a +b 2. 32 +2y 3. 2C ––6d.” a —b 3a –2/ 2c — 6d? a2+-ab 92%-H6ay 4c”—H·12cd.” — ab–b% –6ay–4y. — 12cd2–3664 O2 — b% 9x2 –4y” 4C2 —36d4 4. 2d-H5b 5. ab–H c 6. 32+2y2 2a – 5b ab–c & 32–2y2 In each of the exercises above, one of the factors is the sum and the other is the difference of the same two terms. The following conclusion can be drawn as to the product: The product of the sum and difference of two terms is equal to the difference of their squares. EXERCISES Use the formula (a+b)(a-b) = a”—b” to find the products. 1. (ax+y) (a — y) = 17. (3ry–4b*) (3ry-H4b*) = 2. (a+3) (a –3) = 18. (cºd-H6m*) (cºd–6m*) = 3. (m+7) on-º- 19. (1+2") (1–22) = 4. (p+9) (p-9) = 20. (32–2y) (33.--2y) = 5. (r-H10) (r-10) = 21. (6aw-Hy) (6aac – y) = 6. (8+b) (8–b) = 22. (6a+b) (6a–b) = 7. (11+c) (11–c) = 23. (r-12s) (r-H12s) = 8. (m-2a) (m+2a) = *24. (cº-i-y”) (c.”—y”) = 9. (6–b) (6+b) = *25. (aft-Hy?) (23–y}) = 10. (20–c) (20-Hc) = *26. (c2a+y?”) (cºe—yº) = 11. (m+5a) (m. – 5a) = *27. (2a2+1) (2a2–1) = 12. (4a:--y) (42–y) = *28. (23-1-y?) (23–yś) = 13. (6a+7y) (62–7 y) = *29. (32–2y) (2/-ī-3a) = 14. (a.”--y) (c”—y) = *30. (mºn?--g?) (m?n?–q2) = 15. (2x3-H1) (22°–1) = *31. (;a;++y}) (#3:} – }y}) = 16. (1–y”) (1+y}) = **32. (#a3++b+) (#ał–4b*) = — 95 — FACTORING THE DIFFERENCE OF Two SQUARES The difference of two squares may be factored by taking the sum of their square roots as one factor and the difference of their square roots as the other factor. The algebraic expression for this is a”—b% = (a+b)(a-b). EXAMPLES 1. Find the factors of 42°–25g”. The square root of the first square is 2a and the square root of the second square is 59. Therefore the factors are 2x+5y and 22–5). 2. Find the factors of a 4–y". The square root of the first square is ac” and the square root of the second square is y”. Therefore the factors are ac”—Hy” and ac”—y". Factor. EXERCISES 1. a”—b% = 12. 9r?–4s?== 2. a.”—y?= 13. 1 — 16a’b%= 3. a 9–9 = 14. 4m.”—9n” = 4. 16 – c’ = 15. 49 — a 2b% = 5. r?— 100 = 16. 1672–25s?= 6. d?— 121 = 17. 9m2n.2 — 250?b2 = 7. 9— w?= *18. .04a:2–.09b?= 8. 92?–y?= *19. #C2–#d?= 9. 49e?—fº = *20. A*—Tº B*= 10. y?–36a”= *21. e4—fº- 11. 64e?–4q2 = *22. 3 sa”—#b% = MIXED EXERCISES Factor each of the following. 1. az-Hab = 10. 4m?--4mm-H m?= 2. ac”—y?= 11. 1 — 16m-H64m”= 3. c2–200+-d” = 12. 25—t?= 4. 3aac-H3ay = 13. R”—# = 5. m?–H2m n+m2 = 14. 36–12B-H B2= 6. 44%–9y4 = 15. 3aac-H6a?ac”—H·9aºz8 = 7. a”ac2+-ac2 = 16. a.”y?–9 = 8. 1622–1 = w 17. a2b%2+b^c”—H·c?d= 9. 4m?–4 = 18. a”b?–H2abc-—c?= — 96 — THE PRODUCT OF Two BINOMIALs EXAMPLES 1. a +3 Observe the following: a — 4 (a) The product is made up of three terms. ac2+3a. (b) The first term of the product is the product of the first terms of the given bi- — 4a – 12 nomials. ac”— a -12 (c) The second term of the product is the sum of the two cross products (3a; and —4a). (d) The last term of the product is the product of the last terms of the two given binomials. The example above can be written as shown at the right. The cross product terms are represented by the arrows. (£160-Y)-2-4-12 2. (a –5)(a —6) = azº (product of the first terms) – 11ac (sum of the cross products)+30 (product of the last terms). 3. (22–3)(32–2) = 64!” (product of the first terms) – 13a; (sum of the cross products)+6 (product of the last terms). EXERCISES 1. (a +1) (ac-H2) = 2. (c-H1) (a +3) = 3. (a +3) (a;+4) = 4. (a +5) (ac-H6) = 5. (a;+2) (a +7) = 6. (a – 1) (a —2) = 7. (a –3) (a —4) = 8. (a —6) (a —7) = 9. (a —2) (a – 11) = 10. (a —9) (a — 10) = 11. (c-H1) (c.—2) = 12. (c.—1) (c-H2) = 13. (y—4) (y–H5) = 14. (3+a) (2–a) = 15. (5+y) (6+y) = 16. (y–9) (y-H 11) = 17. (3—a) (2—a) = 18. (4–b) (3–b) = 19. (6–2) (7+ r) = 20. (a+3b) (a+4b) = 21. (ax+4) (aa – 3) = 22. (3c-H1) (4c—3) = 23. (a+3b) (a —2b) = 24. (2–12) (c.—10) = 25. (r–9s) (r-i-10s)= 26. (u-H3v) (w-7v)= *27. (4a–H5) (2a–1) = *28. (3b-i-4) (45–5) = *29. (2a–H3b) (3a–H2b) = *30. (2:c-3c) (52–7c) = PIERRE DE FERMAT (1608–1665) Fermat was a lawyer and a holder of public of- fice, with algebra as his hobby. He had become in- terested in the subject through reading some of the works of Diophantus. Fermat made a number of mathematical discoveries, which he stated (without proofs) in the margins of his books. Although mathematicians have worked on it for almost three centuries, one of the statements Fermat made has not yet been proved. Given a 2–Hy” = 2*, we can find any number of sets of values for ar, y, and 2 which satisfy the equation. Examples are: a =3, y = 4, 2 = 5 or a = 5, y = 12, z= 13. Given ac"+y”=2”, where n is greater than 2, no sets of integral values have ever been found to satisfy the equation. Fermat said there were none, and this statement is known as “Fermat's Last Theorem.” It is still one of the unsolved problems of mathe- matics. QUADRATIC TRINOMIALs When two binomials, as a -i-2 and a H-3, are multiplied together, their product, a 2+52+6, is known as a quadratic trinomial. On this page, you will learn how quadratic trinomials are built from binomial factors, and how they are broken into binomial factors. EXERCISES Multiply. 1. a +3 2. a -3 3. a +2 4. a -3 a;+2 ac-H 5 a – 7 a – 7 Each of these products is a quadratic trinomial. Observe that in each the first term is acº, the middle term is the algebraic sum of the second terms of the binomial factors times a, and the last term of the tri- nomial is the algebraic product of the second terms of the binomial factors. Use these observations to write the following products. 5. (a +3) (a +4) = 9. (2-H2) (ac-H6) = 6. (a –3) (a +4) = 10. (2–2) (ac-H6) = 7. (a +3) (a —4) = 11. (a;+2) (a —6) = 8. (a –3) (a —4) = 12. (c.–2) (a —6) = Be sure you are able to explain each term and each sign in the products you have written. The reverse of the problem presented in the exercises above is to find the binomial factors when the product (quadratic trinomial) is given. In many cases this can be done by inspection. With the binomial factors of exercises 5–8 covered, study the products you wrote. By reversing the reasoning you used in finding the products, you should be able to determine the factors. The inspection method of breaking a quadratic trinomial into its binomial factors is: (a) Make blank parentheses as are shown in the examples below. (b) Inside these parentheses build up the binomial factors whose product is the given trinomial. EXAMPLES 1. We select a as the first term of each binomial fac- tor. This checks, for a times a is wº, the first term of the ºf 9r-F20=( ) ( ) trinomial. The product of the second terms of the binomials must equal 20. Thus, these terms could be 1 and 20, or 2 and 10, or 4 and 5. However, since their sum times a must equal the middle term of the trinomial, 9a, we must select 4 and 5 as the second terms of the binomials. 2. The first term of each binomial is ac. The product of the second terms is – 10. Thus, these second terms could a 2–33 – 10 = ( ) ( ) be — 1 and 10, or 1 and — 10, or –2 and 5, or 2 and – 5. Since their sum must equal –3, the coefficient of the middle term of the trinomial, we must select 2 and –5 as the second terms of the binomials. In the blank parentheses above, write the binomial factors of the given quadratic trinomials. EXERCISES Break the following quadratic trinomials into their binomial factors. 1. ac”—H 102–H 16 = 3. a.”—4a – 12 = 2. ac”—7a;+10 = 4. ac”—H·63 – 16 = — 98 — EXERCISES When two binomial factors are given, find their product; when a quadratic trinomial is given, find its binomial factors. - 1. (a +4) (a;+6) = 11. ac”—92–H 18 = 2. (a —3) (a;+7) = 12. a 2–3a – 18 = 3. ac”—ac— 12 = 13. ac”—H·3a – 18 = 4. ac”—Ha: – 12 = - 14. a 2–H92- 18 = 5. ac”—H·4a: — 21 = 15. (a – 1) (a —8) = 6. 29-–32–10 = 16. (ac-H3) (2–5) = 7. (ac-H2) (a —5) = 17. (2–3) (a —7) = 8. (ac-H5) (a —2) = 18. a 2–H 10a;+24 = 9. (ac-3) (2–6) = 19. ac”—a – 20 = 10. (ac-9) (a;+3) = 20. (a —3) (a +3) = EXERCISES Find the binomial factors. 1. ac”—H·3a;+2= 19. 3%–2–90= 2. **º-s- 20. r?--3r—28 = 3. a”--5a–H6 = 21. f*–13f–H36= 4. m?--7m-H 12= 22. g”–8g–20= 5. ac”—3a;+2= 23. 22–H 102 — 11 = 6. c”—6c—H.8 = 24. a 9–62 – 7 = 7. m”—5m-i-6 = *25. rº–H4a:y--3y2= 8. p?–7p-H 12= *26. a”— 7ab – 18b2 = 9. ac”— 12a:--36 = *27. y”—ay–12a” = 10. b%— 10b-i-16 = *28. a7+19.xy—20y?= 11. d2–9a–H20 = *29. c2 – cal–12d2 = 12. e”— 15e–H 14 = *30. p?--5pa;+63:?= 13. b%–3b —4 = *31. d2+-de-20e2 = 14. d”—3d— 18 = *32. q*-ī-7mq+12m2 = is. g”—y–20= *33. wº—3ry-H2y2 = 16. y”-Hy—20= *34. a 2–ac-i-4 = 17. a 2–Ha:–2= *35. a 4–H4+2+3 = 18. a”--9a–10 = *36. b3–3b3–4 = —99 — *FACTORING MORE DIFFICULT QUADRATIC TRINOMIALs Be able to explain the reason for the value and the sign of each term in the quadratic trinomials in ex- ercises a-d. Write the quadratic trinomials in exercises e—h. (a) (2a–H3) (3a;+2) = 64!?--13a;+6 (e) (52+3) (3a;+2)= (b) (2a–H3) (32–2)=63:?-- 52–6 (f) (52–3) (3a;+2) = (c) (22–3) (3a;+2)=63:?– 5a–6 (g) (5a-H3) (32–2) = (d) (22–3) (3r—2)=6a’–13a;+6 (h) (52–3) (3r—2) = FACTORING QUADRATIC TRINOMIALS BY INSPECTION Example: 62%–25a;+4 = ( )( ) Fill the blanks. From the first term of the trinomial, one reasons that the first terms of the binomial factors may be 6a: and ac, or 3a and 22. From the last term of the trinomial, one reasons that the second terms of the binomial factors may be 4 and 1, -4 and – 1, 2 and 2, or –2 and –2. These must be selected and combined in such a way that the cross product term is —25a. By mentally listing various possibilities and excluding those which do not fit, one finds the binomial factors to be (62–1) and (a —4). EXERCISES 1. 2a:”—H·7a;+3 = ( –H1) ( --3) *7. 3c2–4c — 4 = 2. 22%+5a–H3 = ( –H1) ( --3) *8. 6y?--y–5= 3. 2a:”–7a;+6= ( –3) ( –2) **9. 922–1–302–1–16 = | . 2a:”–7a;+5 = ( – 5) ( – 1) **10. 642-1-13ay-H6/?= 5ac--2= **11. 2a2+-5aac —33:? = *6. 23:?--a – 6 = **12. 6142–Hwy—50% = | The quadratic trinomials given above are of the form Aa2+Bæ--C, or Aaº-HBay-HCy”, where A, B, and C represent the numerical coefficients. For every trinomial we have factored, the quantity Bº-4AC has been a perfect square. This test as to whether or not a quadratic trinomial can be factored should be learned: If B2–4AC is a perfect square, the trinomial can usually be factored by inspection; if Bº-4AC is not a perfect square, the trinomial cannot be factored by inspection. PROBLEMS USING FACTORING 1. The radii of two circles are 16 and 5. Find the difference in their areas. Solution: The areas of the circles are T162 and 15°. The difference in their areas is T16?– TB* = Tr(16°–52) = Tr(16+5)(16–5). Use 4” as the value of T. Then the difference in the areas of the circles is 4%(21)(11)=726. *2. The radii of two circles are 10 and 4. Find the difference in their areas. *3. Four circular holes, each with radius .3 inch, are stamped out of a circular disc whose radius is 3.4 inches. Find the area of the disc that is left. Hint: Use tr(R*—4r”). — 100 — ExAMPLES OF ExPRESSIONS HAVING MoRE THAN Two FACTORS ... a 4–2 = a (22–1) = a (x+1)(a:–1) . a”b-H4ab-H4b = b(a”--4a–H4) = b(a+2)(a+2) ... a 4–1 = (a,”--1)(x2–1) = (a,”--1)(x-H1)(x-1) . 4a/b2+20ab?–96b%=4b*(a^+5a–24) = 4b*(a+8)(a –3) : The polynomials above have been factored completely. To factor completely means to break the poly- nomial into its prime factors, with the exception that monomials are to be left in the form 4b* (above) rather than 2. 2. b. b. A number or a literal expression is prime if it cannot be factored. In arithmetic such numbers as 3, 5, 7, 11, 13, 17, . . . are prime, and such numbers as 4, 6, 8, 9, 10, 12, . . . are not prime. In algebra such ex- pressions as a +y, a 2+y”, a 2+a;+1, . . . are prime, and ac”—y”, a 2+3++2, . . . are not prime. SUGGESTIONS FOR FACTORING COMPLETELY I. Look for a common monomial factor. If there is one, divide by it. II. If the quotient is a binomial, is it the difference of two squares? III. If the quotient is a trinomial, is it a trinomial Square, or may it be factored by the cross-product method? IV. Be sure to write all the factors in the final result. EXERCISES Factor completely. If the expression cannot be factored, write “prime” after it. 1. a”— a = 21. 20%–H 12a+18 = a” — a = 22. 20:8–10a2+12a: = . 169°–1 = 23. d?--d-- 1 = . 4c.—4cd}= 24. 32b%–48b+18 = 2a:3—24: = 25. aa’--8aa’y-H16airy” = 26. b%+4ab–45a?= C 5 * C := . 9a2b3–16bé = 27. ar”—4a:--5 = ... ac”—H·1 = 28. a”ac”—H·4a2a: —4aa’ = ... ab?–Ha*b = 29. 4m.”— 16m – 20 = 10. yº–25= 30. a.”y?--6xyz+92* = 11. a 4–16 = 31. 52%–H3ry-H2y2 = 12. a 4-H.32 = 32. a 4-i-2a2+1 = 13. a 4–4a: = 33. a 4–1–2a:8–Har?= 14. 32a: — 23:8 = 34. yº-H7y?--12y = 15. 50°b?–5c2 = 35. a 3–74 —H·8 = 16. a.”y–25) = 36. cº–8c”-- 10c = 17. 22°y–32y = 37. a 2–13a;+42= 18. 36–y?= 38. a+–15a-2–H56a = 19. 9–ac”y?= 39. 14—92-Haº = 20. 1622–25 y?= 40. 20–8a; – ac” = — 101 — You have now used the following special products: (a) The square of the sum of two terms (b) The square of the difference of two terms (c) The product of the sum and difference of two terms (d) The product of two binomials (using cross products) A REVIEW OF SPECIAL PRODUCTS EXERCISES Within the square before each of the following, place a, b, c, or d to indicate the type; then find the product. 1 2 3 4 5 6 7 8 9 10 1 1 12 ... [] (r-i-2y)*= ... [T] (2-H2a) (2–2a)= ... [T] (r-i-4) (2–2)= . [] (3a–2b)?= ... [T] (3m-H4n) (3m–4n)= . [] (22–1) (r-1-2)= ... [I] (22–3c)?= ... [I] (2d—7) (2d--7)= . [] (1+4c)?= . [] (m+35) (m–1) = . [...] (36–7aa) (3c--7aa)= ... [I] (7–6bc)*= 13. [T] (1+2cy) (1–2xy)= 14. [...] (3m–10n)?= 15. [] (q-7) (24+5)= 16. [T] (32?--2) (322–2)= 17. [I] (2n+5) (n–7)= 18. DT (2bc-11a)*= 19. [I] (b?--7) (b?–71)= 20. [I] (cº-y”) (cº--yº)= *21. [T] (#ab-i-4cd)?= *22. [ ] (++-abc) (#–abc) = *23. [I] (.1b-i-4) (.1b–4)= *24. [] (32–3)?= TIMED PRACTICE TEST You should be able to multiply the following accurately and rapidly. If you cannot, you should review. 1. 2. 3. 10. (3a-Hb)*= (2–H3y) (2–3y) = (y—6)*= . (a+6) (a+3) = . (2C-5d) (2C+5d)= . (2e –36)*= . (c-H4d) (c.—2d)= . (6b – 1)*= . (32-H4)?= (3ry–2)*= 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. (2–10) (2-H 9) = (92–5y) (92–H59) = (2C-3) (c.—4) = (e?–2ſ)?= (4a:--3b)?= (ayz+1) (a:yz–1) = (3n+11m)*= (5–86) (5-H8c) = (9g–10h)?= (2a–7b) (5a-H6b) = My time was — 102 — A REVIEW OF FACTORING You have now used the following types of factoring: (a) The common monomial factor (b) The trinomial square (c) (d) The difference of two squares The quadratic trinomial Within the square before each of the following, place an a, b, c, or d to indicate the type; then factor. 1 2 3 4 7 8 9 10 11 12 ... [T]3+2+3+ = . [I] w”--8w-i-16= . [] a”--5a-H6= . [T]4b2–20bc-1-25cº- ... [I] 4cº-9d” – . [I] 16cº–24cd+9d?= . [...] 1–9m?= ... [T]92%–wy= . [ ] b”—3b–54= ... [T]49a2–952– . [T] rº–7r-18 = . [T]64cº–20cd+d?= 13. [I] 162°y–24ty” – 14. D. 81e4+72e--16= 15. [T] c”—3c.— 28 = 16. D25a?–1= 17. [T] a’--acy—20y”= 18. [] a2+-a8–aft = 19. [T] 49m?--70mm-H25m?= 20. [T] a”–12a–85= *21. [] 162°y°–40aay-H25a”= *22. #3:6–9– *23. [I bº-Šb—#= *24. […] 12a3–5a–2= TIMED PRACTICE TEST You should be able to factor the following accurately and rapidly. If you cannot, you should review. 1. 10. 7a.2b–21qb” = . 9a2+-6ab–H b% = . 9a.”—y?= ... y”— 12y+36= a?--9a–H 18 = . 4c”—256*= 4e?–12ed--9d” = . c2–260–80% = . 36b?–H 12b-i- 1 = ac”—H·8a;+16= 11. ac”—a —90 = 12. Tr?--Trh = 13. 81a.”—259% = 14. 1622–24ba;--9b% = 15. a.”y”2?–1 = 16. a 2-H42–H3 = 17. 121 — t? = 18. 2.c4–62°y?= 19. P-H Pri = 20. a”— 11a–H28 = My time was — 103 — INVENTORY TEST 1. Factoring is the reverse of 2. In multiplication we have the two factors given and we find the OT (a +6)(a – 11) = 3. In division we have the product and one factor given and we find the other , Or (a.”—y”) + (a —y) = 4. In factoring we have the re given and we find the - * , Or 2°–ac – 12 = - 5. (a) (42%)?= (c) (–6+)-- (e) Gº- (b) (–2a)*= (d) (–3a*b)?= (f) (.32%)?= 6. Find the area of a square when one side is: (a) a +2 (b) 22–4 (c) 3a–2b 7. In squaring a binomial one obtains a whose middle term equals the product of the and - terms of the binomial. 8. Check each of the following which is a trinomial square. (a) ac”—H·8a;+9 (b) 25–80a-i-64a.” (c) c”–4c-H9 9. Find the areas and perimeters of the rectangles whose bases and altitudes are given. Base Altitude Area Perimeter (a) 3a;+2y 32–29 (b) d–H6 d—7 (c) 2. +3 3v – 1 (d) 2m+1 2m+3 (e) 6a – 7 - 6ac—7 10. If the area of a rectangle is a "+62+8 and its base is a -i-4, its altitude is 11. If the area of a rectangle is 42°–9, its base may be — and its altitude — 12. (a) (32–4) (2a–H5) = (b) (62–7 y) (92-Hy) = 13. (a) (c2+10a;+25)+(a +5) = (e) (22–36) + (2–6) = (b) (cº--7a;+12) + (x+4)= (f) (4a:”–9y”) + (2a–H3y) = 4–4a–Ha” ac”--a;+} () ---- - @-T-- 402b:3–603b2 # – 53:-Ha!” G2b% 2} – 3: — 104 — CUMULATIVE REVIEW . Subtract. a 2–2a:y-Hy” 3.c2–4y 6a–27 –4acy-Hy”–4 —62%--7y--82 — 170 — 27 . Subtract by removing parentheses and combining like terms. 7t–(6t-H2) = 3a–7–(–7a-H3) = . Multiply. (–3a.”y)(+4a:*) = (–6a’b–7b3)(a^b%) = . Divide. 802b% –72°y –25a5b” abºº gy. T –5aºbº . Using parentheses when needed, write algebraic expressions for: 4 less than a – 3a–H4b multiplied by 4a a less than 4 — 4 less the sum of a and b The sum of m and n multiplied by their difference . If a = 2 and y = —3, find the value of: a 3-H2a:”y-Hyº- (2a–Hy)(22–y) = . Write the coordinates of the following points: 2 units to the right of the y-axis and 6 units below the z-axis. 5 units to the right of the y-axis and 3 units above the 2-axis. 4 units to the left of the y-axis and 9 units below the ac-axis. . Write in their simplest form: a2+2a2–7a?= c2 . C3 = c-Hc-Hc—d—d = (z+y) (c-Hy) = a a a a y : y = — 10m?--5m?= — 105 — UNIT 7. LINEAR EQUATIONS IN ONE UNKNOWN, AND PROBLEMS JAMES CLERK MAXWELL (1831–1879) The rays which cause light, the rays which carry radio messages, and the rays of the X-ray are all bound together in the same mathematical equa- tions. These equations were set up by James Clerk Maxwell many years be- fore any one had ever heard of a radio or an X-ray. Physicists have said that our whole universe is immersed in a something which they call the ether, and that waves of different lengths continually race everywhere in this ether. Some of these waves cause light. No one has ever seen the ether or an ether wave, but since the beginning mankind has known light and darkness. The ether was invented to explain, and Maxwell set up equations to show, how light travels. Max- well's equations did far more than merely describe the motion of light. He read from them the possi- bility of ether vibrations whose wave lengths are different from those of light. Hertz, a German JAMES CLERE MAXWELL, physicist, set up apparatus and discovered experi- mentally longer waves than those of light. Maxwell had predicted these. Marconi showed how these longer waves may be used to carry ra- dio messages. The X-ray, used by the scientist, surgeon, and dentist, to see through opaque substances, also obeys Max- well's mathematical equations. The wave lengths of the X-ray are less than those of light, but the waves move in the same ether according to the same law as do light waves and radio waves. Within comparatively recent years, man's life and more especially his way of making a living have been revolu- tionized by science. The development of the radio and the X-ray are typical of the ad- vances made in many fields. The advances have been the joint work of the mathematician, the ex- perimental scientist, and the inventor. Every Mar- coni has been preceded by a Hertz and a Maxwell. LINEAR EQUATIONS IN ONE UNKNOWN A linear equation in one unknown is an equation which, when reduced to simplest form, has only one unknown and the unknown is raised to the first degree. Examples: 4-Ha = 6; 3a = 12; }a = 6; 7/- 1 = 13. EXERCISES Place a check mark before each of the following which is a linear equation in one unknown. 1. 3a – 1 = 2 5. a.”-Ha = 7 2. a 2 – 7 6. 8a;+4a: = 12 3. a-H2y=6 7. w–62 = 16 4. ac-H4 = 24 8. 32 = 2–4 9. 4r–H3S = 8 10. —6m-H2 = 4 11. 5m — 7 m = – 6 12. a.”--y”=4 The equation cº-i-44 = 42–10 does not look like an equation of the first degree in one unknown, but when it is simplified by subtracting arº from each member, we see that it is such an equation. — 106 — SOLVING LINEAR EQUATIONS The Addition Law: The same number may be added to each member of an equation without destroying the equality. Use the addition law to solve the exercises. 1. ac-7 = 3 2. ac-4 = —2 3. a -1 = – 7 The Subtraction Law: The same number may be subtracted from each member of an equation without destroying the equality. Use the subtraction law to solve the exercises. 1. ac-H4 = 9 2. a +3 = –1 3. ac-H 1 = – 5 The Multiplication Law: Both members of an equation may be multiplied by the same number without destroying the equality. Use the multiplication law to solve the exercises. 1. a = 4 2. #2 = — 3 3. # # # (U = 6 The Division Law: Both members of an equation may be divided by the same number without destroying the equality. Use the division law to solve the exercises. 1. 32 = 9 2. 5a = –10 3. – 2a: = 4 It is usually easier to solve an equation in the form 32–2=4 than in the form 4 =3a –2. This suggests the need of a new law: The members of an equation may be interchanged. The truth of this is evident if we think of the equation as scales; the places of the weights may be interchanged without destroying the balance. Change the following equations so that the unknown letter will appear in the first member. 1. 4 =b+2 3. 11 = 2b–7 5. 5 =4y–7 2. – 6 = 2a: 4. 16 = }ac-H2 6. 10 = 2+16 In solving an equation, it may be simplified to the form —a =3. This is not finished. To solve is to find the value of +2. This is done by multiplying or dividing both sides of the equation –a =3 by — 1. Given the equation, –32 = 6, one divides each member by to solve for +2. Given the equation, – #2 = 5, one multiplies each member by to solve for +2. EXERCISES Solve, indicating what you do in each step. The first has been done correctly. 1. 5a — 7 = 5-H3a. 2. 2.):--5 = 3a – 5 3. #2+1 = a – 1 2a – 7 = 5 (S3a) 2x = 12 (A7) a = 6 (D2) The goal to work toward is to be able to solve mentally simple linear equations in one unknown. — 107 — EXERCISES Solve mentally or on the margin of the page. Write your answers and checks in the spaces provided. Equations Answers Check 1. ac--11 = 7 2. 15 = y–H21 3. 22–4 = — 10 - 4. 25 = 17—w 5. 4h — 14 = 10 6. m = 9m-H 64 7. 6k-H9 = — 15 8. 2a:–H3 = 4a – 5 10. To ac-H 1 = 0 11. 0 = 3m-H 6 12. 99–2=2g–4 13. ——H·8 = — 11 14. y–7=8–y 15. m—# = 2m – } 16. a. —3% = 4a – 1% 17. #b-H 10=0 18. 19pm–H7 = — 31 19. 3b — 4 = 2b–7 20. 2b–4 = —4% *21. 3.5c+27 = —8 *22. —4+}a = —#a *23. .1ac – .4 = .6 *24. 4.3y–2.2=15 *25. .62+..6= 6 — 108 — PROBLEMS Solve the following problems and check. 1. Six times a number equals the number in- creased by 15. Find the number. Check: 2. Mr. Smith's electric light bill for April was $1.28 less than for March. The bill for both months was $8.68. Find the bill for each month. Check: 3. A rectangular lot is three times as long as it is wide. Its perimeter is 392 feet. Find its length and width. Check: 4. The perimeter of a triangular lot is thirty-six rods. The second side is twice the first less four rods and the third side is three times the first less eight rods. Find the sides. Check: 5. Find three consecutive numbers whose sum is 171. Check: 6. The sum of two angles is 180 degrees, and one is six degrees more than half the other. Find the angles. Check: — 109 — PARENTHESES Parentheses preceded by a plus sign may be removed without changing the sign of any term inside the parentheses. Recall that when no sign precedes a number or a parenthesis, the sign is understood to be plus. Remove the parentheses below. 1. (+3a –2) = 3. (–52-H2) = 2. –H (52+3) = - 4. (–2a –6) = Parentheses preceded by a minus sign may be removed by changing the sign of each term inside the parentheses. Remove the parentheses below. 1. – (+2a:–3) = 3. – (–5a-H2) = 2. – (3–2a) = 4. – (– a –4) = When a parenthesis is preceded by a number, each term inside the parentheses should be multiplied by the number when the parentheses are removed. The first and fourth exercises below have been done correctly. The student will do the others. 1. 3(2a –4) = 63 – 12 - 4. –2(2–a) = —4+2a: 2. 3(–4a – 6) = 5. – 4(2a:--5) = 3. 5(2x+10) = 6. –3(–3a – 5) = EXERCISES Remove parentheses and combine like terms. The first has been done correctly. 1. –2(a+b) –7(–3a–H2b) = —2a–2b-H21a–14b = 190–16b 2. 7(a:–3)–2(a;+6) = 3. —7+3a–H3(a+6) = 4. –2(a:—y)–6(2a –4y) = 5. 7c.—3–2(3-H2c) = *6. a.”— (a;+3)(a —3) = *7. (a +3)(a —2)—3r”--6 = *8. 3a(a+3)–3(a”--2) = *9. —6c”—3(2c-H3)(c.—2) = *10. (a+2)(a —2) — (a+3)(a —3) = *11. – 11a:(ac—3)+3+(a —2) = EQUATIONS OF LONG AGO Ahmes, the Egyptian scribe, probably wrote his one of his problems reads: “Heap its half, its mathematical papyrus about a century before the whole, it makes 16.” The word heap was used to birth of Moses. In this mathematical writing, we indicate the unknown. Translated into an algebraic find a number of simple problems and simple equa- e 70, - tions stated and solved. Translated into English, equation the statement becomes 2 +m = 16. — 110 — SOLVIN G EQUATION S CONTAININ G PARENTHESES EXAMPLES Study carefully. State exactly what is done in each step. 1. (y–3)(y–4) = y”–2 g”–7y--12=y?–2 –7y-H12= —2 –7y = –14 g = 2 Check: (2–3)(2–4)=2°–2, or (–1)(–2) = 4–2, or 2 = 2. 2. 23:2–2(a;+3)(x-H4) = 4 22°–2(~2–H7a;+12) = 4 2a:”—2a:”— 14a – 24 = 4 – 14a – 24 = 4 — 14a: = 28 a = — 2 Check: 2C-2)” –2(–2+3)(–2+4) = 4, or 204)–2(1)(2) = 4, or 8–4 = 4, or 4=4. EXERCISES Solve and check. Do as much as you can mentally. 1. a, –3(a;+4) = 2 4. 3(y–2) = 2(y–6) 2. 7m – 28 = – (m-4) 5. 3–3(c–2)=0 3. –2(c.—2) (c-i-3) = -2a:” 6. (q+2) (q+3)–(q—2) (q—3)=0 — 111 — EXERCISES Solve and check. 1. 6v°–2(3v-2) (w-)=0 5. (y-H9) (y-H5) =y?--3 2. 4(62-H2)–3(7a;+3)=0 6. (2+1) (a +2) = (2–1) (2–3)+2} 3. (c-H3)”— (a;+4)”= –14 7. y”--3– (y-H1) (y-H2)=0 4. (3b-H1) (b–1)–3b*=0 • *8. (2–2)?— (2+1)*= —21 — 112 — USING PARENTHESES IN SETTING UP EQUATIONS IN PROBLEM SoLVING COMPLEMENTARY AND SUPPLEMENTARY ANGLES Two angles are complementary if their sum is 90° (a right angle). Angles a and b are complementary, each being the complement of the other. Two angles are supplementary if their sum is 180° (l (a straight angle). Angles c and d are supplementary, b each being the supplement of the other. c / d 1. Write the complement of each of the angles listed: (a) 10° (b) 20° (c) 45° (d) 1° (e) wº (f) 3y° 2. Write the supplement of each of the angles listed: (a) 90° (b) 45° (c) 150° (d) 19° (e) wº (f) (y-H2)* EXAMPLE Let a = the number of degrees in the angle. Then 90—a =number of degrees in its complement, and 2090–ar) = number of de- grees in twice its complement. Then a = 2(90–a) = 180–2a. 3a = 180, or a = 60 = number of degrees in the angle. Find the angle which is equal to twice its comple- ment. EXERCISES 1. Find the angle which is equal to three times its supplement. 2. Find the angle whose complement and sup- plement together equal 120 degrees. 3. The supplement of an angle is equal to three times its complement. Find the angle. 4. A certain angle less its complement equals ten degrees. Find the angle. 5. An angle less four times its supplement equals five degrees. Find the angle. 6. The supplement of an angle plus twice its complement equals 120 degrees. Find the angle. — 113— SOLVING EQUATIONS HAVING FRACTIONAL CoEFFICIENTS EXAMPLES Study carefully. State exactly what is done in each step. 1. +++ - 25 2 3 Q: Q. *(i)+(−)-sº 3.c-H2a: = 150 5a = 150 a = 30 checkº-25, or 15+10=25, or 25 = 25 2. .3a;+.4a: = 1.4 3a;+ 4a: = 14 7a: = 14 a = 2 Check: .3(2)+.4(2) = 1.4, or .6+.8 = 1.4, or 1.4 = 1.4. EXERCISES Solve these equations. You need not write the explanation of each step. 1. +4-6 3. .3s = .7–.2s 2 2 H-3, 4. .39–4 = 2 QU) QU) = –1 3 4 6. .5(c.— .3) = .75 SIMON STEVIN (1548–1620) No one person can be given credit for having invented decimal fractions. Reference has been found to their use as early as 1492. Simon Stevin was the first to discuss the underlying theory. In 1585 he published a pamphlet with the title, La Disme. He introduces it like this: “Teaching how all computations that are met in business may be performed by integers alone with- out the use of fractions. Written first in Flemish and now in French, by Simon Stevin of Bruges. “To Astrologers, Surveyors, Measurers of Tapes- try, Gaugers, Stereometers in General, Mint-mas- ters, and to All Merchants, Simon Stevin Sends Greeting.” — 114 — Solve and check. 1. *4-6-0 4 EXERCISES 2 6. 30–33– 4 9. — ac— 1 = 6 5 $/ , !/ , !/ *10. —–H––––– = — 13 2 3 4 ac-H2 ac-H3 1 g–H–- *11. 2. *12 *, * 14 ... ——H·- = QL) — 3 4. Q. *14. 5a — 19 = — 2 *15. .152 = .25(2–4) — 115 — USING FRACTIONAL EQUATIONS IN PROBLEMS 1. A radio was marked down # and then sold for $60. What was the original price? Check: 2. Mr. Brown sold a house for $5,600 and made a gain of 25%. What did the house cost him? Check: 3. When asked his age, James replied, “The dif- ference between my age and # my age is 3.” Find James’ age. Check: 4. A farmer planted # of his land in potatoes and. # in corn and had 20 acres left. How much land had he'? Check: 5. One third of a certain number is 4 more than one seventh of it. Find the number. Check: 6. One sixth of a number is three less than one fourth of the number. Find the number. Check: — 116 — SoLVING LITERAL EquaTIONS The equations we have solved thus far have had only one letter, which we have called the unknown. In Solving these equations we found the value of this letter, as a positive or negative whole number or fraction. In Unit 1 we studied formulas which were made entirely of letters. We shall now learn to solve such an equation (formula) for any letter in terms of the other letters. EXAMPLE Given the height and base of a rectangle, we can find the area by the formula, A =bh. But if we have the area and base given and wish to find the height, this formula does not apply so easily. We can divide through by b, thus solving the equation for h. The resulting equation is: h = A/b. In this, h is the unknown, for we know the values of A and b. An equation in which the known quantities are represented by letters is called a literal equation. Given the formula A = hb, explain how you would solve for b in terms of A and h. EXERCISES Write the appropriate formula (A =bh, or h = A/b, or b = A/h), substitute the values given for the knowns, and determine the unknown. Given Formula Make substitution here Answer 1. h = 5, b = 4 2. A = 30, b = 6 3. A = 32, h = 4 4. h = 1.5, A = 9 DERIVING NEW FORMULAs 1. The formula for the perimeter of a rectangle when b is the base and h is the altitude is P=2b-i-2h. (a) Solve this formula for b. (b) Solve this formula for h. 2. The formula for interest in terms of principal, rate, and time is I = pri. (a) Solve this formula for p. (b) Solve this formula for r. — 117 — SELECTING AND EVALUATING FORMULAS Write the appropriate formula (P-2,+2, or b = P–2h P–2b for the known letters, and determine the unknown. , or h = 2 ) Substitute the values given Given Formula Make substitution here Answer 1. P= 16, h = 3 2. P=26, h = 4 3. P=32, b = 5 DERIVING NEW FORMULAs Solve for the indicated letter. In each case you derive a new formula. 1. P= 4s S = 2. P=3s-Hb s = 3. A = #bh h = 4. A =#bh b = 5. d = ri r = 6. C = 2irr r = 7. S=#at” a = *8. A = p +prt t= *9. A = p +prt p = *10. A = }h (b+b') b = *11. s=} (a+l) l- *12. C=#(F–32) F= — 118 — LITERAL EQUATIONS Solve each of the following literal equations for a:. Q. 1. aa: =b 6. — = C 0. b 2. ba’ = –1 7. —=— O, C (C 3. cac-H d =e 8. †-c-d Q. 4. – 30a = 6 9. ++b=-3 ac—b 5. ac-b = 4 10 = C C USE OF LITERAL EQUATIONS BY HEATIN It has been found by advanced mathematics T1 – T., that y_*(Ti-") Q. amount of heat that escapes through the walls of a steam pipe and is wasted, or of the amount of heat that penetrates the walls of a refrigerator and wastes ice. The engineer can compute W when he knows T1 and T2 (the temperatures on the warmer and the cooler sides of the wall), w (the thickness of the wall), and k. The material of which the wall is made determines k; k is small for asbestos or ground , where W is a measure of the 11, 2a – b = 0 3a – a 12. - 3 Q QC Ol. 13. — — — = — 3 2 6 Q: Q. *14. — — — = 2 a b (130 *15. ——a = c 2b G AND REFRIGERATING ENGINEERS cork, and large for metals. The theory of heat conduction was not so im- portant a hundred years ago as it is today. Living habits were different then. A large part of the food we eat (meat, milk, fruit, and vegetables) has been in cold storage, or has been shipped in refrigerator cars. Our homes and public buildings are kept warm in winter by heat which comes through pipes. De- signing the equipment to do this and working out the underlying theory are the work of the heating and refrigerating engineer. — 119 - PROBLEMs 1. Find two consecutive integers so that the first added to twice the second gives 113. Check: 2. On a trip of 397 miles, a father drove the car twenty-seven miles farther than his son. How far did each drive it? Check: 3. Solve the equation d =rt for t. At an average speed of fifty-one miles an hour, how long will it take an automobile to travel 170 miles? Check: 4. Find two consecutive even numbers whose sum is 130. Check: 5. A man invested half his money at five per cent and half at six per cent. He received $71.50 interest. Find the total amount invested. Check: 6. If half a number is subtracted from 60, the result equals one third of the number less 5. Find the number. Check: — 120 — PROBLEMs 1. A baseball team played 154 games and won twenty-four more than it lost. Find the number of games won. Check: 2. The 1930 census showed that a certain town had increased its population 12% per cent during the ten-year period. The 1930 population was 46,503. Find the 1920 population. Check: 3. A carpenter wishes to divide a twelve-foot board into three parts such that each part is six inches longer than the preceding one. Find the length of each part. Check: *4. The difference of the squares of two con- secutive odd numbers is 64. Find the numbers. Check: *5. One eighth of the supplement of a certain angle exceeds one fifth of the angle by five degrees. Find the angle. Check: *6. In the triangle ABC, the angle A equals one half of angle B, and angle B equals one third of angle C. Find the angles of the triangle. Check: — 121 — *THE COURIER PROBLEM Before the time of mail and telegraph service and the radio, messages were sent by couriers. This practice was the basis for a multitude of problems which have appeared in the algebra books of all countries for many centuries. In olden times, the problem-story was of messengers on foot, or on horse or camel, or of a dog pursuing a hare; in modern times the problem-story has involved motorcycles, automobiles, trains, and air- planes. The clock problem, where the minute hand pursues and overtakes the hour hand, belongs to the same family of problems. In solving a courier problem, express distance in terms of rate and time, d = r. t. Let d be the distance given in the problem, di the distance traveled by the From a painting by W. H. Jackson. Courtesy Journal of the National Education Association PONY EXPRESS AT A RELAY STATION first courier, and dº the distance traveled by the second courier. Similarly, let ri and ti be the rate and time of the first courier, and let rø and tº be the rate and time of the second courier. The solution of a courier problem depends on one of these formulas: di-H dº = d di—d, = d. di = d2 EXAMPLE A and B start at the same time from two cities 288 miles apart and travel toward each other. A travels eight miles an hour faster than B, and they meet at the end of six hours. What is the rate of each? Use the formula, di-H dº = d. The problem gives r1=ro-H 8, and t = 6. Substi- tute these values in the formula, (r2+8), 6––rº 6 = 288. Solving, r2 = 20 and r1=28. Answer: A travels 28 miles an hour, and B trav- els 20 miles an hour. PROBLEMS 1. A courier starts from P toward Q at the rate of twenty-six miles an hour. Four hours later a sec- ond courier starts from P toward Q along the same road, and overtakes the first at the end of eight hours. What is the rate of the second courier? Use the formula, di-dº. 2. A messenger leaves X and travels toward Z at the rate of twenty-five miles an hour; at the same time a second messenger leaves Y and travels to- ward Z at the rate of twenty miles an hour. It is 60 miles from X to Y. How many hours will it take the first messenger to overtake the second? Use the formula, di-dº - d. X Y º 3. A travels twice as fast as B. They start to- ward each other at the same time from cities 180 miles apart, and meet in exactly five hours. What is the rate of each? – 122--- *MIXTURE PROBLEMS 1. The government bought 180,000 acres of land for park purposes. Some cost $2 an acre, and some cost $3 an acre. The total cost was $475,000. How many acres of each kind of land were bought? Let a = number of acres at $2 an acre. Then 180,000—a = number of acres at $3 an 2,CI’é. * Then 2a+3(180,000—a)=475,000 2. An importer mixes two kinds of tea, worth 63 cents and 48 cents a pound, to make 100 pounds worth 54 cents a pound. How many pounds of each kind does he use? 3. A grocer mixes enough pecans at 60 cents a pound with 80 pounds of English walnuts at 25 cents a pound so that the resulting mixture is worth 37.5 cents a pound. How many pounds of pecans does he add” 4. A bronze bell weighing 572 pounds is to be melted, and enough pure copper added to the alloy so that the copper content is increased from 70 per cent to 75 per cent. How many pounds of pure cop- per should be added? 5. A sirup manufacturer wants a product that is 12%% maple and 87%% cane. How many gallons of maple sirup need he add to one hogshead (63 gal- lons) of cane sirup to make the mixture he wants? 6. A seed dealer wishes to make a lawn mixture to sell at three pounds for one dollar. How many pounds of clover seed at 35 cents a pound should he add to each 100 pounds of bluegrass at 28 cents a pound? INVENTORY TEST 1. Solve mentally. (a) #2 = −4 a = (d) #w = } w = (g) a = −3y y= (b) ac-3=7 a = (e) az = b z = (h) b =ac – a z = (c) 6y = 15 y= (f) #m = —b m = (i) c-d—y y= 2. Write three equations of the first degree in one unknown. (a) (b) (c) 3. Remove the parentheses and combine like terms. (a) 3a–6(a+b) – b = (b) —a;+3(a —y)+6y = (c) 201–29)—3(u–3v)+4w = 4. Write the equation for each of following, letting n represent the unknown. Do not solve. (a) The difference between one half and one fifth of a number is 7. (b) If 6 is added to three times a number, the result is 39. (c) If 5 is subtracted from a number, the result is a. 5. Write the formulas. (a) The area of a rectangle with base a-Hb and height a -b. (b) The sum of the areas of two squares whose sides are a and b. (c) The volume of a rectangular box with dimensions 3, ac, and a —b. 6. Write a T or an F before each of the following according as the statement is true or false. (a) ar”–9=0 is an equation of the first degree. (b) y= c-d is a literal equation. (c) The supplement of an angle is greater than its complement. (d) a”–2+2a: = 2* is an equation of the first degree. (e) The members of an equation may be interchanged. (f) When solved for h, the formula A =bh/2 becomes h =2A/b. (g) An angle of 10° is supplementary to an angle of 80°. — (h) Linear and literal mean the same in describing equations. (i) An angle of 40° is complementary to an angle of 50°. 7. Fill blanks so that resulting statements are true. (a) Parentheses preceded by a sign may be removed by changing the sign of every term inside the parentheses. (b) Two angles are *-m. if their sum is 90°. (c) In a equation, the knowns are represented by letters. c—b **8. Solve for the letters indicated when it is given that s = * (a) c- (b) a = (c) b = — 124 — CUMULATIVE REVIEw 1. Above the dash, write the exponent which will make each equation true. (a) a a a = a – (b) 2—-4 (c) a 9-4-2% = a – (d) ac"--a!" = a – Multiply and check. (a) 32–2y 22–H5y . Divide and check. (a) ac—y)a?--2a:y–3y” 4. Find the products mentally. 5. (a) (a +2y) (c.–2y) = (b) (3a–5b)?= (c) (6a–b) (a-5b) = Solve for a and check. (a) (c-i-3) (2–4)–(2–1) (a –2)=0 6. Factor completely. (a) 27aa”–48ay” = (b) 2a3+2a2–60a = (c) aa:4–16a = (e) a a a a = a- (f) 3–=27 (g) 5–=125 (h) a--a+a+a = 4a– (b) a2+-ab-i-b” a?— ab–H b” (b) a 4-y)a?--y” (d) (6m2–1) (2m2+3) = (e) (622–1)*= (f) (2–1) (c-i-1) (rº-H1)= (b) (r-1) (c.—3) = z*-19 (d) ama”–6ama;+9am= (e) aºy—ay” = (f) 629–1929-–10y” = — 125 — UNIT 8. FRACTIONS THOMAs JEFFERSON (1743–1826) Thomas Jefferson—writer of the ometry. This has proved the only Declaration of Independence, Presi- dent of the United States, founder of the University of Virginia! We know Jefferson as a great statesman, but most of us do not know him so well as a patron of learning. During the latter part of the 18th century, France was going through a great cultural and political upheaval. Jefferson was influenced by this movement, and he transplanted as much of it as he could in this coun- try. The French discarded their old significant and permanent revision of elementary geometry since the time of Euclid (300 B.C.). Jefferson succeeded in having the geometry of Legendre made a part of early American education. Jefferson used his influence on a number of occasions in bringing dis- tinguished European scholars to this country. They served as professors of mathematics at West Point, and at the University of Virginia, and as directors of the Bureau of Standards and of the Coast and Geodetic Sur- system of weights and measures and adopted the metric system. Jeffer- son sought to have the newly formed United States follow the example set by France. He succeeded insofar as our money system is con- cerned. Legendre, of France, had just recast Greek ge- Paul's Photos THOMAS, JEFFERSON vey. When, as an old man, Jefferson was interested in the education of his grandson, he wrote to a friend, “Having to con- duct my grandson through his course in math- ematics, I have resumed the study with great avidity. It was ever my favorite one.” ALGEBRAIC FRACTIONS A. The expression h is called an algebraic fraction. A is the numerator or dividend, and his the denomi- nator or divisor. The numerator and the denomina- tor are the terms of the fraction. Algebraic fractions are necessary in solving formulas. Example: Given A. A =bh, one can solve and get b–º or h-i- Algebraic fractions also furnish a convenient way to write a quotient. (c-Hy) + (x-y) may be a;+ written *-i-9. Notice that the line separating the a -y terms of the fraction takes the place of both the division sign and the parentheses. In an algebraic fraction, the numerator as a whole is divided by the denominator as a whole. REDUCTION OF FRACTIONS In algebra, as in arithmetic, it is often desirable to change the form of a fraction without changing its value. This is done by multiplying or dividing the numerator and the denominator by the same number. EXAMPLES 6 3 1 3 2a: a. 3, 2a: 1. – = — 3. – = — ... — = — 7. — = — 8 4 2 6 6/ 3/ y 2y 14 2 2 4 33-2 3 (I (IJC 2. – = — 4. — = — 6. – = — 8. — = — 21 3 5 10 a 3 a. b ba: EXERCISES Indicate what was done to the first fraction in each of the examples above to get the second fraction. The answers to the first and third are correct. 1. D2 3. M3 2. — 4. — REDUCTION OF ALGEBRAIC FRACTIONS TO LOWEST TERMS A fraction is in its lowest terms when the numer- ator and the denominator have no common factor 5 a. except 1. The fractions 7' … ' and lowest terms. To reduce a fraction to its lowest terms, divide 152°y 1. Reduce 10ay 3a. 15:rº/ 3r 19xy 2 2 Reduce to lowest terms. 1. 10. 11. 12. 13. 14. Olº, ba: 6a?b 3abº - —35ac? 523 b2C bC2 — 16b2C —- - —84pm? –21rºm abc a—-- * cba 3a;+4y Ayº –3a a-Hb - Q/ ac-H ac-y are in both numerator and denominator by all common factors. When the numerator and the denominator are monomials, the common factors can be cancelled by inspection. When they are not monomials, it is usu- ally best first to factor, and then to cancel the fac- tors which are common to both. EXAMPLES 52%–5/? a 4–2a:y--y” 50-7(t+u) 5(z+y) 2. Reduce 2 — a 12 3. Reduce 22-H2y (r-t)(x-7) r-y (z=y)(x-y) a — y 2(z+y) 2 EXERCISES 15. 16. 17. 18. 19. 20. 21. *22. *23. **24. **25. ac”—H·6ac-H9 ac2–9 5c — 5d cº–2cd+dº 2a – 4 6a–24 (r-y)* a’–2ay--y” aac-H bac-H ca: ay-Hby-H cy cº–3c-H2 cº-i-2c-8 4b2+8b+4 2b-i-2 2c2–11cd+12d? 2C2 — cal–3d? - a 4–yº a 4–2.cºy?--yº * 62%-H 11acy—10y” 92°–12ay-H4y” - 2b?–H2b–24 4mb%–36m. — 127 — THE LAW OF SIGNS OF FRACTIONS There are three signs to consider in a fraction: (1) the sign of the fraction, (2) the sign of the numerator, and (3) the sign of the denominator. These three signs can be arranged in eight different ways. +N — N — N —HN +N — N +N — N +—— +— +— –H– tºº smºmºmº sºme ºm---sºme tº-E *m-º. *E= a-mºme mºme-mas +D — D +D — D +D +D — D — D If all three of the signs associated with a fraction are positive, or if one is positive and two are negative, the fraction can be simplified into a positive expression. +++2 +---2-42 tº--(-2-42 +++2 +2 +2 -- –2 - –2 If all three of the signs associated with a fraction are negative, or if one is negative and two are positive, the fraction can be simplified into a negative expression. –4 –4 +4 +4 ——= — (+2) = —2 +— = —2 +— = —2 —— = — 2 — 2 +2 — 2 +2 Usually only the minus signs are written. Wherever no sign is written, a positive sign is understood. Law of Signs. The value of a fraction is not changed by changing (1) the sign of the fraction and the sign of the numerator, (2) the sign of the fraction and the sign of the denominator, or (3) the sign of the numerator and the sign of the denominator. The value of a fraction is not changed when any two of its signs are changed. 4 — 4 4 –4 2 2 –27–2 When the law of signs is applied to a fraction whose numerator or denominator or both are polynomials, remember that to change the sign of a polynomial it is necessary to change the sign of each term of the poly- nomial. a —b —a-Hb a – b – a--b c-i-dº c—H·d –c—d -c-d Two fractions are equivalent if they have the same value although their form is not the same. 1 3 2a: (C ac—y — ac-Hy 2 6 2(ac-Hy) a +y – a – b a-Hb EXERCISES Write as equivalent fractions in three different ways by applying the law of signs. QC —a-Hb 1. — = 4. — F Q/ a –y Ol. — Q, 2. –— = 5. F b c—d 3. —— = 6. – = d b— a Reduce to lowest terms. cd –d? 5a–200b.” 7. — - 10. — = c—d 6b — 3 ?— a?—b% 8. — *T*9__ *11. -: — (a —y) a —b ë 2 — a 12 a8+-b% 9. _*-9 = *12. = gy—a: a-Hb — 128 — MULTIPLICATION OF FRACTIONS Fractions are multiplied in algebra as in arithmetic. EXAMPLES Arithmetic Algebra 1 3 11 3. 11 33 3 Ol C 0 ° C (10 5 7 5.7 35 b d b d bd 1 1 2 1 2 6 7 1 1 1 4 2a-H2y a _2(x+y}- -a- 2. 1 2 7 12~ 1.2 2 a 24-y la” ---- a 1 a 1 2 Ol 1 9 EXERCISES Find the products of the following fractions. When the numerators or denominators can be factored, first separate them into their prime factors and then cancel any factor which is common to any numerator and any denominator. 3 8 4a:” m—H·m. 1. — — = 10. * = 4 15 n–H m 8a.” —2 14 ab a”—b? 2. — — = 11. —. = 12 – 24 a-Hb 1 3 1 8 12 42%+4a:--1 a-H2b 2 4 a”--4ab+45° 2x+1 6a 3 a?– a – 6 3 4. — — = 13, * - 4 ac” 3a – 27 a -3 – a – b” O.Q. b2 5. —. = 14. +ay. == b? a ab bac-H by 6a 3 9—aº 2(c.—d) 6. — — = 15. sº = 4 2a: c—d ac-H3 c—Hd c–d 63 a 2–4 7. & -: 16. & :- c—d a a:4 – 16 45 ac”— y” a-Hb *. a–H3 3a – 18 8. $/ 17 * 2–5 º' cº-36 ab-I-3b a’—ary–2y” y–2a: 6(a — d)2 (a-v) (o-Fā) *18. c—Hd a —y 2a –y a -2J — 129 — DIVISION OF FRACTIONS Fractions are divided in algebra as in arithmetic. EXAMPLES Arithmetic Algebra 3 7 3 11 33 3. * * a b ba: 5 11 57 35 y b y a dy 4 8 2 3 3 2 1 2. —--— = — — = 1 4. 2r-2ſ, r-y_2-25-0--& 2 12 3 #2 a”—bº a+b a?—b”-a-y- a-b a – b 1 To divide one fraction by another, invert the divisor and multiply. EXERCISES Perform the indicated divisions. 2 2 1. –––2= 10. 8+– = 3 $/ a? 2 4+–= § 11 —--a” = b2 2 4 a8 — a” 3. —--—= 12. —--—= 6 12 C3 b? 3 Q. QC 4 ac-Hy y 4 7 14 5. 8+-- = 14. ––– = 11 a;+y a 2–y” 7 14 Ol. a2b 6. —--— = 15. —H. - 23 46 b—c b? — C2 2d 4a c”—2Cd–H d” c – d 7. —-- —– = 16. —H - 3b 3b a;-y ac-y , a. 7b b2 8. +++= 17. + = Q/ ?/ a — 5 2a – 10 1 O; 18 a?– a – 6 a –3 9. 1 +– = º –H F. b 2b b2 — 130 — REVIEW IN MULTIPLICATION AND DIVISION OF FRACTIONS Perform the indicated operations. 1 a;+y 2 4 2+y 52%/ 5 2. * *-- 10 acy” c—d 2(c—d 3. —- ( !- 4 3 a?—b” a +b 4. ————— = 7??, 7??, 44.2 Q2 5. — — = ab 2a: m?n m? 6. —–––– = 3y” y 2a:--8 3 32–9 r-H4 8. 3ay--—= a;2 ac3 11 12. (a — a)” c-H d -º- o (c-i-d)? z-a ab-Hac e b-Hc - e bm-Hbn. - n–H m. 27 a-H9 a?—a —90 7 7a-H2 Oś 13. —H — = 9— a” 3+a ac—3b 14. --a –3b = ac”—H·6ac—H.9 ac-H5 15 - ºToy F.25 ºr " 1 a.”—y” a -1 16. - —H – ac-Hy a Ol. *17. ––––. — = *18. —--—--—= — 131 — ADDITION AND SUBTRACTION OF FRACTIONS Fractions are added and subtracted in algebra as in arithmetic. In algebra as in arithmetic, fractions cannot be added or subtracted unless their denominators are equal. EXAMPLES 4 3–6 4–(3—a) 4–3–Ha 1+a 1. ———–H–=— = 1 3. ——— = - - 5 5 5 Ol. Ol. Ol. 0. Ol, 4 2a+b a-Hb 20-Hb — (a+b) • ac-Hy ac-Hy ac-Hy 3 2 4 5 20-Hb – a – b a 2. –––––– = — F - Q: Q. Q. Q2 ac-Hy ac-Hy EXERCISES Add. Express each result in its simplest form 1 3 2 5 a -5 – -º- – - - = 7. + - 4 4 4 a-Hb a-Hb 1 3 4. a; ac-Hy 2. —–H–—— = 8. - – = 5 5 5 c—d c–d 3 4 6 a2+-y” a 2–y” 3. ———–H–= 9. - F. 7 7 7 n–H n m-H n 3 6 4. a2+2ab-Hb” a”–2ab–H b” 4. —–H–——= 10. - - 10 10 10 ac-y a;-y 20, 3a 4 a2+y” —aº-Hy” 5. ————H-- = 11. - - b b b a — ab-Hb a-ab-Hb a? b2 C2 6 7 - 6 a 12. - - ac-y ac-y 2-y — 132 — LEAST COMMON DENOMINATOR When the given fractions do not have the same denominator, they must be changed into fractions which do have the same denominator before they can be added or subtracted. The least common denominator (L. C. D.) is the smallest number which exactly contains each of the de- nominators as a factor. Thus, the L. C. D. is the least common multiple (L. C. M.) of the denominators. L. C. D. of ; and # is 6; L. C. D. of #, #, and # is 12; L. C. D. of #, #, and # is 42. In algebra, when the denominators are monomials, the L. C. D. can be found by inspection. L. C. D. of 1 and 1 is ab; L. C. D. of 1 and 1 is a 2; L. C. D. of 1 and 1 is a 3b3. Ol, b ac2 Q. 03b Ob3 If, however, we need to find the L. C. D. for more difficult fractions, we may proceed as follows: Factor each denominator into its prime factors. The L. C. D. is the product of all the different prime factors, using each prime factor the greatest number of times it occurs in any one denominator. EXAMPLES 3 5 7 Ol. b C € f 1. – ; – ) and — 2. y j 3. y 8 6 12 a?–y” ac-Hy da —dy ac”—y” a 2–33-y-H2y” 8 = 2.2.2 ac”—y”= (a+y) (a —y) ac”—y”= (a+y) (2–y) 6 = 2.3 ac-Hy=ac-Hy ac”--3ay-H2y2 = (x+y) (2+2y) 12 = 2. 2.3 da – dy=d(x, -y) L. C. D. is (2-Hy) (c.—y) (2-H2y). L. C. D. is 2.2. 2. 3. L. C. D. is (a +y) (a —y) (d) In reducing fractions having different denominators to fractions having the same denominator, one may proceed as follows: Find the L. C. D. Multiply both the numerator and the denominator of each of the fractions by such a number that the denominator of each of the new fractions equals the L. C. D. Example: In Example 1, above, the L. C. D. is 24. Multiply the numerator and the denominator of the first fraction # by 3, getting the equivalent fraction #; multiply the numerator and the denominator of ; by 4, getting the equivalent fraction #; multiply the numerator and the denominator of 13 by 2, getting the equivalent fraction #. The given fractions thus become #1, #4, and #. EXERCISES Find the L. C. D. of each set of fractions; then reduce each fraction to an equivalent fraction having the L. C. D. as denominator. The first has been done correctly. 0. 1 3 2 1. Given y 3. Given — , — , and — a 2–y” a 2–2xy--y” 2 9 ac”—y”= (a, -y) (a +y) a 2–2a:y-Hy”= (a, -y) (a –y) L. C. D. is (a –y) (2–y) (2+y). Multiply the numerator and the denominator of the first fraction by (a —y), and of the second fraction by (ac-Hy). ac—y a(a +y) (2–y)(a –y)(a +y) (2-y)(x-y)(2+y) 1 1 & a b e - - --> 4. G — , — 2. Given Ol y ab IVéI). 30 l/ — 133 – EXERCISES Find the L. C. D. of each set of fractions; then reduce each fraction to an equivalent fraction having the L. C. D. as denominator. 3 4 1. – ) — c d 3 7 2. – º – 3:2 p3 a b 3. – º – acy y” 1 1 1 4. — , — » — aſ b bº 1 1 5. 1 — a - y Q. (I. b 6. y a – b a +b 2 3 7. y - c—d c – 2d 1 8. — , Q. 10. 3a. 11. ar—H 1 ºrs' 1 a-Hb' 1 C Q. - 1 Q. - 2 2y º 42 a”–3y-H2y” a — 4 12. y ac” – a – 6 a”–4a:–H3 SIR ISAAC NEWTON (1642–1727) No other person ever made such im- portant advances in pure mathematics, in physics, and in astronomy as did Sir Isaac Newton. It has been well said that “No one ever left knowledge in a state so different from that in which he found it.” The story is told of a distinguished Hindu visiting in this country. In parts of India agriculture, medicine, and in- dustry are carried on much as they were many centuries ago. The Hindu was asked, “What is the basic differ- ence between this India and America?” International News Photos - SIR ISAAC NEWTON — 134 — He replied, “Sir Isaac Newton.” He made Newton the symbol for the sci- entific and technological progress made in Europe and America during the past two or three centuries. Newton, however, wrote modestly of his own achievements. “I know not what the world will think of my labors, but to myself it seems that I have been but a child playing on the seashore; now finding some prettier pebble or more beautiful shell than my compan- ions, while the unbounded ocean of truth lay undiscovered before me.” ADDITION AND SUBTRACTION OF FRACTIONS HAVING MonoMIAL DENOMINATORS EXAMPLES 1 1 1 4 3 2 9 3 1 1 b a b-a 1. + + E + + = – = - ... — — — = — — — = 3 4 6 12 12 12 12 4 a b ab ab ab EXERCISES 1 **- 12 ac—b 6 3 12 2c "c 2 1 1 13 * +* 6 7. T 5' 4"T 1 1 1 QC (C 3. ————H–= 14. ——— = 2 3 4 5 8 1 1 2a, 3 4. ——— = 15. # Lº- a b 3 4 a b 3 4 5. —--— = 16. ——— = a y 2 y 1 1 9 4 6. ———= 17. —–H– = c 26 ac2 a. a b 6 7 7. —–H– = 18. ——— = b a ab b 1 1 a b c 2a: ac a ac2 acº a—-b b 5 2 9 –H – = 20. ——— = 2 4 2a, 3a; 2 m. 2 3 2 10. — — — = 21. ————H--- m 2 a? aſy y” Q. C 1 2 3 11. *-*- - 22. ———–H– = Ol d 22° 3ry 4y” — 135 — ADDITION AND SUBTRACTION OF FRACTIONS HAVING PolyNoMIAL DENOMINATORS EXAMPLES a 1. b a(a-b) —l- b(a+b) a F5"a-b (gºb)(a-b)" (a+b)(a-b) _dº-ab--ab-i-bº a2+-b” (a+b)(a-b) (a+b)(a-b) EXERCISES Ol, + b --> 7 º a;+y ac-Hy O 1 b º •- – - 8. a;+y ac-y b 1 - - - - 9 b°–4 b-H2 2 —l- 3 tº- a+b | ac-i-cb" 10. 1 1 wº- C e-Ef 11. Ol. 1 • —-H- = 12. a-Hb a 1 * **u Pº-9) 2–y 24-y (c-Hy)(x-y) (2+y)(e–y) _24-y–2x+2}_ —a;+3y (ac-Hy)(ac—y) (2+y)(a:-y) 1 1 ac”— 4 -e ac”—3a;+2 = 2a: ac—2 2 a-H1 ..I;" Q. 1—a: 5 1 — ac” 1 gº-bº" 3y y?–4 - g°–4y-H4 *- — 136 — ADDITION AND SUBTRACTION OF FRACTIONS EXAMPLES 1 2 1 —2 – 1 Oſ, b 0, b a-Hb . —–H– = —–H–= 2. ——— = —–H– = ac-y y–a, a -y ac-y ac-y b–4 4–b b–4 b – 4 b–4 What change was made in the fractions above before they were combined? Answer for— Example 1: Example 2: EXERCISES 3 6 3 b 1. + -: 8. - - b–2a: 22–b g?–9 9–y” 7 6 02 d? 2. - - - & 9. -m – b – c c-b b”— a” a”—b? Ol, b 4 4 –4 3. + F 10. ———–H–= n–2m 2n —m. - O. O. — O, a b b 3 4 4. —–H–——= 11. F — QC (C — 4. 7?? - ??, Olſº, - O/Y, 1 2 s. * * –––– 12. - – - 20 — 20 – 2d. g°–4 2–g 4b 6b d” d-He 6. — — — = *13. +— = 2b — 1 1–2b d?—e” e—d a;+ 1 7. 3*__ 7” *14. * - – ac—4 4–a: ac-y y–º — 137 — CoMBINING ALGEBRAIC WHOLE NUMBERS AND FRACTIONS EXAMPLES +! *11 ab 1 ab–H1 O, —s = —- –s = - b 1' b bº' b b 1 c 1 col 1 cal–1 3. EXERCISES 10. •= — =: a;-y a–25T *11. *12. *13. *14. *15. *16. a-H2— a-H3 a-H3 a-H3 _a^+5a--6–1 * = --> a2+-5a–H5 g a +3 * 6 b——= b–2 6 ac-H4— :=: a;+3 2 —20—H -> C c–3d 1 ac-H5y-H = ac—y 3. 2— ab–H b%— -: Ol. a-Hb q/3 ac2+-acy-Hy?--—= a;-y g” Q. viziº 1 (a+2)(a+3) 1 a-H3 — 138 — COMPLEx FRACTIONS Arithmetic The division of fractions may be indicated as in either of the forms Algebra 1 1 at the left, or as in either of the forms at the right. The first form in 1 1 — — — — each case is the indicated quotient of two simple fractions. The second — —H· — 2 8 form in each case is called a complex fraction. We may simplify a a b 1 complex fraction by writing it as the quotient of two simple fractions 1 2. and carrying out the indicated division. This is shown in the examples am- OI’ below. OI’ Ol, 1 1 8 b EXAMPLES 1 a-Hb 1 2 2 3 (I, a-Hb a – b a-Hb aſ _a-Hb 3 T 2 3 3 a-b a a a a-b a-b 4 Q, 7 2 acº — 2 2–1. $/ d4 c6 – dé (C Q. a’—y” a +y c”—— 2. + F –H 4 C C cº–d4 c – d. - Q. Q} Q: - - —H — — 1++ !/ c—d c—d C 1 Q. (U (r—y)(3:4-g) -a- (cº-H ca-H d”) (c.—dºr 1 - c”—H·cd+-d? = _2~ - 2-r a — y C _c=-d- C EXERCISES 1 ac?--5ac—H6 s (l, 1. — = 6. — = 6 a +3 7 a? 1 3C 2. - 1 (C 1 1 a 7 a b 3. ––– 1 IT a a b b 1+ d 4. d–H c 7??, 4 v-7 1 -: w—— 8 2 5 Q. 1–H– a;+1 !/ 3C — 139 — Perform the indicated operations. 1. REVIEW IN ADDITION AND SUBTRACTION OF FRACTIONS 1 1 ——H--- 2 5 1 1 a b a: y Q/ Q} 10. 11. 12. 13. 14. 15. 16. Ol. b - c–1 JET 20. 3a a-Hb 2a+25T —--1 º-Egº Toº — 140 — INVENTORY TEST 1. Find the value of each of the following expressions when a = 1 and b = -2. © . () -- © 1+” o, ºn tº 8, b2 2a b C a b–a 2. Reduce each of the following to lowest terms. abc - (a) bedſ c—Hd cº–d? (b) a”— 4 (c) a”— 50–H6 -: e”—3e — 10 d) — = (d) e? – 10e–H25 3. Write each of the following as an equivalent fraction in three other ways by applying the law of signs. — 1 (a) == 4. Multiply as indicated. ac2–9 a – 5 g-gº-E20 º-H3T (a) 5. Divide as indicated. ac”—ay a 2–y” (a) —H· = G2. Ol, 6. Add and subtract as indicated. (a) 6a, 2. Q. a, *mº ºn wºmºsºm-º. -- sº- 5 6 15 3d, a +2 (b) —–H–= 4 Ol, 7. Simplify the following. 1 (a) 4-H - = (b) ac-— = Ol. Q. 8. Perform the indicated operations. a” — 16 (a) → 8, = a — 4 3C — QC (b) —= E - —y a-Hb a”—3ab-i-2b% a — b a”--2ab–H b% (b) a — 3 3–ac (b) —--—= QU — Q, (c) 4 3 C Ex- == a — 3 ac-2 (d) Ol. b 3 — a a–3 © 2-14– C ) a - ~ a;+ 1 b2 1 —— Q? b > (b) b 1–H– O} — 141 — CUMULATIVE REVIEW 1. (a) Solve the formula V = Tr”h for h. 2. Remove the parentheses and combine like terms. (a) 6a+4a – (3a–H4b) —6b = 3. Solve for a and check: 22–9(2–2) = –3(2–2)+4 4. Factor completely. (a) y”—y–20= (b) aa:4–16a = 5. Supply the missing numerators and denominators. Q. –– = + a —b 6. Divide the following mentally. (a) (ac”—y”) + (a —y) = (b) (b?–15b-i-56) -- (b–8) = 7. Perform the indicated operations. 2 . () (a) +++ - a b am-º º -º b a a-Hb (b) 8. 6ay?(wy°–4a:--2a.”y–6) = 9. W = }Tr”h. Solve for h. 10. If b apples cost twenty-five cents, then one apple will cost — cents, and a apples will cost — 02 centS. (b) Find h when W = 1452, r = 11, T =%. (b) 23:9–2(2–3) (a +4)+23 = (c) acºy?--ac”y–12a = (d) 100–20b-i-b” – (c) (9a”–30ab-H25b”)+(3a–5b) = (d) (42°–9) + (2x4+3) = ac”—62–H.9 ac”—9 (c) ac-i-3 ac—3 - *(d) -: cents, five apples will cost — 142 — UNIT 9. FRACTIONAL EQUATIONS THREE Construction PROJECTS MADE POSSIBLE BY ENGINEERING MATHEMATICs Paul’s Photos LEFT.-The bridge to span the Gol- den Gate is the longest suspension bridge ever built. There is a clear span of 4,200 feet between the two main towers; these rise 746 feet above the water level. The roadway, at the height of a twenty-story building, is supported by steel cables more than a yard in diameter. The bridge is designed to accommodate 16,000 vehicles an hour. CENTER.—The Empire State Building, which rises 102 stories above Fifth Avenue, New York City, is the tallest structure ever built. Its floor space for office use exceeds two million square feet, mak- ing it the largest office building in the world. It has seven miles of elevator shafts, and uses as much electricity as the city of Albany. RIGHT.-In height of dam (730 ft.) and in area of the reservoir formed, Boulder Dam is the greatest project of the kind ever undertaken. Its purpose is threefold: (1) to protect valuable farm lands from the floods of the Colorado River; (2) to produce electrical current for the Southwest; (3) to store wa- terfor irrigation purposes. Its capacity is 30,500,000 acre-feet. This is about enough to cover New York state to a depth of one foot. FRACTIONAL EQUATIONs An equation in which the unknown letter or letters appear in the denominator of the fraction is a frac- tional equation. The following are fractional equations. 1 1 2 — = 6 ––– J. Q: rº-IT a y SoLVING FRACTIONAL EQUATIONs The best way to solve a fractional equation in one unknown is to multiply through by the L. C. D. This is called clearing of fractions. One thus obtains an equivalent equation which is not fractional. EXAMPLES 2 1. Given the equation — = 4 Q’ Multiply through by a and get 2 = 4a. Solve this equation: a = }. 7 2 1 2. Given the equation –––=– y 3 2 Multiply through by 6y and get 42–4y =3y. Solve this equation: y = 6. EXERCISES Solve. 5 1 2 1 1. ––– = 2 2. = a; 2 ac—2 ac—3 — 143 — EXERCISES Solve and check. 1 2 1 1 1 1. —=— 9. ———=— g 4 gy 6 3 2 2 1 10 ++ 1 a; 8 'm 6 3-ºxº 4 1 1 1 3. —= 2 11. ———=— 2. 2 4 4 4 5 12 1 + 1 1 770, a 's T 4 8 1 2 1 1 .5. —=— 13. — — — = — 2w 3 3a; 3 6 3 1 1 1 2 6. — = — 14. —–H–=— 7r 2 2a, 3 3 1 3 2 3 2 7 *-* = −e 15. - - - - -e 2 a. gy 5 5 2 3 3 8. •- - -ºs 16. —— 1 = 4 7 y 4n. – 144 — Solve and check. 1 2 1. — = g–1 y–2 6 2 * = * EXERCISES 7. * * * _3^+6 ac—2 ac-H2 ac”— 4 9 J__y=F20 g-H2 y–2 y?–4 ac—6 a – 7 3 7 4 5. —–H–=— 42 16 32 4a: 2a: 9. = +1 2a – 1 2a:--1 — 145 – Solve and check. 1 32–6 5a. 32–5 52+1 2 3s-H4 s–H3 6s–2 2s-H1 3 — 1 2m – 5 == 2m – 1 a;+3 ac-H7 a;+2 ac-H5 24-1 2–2 2-H2 z+3 EXERCISES 2a – 1 2a: & ac—3 6 2 3 ++7. •= — —- sº g”–1 y–1 y–H1 h?–2 2h-H3 **8. = h–H2 2 — 146 — NUMBER PROBLEMS USING FRACTIONAL EquaTIONs Solve and check. 1. A number divided by 3 less than itself gives a Quotient of #. Find the number. Check: 2. The sum of a certain number and 12 is di- vided by 3 times the number. The result is #. Find the number. Check: 3. What number must be subtracted from both the numerator and the denominator of the fraction # to double its value? Check: 4. Separate 40 into two parts so that one part will be two thirds of the other. Hint. Let a = one part. Then 40—a = the other, and the equation is: a = #(40–ac). Check: 5. Separate 75 into two parts so that their quo- tient will be 1%. Check: 6. Find two consecutive even numbers such that their quotient is #. Check: — 147 — PROBLEMs 1. The sum of #, #, and # of a certain number is 124. Find the number. 2. Find the number whose double exceeds its half by 54. 3. One sixth of a certain number exceeds one eighth of the number by 3. Find the number. 4. Divide 80 into two parts such that twice the Smaller equals one half the larger. 5. Divide 72 into two parts such that % of the larger part less 5 is equal to # of the smaller part plus 4. 6. The difference of two numbers is 72. The larger number divided by the smaller gives a quo- tient of 4 and a remainder of 12. Find the numbers. — 148 — WoRK AND TANK PROBLEMs 1. If a man can do a piece of work in six days, he can do what part of it in one day? 2. If a boy can do a piece of work in ten days, he can do what part in one day? 3. If it takes John a days to do a piece of work, he can do what part in one day? 4. If William can do a piece of work in five days, he can do of it in one day. If James can do of it in one day. Working together they can do the same piece of work in four days, he can do plus in one day. Many problems use the ideas involved in the questions above. EXERCISES 1. William can do a piece of work in five days and James can do the same piece of work in four days. How long will it take them working together? Let n = the number of days it takes both working together. 1 Then — = the part of the work they can do together in one day. 70, # = the part of the work William can do in a day. # = the part of the work James can do in a day. Th 1 *...* 1 9 9m = 20 23 d en — =–––H– ; – = − ; 9m = ZU; n = 3.V.S. n 5 4 n 20 # Clay Both working together can do the work in 23 days. 2. A tank has two inlet pipes. One can fill it in twenty minutes and the other in thirty minutes. How many minutes will it take both pipes together to fill the tank? Let m = the number of minutes it will take both pipes to fill the tank. 1 Then — = the part that both pipes will fill in one minute. 70, # = the part that one pipe will fill in one minute. # = the part the other pipe will fill in one minute. 1 1 1 Then — = −–H–. Complete the solution. n 20 30 3. A man and a boy can do a piece of work in six days. It would take the man alone nine days to do the work; how long would it take the boy working alone? — 149 — PROBLEMS 1. In three days a tractor can plow a field which it takes a team of horses eleven days to plow. How long will it take both working together to plow the field? Check: 2. A man can dig a ditch in eight days and a boy can dig it in fourteen days. How long will it take them to do it together? Check: 3. A tank has two inlet pipes. One can fill it in forty minutes and the other in one hour. How long will it take the two pipes together to fill the tank? Check: *4. The inlet pipe can fill a tank in thirty min- utes, and the outlet pipe can empty it in twenty minutes. If the tank is full and both pipes are opened, how long before the tank will be emptied? Check: *5. Two bricklayers together build a wall in seven hours. If one could do the same work in twelve hours, how long would it take the other to do it? Check: *6. A paper mill installed an improved machine which, working with the old machine, did a certain amount of work in fourteen hours. The old machine alone had taken forty hours to do the work. How long would it take the new machine to do the work alone? Check: — 150 — SoLVING AND EVALUATING FORMULAs One of the important applications of algebra is the formula. You should be able to solve a formula for any letter in it and to find the numerical value of any letter when the values of all the other letters are given. EXERCISES Solve each of the following formulas for the indicated letter. k 7. S=#Tr”h Solve for h. 1. A = — Solve for k. d? l2 *8. d = — Solve for w. W1 d2 8s 2. — = — Solve for d1. W2 di *9. V = u-Hat Solve for a, 3. s = #at” Solve for a. rl — a -v- **10. S = Solve for r. W r—l 4. M = — Solve for E. E Or” — O, *11. S = Solve for a, 5. P=2b–I-2h Solve for b. r — 1 6. A = p +pri Solve for r. *12. A = p(1+rt) Solve for t. — 151 — EvALUATION OF FORMULAs 1. Given the formula S = (a+b) find the val- ue of S, when a = 1, l- 27, and n=8. 2. Given the formula A = Tr”, find the value of A when T = ** and r=3}. h 3. Given the formula A = − (bi-H bo), find the T2. value of A when h = 4, bi=4}, and b2 =3%. 4. Given the formula A = P(1+rt), find the value of A when P=600, r=.06, and t=2#. W *5. Given the formula " i "* , find the value 2 I of W1, when W2 = 80, di =8, and d2 = 10. *6. Given the formula I = , find the 7"??, value of R when I =4, E=8, n = 16, and r=2. 77, **7. Given the formula S == (a+l), find the value of l when S=88, n = 4, and a = 1. Trr?h **8. Given the formula V--- find the value of h when W = 49, T = **, and r = 4. — 152– INVENTORY TEST 1. What law of the equation is used in clearing fractions? -- 2. Write in the blank space after each of the following equations the multiplier to be used in clearing it of fractions. @ 4 - © ºr -- *) at a to ° 5–4" bia T2 (b) a 1 (d) QC 1 = 1 O, *=e b T ac”— 7a;+12 2–4 3. Solve for ac. 1 (C a – 2 * (b) F. 2 a;--8 QC 4 - (a) 3, 4. (a) Write a formula for a which will show the relationship between r and a in the following table. (b) Solve the formula you have discovered for a and complete the table. Formula: a = QC 2 || 3 6 9 || 10% 11 | 12# 13 14 Ol. 6 8 14 20 23 O = 5. Write the equation for each of the following problems. Let r represent the unknown number. (a) One half of a number less 4 equals one third of the number. (b) What number added, to both the numerator and the denominator of ; will make the fraction equal #7 (c) A can do a piece of work in five days and B can do it in seven days. How long will it take them working together? 6. In each of the following equations determine by substitution for a whether or not the number or letter given to the right is a root of the equation. Write “Yes” or “No” in the blank space. @*--- do (e) – –a–b b) *) a "...I2 5 / — o, –a (-ob) — 5ac-i-3 2a: —3 1 1 1 b (b) → = --- (3) — 'G = + ( ) 5a;+7 2a: .c a b a +b — 153 — CUMULATIVE REVIEW Supply the missing words or complete the statements. 1. The sum of 3a, 4b, and – 6 is 2. If ac-2 is subtracted from 32, the remainder is 3. If V = abc, then a = , b = , and c= 4. The tells how many times a number is used as a factor. 5. The sum of –3, +4, -6, -7, +2, —4 is 6. (–3a.”)(2ab) = (–8c") + (2C+)= —#2)(–;c2a) = (—#cºd?) -- (-cd2) = (—a)(–b)(–c) = (–4y”–2y) + (2y)= 7. In multiplication we coefficients and exponents of like letters. 8. In division we coefficients and exponents of like letters. 9. By how much does 3a*-H4a-Hb exceed 20°–6a+4? 10. If a = —2, then a 3–2°–2–2= 11. Rewrite – a –y–4b, placing parentheses preceded by a - sign around the last two terms. 12. Write the equation for the following problem. The difference between the numbers 32-H4 and 22–2 is 6. The root of this equation is 13. The formula aº ; a "=a^+” shows that in multiplication we exponents of like letters. Since a formula is true for all values, then *...* = rº, and *r =– ; r; .2% † ———; w" wºe ; a ‘‘a’= ; wſſ a 3 = 14, wa-i-º-º- and cº-wº- y *** = } ac"--at-º- ; a "--a!" = - 15. (–4a)*= ; (3a*)*= ; (-;a)*= 16. If m represents A's age now, then ten years ago his age was and five years from now his age will be 17. The coordinates (2, —3) represent a point which is two units to the of the y-axis and three units the a-axis. 18. By how much does 5 exceed 3a–4y–2? 19. If a man can do a piece of work in n days, he can do part of it in one day and part of it in three days. 20. Multiply mentally. (a) (ac-H4) (ac-4)= (d) (c.--a”) (c.—a”) = (b) (2a–1)*= (e) (22°–a)?= (c) (a —2) (a+3)= (f) (22–3) (3.c-2) = — 154 — UNIT 10. LINEAR EQUATIONS IN TWO UNKNOWNS THREE FAMOUs MATHEMATICLANs AND AstroNoMERs wºn Tººn NICHOLAS COPERNICUS (1473–1543) At the beginning of recorded history, oxen and horses were used as means of transportation. Until less than two centuries ago, no better way was known of traveling from one place to another. As far back as we have any record, messages were car- ried by swift runners or by horsemen. No better way of carrying messages was found till very recent times. A few generations ago, wheat was cut and threshed by the methods used in the time of Moses. The era of science, which began about three cen- turies ago, has brought in a multitude of changes. Scientific method met with great opposition from those who were satisfied with things as they were. The great battle of science was to overcome this op- position. Tradition argued that the rate at which a body falls depends upon its weight. By trying it, Galileo convinced himself that this assumption was false. William Thompson GALILEO GALILEI (1564–1642) William Thompson JOHANN KEPLER (1571–1630) To convince others, he climbed the tower of Pisa and released a hundred-pound cannon ball and a half-pound weight together. They struck the ground together at the feet of those who argued against him, and who continued to insist that he was wrong. Galileo made a telescope through which he could see new planets in the sky. Theologians in- sisted that this was impossible. They refused to look through the telescope. Instead they spent their time devising ingenious arguments to prove they were right and Galileo was wrong, “as if with mag- ical incantations to charm the new planets out of the sky.” Workers in astronomy won this first great battle for all science. Copernicus, Galileo, and Kepler were ridiculed and persecuted, but they successfully laid the foundations for the development of the modern scientific era. LINEAR EQUATIONS IN ONE UNKNowN EXAMPLES 1. ac-H4 = 6 2. 23:--3 =9 3. 2a:-H3 = 6–3. 4. a -4 = 7 a = 2 2a: = 6 3r–H3 = 6 ** = 11 p = 3 3a – 3, or r = 1 a = 22 LINEAR EQUATIONS IN Two UNKNowNs A linear equation in two unknowns has an unlimited number of solutions. - EXAMPLES 1. *-i-2, -8. 2. 2. —30/=6 * | 0 || |2|3 || | | | |s * || 0 || || 2 || 3 || || 5 || || |s y || 4 || 3 || 3 || 2 || 2 | | | | | | | | 0 || 2 || | | || 0 || || 2 |2|s, Any one of these pairs of values for r and y is a solution to the equation written above the table. It is easy to check by substituting the pairs of values in the equation. — 155 — A LINEAR EquaTION IN Two UNKNOWNs Consider a rectangle whose perimeter is 20. The base and the height are related by the equation 2b-H2h = 20. When divided through by 2, this equation becomes b-H h = 10. Below is a table of corresponding values of b and h. These pairs of values are graphed at the right. h axis b 1 2 3 4 5 6 7 8 9 9 h 9 8 7 6 5 4 3 2 1 8 Notice from the graph that— (1) The pairs of points which satisfy the equation lie on a Straight line. (2) Any point on this line, as P, can be used as a vertex to * determine a rectangle with perimeter 20. 3 Every equation of the first degree can be graphed as a 2 straight line. This is a reason for the name linear equation. A linear equation in two unknowns is sometimes called an indeter- minate equation, because an unlimited number of pairs of values satisfy it. b axis EXERCISES 1. Given the equation –2a+y=1, complete the table below by writing the values of y which correspond to the indicated values of a. It is easier to do this after first solving for y. The equation then becomes y = 1+2a:. * |-2 || – || 0 || || 2 || 3 || 4 || 5 || 6 || $/ Each of these pairs of values represents a point on the graph. Plot six points on the graph at the right. Draw a line through the six points plotted. (a) Is it necessary to plot as many as six points to graph a straight line? (b) How many points are necessary? (c) If other pairs of values from the table were plotted, would the points lie on the line you drew through the six points? (d) The point (2%, 6) lies on the line. Does it satisfy the given equation? (e) Will the coordinates of every point on the line satisfy the given equation? 2. Given the equation 2a:-y=3, make a table of pairs of values for a and y, and graph. It is easier to do this after first solving for y. Thus, y= Q. , — 156 — AN EASY WAY TO GRAPH A LINEAR EquaTION IN Two UNKNowNs Two points fix the position of a straight line. are the points where the line crosses the at and y Thus two points are all that it is necessary to plot axes. These are obtained thus: (a) Set a = 0, and in graphing a straight line. As a check, however, it solve for y. This gives the point where the line is usually well to plot an additional point or two. crosses the y axis. (b) Set y = 0, and solve for ar. This Usually the two most convenient points to use gives the point where the line crosses the ac axis. EXAMPLES 1. 22–H39 = 6 2. 4a:-Hy=8 3. 5a –2y = 12 (C 0 3 (U 0 2 (C O 23 $/ 2 0 $/ 8 0 g | – 6 0 When the line crosses both axes at the point (0, 0), it is necessary to find at least one other point. It is also advisable to find another point when the two points at which the line crosses the axes are close together. EXAMPLES 1. 2a:--39 = 0 2. 32–59 = –1 3. 52–29 = 0 Q. 0 3 3U 0 |—|| 3 (U 0 2 $/ 0 | – 2 Q/ # O 2 $/ 0 5 EXERCISES 1. By inspection find two points on each of the lines. (a) a -i-y=4 ( y ) ( y ) (f) a -39 = 4 ( y ) ( ) (b) ac-y = 6 ( y ) ( y ) (g) a +2y=3 ( 5 ) ( j ) (c) a +4y=8 ( y ) ( } ) (h) 23:--7y = 14 ( j ) ( 3 ) (d) 22-y–0 ( , ) ( j ) (i) 32+4y=0 (- - , ) ( ) (e) 62–3y=12 ( -, -— ) ( — , . ) (j) 22–H5g = 1 ( j ) ( ) 2. Find two points on each line, and graph the straight lines. (a) as-Hy-5 ( y ) ( y ) (b) 23:--y= —8 ( y ) ( y ) (c) a —39 = 6 ( y ) ( — » ) y — 157 — FINDING EQUATIONS WHEN PoinTS ARE GIVEN There are an unlimited number of pairs of values which satisfy the equation y=3a. As the value of a changes, the value of y will also change. However, the rate at which y changes is three times as great as the rate at which a changes. When a = 1, y = 3; when a = 2, y = 6; when a = 3, y =9, and so on. The reverse process is also possible. When the values for a and y for several points on a line are given, we are able to find the equation of the line. EXAMPLES 1. From the table at the right, find the equation which de- Q. 2 4 6 8 10 scribes the relationship between a and y. Solution: Each value of a is exactly twice the corresponding $/ 1 2 3 4 5 value of y. Thus, a = 2, is the required equation. 2. From the table at the right, find the equation which de- scribes the relationship between a and y. (C 1 3 5 7 9 Solution: Each value of a is 1 less than twice the corresponding value of y. Thus, æ =29–1 is the required equation. !/ 1 2 3 4 5 EXERCISES From the tables below, find the equations which describe the relationship between a and y. 1. 4. a | 1 || 2 || 3 || 4 5 a | 0 || 1 || 2 || 3 || 4 Q/ 4 8 12 16 20 $/ 1 3 5 7 9 Equation: y = Equation: y = 2. 5. 30 1 2 3 4 5 Q. 0 1 2 3 4 $/ 5 9 13 17 21 g | – 1 1 3 5 7 Equation: y = Equation: y = 3. 6. Q. 3 6 9 12 15 Q. 2 4 6 8 10 º/ 1 2 3 4 5 $/ 1 2 3 4 5 Equation: a = Equation: a = EQUIVALENT EQUATIONS The equation a =2/ may also be written y = }a. Either can be obtained from the other by dividing or by multiplying through by 2. Such equations are called equivalent equations. When two such equations are graphed, they are represented by the same straight line. {/ EQUATIONS REPRESENTED BY PARALLEL LINES The graphs of the equations a = 29 and a = 2iy–4 are shown at the right as two parallel lines. The coefficient of a is 1 and the coefficient of y is 2 in each equation. P If graphed, the equations ac-2y=4 and 2a:–4y = 12 would be repre- sented by two parallel lines. Notice that 2a:–4y = 12 is equivalent to the equation ac-2J = 6. If the coefficients of x and y in one equation are (or can be made) equal respectively to the coefficients of x and y in a second equation, and the two equations are not equivalent, their graphs are parallel lines. — 158 — SOLVING A PAIR OF LINEAR EQUATIONS We have shown that two linear equations such as a = 3y and g =#2, which can be reduced to the same form, represent the same line, and that two equations such as a =3y and a = 39-H1 represent parallel lines. However, you know that usually two straight lines intersect. If you graph the equations ac-Hy=3 and ac-y = 1 on the same axes, as at the right, the lines will intersect in one and only one point. The point of intersection (2, 1) in this case is called the solution of the pair of linear equations, for it is the only point which is on both lines, and whose coordinates satisfy both equa- tions. To prove our solution we substitute it in each equation. To solve a pair of linear equations (sometimes called a linear system) is to find the coordinates of their point of intersection. This point is called the solution or the root. EXERCISES Solve graphically. 1. ac-Hy=6 3. 3r— y = 4 ac—y = 2 - e ac-H3y = -2 2. 2a:-H y = 6 4. 2a:--y=8 a –2\, = —2 4a –y=4 — 159 — EXERCISES Solve the following pairs of linear equations graphically. 1. ac-Hy=5 4. a +2) = —5 a – y = 1 2a – y = 5 2. 2a:-H y = —7 K. *5. ac—2y=0 52–2y = — 4 2a:-H y = 0 3. 2a:-H39 = 4 *6. 2.--2y=4 —ac-H2y=5 ac-H2y=8 — 160 — ALGEBRAIC SoLUTIONS OF PAIRs of LINEAR EQUATIONS A graphical solution of a pair of linear equations is not always satisfactory. For example, the point of intersection of the two lines might be (3}, 24). In such a case it would be difficult to determine the point with exactness. Other and better methods are available. THE METHOD OF ADDITION AND SUBTRACTION The principle involved in the method of addition and subtraction is to eliminate one of the unknowns by adding or subtracting the given equations. EXAMPLES Steps: (1) Add or subtract the given equations, thus 1. *-Hy=4 eliminating one of the unknowns. 2. 4a:--y = 10 a y=2 (2) Solve for the unknown which is left. *-tt- 1 2a: = 6 (3) Substitute the value obtained in (2) in one of the 3a – 9 a: = 3 given equations, and solve for the second unknown. a = 3 Q) = 1 To check, substitute the solutions in the given equa- y = -2 Solution (3, 1) tions. Solution (3, -2) EXERCISES Solve by use of the method of addition or the method of subtraction. Check. 1. 2a:-y=4 2. 33.--2)/ =8 3. a-H 3y = 7 3a –-y=6 a;+2y = 4 a –2/=2 5. 2.-- y=3 6. r–H5)=–3 7. 3r-H2y=4 –2a:--3y = 1 a – y = 3 a – 29 = 4 The idea of representing an equa- tion by a graph, or of finding an equa- tion which expresses a given relation- ship between a and y, was developed by Descartes. He published the basic prin- ciples of this method in a book called La Geometrie (1637). John Stuart Mill called this “the greatest single step ever made in the progress of the exact sciences.” These principles are now in- cluded in a course in college mathe- matics, known as analytic geometry. The greatness of Descartes' discov- DESCARTES AND GRAPHING William Thompson RENE, DEscARTEs (1596–1650) – 161– 4. 2.r-H y = 2 –2++2y=4 8. r–3y=0 2a –3) =3 ery lies in the fact that it has given men a way to make pictures of abstract ideas and relationships. Making pic- tures of animals and other concrete things dates back thousands of years. Such pictures are found in the pyra- mids of Egypt, and in the caves of the cavemen of prehistoric Europe. But prior to the time of Descartes, mankind had not been able to make a picture of an abstract idea, as (a) the sum of two things is 10, or (b) one thing is 3 more than twice another. EXERCISES Solve each of the following by the method of addition or the method of subtraction. 1. ac-Hy=3 5. 3a;+4y = —7 9. 3a;+2y=8 ac—y = 1 62–4y = -2 3a;+4y=7 2. 2a: —y = 5 6. 2a:-Hy= 8 10. 6a+9y = 6 ac—y = 2 3a;+y= 10 82–9y = 1 3. 32-H4y = –1 7. 3a;+5y = –17 11. 2a:-Hy= –1 ac—4y = 5 3a;+3y = – 9 4a –y= —# 4. a7+671 = 4 8. a7+ 2y = 2 12. 33.-- y = } a —2y = -4 32–29 = 2 32–4y = –1 — 162 — USING MULTIPLICATION IN SOLVING LINEAR EquaTIONS The addition and the subtraction methods, as we have used them, are applicable only if the coefficients of a (or of y) in the two equations are numerically equal. If this is not true, as in example 1 below, we cannot eliminate an unknown by adding or subtracting. However, we can, in this case, eliminate a by first multiplying the upper equation through by 2 and then subtracting. Sometimes it is necessary to multiply both of the given equations by suitable numbers in order to eliminate an unknown. This is true in example 2. After multiplying the upper equation by 2 and multiplying the lower equation by 3, we eliminate y by addition. EXAMPLES 1. ac-H2y=3 2a:--4y=6 2. 2a:--3y=7 4a:-H6g = 14 22–H3y=5 2a:--3y = 5 32–29 = 4 92–6/ = 12 !y = 1 13a; = 26 ac-i-2 = 3 a = 2 4–H39 = 7 Solution (1, 1) a = 1 Solution (2, 1) 3y = 3, and y = 1 EXERCISES Solve for a and y, and check by substituting your solutions in the given equations. 1. a, +y=2 22–H3y = 5 2. 2a – y = 3 42–H3y=11 3. a -6 y = 13 4. 33 – 2/=8 2x+y=0 2a-H3 y = 1 *DIOPHANTINE EQUATIONS Given a single equation with two unknowns, one can find an unlimited number of solutions. Two equations in three unknowns also have an unlimited number of solutions. This is true of any system of equations in which there are more unknowns than there are equations. Diophantus (c. 275 A.D.) studied such systems of equations; after him they are called Diophantine equations. Some of the well-known trick problems of today are of this type. Problem: A man paid $100 for 100 animals (calves, lambs, and pigs), paying three dollars for each calf, one dollar for each lamb, and fifty cents for each pig. How many of each sort could he buy? In Solving, let a, y, and 2 represent the number of calves, lambs, and pigs, respectively. Then 3a;+y+}z = 100 (The total cost was $100.) And a +y+ z = 100 (There were 100 animals.) 2a – #2 = 0, or z = 4a: And y = 100–53. (By substituting 2 = 4a in the sec- Ond of the given equations.) Using the equations y = 100–5a, and z = 4a:, let a = 1, 2, 3, . . . and find the corresponding values of y and 2. These are given in the table below. Any one of these sets of values satisfies the condi- tions given in the problem. 3C 1 2 3 4 5 6 7 8 9 10 || 11 | 12 || 13 || 14 | 15 18 19 y 95 90 | 85 80 | 75 70 65 | 60 55 50 45 40 35 | 30 25 | 10 5 2 4 8 12 16 || 20 24 || 28 || 32 || 36 | 40 || 44 || 48 || 52 56 60 | 72 76 Solve for a and y. 1. 2a:–3y = –1 32–29 = 1 2. 22-69 = – 2 2a:--4y = 10 3. 42–7 y=29 32–2y = 12 4. 5. 6. EXERCISES 52–29 =5 4a:--7y = 4 7a;+2y = – 8 2a:--5y = —20 42–7 y = 10 —52–3) = 11 2a+5y=6 *8. —22+4y = 9 72–99 = –1 **9. az-i- by =c ma;+?vy=k —"164 — USING THE METHOD OF SUBSTITUTION IN SOLVING EQUATIONS Solve one of the given equations for either of the unknowns in terms of the other, and substitute this value in the second equation. 1. ... } 2a:--3y =7 a = 29 2(2g)+3y = 7 7ty = 7 gy = 1 a = 29, or a = 2 Solution (2, 1) 1. a = 29 2a – y = 10 2. a = −3y 3a;+4y=5 3. a =y+1 22–30) = 2 > EXAMPLES Given equations. Solve the first equation for a in terms of y. Substitute a value in the second equation. Simplify, combine terms, and solve. EXERCISES 4. ar-H4y = –14 22–5y = 11 5. 62–7 y = –1 2a-H y = 3 6. a--7 y = —6 2a:-H39 = –1 2. º: 3a –4y = 10 1–39 2 º) 3| —— ) —47/ = 10 ( 2 $/ 3–94) —8, = 20 3–17 y = 20 – 17 y = 17, or y = –1 Q. 1–3y T 2 T 2 Q. *7. 2a-H3/= — 5 ac—4y = 14 *8. 2a –30) = 7 ac-H7 y = —5 *9. 3a –77) = 10 2a:--9y = – 7 — 165 — Solve by any method you choose. 1. 2a:-Hy=7 3a –y=8 2. 32–y = -2 2a – y = –1 3. –a;+5y = — 9 ac—7ty = 11 4. a7+2y = 15 22–39 = –19 REVIEW EXERCISES 5. 22– y = —5 32–29 = 4 6. – a – y = 1 –52+6/=27 7. 32— y = 17 a;+3y = 19 8. a7+2y=2 3a –2y=2 9. —2a-H y = 2 6a-i-3y = —6 *10. 22–31) = —2 32–2y = 2 *11. 22-H7/= –14 52–2 y = 43 *12. az-i-by =c da:-ey=f — 166 — REVIEW EXERCISES Solve by any method you choose. Check. 1. 2a:-H y = 1 - 5. 52–99 = 37 9. 9a;+ y = 5 3a;+2y=3 a — y = 5 6a+5y = 12 2. 32–29 = 0 6. ac—2y = 1 *10. 52--39 = 15 42-H2y = 14 2a:--5y=20 32–39 = 9 3. 52--3y = 17 7. 52–3) = 0 *11. 2a:--5y = 9 32–59 = 17 22–4y = –14 32-H2y=30 4. 15a–H y = –15 8. 32–99 =24 *12. — 11a;+ 6y =29 10a;+7y =66 5a-H79 = 40 ac-18y =41 — — 167 — NUMBER PROBLEMs Heretofore we have solved number problems by using one equation in one unknown, but many times they can be solved more easily by using two equations in two unknowns. EXAMPLE The sum of two numbers is 19. Six times the smaller exceeds four times the larger by 4. Find the numbers. Set up the equations: Solve the equations: Let a = the larger number 4a:-- 4y = 76 and y = the smaller number. –4a:-H 6y = 4 Then ac-Hy = 19 10y = 80, or y=8 and 6y–4a: = 4. a;+ 8 = 19, or a = 11 In checking it is better to go back to the given word problem rather than to the equations you set up. It is possible that the equations may be wrong. Check: The sum of the two numbers, 11 and 8, is 19. This checks. Six times the smaller, 48, exceeds four times the larger, 44, by 4. This also checks. PROBLEMS 1. The sum of two numbers is 26 and their dif- ference is 8. Find the numbers. Check: 2. The difference of two numbers is 10 and their sum is 48. Find the numbers. Check: 3. The sum of two numbers is 15. Three times the first number added to four times the second is 55. Find the numbers. Check: 4. The sum of two numbers is 30. Four times the smaller exceeds three times the larger by 1. Find the numbers. Check: — 168 — NUMBER PROBLEMS 1. The sum of two numbers is 100, and twice the larger equals eight times the Smaller. Find the numbers. Check: 2. Separate 70 into two parts so that the larger is five times the Smaller. Check: 3. The sum of two numbers is 64. If the larger is divided by the smaller, the quotient is 15. Find the numbers. Check: *4. One third of the sum of two numbers is 12, and one fourth of their difference is 3. Find the numbers. Check: *5. A certain fraction equals 1 when 2 is added to its numerator and equals 7 when 8 is subtracted from its denominator. Find the fraction. Check: **6. Find two numbers whose sum is a and whose difference is b. Check: — 169 — PROBLEMs ABOUT PLANE FIGURES C G L O E F H K M N A B Facts to remember about the figures above: The perimeter of a rectangle is expressed by the formula The sum of the three angles of any triangle is 180° (Fig. 2). Figure 3 is an isosceles triangle. Two sides, LH and LK, are equal, and two angles, H and K, are equal. Figure 4 is a right triangle. One angle, M, is 90°, and the sum of the other two, N and O, is 90°. Figure 5 represents two complementary angles, R and S. Their sum is 90°. Figure 6 represents two supplementary angles, T and U. Their sum is 180°. PROBLEMS Solve the following, using equations with two unknowns. 1. The perimeter of a rectangle is eighteen inches and the length is three inches more than the width. Find its dimensions. Check: 2. The sum of two angles of a triangle is 120° and their difference is 30°. Find all three angles of the triangle. Check: 3. The perimeter of a triangle is 32 inches. The sum of two sides is 19 inches and their difference is 5 inches. Find all three sides of the triangle. Check: 4. An angle is 3} times its supplement. Find the angle and its Supplement. Check: — 170 — PROBLEMS 1. The sum of two unequal angles of an isosceles triangle is 158°, and # their difference is 57°. Find the three angles of the triangle. Check: 2. The difference of the two smaller angles of a right triangle is 21°. Find all angles of the triangle. Hint: The sum of the two angles equals what? Check: 3. One half of an angle and one third of its com- plement together equal 40°. Find the angle and its complement. Hint: The sum of the two angles equals what? Check: 4. How large is an angle which equals one fifth of its supplement? Check: **5. If the length of a certain rectangle is in- creased by one inch and its width is increased by one inch, its area is increased by thirty-one square inches. If its length is decreased by two inches and its width is decreased by three inches, its area is de- creased by seventy-four square inches. Find its 8,I'628,. Check: — 171 — MIXED PROBLEMS 1. In a collection of cents and nickels there were twenty-three coins amounting to forty-three cents. Find the number of nickels and the number of CentS. Hint: a nickels = 52 cents. Check: 2. When one can buy two pounds of butter and four loaves of bread for $1.16 or one pound of but- ter and six loaves of bread for $1.06, what is the cost of a pound of butter? Of a loaf of bread? Check: *3. Find a two-digit number such that the sum of the digits is ten and the tens digit exceeds the units digit by four. Check: *4. Six years ago a father was three times as old as his daughter, while nine years hence he will be twice as old as she. How old is each now? Check: **5. A man invests $6,000 in two mortgages, one paying five percent and the other six percent. His annual interest from the two investments is $320. How much does he invest at each rate? Check: — 172 — *FRACTIONAL EquaTIONS IN Two UNKNOWNS EXAMPLES Solve for ac and y and check. - - - - *- The subtraction method may be used to elimi- nate a. When the second equation is subtracted 5 5 from the first, we get – - This gives y = 6. If Q/ we substitute y = 6 in the first equation, we get a = 4. 2 3 2 2 1 These values check for —–H–= 1, and ———= −. 4 6 4 6 6 1 1 3 3 + 2 149 a y 10 a. !/ - LTO Multiply the first equation by 2 and add to the 5 Second equation. We get † = a or a = 2. If we QC Substitute a = 2 in the first equation, we get y = 5. These values check, for –––=— , and —-i-— 2 5 5 = 11%. In solving two equations in two unknowns in which the unknowns occur in the denominators, do not clear of fractions. Proceed as in the examples above. 1 1 The fractions — and — are said to be the reciprocals of a and y, respectively. (C !/ EXERCISES Solve for a and y. 3 2 1 6 2 1 1 — 1 2 3 5 a: y 2 a: y 4 a: y 12 30 2/ 12 3 4. 1 1 1 1 1 1 — 1 O ———=~ ; - – `- - 5. ——H–= 1 ; - - - - - a: y 30 a y 6 a 3/ 2 a y 12 3 1 2 6 5 + 1 6. —-H 1 1 1 2 a y 5 a y 2 a $/ T2 " … yT — 173 — PROBLEMS USING RECIPROCALS 1. The sum of the reciprocals of two numbers is #, and the difference of the same two reciprocals is #. Find the two numbers. 2. Five men and eight boys working together can do a job in three days, while one man and eight boys can do the same job in five days. How long would it take one man to do the work alone? One boy alone? 3. One man and one boy can do.3 of a job in one day. The man and three boys can do the whole job in two days. How long would it take the man alone to do the job? How long would it take one boy alone to do the job? 4. The sum of the reciprocals of two numbers is #. Four times the reciprocal of the first plus three times the reciprocal of the Second equals 1. Find the numbers. 5. Solve for a and y when 1 2 5 1 1 1 =— ; ––– a: y 6 2a, 3y 4 6. Solve for a and y when 3 2 – 5 1 3 5 a: y 6 Q} — 174 — INVENTORY TEST 1. Find the coordinates of three points which satisfy the equation a -39 = 0. ( 2 ), ( 2 ), * ). N. y 2. At what point will the line 32+4y = 10 cross the ac-axis? The y-axis? — 3. Can you find a single point that will satisfy a pair of linear equations (a) if they are parallel? –– (b) If they are equivalent? (c) If they intersect? 4. How many points will satisfy the equation ac-39 = 10? 5. How many points will satisfy both of the equations a +y=4 and a -y = 6? 6. By inspection pick out two points which may be used to graph the line for the equation 32+4y=12. ( , ) ( , ) 7. Write the equation of the line passing through the points (3, 1), (6, 2), (15, 5), (60, 20), etc. 8. Solve each of the following linear systems by inspection and place the root in the parenthesis at the right of each. (a) ac-Hy=4 ( ) (b) 23:-Hy=4 ( ) (c) m-H m = 4 ) ac—y = 2 ? a — y = 2 y 4m-H n = 1 9. The three angles of a triangle equal degrees. 10. The complement of a 40-degree angle contains degrees. 11. Write the equations which you would use in solving each of the following problems: (a) The sum of two numbers is 37, and their difference is 3. Find the numbers. (b) Five pounds of coffee and three pounds of tea cost $3.30, and four pounds of coffee and seven pounds of tea cost $5.40. Find the cost of a pound of coffee and of a pound of tea. (c) Find an angle which equals three fourths of its supplement. (d) The sum of two numbers is a and their difference is b. Find the numbers. y 12. If a motorboat can travel a miles per hour in still water, what is its rate (a) upstream, and (b) down- stream, on a river whose current is 3% miles per hour? (a) —--— – (b) 13. How many points will satisfy both of the equations a +y=1 and 22-H2y=2? *14. How many points will satisfy both of the equations ar—Hy = 1 and ac-Hy=4? **15. John has fifteen coins (quarters, dimes, and nickels) whose value totals $1.50. How many of each could he have? Quarters Dimes — — Nickels — 175 — CUMULATIVE REVIEW 1. Simplify by performing the indicated opera- (e) Their product is 7. tions. (a) ac-Ha!-Ha!-Ha! = (f) Their quotient is #. (b) (–33:4) (#25) = (g) The sum of their squares is 25. (c) (–328)?= (d) (ac-Hy) (2–y) = (h) Their product is a. (e) a”. a”, a”. a”= (i) One half their difference is 4. (f) VAmº– (j) They are supplementary angles. (g) (–3)4= (h) (m?--8m--16) -- (m-H4)= (k) They are complementary angles. (i) (–3a*b-i-2a) (ab) = i) —4(–3–Ha:) = (j) ( ) (1) One equals twice the other. 2. Express each of the following in algebraic terms, using a for the unknown number. (m) One equals one third the other. (a) A man's age ten years hence_ (n) The difference of their cubes is —27. (b) A man's age five years ago (c) The complement of an angle (o) Their sum divided by their difference is 15. (d) The Supplement of an angle (e) Six decreased by one half a number (p) Their sum divided by their difference is — 15. (f) The part of a piece of work a man can do in one day, if he can do it all in a days. 4. The graph of the equation ac-H2y = 12 crosses the ac-axis at what point? The y-axis at what point? 3. If a and y represent two numbers with a larger than y, write the equations for the following: 5. The equation 32-H46, =0 crosses the ac-axis at (a) Their sum is 4. what point? — (b) Their difference is 2. a; a 5 6. Is 6 a root of the equation 2 + 3 =—? (c) Their quotient is 3. 2 (d) Their difference is –2. * =s=====s**E=º — 176 — UNIT 11. RATIO, PROPORTION, WARIATION EARLY BEGINNINGS OF MATHEMATICs No one can tell now among what strange people, or in what dim, distant past, mathematics had its earliest beginnings. Those who and time do not easily destroy. From the writings upon these bricks, we have learned the interesting number system used by these know most about very early times say that mathematics had reached rather advanced stages of development by 3500 B.C. in at least three regions—in China, in Mesopotamia, and in Egypt. Of these three, we are least sure about China. About 200 B.C., the emperor had all books and manuscripts burned. Thus, Chinese claims of mathematical antiquity rest upon tradition and upon second-hand knowl- edge, yet there is little doubt that mathematics thrived in China many centuries before the beginning of the Christian GI’8. Explorers in Mesopotamia have dug away great mounds of peoples, and the applications they made of their mathemat- ics. Our knowledge of ancient Egyptian mathematics comes from inscriptions in their tem- ples and pyramids, and from two rather long mathematical writings. These writings are upon long rolls of papyrus. They give us the best idea we have of the advancement of mathemat- ics in Egypt at the time they were written. Each of these peoples used such mathematics as was needed to survey fields, to plan and erect buildings, to care for crops and herds, to pay and collect taxes, and to carry on banking earth, uncovering the ancient W** cities of Nineveh, Nippur, and Babylon. The Sumerians, Baby- RUINS OF ANCIENT BABYLON and commerce. What they knew has come to us; from whom they learned it we do not know. To lonians, and others who lived there in ancient times wrote upon soft clay bricks and then dried or burned the bricks, thus making a record that fire China, to Babylon, and to Egypt of about 3500 B.C. is as far back as scholars have yet pushed the curtain which covers the past of mathematics. RATIO The ratio of one number to another is the indi- cated quotient of the first number divided by the second. Sue is one half as old as John. Sue might be 7 and John 14. The ratio of their ages is +1 or 4. A ratio may also be written 7:14. This is usually read “7 is to 14,” but it may be read “7 divided by 14.” Mary is 8 years old, and Charles is 12 years old. The ratio of their ages is Mary weighs 60, and Charles weighs 96 pounds. The ratio of their weights is John has 18 cents and Sue has 12 cents. The ra- tio is +3 or 18:12. When reduced to lowest terms, this fraction is #. It is helpful to understand that ratio is nothing more than a new name for a frac- tion. In order to have meaning, a ratio must refer to like things. The ratio of 10 books to 15 books is 10:15, or 2:3, but the ratio of 10 books to 15 apples has no meaning. The ratio of 3 feet to 9 inches can be given meaning by writing the feet as inches. The ratio is 36:9, or 4:1. Similarly, the ratio of 3 pounds to 6 ounces is 48:6, or 8:1. EXERCISES Write these ratios as fractions and reduce the fractions to lowest terms. 10 feet:2 yards 2 pounds:8 ounces 9 cents: 15 cents 4 inches: 18 inches — 177 — EXERCISES Simplify the ratios by writing as fractions and reducing to lowest terms. 1. 20 to 25 = 8. (a-b): (a^–2ab-i-b?)= 2. 15a to 21a: = 9. (c-d): (d—c) = 3. 36 in. :4 in. = 10. (–3):# = 4. 1 ft. :3 yd. = 11. #:# = 5. 1.2:4.8 = 12. 22:2y2= 6. $2:15 centS = 13. (4a:*-i-4ay--y”): (y-H2+) = 7. 922: 2724 = 14. 4 qt.:6 pt. = EXAMPLE Find two complementary angles whose ratio is Let 2x =no. of degrees in one angle, and 2:3. 32 =no. of degrees in other angle. Answers: 36° and 54° Then º º-s, PROBLEMS 1. Divide 99 into two parts whose ratio is 8 to 3. Answers: 2. The perimeter of a rectangle is 40 feet. The ratio of its width to its length is 2:3. Find width, length, and area. Answers: 3. Air is composed principally of nitrogen and oxygen in the ratio of 4:1. Find the number of cubic feet of oxygen in a room 20 feet by 18 feet by 10 feet. Answers: *4. The angles of a triangle are to each other as 1:2:3. Find each angle. (1:2:3 means the ratio of the first angle to the second is 1:2 and of the Second angle to the third is 2:3.) Answers: — 178 — PROPORTION 1 12 A proportion is formed by equating two equal ratios. Examples: 3 T36 is a proportion; it may be read “1/3 equals 12/36.” 1:3 = 12:36 is a proportion; it may be read “1 is to 3 as 12 is to 36.” EXERCISES Supply the missing term in each proportion below. 1 3 2 4 6 2 1 Ol. 1. – = — 2. — = — 3. — = — 4. — = — 5. — = 6. — = — 2 3 4 24 3 b TERMS OF A PROPORTION O. C. The proportion º, T., may also be written a b = c :d. The numbers a, b, c, and d are called the terms of the proportion. The terms a and d are called the extremes, and b and c are called the means of the propor- tion. Stated differently, the first and last terms of a proportion are the extremes; the second and third terms are the means. FUNDAMENTAL PRINCIPLE In any proportion, the product of the extremes equals the product of the means. This is immediately O, C evident if we multiply each term of the proportion º, T. by bà. The resulting equation is ad=bc, or the product of the extremes equals the product of the means. This principle furnishes an easy method of Solving an equation written as a proportion. EXAMPLES Solve for ac. * ac-i-2 ac-7 1. 9:4 = ac:8 2. F. 4a: = 72 4 5 a = 18 5ac-H 10 = 4r — 28 a = — 38 EXERCISES Solve for ac. 1. *-*. 6. 4: a = 9:11 3 6 2 4 3 7. – 4:8 = r : —2 a 9 1 8. #:# = x:# 3. — = — 9 81 a –3 at-4 9. a. a = b : c 4. - 8 9 (C 7??, 5. 24; ; 7 = 4:9 10. -; — 179 — PROBLEMS Set up the proportions and solve. The first has been done correctly. 1. The taxes on a home valued at $11,000 are $275. What should be the taxes on a home valued at $9,000? Answer: $225. Let a = the amount of taxes on the second home. Then ac: 275 = 9,000: 11,000, or ac: 275 = 9:11. 11a: = 275.9 = 24.75 a = 225 2. If eight pounds of coffee cost $2.80, how much will twenty-seven pounds cost? Answer: 3. If four yards of cloth cost ninety-two cents, how much will eighteen yards of the same kind of cloth cost? Answer: 4. A man used twelve gallons of gasoline in driving 174 miles. At the same rate, how many gal- lons would he use in driving 261 miles? Answer: 5. A wheel goes 144 feet in eighteen revolutions. How many revolutions are needed to go one mile (5,280 feet)? Answer: 6. An architect makes a building plan so that three inches represents twenty feet. A distance of eight inches on the plan represents how many feet? Answer: — 180 – SIMILAR TRIANGLES Two triangles are similar if they have the same shape. The triangles ABC and DEF are similar tri- angles. In similar triangles, the corresponding angles are equal. This means angle A equals angle D, angle B equals angle E, and angle C equals angle F. In similar triangles, the corresponding sides are A proportional. This means c.f-a: d=b: e. EXAMPLE How high is a tree that casts a shadow of twenty feet when a ten-foot pole casts a shadow of four feet? Answer: 50 ft. / PROBLEMS F 6 d B D f E Let a = height of tree a 20 Then — = — 10 4 4a = 200 a = 50 1. A building casts a shadow of seventy-five feet. At the same time a post eight feet high casts a shadow of three feet. How high is the building? Answer: 2. A post twenty feet high casts a shadow of thirty-five feet. At the same time, the shadow of a tower is 665 feet long. How high is the tower? Answer: 3. The hypotenuse of a right triangle is twenty- four feet and its base is seven feet. Find the hy- potenuse of a similar right triangle whose base is 10.5 feet. Answer: *4. Given similar triangles ABC and DEF, such that AB = 7, BC =4, CA = 6, DE = 10.5. Find EF and FD. Hint: See the figure at the top of this page. Answers: — 181 — DIRECT WARIATION It costs three cents a mile to travel by a certain railroad. Thus, the cost of a trip will depend on the number of miles in the trip, or C =.03m, where C represents the cost in dollars and m represents the number of miles. In this formula, C and m are called variables because they may have a succession of different values. Notice that the value of C depends upon the value of m. For this reason C is called the dependent 10 9 Variable, and m is called the independent variable. 8 The formula P=3s holds true for all equilateral triangles. The 7 value of P depends on the value of s, as is shown in the table. 6 S S 5 S | 1 || 2 || 3 || 4 || 5 || 6 || 7 || 8 || 9 || 10 4 S P = 3S P 3 6 9 | 12 | 15 18 21 24 || 27 || 30 In studying the table above, you will notice: (a) As S increases, P increases; as s decreases, P decreases. (b) P increases (or decreases) three times as fast as s. The ratio between s and P is always the same. When two quantities are related in this way, we say that P is directly proportional to s, or that P varies directly as s. A graph of the linear equation P=3s is shown at the right. 1 2 3 4 EXERCISES In exercises 1 to 4, (a) find the constant ratio of the first variable to the second, and (b) write the equation which expresses the variation. 1. 3. Q. 4 12 16 20 t 10 20 30 40 50 Q/ 1 3 4 5 ?!, 1 2 3 4 5 Constant ratio: Constant ratio: Equation: a = Equation: t = 2. 4. S 1 3 4 5 'll # 1 1% 2 2% 7. 4 12 16 20 t 5 10 15 20 25 Constant ratio: Equation: s = In exercises 5 and 6, (a) supply pairs of values which have the given ratio, and (b) write the equation of the required relationship. 5. Constant ratio is sº P- 1:4. S P Equation: s = Any equation of the type y = ca. represents direct variation. In this equation, a is the independent vari- able, y is the dependent variable, and c is the constant. It is possible to solve this equation for ac. Thus, Constant ratio: Equation: w = 6. Constant ratio is r. S=3:5. r S Equation: r= 1 a = — y. After solving, y is the independent and a is the dependent variable. C — 182 — *INVERSE WARIATION John is to walk from A to B, a distance of twelve miles. Two variables are involved: the rate and the time. Rate multiplied by time gives distance, or r; t = 12. Notice (a) there are two variables r and t, and a constant 12; (b) the product of the two vari- ables equals the constant; (c) as either of the vari- ables increases, the other decreases; (d) the equa- tion is of the second degree. The equation ri=12 can be solved for either r 12 or t. In the form r–7. t is the independent and 12 r is the dependent variable. In the form t = — , r is r the independent and t is the dependent variable. EXERCISES 1. Finish the table below by finding the values of t that correspond to the given values of r in the equation r , t = 12. r | # | 1 || 2 || 3 || 4 || 6 || 8 || 12 | 18 |2. t 2. Plot the pairs of values shown in the table above on the graph at the right. Join your points with a curve. 3. Use the table above to fill the blanks: (a) As r increases, t (b) As r decreases, t (c) As t decreases, r : : 25 20 15 sº |- 1 f | 1 || || 1–1–1– I-1–H–H–1 H 1—1–1–1–1–1–1–1–1–1–1–1–1–1 5 toº---T: '20' ' ' ' 25"T" so 4. Does the graph you have made check with your answers in exercise 3? 5. The product of each pair of values in the table above is DEFINITION OF INVERSE WARLATION If the product of two variables is equal to a constant, then either variable is inversely proportional to the other. We say that one varies inversely as the other. h EXERCISES 1. The area of a rectangle is thirty Square inches. (a) Write the equation which expresses this relationship, using w and h as the variables. (b) Complete the table below. (c) Plot the pairs of values shown in the table, and join the points with a, CUITVe. Equation: w) || 1 || 2 || 3 || 5 || 6 || 10 | 15 20 h 2. An inverse variation is illustrated in the table at the right. Determine the constant product, and write the equation for the inverse variation. Constant product: Equation: – 3. In direct variation, the 30 . : 25 20 * f { | I tº 1 1 1 t t t l—1–1–1–1–1–1–1–1–1–1–1–1–1 I-I-I-I I-I-II IT II * TITUTI s --- io" -"sº-º-º-º-º: go to )" 5 || 6 || 7% 10 | 12 15 l 12 10 8 6 5 4 tion, the of the two variables equals a constant; in inverse varia- of the two variables equals a constant. — 183 — GENERAL DEPENDENCE Algebra deals largely with the relationship be- tween quantities, or the dependence of one quan- tity upon one or more other quantities. We have studied this dependence in formulas, equations, problems, tables, graphs, proportions, and varia- tions. The idea of dependence is, however, much more general than has yet been indicated. Instead of a dependent and an independent variable, there may be a dependent and two or more independent variables. The area (A) of a rectangle depends upon two variables, the width (w) and the height (h); the volume (V) of a rectangular box depends on three variables. In many cases a relationship cannot be expressed as a mathematical formula. DISCUSSION EXERCISES The price of steak depends upon what? : The college one attends depends upon what? The amount of wheat grown in the United States depends upon what? . The fuel cost of keeping a home warm in winter depends upon what? Whether or not one does good work in school depends upon what? * FUNCTIONS Consider such equations as y=3a, y=a^, y=5-2a. In every such equation, the quantities ac and y are related so that a change in one makes for a change in the other. We say that the value of y depends upon the value we give to ac, or that y is a function of ac. We call a the independent variable and y the dependent variable. EXERCISES Write an equation showing the relationship expressed in each of the following. 1. The perimeter (P) of a square depends upon its side (s). 2. The area of a triangle (A) with base 10 depends upon its altitude (a). 3. The area of a circle (A) depends upon its radius (r). 4. The circumference (C) of a circle depends upon its diameter (d). 5. The total cost (T) of thirty automobiles of the same kind depends upon the cost of each (c). 6. The amount of interest (I) depends upon the time (t), the rate (r), and the principal (p). In statements like those above, the phrase “depends upon” can be replaced by “is a function of” without changing the meaning of the Sentence. PROBLEMS 1. The speed of a motorboat in still water is eight miles an hour; the rate of the current is c miles an hour. (a) Write the formula for the distance the boat can go upstream in One hour. D = (b) Write the formula for the distance the boat can go upstream in t hours. D = (c) Find D when c=3 and t = 5. (d) When c=2, how long will it take the boat to go forty miles upstream? t = (e) Write the formula for the distance the boat can go downstream in thours. D= D = (f) When c=2, how long will it take the boat to go forty-five miles down- stream? *2. Mr. Brown rents A acres of corn land from Mr. Jones, agreeing to pay the latter three dollars an acre plus one fifth of the crop raised. Mr. Brown raises C bushels of corn worth d dollars a bushel. Write a formula showing Mr. Jones’ income for the year. — 184 — *THE LAws of THE LEVER The following proportion holds true for all types of levers, W1: W2 = D2: D1, where W1 is a weight, W., is another weight, D1 is the distance of the first weight from the fulcrum, and D2 is the distance of the second weight from the fulcrum. A w KWA Ø {S) º º *—A-3 TYPE I TYPE II TYPE III PROBLEMS 1. John and Mary weigh forty-two pounds and 52% pounds respectively. If John sits five feet from the fulcrum on one end of a seesaw, how many feet from the fulcrum on the other side should Mary sit that they may balance? 2. Two boys carry a load of seventy-five pounds Cn a pole between them. If the load is four feet from one boy and six feet from the other, how many pounds does each carry? 2" 3. What effort must be expended at Wi to lift the weight shown in the figure below. (40 lbs. K. : V W F W2 to W1 =3 ft.; W1 to F=7 ft. **4. A lever is five feet long. Where must the ful- crum be in order that an effort of fifty pounds at One end will lift a weight of 200 pounds at the other end? / **5. A nut cracker is eight inches from hinge to end of handle. What pressure is exerted on a nut 1% inches from the hinge when a pressure of twelve pounds is brought to bear } inch from the ends of the handles? — 185 — INVENTORY TEST 1. Express each of the percents as a ratio, and reduce to lowest terms. (a) 20% = (b) 75% = (c) 163% = (d) 37%% = 2. Supply the missing terms to make each of the following a proportion. 1 Q. ac-y aſ”—y” (a) —=— (b) —=— (c) == 3 12 * 24/ ?/ ac-Hy 3. Solve for the letter indicated. (a) a 3 (d) h;2 = 4:7 h = ( yºtº 3 a, ) — = — = - – QC = 2 = 2 * g)---, 1 y (e) 7: m = 8:9 m = g-H2 3 7 14 g–2 4 4 f) S. c = die s = ſm — 4 9 (c) — = — m = (f) (i) — = — m = "m — 7 n—8 5 4. In two similar triangles, the corresponding 2.Te , and the corresponding al’e 5. Write three different ratios each equal to #. (a) : (b) : (c) 3– 6. Complete the proportions: (a) 3:7 =9: (b) 5:2= .4 (c) m: n = 2m: 7. Given the formula d = 30t. As t increases, d , and as t decreases, d — Therefore d varies as t. 8. When a post twelve feet high casts a shadow of eighteen feet, a tree which casts a shadow of ninety feet must be feet high. 9. In direct variation, the of the two variables is a *10. In inverse variation, the of the two variables is a 11. Jn the equation y = 4a2+3a;+5, y is called a of ac. 12. In the equation v =#gt”, t is the variable, v is the vari- able, and g is a constant. 13. Other things being equal, the number of oranges one buys varies as the money spent for Oranges. 14. In the equation =r j is the independent variable, and is the dependent variable. *15. Other things being equal, the time required to do a piece of work varies 2,S the number of men employed. *16. Given the formula r--- As i increases, r , and as t decreases, r—— Therefore r varies as i. 17. In any proportion the of the means equals the of the 18. Find two complementary angles whose ratio is 1:5. and 19. A ratio is an indicated ; the equality of two ratios is a 20. Two neighbors paid city taxes of $120 and $132 respectively. The assessed valuation of the first property was $5,400. Write the proportion you would use in finding the assessed valuation of the second property. — 186 — CUMULATIVE REVIEW Place on the blank following each first column statement the letter which identifies the Second column statement which correctly completes it. 1. The area of a trapezoid is (a) (4, 0). 2. 2:y may be written (b) 24. 3. (a.”—a —20) + (a –5) = (c) is a proportion. a-Hb 20-H.2b 4. —-i- f — (d) variables. a —b a -b bc 5. An equality of two ratios (e) —. Ol. Q: C - 6. If -=— , then a = — (f) —3. b a *-*- ac-H 1 7. V–27a:* = (g) -> a – 1 ac2–9 1 8. & f — (h) 5, 0. 3 ac-H3 9. The root of an equation is (i) — 15. 10. If a bushels of apples cost y dollars, then 2 bushels will cost (j) ar–H1. 11. (a.”–H52-H4) + (2-H4)= (k) #h (b.1+b2). 12. In multiplication we add (l) ac-H4. a — 3 13. The lines ac-Hy=4 and 22-y-8 intersect at the point (m) 3.T.’ 1 1 14. ( ++)-( –)- **-* (n) the number which satisfies it. Q: Q. 15. When g-a-4, 2 * — (o) exponents of like letters. g b 16. If –=— , then y = (p) #. Ol C 17. In the equation ac-Hy=6, as and y are called Q/2 (q) – dollars. Q} Q. 18. When a = −2, acº—ac”-Ha!— 1 = (r) —. Q/ 19. The line ac-Hy = 5 crosses the ac-axis at the point (s) —3a.”. º ab 20. One root of a 2–H 7a;+12 = 0 is (t) — . C — 187 — Many large companies maintain schools for their employees. The Henry Ford Trade School of Dearborn, Michigan, is an excellent example of this type of education. This school is for boys between the ages of twelve and eighteen years who must support themselves while they learn. The course consists of four years each of English, mathematics, mechanical drawing, and shop theory. In mathe- matics, the boys review (23%)?=2x2.24% = 4x4 parts of arithmetic and study algebra, geometry, and trigonometry. The school works on the principle that young people should learn to do and to make useful things as a part of their education, and not spend their time merely practicing on the making of things which are to be thrown away. The picture shows a class of boys doing school work which will later be used in industry. The company pays them for this work. (3.c)*=34::34: , 3.c-27a.” To raise a monomial to a given power, raise the coefficient to the required power, and multiply the exponent of each letter by the exponent of the power. 1. (–22)*= 2. (20°c)?= 3. (54%)*= (z+y)*=a^+2+y+y” 8. (– : a*b)*= (3m—4n)?=9m?–24mm-H16m* The square of a binomial equals the square of the first term, plus twice the algebraic product of the two terms, plus the square of the second term. 1 . (26–d)*= 2. (44-H 5y)*= 3. (e–ary)*= 4. (3ab-H4)*= 5. (6m – #n)*= UNIT 12. POWERS AND ROOTS Courtesy Henry Ford Trade School SHEET METAL DEPARTMENT, HENRY FORD TRADE SCHOOL PoweRS OF MONOMIALS EXAMPLES (...) aſ a rº 2/T2 2 T 4 EXERCISES 4. (–4y”)*= 33:2 2 s (...)- 4y 6. (ay”)*= 7. (54%)? - *9. (ra)* - PowLRs OF BINOMIALS EXAMPLES EXERCISES *6. +7. **8. **9. **10. — 188 — (a 2–0)?= (a2+ax+)*= (a"—y")*= *a*—b)*= (a^+av)*= ROOTS AND RADICALs When a given number is the product of a second number taken two or more times as a factor, this second number is called the root of the given number. Specifically, the square root of a number is one of two equal factors, and the cube root of a number is one of three equal factors of the number. EXAMPLES 1. 4 = 2.2, or 4= —2. –2 4. y”= y: y : y 2. a.”= a -a, or a:”= —ac. –a: 5. – 8a? = — 20 — 20 —20. 3. 27 = 3.3.3 In example 1, either +2 or -2 is seen to be a square root of 4; similarly in example 2, either +3 or -a. is seen to be a square root of ac”. The cube root of 27 is 3; the cube root of y” is y; the cube root of -8a” is -2a. The radical sign V placed over a number indicates that a root is to be taken. The small number, index, placed in the V of the radical sign indicates what root is to be taken. When the index number is 2, or when no index number is given, the square root is indicated. EXAMPLES V4=2 ~/9=3 ~/8=2 ~/27=3 A radical expression may be preceded by a + sign, as +V4, or by a - sign, as – V4. When no sign is written, the + sign is understood. We observed above that either +2 or – 2 is a square root of 4. In practice, we usually write + V4 when we mean +2, and we usually write – V4 when we mean –2. EXAMPLES +V9= +3 + V16 = +4 — A716 = — 4 — V/ac2 = — ac We cannot now find the square root of a negative number. Why? In higher mathematics a new kind of number is studied, by means of which one can represent the square root of a negative number. EXERCISES Find the indicated roots of the following: 1. V4 = 8. V8 = 15. – V-º- Sºº-ºº: 5 r2014 = 2. –V4 = 9. — V8= 16. v. 25.cºy *17. V-12529– 3. – V25a.” = 10. V-8 = / 1. *18. — i- 4. H- V.9b%= 11. Vas– 36b 3/~8 /SAZ2. 3 * *19. —- sº 5. V/64c” = 12. -- Var* = Q9 :R —“ 8 = 6. – V% = 13, V-27c3= 20. — V81c 3 — 06m &= - **21. == 7. — Vac2c2 = 14. V-º- | 27 — 189 — SQUARE ROOTS OF ARITHMETIC NUMBERS EXAMPLES 2-> Procedure: Divide the given number, 1,369, into two groups of two figures each. 6 9 |37 The first group is 13 and the Second group is 69. Find the largest square contained in the T first group. This is 9. The square root of 9, or 3, is the first answer-figure. Write 3 in the quotient position. Subtract 9 from 13; bring down the second group. This makes the new dividend 469. Double the 3 and write to the left of 469. Find the number of times 6 is contained in 46. This is 7. Annex the 7 to the 6, making 67, and annex 7 to the 3, making 37 in the quotient position. Multiply 67 by 7 and write the product under 469. Subtract. The remainder is 0. The square root is 37. 1. 3 *— 67 i } . *==e 0 0 0 ,--> 82 si |01 T. 69 |1.3 8 1 1 is T : 23 1 8 1 0 0 0 ,-\ 0 0 |1.73+ º 27 } . . *E-º- iº 0 0 343 | . 0 0 gº 2 9 In marking a number off into groups of two figures each, always begin at the decimal point. The following 2- 2-> ,-\, ,-\ ,-\, ,-, 2-, numbers have been marked off correctly: 4 73. ; 4 187; .60; 1 76.3 72 0. Often in finding the square root of a number, the result does not come out even (see the last example above). In such case we shall find the root correct to two decimal places. In finding the square root of a fraction, first write it as a decimal. Example: # is written as .6667. EXERCISES Find the positive square root of each of the following. 1. 8 4 1 4. 1 2 5 4 4 2. 5 4 7 6 5. 9 4 ... O 9 3. 8 6 4 9 6. 2 9 , 1 6 — 190 — TABLE OF SQUARES AND SQUARE ROOTS By use of the table below one can read directly the square or the square root of any whole number from 1 through 99. Examples: The square of 23 is 529; the square root of 23 is 4.796; the square root of 48 is 6,928. - No. Squares i. No. Squares §.G No. Squares §º 1 1 1.000 34 1,156 5.831 67 4,489 8.185 2 4 1.414 35 1,225 5.916 68 4,624 8.246 3 9 1. 732 36 1,296 6.000 69 4,761 8.307 4 16 2.000 37 1,369 6,083 70 4,900 8. 367 5 25 2. 236 38 1,444 6. 164 71 5,041 8.426 6 36 2.449 39 1,521 6.245 72 5,184 8.485 7 49 2.646 40 1,600 6. 325 73 5,329 8. 544 8 64 2.828 41 1,681 6.403 74 5,476 8.602 9 81 3.000 42 1,764 6.481 75 5,625 8. 660 10 100 || 3.162 43 1,849 6. 557 76 5,776 8. 718 11 121 3.317 44 1,936 6.633 77 5,929 8. 775 12 144 3.464 45 2,025 6. 708 78 6,084 8.832 13 169 3. 606 46 2,116 6.782 79 6,241 8. 888 14 196 3.742 47 2,209 6.856 80 6,400 8.944 15 225 3.873 48 2,304 6.928 81 6,561 9.000 16 256 4.000 49 2,401 7.000 82 6,724 9.055 17 289 4. 123 50 2,500 7,071 83 6,889 9. 110 18 324 4. 243 51 2,601 7.141 84 7,056 9. 165 19 361 4.359 52 2,704 7.211 85 7,225 9.220 20 400 4.472 53 2,809 7.280 86 7,396 9.274 21 441 4.583 54 2,916 7.348 87 7,569 9.327 22 484 4.690 55 3,025 7.416 88 7,744 9.381 23 529 4. 796 56 3,136 7.483 89 7,921 9.434 24 576 4.899 57 3,249 7.550 90 8,100 9.487 25 625 5.000 58 3,364 7.616 91 8,281 9. 539 26 676 5.099 59 3,481 7.681 92 8,464 9. 592 27 729 5. 196 60 3,600 7.746 93 8,649 9.644 28 784 5.292 61 3,721 7.810 94 8,836 9.695 29 841 5.385 62 3,844 7.874 95 9,025 9.747 30 900 5.477 63 3,969 7. 937 96 9,216 9. 798 31 961 5.568 64 4,096 8.000 97 9,409 9.849 32 1,024 5. 657 65 4,225 8.062 98 9,604 9.899 33 1,089 5. 745 66 4,356 8. 124 99 9,801 9,950 EXERCISES From the table above read the answers called for below and write them on the blanks provided. 1. Find the square of 31 —; 54 —; 67 —; 71 ; 93 2. Find the square root of 13 —; 37 —; 43 ; 71 — PETER BARLow (1776–1862) Peter Barlow was professor of mathematics at bers. This has been a standard reference for more the Woolwich Military Academy in England. He than a century, playing in its narrower field a role made a number of worth-while contributions to somewhat similar to that played by Webster's dic- mathematics and allied fields. Among these contri- tionary in the field of language. Barlow was also butions was a table of the powers and roots of num- interested in mathematical physics. — 191 — *INTERPOLATION Many numbers of which we may need to find the square or the square root are not in the table on page 191. However, we can find the squares and the square roots of such numbers by use of a method called interpolation. EXAMPLES Find the square of 31.6. 31.6 lies between 31 and 32, both of whose squares are found in the table. The square of 32 = 1024 The square of 31 = 961 Differences 1 63 As the number increases by 1, the square in- creases by 63. By proportion, when the number is increased by .6, the Square will be increased by .6×63, or 38. Find the square root of 67.3. 67.3 lies between 67 and 68, both of whose Square roots are found in the table. The square root of 68 = 8.246 The square root of 67 = 8.185 Differences 1 .061 As the number increases by 1, the square root increases by .061. By proportion, when the number is increased by .3, the square root will be increased by .3X.061, or .018. e 961 ſº 8. 185 Thus the square of 31.6 is +38 Thus the square root of 67.3 is +.018 999 8. 203 EXERCISES Using the examples above as models, find the squares of the numbers in the first column below, and the square roots of the numbers in the second column. 1. 86.3 2. 34.9 3. 49.8 4. 65.8 5. 91.4 6. 34.75 — 192 — THE RIGHT TRIANGLE As far back as we have any record there were Many years later Pythagoras, a Greek mathe- surveyors in ancient Egypt. They called themselves matician, discovered the law that must exist among “rope stretchers.” Like a present day engineer, the sides of a triangle in order that two of the these early surveyors often needed to con- - sides be perpendicular. Stated in words this law is: In any right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides. The law may be written in symbols as cº - a”--b”, where c is the hypotenuse, and a and b are struct a right angle. They needed this in making the corners of temples and pyra- mids square, and in laying off farm lands. The rope stretchers’ method was to tie twelve knots at equal distances in a rope. They then stretched the rope about three the legs of the right triangle. stakes so that a triangle was formed, the sides of Pythagoras stated the general principle; the which were three, four, and five of these equal divi- Egyptian rope stretchers were using a special case sions. Such a triangle is shown in the illustration. of this, namely, 5*=3%+4°. There are a great many The angle opposite the longest side is a right other special cases where the square of one whole angle; the other two sides are perpendicular to number equals the sum of the squares of two other each other. whole numbers. EXERCISES Find the third side of each of the following triangles. 1. - 2. 24 25 b —: —; Using the formula cº-a2+b^ and the table of squares and square roots on page 191, find the value of the side indicated below. 4. Given a =4, b = 7. Find c. 7. Given a = 12.5, b = 15.5. Find c. 5. Given a = 7, c = 12. Find b. - 8. Given b = 4.5, c = 7.5. Find a. 6. Given a = 5, b = 7. Find c. 9. Given b = 11, c = 17. Find a. — 193 — PROBLEMS USING THE FORMULAS FOR RIGHT TRIANGLES c2 = a 2–H b% q2 = C2—b2 b? = c2 — a 2 1. Find the length of the hypotenuse of a right triangle whose other two sides are 48 feet and 20 feet. Answer: 2. Find the diagonal of a rectangle whose sides are 69 feet and 92 feet. Answer: 3. How high up the side of a building will a lad- der 30 feet long reach when the base is set 8 feet from the building? Answer: 4. What is the longest straight line that can be drawn on a rectangular piece of paper whose dimen- sions are 16 by 30 inches? Answer: 5. A baseball diamond is 90 feet square. Find the distance from home plate to second base, correct to two decimal places. Answer: *6. Find the area of a right triangle whose base is three fourths of its altitude and whose hypotenuse is 20 feet. Answer: — 194 — **THE COMPOUND INTEREST FORMULA Mr. Smith lent Mr. Jones $500 for three years, the interest to be compounded annually. At the end of the three years, Mr. Jones settled his obligation by paying Mr. Smith $595.51. The $500 is called the principal, three years the time, six percent the rate, $595.51 the amount, and $95.51 the interest. The formula A = p(1+r)! gives the amount (A) in terms of principal (p), rate (r), and time (t). Use this formula to find A in exercises 1 and 2. When t=2, the compound interest formula becomes A = p(1+r)”. Use t = 2 in solving exercises 3, 4, 5, and 6. EXERCISES 1. p =$200, r=.04, t =3. 4. Find 1+r when A =$112.36 and p =$100. 2. p =$500, r=.05, t =2. 5. Find r when A =$441 and p =$400. 3. Solve A = p(1+r)” for (1+r). 6. Find r when A =$583.20 and p =$500. — 195 — COMPOUND INTEREST TABLE Bankers and others whose business is the handling of money and credit make continual use of com- pound interest. They have compound interest tables which save them a great mass of detailed computation. A very short table of this sort is given below. To use this table, locate the row which identifies the number of interest periods and the column which identifies the rate for a period. The amount of $1 for the given number of periods at the given rate falls in this row and this column. EXAMPLES 1. The amount of $1 for 5 periods at 4% a period is $1.22. 2. The amount of $1 for 6 periods at 3% a period is $1.19. In writing dollars and cents, in an answer, it is customary to discard a part less than one half cent, and to count as an additional cent a part which equals or exceeds a half cent. This was done in the examples above. In example 1, $1.2167 was written as $1.22, and in example 2, $1.1941 was written as $1.19. 70, 1;% 2% 2%% 3% 3#9% 4% 5% 6% 1 1.0150 1. 0200 1.0250 1. 0300 1.0350 1. 0400 1. 0500 1. 0600 2 1.0302 1. 0404 1.0506 1. 0609 1. 0712 1.0816 1. 1025 1. 1236 3 1. 0457 1.0612 1.0769 1.0927 1. 1087 1. 1249 1. 1576 1. 1910 4 1.0614 1.0824 1. 1038 1. 1255 1. 1475 1. 1699 1. 2155 1.2625 5 1.0773 1. 1041 1. 1314 1. 1593 1. 1877 1. 2167 1. 2763 1.3382 6 1.0934 1. 1262 1. 1597 1. 1941 1. 2293 1.2653 1. 3401 1.4.185 7 1. 1098 1. 1487 1. 1887 1. 2299 1. 2723 1. 31.59 1.4071 1. 5036 8 1. 1265 1. 1717 1. 2184 1. 2668 1. 31.68 1. 3686 1.4775 1.5938 9 1. 1434 1. 1951 1. 2489 1. 3048 1. 3629 1.4233 1. 5513 1.6895 10 1. 1605 1. 2190 1. 2801 1. 3439 1.4.106 1.4802 1.6289 1 .. 7909 The number of periods, n, is the number of years if the interest is compounded annually; it is double the number of years when the interest is compounded semiannually; it is four times the number of years when the interest is compounded quarterly. The rate, r, is the annual interest rate when the compounding is done annually; it is one half the annual interest rate when the compounding is done semiannually; it is one fourth the annual interest rate when the compounding is done quarterly. The compound interest equals the com— pound amount minus the given principal. To find the compound amount when the principal is other than $1, read the amount for $1 from the table, and multiply this amount by the number of dollars in the principal. EXERCISES AND PROBLEMS 1. On the blanks provided write the compound amounts of one dollar for the given values of n and r. n = 7, r = 5% n = 4, r = 3% n=8, r =4% n = 6, r =2% 2. The amount of $100 for 5 periods at 3% is $ 3. The amount of $500 for 5 years at 3% compounded semiannually is $ - - 4. The compound interest on $100 for 3 years at 24% compounded annually is $ 5. The compound amount of $400 for 2 years at 6% compounded quarterly is $ 6. The compound amount of $300 for 5 years at 7% compounded semiannually is $ - 7. The compound interest on $100 for 2 years at 10% compounded quarterly is $ 8. The compound amount of $1000 for 2 years at 6% is $ more if it is compounded quarterly than if it is compounded annually. — 196 — IRRATIONAL NUMBERs A whole number or the quotient of two whole numbers is called a rational number. A number which is not a whole number or the quotient of two whole numbers is called an irrational number. EXAMPLES Rational numbers: 1, 4, 6%, 3.5, etc. Irrational numbers: V2, V3, V7, etc. Many indicated roots are irrational numbers. They will never come out even, however far the work is carried. For the V2, we often 1 use 1.414, but this value is only approximate. If carried out further, the more accurate value 1.4142 will be found; this procedure could be continued indefinitely. Though it is impossible to find the square root of 2 exactly, we can construct a line that will represent its exact length. When each leg of \5 a right triangle is 1, the hypotenuse is V2. The V5 is also an irrational number. We can never find its value exactly, but we can draw a line that will represent its exact length. This is done in the lower figure at the right. RADICALS A radical is an indicated root of a number or of an algebraic expression. W; b The expression under the radical sign is called the radicand. The number above and to the left is called the index. In the radical Va.--y, a +y is the radicand, and 3 is the index. The index indicates the root that is to be taken. Where the Square root is to be taken, the index 2 is usually omitted and understood. EXAMPLES v/2 v/9 V; ~/r2-Hy? VT-Ea v/a25 vº PRINCIPAL ROOT The square root of 4 is either +2 or –2, for (+2)*=4 and (–2)?=4. The root preceded by the plus sign is called the principal root. In solving problems, sometimes only one of the two square roots checks with the sense of the problem; sometimes both must be used. You can tell from the problem itself whether to use one or both square roots. If the plus sign or if no sign precedes the radical, the plus root is meant; if the minus sign precedes the radical, the minus root is meant. EXAMPLES +V4 = +2; — V4 = —2; V% = +3; — V9 = –3; V16=4; –VT6= —4; — V-27= +3. TYCHO BRAHE (1546–1601) “Twinkle, twinkle, little star, How I wonder what you are, Up above the world so high, Like a diamond in the sky.” Tycho Brahe, as a boy, wondered what the stars were as he eagerly gazed into the heavens. He continued to wonder, and to study, and to find joy in the stars as long as he lived. While still a youth, Tycho discovered a new star. This brought him to the attention of King Frederick of Denmark. Frederick built an observa- tory on a little island for the young astronomer; Tycho called this Uraniborg, “the city of the heav- ens.” There he spent most of his life, watching the heavens by night and making computations by day, charting the place and the course of every star. Today mathematics finds many common appli- cations in physics, mechanics, and engineering. For centuries before these fields were developed, astron- omers had used mathematics in their work. Tycho Brahe was one of the founders of modern astron- Omy. — 197 — SIMPLIFICATION OF RADICALS The form of a radical expression may be changed without affecting the value. It is often desirable to change the given form to one for which the approximate value can be more easily obtained. EXAMPLES v/8= V4, 2–2 V2 ~/24= V/8, 3–2 V3 Va2b– Voºb-av/b Waßb- Was, b-aw/b In finding the square root of an expression like those above, factor out the largest perfect square. Take the square root of this, and write before the radical. In finding the cube root, factor out the largest perfect cube. Take the cube root of this and write before the radical. EXERCISES 1. V8 = 11. 2v3 = 21. V16= 2. V16= 12. –3vſ18 = 22. Vancº- 3. V18= 13. 4V125= 23. V27x2= 4. V27 = 14. Vancº- 24. 2 V192= 5. V100 = 15. 3 V162 = 6. V75= 16. 3M/a2bº- 7. V32= 17. 3M/27b2– 8. V48= 18. 10 V8.c5 = 9. V63= 19. 3-M5a?= 10. V108= 20, 2N/16a?= 25. —3 W/64a:2y+= 26. V(a–b)ºy = 27. — a W56bºc= 28. —3 W/–125– 29. – 5 W/-64a3= 30, a V-54 cº- LEWIS CARROLL (1832–1898) Charles Lutwidge Dodgson was Lewis Carroll's real name, and he was professor of mathematics at Oxford University, in England. Most of his writings were in the fields of advanced mathematics; schol- ars still consult these. |Millions of children and grown-ups read other books he wrote: Alice in Wonderland and Through the Looking Glass. He wrote these for some little girls he knew, using the pen name of Lewis Carroll. The story is told that Queen Victoria read Alice in Wonderland, and was delighted with it. She asked that a copy of every book the author had written be brought to her. They were A Treatise on Deter- minants, Commentaries on Euclid, and others of like nature. These were not the sort of books the Queen had expected. — 198 — SIMPLIFICATION OF FRACTIONAL RADICALs ! ... . T VI Use the table of square roots to find the value of 2 This may be done as follows: 2 TV5 1 1 707. Dividing 1 by 1.414 is a 1 A bett f finding the value of w/º: = − = — = . (U ( .. LJ1 V1CIII) .4:14: IS a, IOI) gº OTOCéSS. €15136]. Wa,V O 1I] gº the Va,IULé O --- v/2 1.414 g 1 Oy g p y 9, 2 1 º 1 1 1 is as follows: g- †- T.2-g 2-g (1.414) = .707. A division has to be made in this, but the divisor is the easy number 2 rather than the hard number 1.414. EXERCISE 1 Find the value of A/3 by each of the methods above. Compare the difficulty of the two methods. EXAMPLES 4 W. 4 2 Vº 6. 1 -— = := 4/ — 5 = —V5 — 4/1/ — = — 4 He -44/ — 6 = — V6 W: 25 25 ;V5 8 16 16 v/6 3. W; 1 1 W. b2a, 5- 4/ * - W -8-ºvº --a/-a/; a-ºva RULE FOR SIMPLIFYING FRACTIONAL RADICALS (SQUARE ROOTs) Multiply the numerator and the denominator by the least number that will make the denominator a perfect square. Discover the greatest fractional factor that is a perfect square. Take the square root of this factor and write it before the radical. This method may also be used for cube roots by making the denominator a perfect cube. EXERCISES 1. Vº- 8. Vºji= 9.4/1. 2. Vä- W = 5 # = 12 -- - Q/ AT 6. Vă = 13. A/ — = 7??, 3. 7. V} = 14. M/ — = acă — 199 — ADDITION AND SUBTRACTION OF RADICALS Similar radicals are those having the same index and the same radicand. Examples: 3V2, —4V2, and av2z are similar radicals; 2v3a, 7 vº3a, and cy/3a are similar radicals. Ol. Dissimilar radicals can sometimes be made similar. Example: W; can be made similar to V20 by ap- tº a 20. 1 1 plying the method used on page 199: g- T- +2a-2 v 20. Similar radicals or radicals which can be reduced to similar radicals can be added or subtracted. EXAMPLES 4V2–2V2+7V2=9v/2 Vº-V8=}v/2–2V2= –14 V2 3 b 3/— 3 3 3 ava-i-bva - (a+b) Vº 2a º-Wab–2Val-Val-Val 0. |EXERCISES Combine terms which are similar or can be made similar. 1. 5V2–6V2+4v/2= 10. VT6–V54 = 2. 4 V2–5V2+8V2= 11. W aacº–H Yaº- 3. avº--bv/2+cv/2= 4. V8–V2= 12. V2+ Vº-HVº- 5. V12–H V27 = 6. 2 V10— V40 = W; T *13. ——-6 s-Wºr 2 * V8 0, 3 7. V3–V75= 8. Vº-V50= T *14. W. vºivº- 9. 2 VF--V20= — 200 — MULTIPLICATION OF RADICALS To multiply radicals of the same order, multiply their coefficients together and multiply their radicands together. Simplify the results whenever possible. EXAMPLES v/2. V3 = V6 ~/2. ~/12= V/24 = V/8, 3–2 V3 –2V3.4 V6- —8VT8 = -24V2 V; V = V, - Vºſs; V2 EXERCISES Multiply the following. Simplify the results when possible. 1. V3. V2= 2. V2. Vº2= 3. V6, V7– 4. Va. V5+ 5. V2. V8 = 7. 2 V2.3V3= 8. 2 Vö.4V6= 9. 10. 11. 12. 13. 14 15 16 ~/a2. Va2– V. Vº- (w/º)?= **Simplify so that no radical occurs in the denominator. (Hint: Multiply the numerator and the denomi- nator of the first exercise by V/3+ V2, and of the second exercise by V/5–2). 10 17. — V3–V2 v/5–2 18. — V5+2 DIOPHANTUS OF ALEXANDRIA (ABOUT 250 A.D.) There have come down to us portions of a book called Arithmetica of which Diophantus was author. Much of this book is what we now call algebra. In his Arithmetica, Diophantus introduced a better symbolism, that is, a better method of translating word statements into algebraic expressions, than had been used before that time. Little is known of the life of Diophantus: when and where he was born; where he was educated. He did his work at Alexandria in North Africa. A curious problem about him appears in a manuscript written about sixteen centuries ago. It reads: “The boyhood of Diophantus lasted # of his life, his beard grew after tº more, after # more he married, 5 years later his son was born, the son lived to half the father's age, and the father died 4 years after the son. How old was Diophantus when he died?” The answer is 84 years. Before you have finished this course you should be able to Solve problems of this sort. — 201 — DIVISION OF RADICALs To divide radicals of the same order, divide their coefficients and divide their radicands. Simplify the results whenever possible. EXAMPLES V14-i-V7= V2 V;--Vº-V3 = }V6 4V6+2V3=2V2 ~/256+ V4 = V/64 = 4 EXERCISES Divide the following and simplify when possible. 1. V12-i-V6= 11. 5V2I--V7 = 2 35-i-V5= 12. V3+ V6= 13. V3 + V2= 3. V75+ V15= V; +V2 14. Vº--Vă = 4. V26+ V13 = 15. V24+ V% = 5. 4V10-i-2V5= w/º: W. 16. — — — — = Ol. b 6. V10a2+ V2a = 17. Wºº-i-Wła 7. ~/4-i-V2= 18. #v}+}Vº- 8. W16+ V2= 1 / 5 1 /10 19. — M/ —--—M/ — = 9. W81.c4+ V/32 = 5 (tº 10 a? --- 2/3 4/4 10. V52a2+ V13= **20. 3 * º — 202 — The square root of a number or of an algebraic expression may be indi- } cated either by a radical sign, or by a fractional exponent. Both have the v/7=7 Same meaning; sometimes it is more convenient to use one and sometimes 'a-E25= (a+2b)} the other. a-H (a+2b) FRACTIONAL ExPONENTS Indicate the square root of each expression below both by a radical sign and by a fractional exponent. 1. 2. The cube root of a number or of an algebraic expression may be indicated 3 either by a radical sign, with index 3, or by a fractional exponent. Similarly, w/a-ał the fourth root or higher roots of numbers may be indicated either by radicals or by fractional exponents. 11 35 3. ac-39 4. a”-H2ab – Wa2+b= (a^+b); Wry-Fº- (xy+2})* In multiplying and dividing, fractional exponents follow exactly the same laws as whole number expo- nents. To multiply numbers of like base, add their exponents. To divide numbers of like base, subtract their exponents. Perform the indicated operations. 5. a. a”= 8. a?: a*= nº- 13. 10. a. a' = 14. 11. a”. a”= 15. 12. as: a*= 16. 17. a”---a = 18. aš--a% = 19. aš--a'- 2 1. 20. O." -- aſ” = It is often necessary to simplify a number with a fractional exponent. Study examples 21, 22, and 23, and see if you can determine the principle used. Use this principle with exercises 24–32. 21. 22. 23. 24. 25. 26. (c)3=26 (32)4=38 (9)*=(3?)?=3 (27)*= (16)*= (81)*= (24)2=28 (53)2=56 (16)*=(24)*=2 27. (8)*= 28. (36)*= 29. (32)*= (28) = x2 (38)}=34 (r.10)}=25 (56)? - 52 (125)*=(53)*=5 30. (49)*= 31. (64)*= 32. (25)*= — 203 — *RADICAL EQUATIONS It is often necessary to solve an equation in which the unknown occurs under a radical sign. The first step in solving such an equation is to square both sides of it so as to remove the radical sign or signs. EXAMPLES Square to remove the radical sign. v/32–2=2 * * * * e º vº-F63–6–3, 3a –2 = 4 Simplify and solve the resulting equation. ac”—H·62 =36–12a:--ac? 32 = 6, or a = 2 ;4-- ~ + v. As e 18a;=36, or a = 2 Check: V6–2=2 * Check by substituting the answer in the Check: V4-FT2=6–2 - given equation. * 2 = 2 4 = 4 — EXERCISES Solve for a and check. The first two have been done correctly. Given Equations Solutions Check 1. Vac-H2=3 a;+2=9, or a = 7 v/7+2=3, or 3 =3 2. V2s + Vº-H4 2a:= a +4, or a = 4 v/24=V4-F4, or V8–V8 5. V2s-T = Vº 6. 2.v/T-a- V8 7. V32-FT= V10 8. A/52–2= V/6+a; *mºm- 10. Vac”–25 = 12 11. (25–2%)} = 4 12. (~2-1-52–9)} = (5a)? 13. (~2–7)} = (–6)? 14. (cº-i-82–13)} = (8a;+3)? — 204 — 1. Any square is the Of equal factors. 2. Any cube is the of equal factors. 3. The square of a positive number is a number. 4. The square of a negative number is a number. 5. The cube of a positive number is a number. 6. The cube of a negative number is a number. 7. The formula for the square on the hypotenuse of a right triangle is 8. The index of the radical 3Va is ; the coefficient is ; the radicand is 9. The Square root of a number is one of the equal of the number. 10. The two square roots of 25 are and The principal Square root is — 11. (a) V3–– (d) V, -— (b) - V.9–– (e) Vicº–20 g-H100 = (c) - V162°= (f) — V92?--42a-a-i-49a2= 12. The cube root of a number is one of the equal of the number. 13. (a) (3)*= (b) (–3)*= (c) (22%)*= (d) (–2a)*= 14. The cube root of a positive quantity is , and the cube root of a negative quan- tity is - - -*- 3/II 15. (a) V27 = - (b) V-82°= (c) W27-y?– (d) Wºº-- 16. Simplify the following. (a) V32= (c) –2 vſ 16aºyº – (b) V32= (d) Vººr' = 17. Solve for S. A = s”. 18. INVENTORY TEST Solve for r. A = wr”. — 205 — CUMULATIVE REVIEW Perform the indicated operations. 1. 10. 11. (42°-H6++4)+(2a2+22–2) = . (4a2+62-H4)–(2a2+2a:-H2)= . 3+4.2–6–4–3–H4 = 3(–4)+6+ (–2)+3= . 3a–2(4a–H1)–6= . (3a;+4) (32–4)= . (2d —3b)?= (a;+4) (ac—7)= . (20–3) (3d-H4)= Factor completely. 3a*—27b2 = 12a2–180b — 125° = aac”—aac – 12a = 16b2–81.c2 = After each true statement write “T”; after each false statement write “F.” 1 | 1 1 Q. I b a+b 1 + 1 z-Hy 20 y 24/ ac”— 2 = a +y a – y c—Hd c – d c—d c-H d a?—b? (a-b) *- ac – y y–a: O, — Q, 12. Perform the indicated operations. 13. 1 1 cd c cº-c-20 c–4 Perform the indicated operations. (–32)?= (3ab%)3 = 7mºn?-- m3 = — (5a)*= acq. acb = ac"--ac” = (wº)" = (a —y)*= 14. The rate of a freight train is three fourths that of a passenger train. The freight requires one hour longer than the passenger train to make a run of 150 miles. Find the rate of each. — 206 — UNIT 13. QUADRATIC EQUATIONS BENJAMIN PEIRCE (1809–1880) A measure of the teacher's worth is the influence he has on his pupils. Few teachers anywhere have exerted greater influence upon those with whom they came in contact than did Benjamin Peirce, professor of mathematics in Harvard University. Many of his pupils attained high place in life; they attributed much of their success to the teaching and the inspira- tion of Professor Peirce. One of his pupils, Charles W. Eliot (President, Harvard University, 1869– 1909), referred to him as “a very in- -- º spiring and stimulating teacher. He #. F.T. dealt with great subjects and pursued abstract themes . . . nevertheless filled them [the students] with admiration and rever- ence.” Another of his pupils, A. Lawrence Lowell (President, Harvard University, 1909–1933) wrote: “I never admired the intellect of any other man as much as that of Benjamin Peirce. I took every BENJAMIN PEIRCE course that he gave when I was in college, and whatever I have been able to do intellectually has been due to his teaching more than to anything else.” - Another of his pupils, Edward Everett Hale (author of “The Man Without a Country”) wrote: “The classical men made us hate Latin and Greek; but the mathematical men made us love mathematics, and we shall always be grateful to them. . . . I had but four teachers in college– Channing, Longfellow, Peirce, and Bachi. The rest heard me recite but taught me nothing.” Many pages could be given to for- mer pupils who thus paid tribute to this great teacher. Upon Professor Peirce's death, Harvard Uni- versity inscribed in its records: “As a teacher, he inspired young minds with a love of truth, and touched them with his own enthusiasm.” QUADRATIC EQUATIONS A quadratic equation has two answers. Consider the examples. 1. a 2–3++2=0. The two answers are a = 1, and a = 2. Each of these will check when substituted in the equation. (1)?–3(1)+2=0, and (2)?–3(2)+2=0. Test to make sure that both check. 2. rº–H5++6=0. The two answers are c = -2, and r = –3. Each of these will check when substituted in the equation. (–2)2+5(–2)+6=0, and (–3)*-i-5(–3)+6=0. Test to make sure that both check. GENERAL AND SPECIAL QUADRATIC EQUATIONs The general form of the quadratic is ar” +b++ c = 0, where a, b, and c may be any con- stants. This means that in a quadratic the variable occurs to the second power and to no higher power. When the quadratic has no a term, that is, b = 0, it may be written cº–k = 0, or cº-k. The two an- swers are a = vſk, and r = – VR. Notice that these two answers have the same numerical value, but have opposite signs. When the quadratic has no constant term, that is, c=0, it is written aaºº-H br=0. The two answers are a = 0, and a = —bſa. — 207 — QUADRATIC EQUATIONS EXAMPLES 1. ac”—4 = 0 2. ac2 = 7 3. 5a-2-1-82 = 0 (2–2)(ac-H2)=0 a = + V7 ac(52+8) = 0 a –2 =0, or a = +2 a: = +2.646 (See table, p. 191.) a = 0 a;+2=0, or a = -2 a: = —2.646 52+8=0, or a = —# You already know that the product of any number and 0 is 0. It is also true that when the product of two factors is 0, at least one of the factors must be 0; when k times m equals 0, then either k or m (or both) must be 0. This is the principle used in the third and fourth steps of examples 1 and 3, above. When a –2 =0, or when ac-H2=0, the equation given in example 1 is satisfied; it cannot be satisfied otherwise. EXERCISES Solve for ac. 1. ac2 = 16 8. ac2 = 10.24 15. 34:2–H5a: = 0 2. ac” = 144 9. a.” =# 16. 4a2–15ac = 0 3. ac2 = 400 10. a 2–49 = 0 17. ap2 = 27 4. ac2 = 18 11. 4a2–81 = 0 18. a.”—# = 0 5. ac2 = 30 12. a 2–4 =0 19. 9a:” – 16 = 0 6. ac2 = 54 13. a 2-H42 = 0 20. 44%–100 7. ac2 = 1.21 14. 2a:”—3a; = 0 21. 3a.”—21a;= 0 —208 — SoLVING QUADRATICS BY FACTORING Whenever the a, b, and c of the general quadratic are such that aw”-Hba;+c can be factored by inspection, the equation can be easily solved by factoring and setting the factors separately equal to 0. EXAMPLE ac?--9a;+20 = 0 Solve: (z+4) (ac-H5)=0. Set (ac-H4)=0 and (ac-i-5)=0. Then a = —4 or —5. Check: (–4)2+9(–4)+20=0, or 16–36–H20=0 a = — 4 a: = — 5 (–5)2+9(–5)+20 = 0, or 25–45+20=0 EXERCISES 1. ac”—ac — 12 = 0 Solve: Check: QC = ac = 2. cº-i-32–10=0 Solve: Check: 3. 4a:%–4a:--1 = 0 Solve: Check: 3C = QC = 4. a 2–5a;+6=0 Solve: Check: QC = (C = 5. a 2–Ha:—20 = 0 Solve: Check: Q = 3C = 6. a7+2a – 15 = 0 Solve: Check: — 209 — EXERCISES Solve by factoring in the spaces provided, and check mentally. 1. a 2–H 122–64 = 0 4. 29–302–400 = 0 7. ac2+40a2+-200 = 425 Hint: Subtract 425 from each member of the equation. 2. y”–10y-H9=0 5. y?–24y–81 =0 8. ac2–8w-H 15 = 99 3. a 2–H 14a: — 51 = 0 6. 22–H222–23 = 0 9. a 2–H 12a:--27 = 40 ADDITIONAL EXERCISES * ... a 2–14a:-H40=0 9. 2%–42–5 = 0 17. d?--240 = — 144 1 2. a 2–142-H48 = 0 10. m?--2m-255 = 0 18. d”—26d = –144 3. a 2–H 14a – 51 = 0 11. n2+3m — 154 = 0 19. w”--1100–H24 = 0 4. y?--5y–36=0 12. n”—H·n — 420 = 0 20. w”–H300–54 = 0 5. y?–59–14=0 13. k?–26k-H 169 = 196 21. w?–5uy—24 = 0 6. y?--5y-H6=0 14. k”—2k+1=25 *22. 2w?—u—21 = 0 7. 22–H52–6=0 15. k?--50k+625 = 1600 *23. 30.4%—u—2 = 0 8. 2”—62–H5 = 0 16. d”–28d–H 196=0 *24. 5u%+27u — 18 = 0 — 210 — SoLVING QUADRATICS BY COMPLETING THE SQUARE The trinomial ac”–4a:--4 is a perfect square; its factors are (a —2)(ac-2). Similarly ac”–62+9 is a perfect square; its factors are ( )( ). Finish the trinomials below so as to make each a perfect square. (a) a”–8a;+ (b) a 4–10a;+ (c) aſ”–H122-- (d) a 2–H52+ In each case above, the number you wrote should equal the square of half the coefficient of ac. Check by testing for this requirement. To complete the trinomial square x2+mz+ (m/2)” where the blank occurs. , or the trinomial square x*— m3+ , place EXERCISES Solve each of the following by completing the square. The first has been done correctly. 1. ac”—4a: —32 = 0 Step-by-step procedure: (a) aſ”—4a: = 32 (a) Add 32 to each member of the equation. (b) a 2–42-H4 = 32+4 (b) Make the left member a perfect square by adding the proper number; add this same number to the right member. (c) (2–2)?=36 (c) Simplify. (d) (a —2) = +6 (d) Extract the square root of each member. (e) a = 2+6 (e) Add 2 to each member of the equation. (f) a = 8, or —4. (f) Write the answers. 2. ac”—H·10a; – 11 = 0 5. y?–10y--16=0 8. 29–52–H4 = 0 3. ac”—62–40 = 0 6. y?–12y+11=0 9. 22–172–H 16 = 0 4. ac”—4a – 12 = 0 7. y”–14y--24=0 10. 22–72--3}=0 — 211 — EXERCISES Complete the solutions. 1. a 2–H.8a. —20 = 0 3. a 2–3a – 70 = 0 ac”—H·8a;+ = 20-H. ac”—3a;+ =70–H a’–6a-H = 16-H ac-H = + 3C — = + 30 T — F iſ 3C = OT (C = Or 2 = OI’ 2. a 2–H92-H20=0 4. a 2–Ha: — 42 = 0 6. a.”—52 – 24 = 0 ac?--9a;+ = —20—H ac”—Ha!-H =42–H *–5r-H =24+– ac-H = + a;+ = + 2-— = + 3C F —— OT (C = OT Q = OI’ It is sometimes necessary to solve a quadratic by completing the square when the coefficient of ac” is different from 1. The method for carrying this through will be made clear in the exercises below. The first has been done correctly. EXERCISES 1. 3a;2–7a;+2 = 0 4. 5a-2–92–2 =0 3a;2–7a: = —2 7 —2 ac”—— a = — 3 3 7 49 – 2 49 25 ac? ac-H–=——H--— 3 36 3 36 36 ac—# = + š and a =#-E: a = 2 or ; 5. 622–13a;+6=0 2. 62%-Ha!— 12 = 0 3. 5a-2–H 7a:—6=0 6. 10a;2–21ac – 10 = 0 — 212 — **USING THE QUADRATIC FoRMULA By the use of a formula, one can solve the general quadratic equation, aa”--ba-HC =0, without factoring —bH Vb2–4ac 2a We shall prove the formula on page 214, but we shall use it now. Study the examples carefully. and without completing the square. This formula, which gives the values of ac, is a = EXAMPLES Use the formula above to solve for ac. 1. 3a*-H 10a;+3=0. In this, a =3, b = 10, and c=3. Substitute these values in the formula. — 10+ \7100–36 — 10+8 –2 – 18 — 1 6 = 6 -º-o: Then z-a-orz=-3. QC = 2. 2a:”--ac—3=0. In this equation a =2, b = 1, and c=-3. Then — 1 + V1+24 — 1 + 5 4 —6 –3 - =– or —. Then a = 1 or a = —. 4 4 4. 4 2 EXERCISES Solve by use of the quadratic formula. 1. 2a:”—3a;+1 = 0 3. 62%— 13a;+6 = 0 4. 2a:”— 13a;+ 15 = 0 6. 6ac”— 7a: — 20 = 0 — 213 — *DERIVING AND USING THE QUADRATIC FoRMULA 2a2+3a – 5 = 0 Compare the examples on the aa.”—H·ba;+ c = 0 2a2+3a; = 5 left and right, step by step. Be sure aac”—H·ba: = —c you understand exactly what is done ac”——— 2-4 in each step. ac”–H– z--" 2 Since the a, b, and c in the exam- (1. Ol. 9 5 9 ple on the right may be any num- b? – c b2 2–1–– – = – = - - e 2–1–– ºr— *=º | ac2+ 2 *H, 2 "10 bers, the solution found may be used ac?-- Ol. a:- 40° a '4a: as a formula to Solve any quadratic. You have already used this formula in solving the exercises on page 213. The quantity under the square (**) —40c-H b” QC 20. *º- 40% ac-H *\_ +". root sign is sometimes called the (U b. == Vb%–4ac 4 T 4 discriminant. When it is not a per- 20. 2a fect square, refer to the tables on 7, 2 TAT2 . QC = — 3 + 7 page 191. Write such answers cor- (C = – b + Vb%–4ac 4 rect to two decimal places. 20. EXERCISES 1. 4ac”—3a – 1 = 0 4. 2a:”—7a;+ 4 = 0 5. 8a;?–17a;+2=0 2. 9a2+ 13a;+4 = 0 6. 3a;2–H52-H2=0 3. 322-H2a –4 = 0 —214 — *USING THE QUADRATIC FoRMULA To SoLVE EQUATIONS EXERCISES 1. a,”—ac—42=0 QC = 4C = 2a:–3 ac-H 1 5. -: 3C = ac—1 Ø 2a:-H1 2. 2a2–6ac-H1 = 0 Q = Q = 6. az-H1 = QC = QC 1 1 1 3. a 2–8a;+16=0 ac= Q = . —- = Q = a; ac-H1 at-1 3a –2 (C 8 QC 1 *- := sº 3 = Q = e := - re 3C 3a;+1 a;+6 a. 3. (Simplify and get the equation in standard quad- ratic form.) — 215 — *PROBLEMS INvolving QUADRATIC EQUATIONS A quadratic equation has two roots. In many word problems only one of these roots is used. Although the other checks when substituted in the equation, it is unreasonable as an answer to the word problem. This is illustrated in some of the problems below. For example, in problem 1, it does not make sense to have – 15 inches as the base of a triangle. Such unreasonable roots are discarded as answers to word problems. PROBLEMS 1. The altitude of a right triangle is 7 inches longer than its base. The hypotenuse is 17 inches. Find the base and the altitude Let a = length of base. Then a +7=length of altitude, and a 2+(z+7)?=289 2a2+14a: —240 = 0 a 2–H7a: — 120 = 0 (a:-H15) (a —8) = 0 a = –15 (unreasonable; discard) a = 8 (length of base) ac-H7 = 15 (length of altitude) 2. Two square flower plots together contain 130 square feet. The side of one is 2 feet longer than the side of the other. Find the dimensions of each square. (Hint: Let a be the side of the smaller Square, and ac-H2 be the side of the larger square.) 3. One leg of a right triangle is twice as long as the other. The hypotenuse is 10. Find lengths of the legs. 4. The perimeter of a rectangular field is 144 rods, and its area is 1,280 square rods. Find its length and its width. (Hint: Let a be the width, and 72–a: be the length.) 5. The sum of a certain number and its recipro- cal is 21's. Find the number. 6. The sum of the areas of two squares is 225 Square inches. A side of the first added to a side of the Second makes 21 inches. Find the length of the side of each square. — 216 — **THE GOLDEN SECTION In order that a rectangular picture frame or window may look best, it is necessary that it be rightly pro- portioned. If it is too long for its width, it will look “skinny”; if it is too wide for its length, it will look “dumpy.” The Greeks developed the proportion of width to length which is considered most pleasing to the eye. We call this the “Golden Section.” In the Golden Section, width (w) and length (l) are related according to the formula below. See if you can derive (2) and (3) from (1). ! (1) *_ or (2) w?--lw—l?=0, or (3) l’—lw—w”=0. l w—H·l EXAMPLE Find the length of the best proportioned picture whose width is six inches. Solution: Substitute was 6 in formula (3). lº–6l–36=0, and l-3--3v3. Use V5=2.236. Then l=3–H 6.708 = 9.708. PROBLEMs 1. Find the width of the best proportioned win- dow whose length is four feet. 2. The perimeter of a rectangle is to be sixty inches. How wide and how long must it be that its proportions be the best? Hint: Let w = 30—l. **THE TANK PROBLEM Dating from ancient times, cisterns and tanks have been filled through pipes and emptied through pipes. This has suggested a family of interesting problems. The family has grown to include problems in no way connected with tanks, though the same mathematical principle runs through them all. EXAMPLE Two pipes together fill a tank in six hours. The larger pipe alone can fill the tank in nine hours less time than the Smaller pipe alone. How long does it take each pipe alone to fill the tank? Let a = the number of hours it takes the larger pipe to fill the tank. Then ac-H9= the number of hours it takes the smaller pipe to fill the tank. 1 Then –=the part of the tank filled by the larger pipe in one hour Q. 1 And the part of the tank filled by the smaller pipe in one hour. Q} 1 1 1 * Then —-H− = − the part of the tank filled by both pipes in one hour. a ac-i-9 6 By multiplying each term of the equation above by 62(a +9) and solving the resulting equation, you find a = 9. Then ac-H9 = 18. PROBLEM A man and a boy working together can do a piece of work in six days. The man working alone can do the work in nine days less time than the boy working alone. How long does it take each to do the work? — 217 — QUADRATIC EQUATIONS Solve by any method you choose. Check. 1. a 2–H 17a;+60 = 0 5. a 2–7a;+6 = 0 *9. 23:2–5a;+3=0 2. a 2 – 53: — 36 = 0 6. a 2–182+72=0 *10. 53:?--2a: —3 = 0 3. a 2–H52–66 = 0 7. a 2–83 —9 = 0 *11. 4ac”— 7a: – 2 = 0 4. ac”— 11ac — 26 = 0 8. ac”—4a: —32 = 0 *12. 32°–H.82 –3 = 0 THE HINDU METHOD OF SOLVING QUADRATICs A trick method of Solving quadratics came from India; it was invented almost a thousand years ago. The example at the right is solved by this Hindu method. The steps are: (a) transpose the constant term to the right member; (b) multiply the equation through by 4 times the coefficient of acº; (c) square the coefficient of a of the given equation and add to each member; (d) extract the square root of each member; (e) transpose the constant from the left to the right member; (f) divide through by the coefficient of a. In words this sounds complicated. In actually doing it one finds no diffi- culty. It is easy and interesting. Try this method on some of the exercises above. 2a:” – 53:--2 = 0 (a) 23:?–5a: = —2 (b) 1639–40a = –16 (c) 1639–40.c-i-25 = 9 (d) 4a – 5 = +3 (e) 4x = 5+3 = 8 or 2 (f) a = 2 or ; — 218 — How a CAME INTO ALGEBRA A wag has defined algebra as the search for a . The French say “étre fort en ac” when they mean be- ing strong in mathematics. The reason is evident. Let a stand for the unknown, and proceed to find the value of ac. This procedure is the heart of prob- lem solving in algebra. As we have already seen, Vieta was the first to use letters to represent quantities. He used vowels for unknowns and consonants for knowns. René Descartes, another Frenchman, in 1637 introduced a variation to Vieta's system. He used the first let- ters of the alphabet for knowns and the last letters for unknowns. When a single unknown was used in a problem, Descartes represented it by ac, and it has been so represented ever since. When two unknowns are used, the second is usually represented by y. PROBLEMS : . The sum of the squares of three consecutive numbers is 110. Find the numbers. . The square of a number exceeds the number by 156. Find the number. . The sum of two numbers is 16, and the sum of their squares is 130. Find the numbers. . The difference of two numbers is 6, and the difference of their squares is 120. Find the numbers. . The product of two consecutive even numbers is 440. Find the numbers. . The product of two consecutive odd numbers is 399. Find the numbers. . The longer leg of a right triangle is five inches longer than the shorter leg, and the hypotenuse is five inches longer than the longer leg. Find the hypotenuse and both legs. 8. A piece of wire 40 inches long is bent so as to form a right triangle whose hypotenuse is 17. Find the lengths of the legs of the triangle. 9. The perimeter of a rectangle is 98, and its area is 600. Find its base and its altitude. Solutions — 219 — *QUADRATIC EQUATIONS IN Two UNKNOWNS EXERCISES Solve for a and y and check. The first one has been solved correctly. 1. a.”-Hy”= 13, and 22–H3y=0. 4. y=a^-5a, and y = a –8. (a) a = —#y (From the second given equation) (b) #y?--y?=13 (Substitute (a) in the first givenequation.) (c) *y”= 13, or y?=4 (Combine like terms of (b).) (d) ... y = + 2 (Extract the square root of (c).) (e) ... a = -F3 (Substitute (d) in the second given equa- tion.) 2. a 4-Hy?=169, and ac-i-y=17. 5. y = —a’–H5a, and y = a – 12. 3. a 4-Hy”=58, and a —y=4. 6. y = 2a:”–12, and y = 2a:. — 220 — **QUADRATIC EQUATIONS HAVING Two UNKNOWNS Two equations are necessary in solving for two unknowns. The method is shown in the first problem be- low, which is solved correctly. The student will set up two equations in two unknowns for each of the other problems, and solve the equations. PROBLEMs 1. The difference between two numbers is 2, 3. Find two numbers whose difference is 8 and and the sum of their squares is 202. Find the num- the sum of whose Squares is 370. bers. Let a = the larger number and y = the smaller number. Then (1) ac-y = 2 and (2) a 2+y”=202. From the first equation (3), y=a:-2. Substitute (3) in (2): a 2+(2–2)*=202. 2a2–4a: — 198–0 a 2–2a – 99 = 0 (ac—11) (a;+9)=0 Then a = 11 (larger number), and y = 9 (smaller number), or a = −9 (larger number), and y = –11 (smaller number). 2. One number is two greater than twice an- 4. Find two numbers such that the square of other, and the difference of their squares is 84. Find the first is 44 greater than the second, and twice the numbers. the first is 4 less than the second. A quadratic equation with one unknown has two solutions, that is, there are two values of the unknown which will satisfy the equation. A quadratic equation with two unknowns has two pairs of solutions when it is solved simultaneously with a linear equation. For example, in problem 1 above, both equations are satisfied for a = 11, y =9, and for a = −9, y = –11. Are the two equations satisfied for a = 11, y = – 11? For a = —9, y =9? — 221 — 1. Graph y = x*–4a. **GRAPHING QUADRATICs EXAMPLES Find the value y has for successive values of a, as 0, 1, 2, 3, 4, . . . a = 0, y = 0; when a = 1, y = –3, and so on. Here is a table of such corresponding values. and – 1, -2, –3, c . . . Thus when Q’ () 1 2 3 || 4 5 – 1 — 2 {/ 0 –3 | — 4 — 3 || 0 5 12 5 12 These pairs of points are plotted and joined with a smooth curve in the graph at the right. On this graph a vertical interval represents two units, and a horizontal interval represents one unit. 2. Graph a 2+y}=25. There are two values of y for each value of ac. This is made evident by transposing ac” to the right member and extracting the square roots of both members. g?=25—acº, and y = + V25—a.” Here is a table of values of y = + V25–2%. ar" 0 + 1 + 2 + 3 + 4 + 5 5 4, 9 4.6 4. 3 0 (U Here is a table of values of y = — v. 26-a%. (C 0 + 1 + 2 + 3 + 4 + 5 * — O — 4.9 —4.6 — 4 — 3 0 These pairs of points are plotted at the right and joined with a smooth curve. EXERCISES Make a table of corresponding values and a graph for each of the following. 1. y = a 2+4a: 2. ac2+y}=16 — 222 — *Solving EQUATIONS GRAPHICALLY On pages 220 and 221, linear equations and quadratic equations were solved together by eliminating one of the unknowns and solving for the other. Another way to solve is to graph the linear and the quadratic equation. The coordinates of the points of intersection of the two graphs give the values of a and y which satisfy both equations. EXAMPLE Solve graphically: y = a 4–6, and 2a:—y=3. Table of values for y = a 2–6 Q. 0 + 1 + 2 + 3 + 4 $/ — 6 — 5 — 2 3 10 Table of values for 2a:—y = 3 Q. — 1 0 1. 2 3 4 $/ — 5 –3 — 1 + 1 +3 +5 The coordinates of points P and Q give the values of a and y which Satisfy both equations. & T — 3 = P Q g = gy = EXERCISES 1. Solve graphically: y=a^-32–1, and y = 2a:–1. Tables here— 2. Solve graphically: a*-Hy”=36, and y = a +6. Tables here— — 223 — INVENTORY TEST 1. Following each of the quadratic equations are four numbers. Put circles around the two which satisfy the equation. (a) ar”–7a;+12=0 2, 3, 4, 6 (c) a 2–a – 12=0 –3, 3, 4, 6 (b) a 2–Ha:–12=0 –3, 3, −4, 4 (d) a 2–8a;+12=0 –4, 2, 3, 6 2. Following each of the pairs of equations are four pairs of numbers. Put circles around the two pairs which satisfy the equations. (a) aº-Hy?= 169, and ac-Hy=17. (10, 7), (12, 5), (–12, 5), (5, 12) (b) g = a 4–3a, and g = 2a:–4. (0, 0), (1, –2), (4, –4), (4, 4) 3. The general quadratic equation has the form: aa”--ba;+c=0. Give the values for a, b, and c for the quadratics below. (a) 23:2–3a;+4=0 a = b = C = (c) a 2–H5a = 4 Ol. F b = C = (b) 8–22–a’=0 a = b = C = (d) 23:?–52 = –2 a = b = C = 4. Write the solutions for— (a) ac” =49 a = and (c) a 2–.25 Q = and (b) a 2–18 a = and (d) a”—.36=0 a = — and 5. Factor and write roots of the quadratic equations. (a) a 2–10a;+24=0 QC = and (b) a 2+2a:–24=0 **— and — (c) 328–1–4a – 15 = 0 a =– and - (d) 62°–11a;+3=0 w Q = and — 6. Solve by completing the square. (a) 24-22–8–0 QC = and (b) a 2–32–18–0 (C = and *º-º-º: 7. Solve by any method you choose. (a) 32°–4a – 15 = 0 a = and — (b) 62°–7a;+2=0 QC = and * *8. Solve by use of the quadratic formula. (a) w”–72–4 =0 Q = and (b) 9a2+13++4=0 a = and — 224 — CUMULATIVE REVIEW 1. Following each pair of equations below are three pairs of numbers. Circle the pair of numbers that is a solution to the pair of equations. (a) ac—39 = 10 *- a;+2y= 5 (13, 1) (1, 2) (7, − 1) (b) 3a;+2y = 13 2. A man can do a job in ten days and a boy can do the same job in fifteen days. Working together they can do the job in days. 3. Reduce the fractions to lowest terms. 9a2–1 a 3–acy” -— = (b) - 6ac”— 11a;+3 a:8–H2:r”y--acy” (a) 4. The sum of three consecutive even numbers is 258. Set up the appropriate equation and find these numbers. 5. Factor completely. (a) 8w8–2uv? (b) 4a:4–ay? 6. Remove parentheses and combine like terms. (a) 3a– {5a — (a —a)+3a) = (b) 5u-Hſ2y+(w–2v)–(v–w)]= " 7. Write the products. (a) (ac-Hy)*= **(c) (r-i-y-H2)(x-1-y—z) = (b) (ac-Hy)(a —y) = **(d) (2+y+2)(a —y-Hz) = 8. Fill the blanks to complete the following statements. (a) The product of two numbers with like signs is (b) The quotient of two numbers with unlike signs is (c) In subtracting, think of the sign of the - 8,S and proceed as in addition. 9. Solve the fractional equations. ac-i-3 at-i-7 (a) - a;+2 a +5 6 2 4 (b) g-H3 re-2 ºff 10. Solve for a and y, and check. 9a2+4y2=36, and 3a;+2y=6. — 225 — *UNIT 14. NUMERICAL TRIGONOMETRY ALBERTA, MICHELSON (1852–1931) From early manhood until almost the day of his death, Professor Albert A. Michelson (1852–1931) gave his best efforts to determining, with exactness, how fast light travels. Some one once asked him why he worked so hard on this problem. He could have given many excellent scientific reasons, but his only answer was, “Because it's such good fun.” Light travels at a rate which would permit it to encircle the earth seven times at the equator in one second. In jº. other words, light travels 186,284 miles a second. Determining the speed of A. A. MICHELSON light is an example of getting the meas- ure of something which cannot be measured directly. Many other examples could be given. How far is it to the moon? As- tronomers have measured, and can tell you. Of course, they did not measure this distance directly as you would measure the length of a table. One can- not layoff the distance from here to the moon with a yardstick. All such meas- ures as these have been obtained by in- direct measurement. Professor Michel- son spent a lifetime determining, by in- direct measurement, the speed of light. TRIGONOMETRY Trigonometry has been called the science of indirect measurement, the science of getting the measure of something which, because of its nature or its inaccessibility, cannot be measured directly. Only those who know a great deal of mathematics can hope even to make a beginning at such prob- lems as determining the speed of light or meas- uring the distance to the moon. However, with only a few lessons in trigonometry, you will be able to solve many interesting problems by use of in- direct measurement. Tools Used in Trigonometry The word trigonometry comes from two Greek words which mean “to measure a triangle.” A tri- angle has sides and angles. To measure the length of a side directly, one can use a ruler, a yardstick, a tape, or another measure of distance. A protractor is the simplest means of measuring an angle di- rectly. Surveyors use an instrument called a transit for measuring angles. This is but a combination of two or more protractors arranged so as to measure both horizontal and vertical angles. Shadow Reckoning Thales (about 600 B.C.) is honored as one of the “seven wise men” of ancient times. One of his ac- complishments was to find the height of a pyramid by measuring the height of its shadow. This is an example of indirect measurement, and at that time it was considered a wonderful achievement. Today many high school pupils discover Thales' method for themselves and use it in measuring the heights of trees and buildings. EXAMPLE The flagpole in a school yard casts a shadow 48 feet long at the same instant that a 3-foot pole casts a shadow 4 feet long. Find the height of the flagpole. Solution: Let a stand for the height of the flagpole. Then a: 3 3 — = — ; Or *-i-48– 36. Thus the flagpole must be 36 feet high. 48 4 21 EXERCISE A man whose height is 5 feet 9 inches casts a shadow 8 feet long. How high is a telephone pole that casts a shadow of 32 feet at the same instant? — 226 — EXERCISES 1. John measured the length of the shadow of a tall poplar and found it to be 108 feet. At the same time he found that a four-foot stick cast a shadow of six feet. How tall was the poplar? 2. Later John measured the length of shadow of a cherry tree and found it to be forty-two feet. At that time his four-foot stick cast a shadow of seven feet. How high was the cherry tree? SIMILAR TRIANGLES Being able to solve triangles as in the exercises above depends on the principle that A the corresponding sides of similar triangles are proportional. The right triangle formed by the tree AC, its shadow CB, and the line AB from the top of the tree to the end of the /* D shadow is similar to the right triangle formed by the stick DF, its shadow FE, and the line I AC DF ^ DE from the top of the stick to the end of its shadow. The proportion is ECTEF B C E F TANGENTS OF ANGLES AC The value of the ratio BC depends on the size of the angle B. In trigonometry, this AC ratio is called the tangent of angle B. It may be written: tan B TBC , or AC = BC. tan B. Using the figure at the right, determine the value of the tangents of the angles shown. The value can be found by dividing the number of blocks in the height of the triangle by the number of blocks in its base. The first exercise below has been done correctly. (a) tan 10°=# = .18 (d) tan 40°= (b) tan 20°= (e) tan 50°= (c) tan 30°= (f) tan 60°= Notice that the value of the tangent depends on the size of the angle, and that the tangent increases as the angle increases. By more accurate methods than you have learned, mathematicians have found the values of the tangents of angles with great exactness. A table of these values is given on page 238. Consult this table and find how closely the values you have just found agree with the table. SoLVING A RIGHT TRIANGLE EXAMPLE Given angle B = 20° and A Solution: AC = BC tan B BC = 100 ft. Find A.C. _T 100(.364) = 36.4 B C Answer: 36.4 ft. EXERCISE Given angle B = 30° and BC = 50 ft. Find A.C. Answer: — 227 — SoLVING RIGHT TRIANGLES EXERCISES AND PROBLEMS 1. Given angle B =46° and BC = 25 feet. Find A.C. Answer: 2. Given angle a = 53° and base b = 63 feet. Find the height. Answer: 3. Find the distance QR across a swamp, when it is given that an- gle P=41° and PQ = 1,275 feet. Answer: 4. A guy wire to the top of a der- rick makes an angle of 78 degrees with the ground. The distance from the base of the derrick to the point where the guy wire is fastened to the ground is twenty-two feet. Find the height of the derrick. Answer: *5. A ladder leans against a Ver- tical wall. It is impossible to meas- ure directly how high on the wall it reaches. What measurements could you get from the ground which would enable you to find the height of the top of the ladder? Explain. *6. From a point at the edge of the water on the south side of a lake, a tower can be seen rising from the water's edge on the north side. The distance from the point to the base of the tower is known to be exactly one mile. What measurement could be taken without moving from the point on the south side that would enable one to compute the height of the tower? Explain. the student.) A (Figure to be supplied by the student.) (Figure to be supplied by — 228 — TRIGONOMETRIC FUNCTIONS The tangent of an angle a depends on (is a function of) angle a. This simply means that the value of the tangent changes as the value of the angle changes. There are two other simple trigonometric functions which we need to know. They are called the sine and the cosine. The sine, cosine, and tangent are defined below. e a opposite side b adjacent side sin A = — = — cos A = — = — C c hypotenuse c hypotenuse * b opposite side a adjacent side b sin B = — = — COS B = — = — (! c hypotenuse c hypotenuse a opposite side b opposite side tan A =–= ºp º tan p_p_oppºnense A C B b adjacent side a adjacent side VALUES OF THE SINE AND THE COSINE Using the figure at the right, find the sine and the cosine of each 90° O of the angles shown. The hypotenuse equals 10 for each triangle; the height or the base may be found by counting blocks. Record your results as decimals, correct to two places. 0° | 10° 20° 30° | 40° 50° 60° 70° 80° 90° sin COS GRAPHS OF SINE, CoSINE, AND TANGENT The way in which one quantity depends upon another is clearly shown by a graph. The ac-axes below are marked off into intervals representing 10°, and the y-axes into tenths. Referring to the tables you have made, or to the table on page 238, plot the necessary points and make graphs of the sine, the cosine, and the tangent. A part of the sine graph has been done correctly. ) | .9 .9 10o 200 300 400 500 600 700 800 909 *Oo 200 300 400 50o 600 700 800 909 100 200 300 400 500 600 700 800 90o SINE GRAPH CoSINE GRAPH TANGENT GRAPH — 229 — USING THE SINE AND THE COSINE EXAMPLE Given angle B = 57° and AB = 20. A Solution: Find AC and BC. AC AC = 16.8 sin B = B’ or AC = A.B. Sin B |BC = 10.9 AC=20(.8387) = 16.774 BC cos B = B’ Or BC = A.B. COs B B421 C BC =200.5446) = 10.892 EXERCISES 1. Given angle B = 41° and A AB = 55. Find AC and BC. AC = |BC = B C 2. Given angle P=27° and PQ = 78. Find PR and QR. PR = QR = Q P 41° 90° Ič 90° 279 L 90° 61° 3. Given angle M = 61° and MN = 65.7. Find LN and LM. LN = LM = M N 4. Given angle D=53°, angle E=28°, DF=10, and FE = 17. Find FG and D.E. F FG = DE = 53° 90° 28° D G E — 230 — DROBLEMS 1. A ladder 27 feet long leans against a building so as to make an angle of 72 degrees with the ground. How far from the base of the build- ing does the foot of the ladder rest? Answer: 2. A straight walk 400 rods long runs diagonally across a rectangular park, making an angle of 39 degrees with the longer side of the park. Find the dimensions of the park. Length: Width: 3. One end of a 68-foot guy wire is fastened to the top of a smoke- stack, and the other end to the ground. The wire makes an angle of 74 degrees with the ground. Find (1) the height of the smokestack and (2) the distance from the ground end of the wire to the base of the Smokestack. Answers: (1) (2) 4. Given angle B = 13 degrees, and AC = 225. Find A.B. AC sin B Answer: AB = Hint: AB = C 5. Given angle B = 59 degrees, and BC = 515. Find AB and A.C. Answers: AB = AC = A — 231 — INTERPOLATION The table on page 238 gives directly the values of the sine, the cosine, and the tangent of angles with a whole number of degrees. By interpolation we can find the sine, cosine, and tangent of angles made up of degrees and parts of a degree. Study the examples carefully. EXAMPLES 1. Find sin 46° 20'. 1A sin 46° = .7.193 20 . 7193 sin 47° = . 7314 Go” . 0121 = .0040 .0040 Difference = .0121 . 7233 B C Therefore sin 46° 20' = .7233 2. Find angle B. A 10 O tan B === 1.2500, tan 51° = 1.2349 - |- 10 tan 519 – 2349 " " -lº" tan 52° - 1 - 2799 Difference e 0.151 Difference 0450 151 f 1-ºx60–20 º In or 1 =1.5×ov = B S C Therefore B = 51° 20' 3. Find angle Q. 9 P sin Q === .7500 Sin B = .7500 in 48° º sin 48° = . 7431 SII). = . sº wº- 12 9 sin 49° = . 7547 Difference © 0069 Diff . 0116 — X 60' = 36' 1.TIGI’éI) Cé 0 TIG’ Therefore Q = 48° 36' Q R EXERCISES 1. Sin 18° 40' = 4. tan B = .7600 B = 2. tan 33°15' = **5. COS 43°45' = 3. sin B = .5431 B = **6. COs B = 4532 B = — 232 — ANGLES OF ELEVATION AND DEPRESSION If we sight from point P, anglea, is called the angle of elevation of point S, and angle y is called the an- gle of depression of point Q. Notice that the angle of elevation is measured upward from horizontal to the line of sight, and that the angle of depression is measured downward from horizontal to the line of sight. PROBLEMS 1. A kite string is 100 yards long. Find the height of the kite when its angle of elevation is 38 de- grees. Answer: 2. A flagpole 51 feet high casts a shadow 100 feet long. What is the angle of elevation of the Sun? Answer: 3. From a second-floor balcony, the angle of elevation of the top of a telephone pole is 8 degrees, and the angle of depression of its base is 16 degrees. The balcony is 50 feet dis- tant from the telephone pole. Find © the height of the latter. Answer: 4. A flagstaff stands on top of a tower. From a point on the ground, at a distance of 58 feet from the base of the tower, the angle of elevation of the top of the flagstaff is 77 de- grees and the angle of elevation of its base is 63 degrees. Find the height of the flagstaff. Answer: *5. The angle of elevation to the top of a tower 175 feet high is 31 degrees. How far from the base of the tower was the sight taken? Answer: — 233 — TRIGONOMETRIC FoRMULAs Using the formulas for sin A as patterns, write the corresponding formulas for cos A, tan A, and sin B. Oſ, 1. sin A =– , or a = c-sin A, or c=- 3. COs A = C sin A 2. tan A = 4. sin B = EXERCISES 1. A = 28°20', c = 148. Find a and b. 5. A = 18°36', a = 309. Find c. 2. B = 65°24', c = 652. Find a and b. 6. A =65°18', b = 55. Find c. 3. B=73°45', a = 120. Find b. 7. b =891, c = 1,000. Find A. 4. B = 67°54', b = 111. Find a. 8. a = 788, c = 1,000. Find A. — 234 — PROBLEMS 1. A vertical cliff rises from the edge of a lake. From a boat 1,500 feet out, the angle of elevation of the top of the cliff is 6°16'. How high is the top of the cliff above the water's edge? Answer: 2. A mountain road a mile long makes an aver- age angle of 17°30' with the horizontal. How many feet higher is the top end than the bottom end of the road? Answer: 3. A rafter 18 feet long reaches from eaves to ridgepole, making an angle of 24 degrees with the horizontal. How much higher is the ridgepole than the eaves? Answer: 4. Two towers of equal height stand on a level field, 150 feet apart. The angle of elevation from the base of either to the top of the other is 33°25'. How high are the towers? Answer: 5. In finding the distance AC across a river, line CB is drawn on a bank and perpendicular to A.C. By measurement it is found that CB is 152 feet long, and that angle CBA equals 64°33'. Draw the figure and solve for AC. Answer: — 235 — INVENTORY TEST measurement. of the side 1. Trigonometry has been called the science of 2. The tangent of an angle is defined as the divided by the side. 3. When the base and the angle of elevation are given, one finds the height by use of the (sine, cosine, tangent). 4. When the hypotenuse and the angle of elevation are given, one finds the base by use of the 5. In a right triangle, the sine of angle A is equal to th of angle B, and the sine of angle B is equal to the Of 6. Recall the graphs of the sine, the cosine, and the tangent. (a) As the angle increases from 0° to 90°, the sine (b) As the angle from — to to (c) As the angle increases from 0° to 45°, the tangent to , the cosine increases from from to from 7. Trigonometric functions of angles of 30°, 45°, and 60° can be easily found by getting the ratios of the sides of appropriately constructed triangles. Such triangles are given on the right. Find the sine, W3 the cosine, and the tangent of these angles, first as 1 common fractions and then as decimals. The sine 45° of 45° has been found correctly. Recall that 1 W3 V2=1.414, and V3=1,732. - - Sine Cosine Tangent 30° 1 V2 1.414 45° —== — = — = .707 v/2 2 2 60° 8. Using the table of trigonometric functions given at the right, insert the correct value wherever a question mark (?) occurs below. 1 2 3 4 5 sin COS tan A. 479 79° 30° 18° .3090 .9511 .3249 B 18° 54° 30° .5000 | .8660 . 5774 0. 7 º 10 10 47° | . 7314 | .6820 | 1.0724 b 2 2 10 54° .8090 | . 5878 | 1.3764 C 10 100 79° | .9816 | . 1908 || 5. 1446 — 236 — CUMULATIVE REVIEw 1. Name or give examples of three ways of indicating that one quantity depends upon another. (a) (b) (c) 2. Make tables of values and graph each of the following. Notice that there will be two graphs on each pair of axes. (a) y=a^*–4 * | | | | | | | | | , TTTTTT ºf , | | | | | |T| The two graphs intersect in the points ( ) and ( y ). y (b) a 2+y}=25 * | | | | | | | | | , | | | | | | | | | The graphs intersect in the points ( ) and ( y ). y 3. Write a simple problem (make it up yourself) as a word statement, and then translate the word state- ment into an algebraic equation. Word statement: Algebraic equation: 4. Is it possible for the same algebraic equation to represent more than one word problem? Give an example. 5. Set up the proper equations and solve: (a) Robert is eight years older than Henry, and the sum of their ages is twenty-two years. Find the age of each. (Let a = Henry's age.) (b) Two hucksters sold twenty-two dollars' worth of produce. The owner of the business re- ceived eight dollars more than his assistant. How much did each receive? (Let x = assistant's share.) VALUES OF THE TRIGONOMETRIC FUNCTIONS Angle Sin Cos Tan Angle Sin COS Tan 19 . 0175 .9998 . 0175 469 . 7193 . 6947 1. 0355 2 . 0349 .9994 .0349 47 . 7314 . 6820 1.0724 3 ... 0523 .9986 ... 0524 48 . 7431 . 6691 1. 1106 4 . 0698 . 9976 . 0699 49 . 7547 . 6561 1, 1504 5 0872 .9962 . 0875 50 , 7660 . 6428 1. 1918 6 , 1045 . 9945 ... 1051 51 . 7771 . 6293 1. 2349 7 1219 . 99.25 . 1228 52 . 7880 . 6157 1.2799 8 . 1392 .9903 . . 1405 53 ... 7986 . 6018 1. 3270 9 . 1564 . 9877 . 1584 54 .8090 . 5878 1. 3764 10 1736 .9848 . 1763 55 . 8.192 . 5736 1. 4281 11 . 1908 . 9816 . 1944 56 . 8290 . 5592 1.4826 12 . .2079 . 9781 . 2126 57 . 8387 . 5446 1. 5399 13 . 2250 .9744 . 2309 58 . 8480 . 5299 1. 6003 14 . 2419 .9703 . 2493 59 . 8572 .5150 1.6643 15 2588 .9659 . 26.79 60 . 8660 . 5000 1. 7321 16 .2756 .9613 . 2867 61 . 8746 . 4848 1. 8040 17 . 2924 . 9563 . 3057 62 .8829 .4695 1.8807 18 .3090 . 9511 .3249 63 . 8910 .4540 1.9626 19 3256 .945.5 . 3443 64 . 8988 .4384 2.0503 20 . 3420 | 9397 . 3640 65 .9063 . 4226 2. 1445 21 . 3584 . 9336 . 3839 66 . 9135 .4067 2. 2460 22 . 3746 . 9272 .4040 67 . 9205 . 3907 2. 3559 23 . 3907 . 9205 . 4245 68 . 9272 . 3746 2.4751 24 . 4067 . 9135 . 4452 69 . 9336 . 3584 2. 6051 25 . 4226 .9063 . 4663 70 . 9397 . 3420 2. 7475 26 .4384 . 8988 . 4877 71 .945.5 . 3256 2.9042 27 .4540 . 8910 . 5095 72 . 9511 .3090 3.0777 28 .4695 .8829 . 5317 73 . 9563 . 2924 3.2709 29 . 4848 . 8746 . 5543 74 .9613 . 2756 3.4874 30 . 5000 .8660 . 5774 75 .9659 . 2588 3. 7321 31 . 5150 .8572 . 6009 76 .9703 . 2419 4. 0108 32 . 5299 . 8480 . 6249 77 .9744 . 2250 4.3315 33 . 5446 .8387 . 6494 78 . 9781 .2079 4. 7046 34 . 5592 . 8290 . 6745 79 . 9816 . 1908 5. 1446 35 . 5736 ... 8192 . 7002 80 .9848 . 1736 5. 6713 36 . 5878 .8090 . 7265 81 .9877 . 1564 6. 31.38 37 . 6018 ... 7986 . 7536 82 .9903 . 1392 7. 1154 38 . 6157 . 7880 . 7813 83 . 99.25 . 1219 8. 1443 39 . 6293 .7771 . 8098 84 . 9945. ... 1045 9.5144 40 . 6428 . 7660 . 8391 85 . 9962 .0872 11. 4301 41 . 6561 . 7547 . 8693 86 .9976 . 0698 14.3007. 42 . 6691 . 7431 . 9004 87 .9986 .0523 19.0811 43 . 6820 . 7314 .9325 88 .9994 .0349 28. 6363 44 . 6947 . 7193 . 9657 89 .9998 . 0175 57. 2900 45 . 7071 . 7071 1.0000 90 1.0000 .0000 | . . . . . . . — 238 — SUPPLEMENTARY AND REVIEW WORK FOR UNIT 1 1. Supply the item or items missing from each column below. Term 3a;2 523 º º 7 9;3 4748 5m. 7 7 Numerical coefficient 7 º 4 12 7 º ? º 1 2 Literal part º º QC Q/ 2 º 7 º QM, ?) Exponent º º 5 2 3 º º º 2 1 2. A certain rectangle is three times as long as it is wide. Let a stand for its width and write the formulas for its length, its perimeter, and its area. Evaluate these formulas for 2 = 7. - 3. The longer base of a certain trapezoid is twice the shorter base and its altitude is 4. Let a stand for the length of the shorter base, and write the formulas for the length of the longer base and the area of the trapezoid. Evaluate these formulas when a = 6. 4. Evaluate the following expressions when a = 2, y = 5, and z = 6. (a) ac”—Hacz (c) y” —acz (e) z*—y? —2% (g) a.”--y” —22:y (b) acz — y (d) a 4-Hy? (f) ac” + acy +y” (h) (2” —y?)a. 5. Write as algebraic expressions with a representing the unknown. (f) Twice a certain number increased by 46 (g) 18 increased by 3 times a certain number (h) 34 diminished by 7 times a certain number (i) A certain number squared less a” (j) k times the square of a certain number increased by 19 (a) A certain number increased by 13 (b) Twice a certain number less 7 (c) 8 plus the square of a certain number (d) 9 minus the square of a certain number (e) The square of a certain number diminished by 6 times the number 6. Translate each of the algebraic expressions into a word statement. (a) 2 –34 (c) 3a;+14 (e) 17–2a: (g) ac” +45 (b) 8+3a. (d) 53 —a.” (f) 65+2% (h) 53 +32 7. Given the formula C =#(F–32). Let F, in turn, equal 41, 104, 95, 77, 50, 86, 68, and 59. Find the corresponding values of C. 8. Given the formula A = #h (b.1+b2). Find A when h, bi, and b2 are given. (a) h = 6, bi = 13, b2 = 6 (b) h = 3, bi =9, b2 = 5 (c) h = 7, bi = 12, b2 = 4 9. Evaluate when u = 2, v = 3, and w = 4. (a) w?--v?--w” (b) wy --ww-Hvv (c) w” +2uv +v? (d) v2 +2ww-Hw” **10. Canals are dug and mountains are moved one shovelful at a time. What seems impossible when looked at as a whole often becomes simple when taken one step at a time. There is a hard formula in geometry, known as Hero's formula, for finding the area of a triangle when the three sides are given. Let a, b, and c stand for the sides of the triangle; let S stand for one half the perimeter; and let A stand for the area. We have the formulas: +-b s-ºttº and A* = S(s—a) (s–b) (s—c). Use the formulas above to find the values of the items missing from the table below. Ol. 3 6 5 10 8 9 10 4.5 17 9 17 13 b 4 8 12 24 15 12 17 6 39 40 25 14 C 5 10 13 26 17 15 21 7.5 44 41 28 15 S * º ? ? º ? º º º º º * S - Q. 7 . º 7 7 ? º * º º º º * s — b º º º º º º º º 7 º º 7 S – C º º º º º º º º º 2 º 7 A? 7 º º º º * ? º º º º 7 A. 6 7 30 7 ? º º º 7 º º º SUPPLEMENTARY AND REVIEW WORK FOR UNIT 2 1. Translate each of the following word statements into an algebraic equation. (a) A certain number plus 4 equals 13. (e) Twice a certain number less 2 equals 28. (b) A certain number subtracted from 17 leaves a re- (f) One half a certain number added to 8 gives 34. mainder of 5. (g) A certain number is 7 less than 29. (c) A certain number is 8 more than 20. (h) A certain number equals the sum of 33 and 46. (d) A certain number is equal to the difference be- tween 37 and 19. 2. Translate each of the following algebraic equations into a word statement. (a) a +3 = 14 (c) #2 +2 = 7 (e) 32 +2 = 17 (g) a = 19–16 (b) 14 — a = 6 (d) 31 – $2 = 27 (f) 15–32 = 3 (h) a = 31-H14 3. Solve the equations and check. (a) a +7 = 10 (f) .22 – 5 = 7 (k) 5 = 2.8a. -- .2a: (p) 1.5a — .32 = 24 (b) 23:-H7 = 10 (g) 1.3a; +5 = 44 (l) .5a. --10 = 1.5a. (q) 4.8a. = 12 (c) 3a – 6 = 15 (h) .9ac-H .2 = 2 (m). 62 = 3 — . 1a: (r) 4.1a: + .22 = 8.6 (d) #2 +3 = 4 (i) 3a – 5 = 3c (n) 7a;+5 = 12a: (s) 2.05 = }a — .45 (e) .5a. — 1.5 = 3 (j) 1.7a;+3 = 2a: (o) 17 = 32 – 1 (t) 72 –3.5a. = 42 4. A father is seven times as old as his son; in six more years, the father will be only four times as old as the son. How old is each now? 5. A mother is three times as old as her daughter; six years ago, the mother was nine times as old as the daughter. How old is each now? 6. Henry is twice as old as George. In twenty years, Henry will be only nine-sevenths as old as George. How old is each now? 7. Anne's age is eight years more than twice Sue's age. The sum of their ages is twenty. Find the age of each. 8. In five years, John will be four-thirds as old as he is now. What is his present age? 9. Mr. Simpson invests part of his money at four per cent and part at six per cent. The latter amount exceeds the former by $500. The total annual income is $230. How much has he invested at four per cent? At six per cent? 10. A man invests $5,000, part at four per cent and part at five per cent. The total annual income is $218. How much has he invested at four per cent? At five per cent? 11. Two investments, one at three per cent and the other at five per cent, yield the same annual return. The amount at three per cent is $2,000 greater than the amount at five per cent. Find each amount. 12. The sum of two numbers is 68. The larger is 6 greater than the smaller. Find the numbers. 13. The sum of four consecutive odd integers is 80. Find them. 14. Three consecutive even integers are such that the smallest plus one half the second plus one third the third equals 28. Find the three integers. 15. Two integers differ by 7, and are such that 4 times the smaller equals 3 times the larger. Find them. 16. The perimeter of an isosceles triangle is 47. The sum of the equal sides exceeds the base by 5. Find each side. 17. The perimeter of an isosceles triangle is 63 inches. The base is 3 inches greater than either of the equal sides. Find each side. *18. The length of a rectangle is 9 more than its width. By doubling the width and halving the length, a new rectangle is formed whose perimeter is 4 more than the perimeter of the given rectangle. Find the length and the width of the given rectangle. *19. A rectangle is ten feet longer than it is wide. By increasing its width by twenty per cent (the length remaining unchanged), the perimeter is increased by six feet. Find the original dimensions. *20. A rope hangs over a pulley so that the length of the longer end is three times that of the shorter end, When fifteen feet of rope has passed through the pulley, the two ends are equal. How long is the rope? — 240 — SUPPLEMENTARY AND REVIEW WORK FOR UNIT 3 1. Add. –3 5 —7 6 — 5 7 5 –4 17 — 16 18 –24 2 2 –4 4 4 9 1 3 – 6 — 13 —9 36 –4 -3 -2 3 -6 -3 –2 -2 4 7 3 - 12 2. Add. 164: 7a: — QC 2y –6/ 22 2y2 422 – 6p 6r 5s 9m. — QC 3a; –2a: 7 y –9y – 2 –7 yz 1222 –2p –67. 2s — 57m. —23. –2a: 3a; —39 3y 52. —30/2 —622 –5p 12r — 6s 4m 3. Subtract. 14 17 6 7 8 —8 5ac 52 — 122 —4c 16," –9y –3 3 16 -4 9 17 112 - 112. 82 12c —6r —2/ 4. Multiply. –4 — 10 8 —6 12 — 5 3a; — 5a, 4a: 322 $/2 –392 6 5 4 -3 -2 15 2 5 −5 –2 – 10 – 12 5. Divide. 81 by 3 60 by — 15 —124 by 31 –96 by — 16 —81 by 27 34 by —2 6. Divide. 52 by — ac 15uv by —3vv 72t by 24 –96m by —2 —54r by 3r 48#2 by — 12t” 7. Use the directed number scale to supply the missing words. (a) To add 4 to 5, begin at and count units to the – (b) To add 3 to —6, begin at and count units to the (c) To subtract 5 from 6, begin at and count units to the (d) To subtract 2 from —4, begin at and count units to the (e) To add –3 to 4, begin at and count units to the (f) To subtract —3 from —5, begin at and count units to the 8. (a) The product of two numbers with like signs is 4 (b) The product of two numbers with unlike signs is (c) The quotient of two numbers with like signs is (d) The quotient of two numbers with unlike signs is 9. Use the race track at the right. (a) Multiply each number on the track by —2; (b) Add to each number on the track +8; —5. Q (c) subtract from each number on the track Start t –4; +9. 10. Find the sum of each row and each column of the squares below. 12 | – 7 || – 6 11 || – 10 10a || – 9a || – Sa: 9.c | — 12.c —8 || – 2 10 5 0 –9 — 3 4 || – 1 Q – 11a: | – 53: 2a || – 33: 73: — 7 3 9 — 1 –8 2 0 | – 2 8 — 10a: 0 — 23 — 40: 63: 13 — 6 || – 9 — 5 12 — 5 1 | — 4 3 5 — 7a: — a — 60: QX 3a; — 4 2 6 — 3 4 10 7 6 — 11 | – 12 8a; 5a: 4.x: — 13a; – 14a: 11 8 || – 11 — 10 11. Find the product of the two numbers of each exercise on pages 45 and 47. 12. Divide each of the following numbers by 2, and then divide each by —2. 2 +6 –22 8 – 14 +28 — 18C —26ab — 1802b3 0 — 4 — 10 16 +4 64 16a +84a: 48c.” 32m. – #g — 241 — SUPPLEMENTARY AND REVIEW WORK FOR. UNIT 4 1. Air is composed of nitrogen 78%, oxygen 21%, carbon dioxide, water vapor, argon, and other elements 1%. Show this (a) in a circle graph, (b) in a rectangular graph. 2. The average retail prices of ten common foods, for certain years, are given in the table. Round POrk Bacon Lard Eggs Butter Milk Flour | Potatoes | Sugar steak chops Lb. Lb. Lb. Lb. Doz. Lb. Qt. 10 lb. 10 lb. 5 lb. 1900 13.2 11.9 14.3 9.9 20.7 26.1 6.8 25.0 14.0 30.5 1903 14.0 14.0 18.2 12.0 25.9 28.5 7.2 27.0 17.0 28.0 1906 14.5 15.2 19.6 12.1 27.8 30.4 || 7.4 29.0 17.0 28.5 1909 16.4 17.4 22.4 14.2 31.9 34.5 8.1 36.0 19.0 29. 5 1912 19.9 19.2 24.4 14.8 34.1 37.4 8.7 35.0 22.0 31 .. 5 1915 23.0 20.3 26.9 14.8 34.1 35.8 8.8 42.0 15.0 33.0 1918 36.9 39.0 92.9 33.3 56.9 57.7 13.9 67.0 32.0 48.5 1921 34.4 34.9 42.7 18.0 50.9 51. 7 14.6 58.0 31.0 40.0 1924 33.8 30.8 37.7 19.0 47.8 57.7 13.8 49.0 27.0 46.0 1927 37. 1 36.8 47.2 19. 3 45.2 55.6 14. 1 55.0 38.0 36.5 1930 41.2 35.9 42.3 17.0 41.0 46. 1 14.0 47.0 36.0 31.0 1933 25.2 19.6 22.3 9.0 26.1 27.3 10.6 39.0 23.0 27.0 (a) Make a broken line graph for each food as the price varies from year to year. Be able to tell from your graphs the year in which the price of each food was highest, and the year in which the price was lowest. (b) Make a bar graph for each food as the price varies from year to year. 3. The table below gives the composition of some common foods. Make a circle graph of the constituents of each food listed. Protein Fat, Carbohydrates Ash Water Cheese 29 % 36 % .3% 3.2% 31.5% Dried Beans 22.5% 1.8% 59.6% 3.5% 12.6% Roast Beef 22.3% 28.5% 0 % 1.2% 48.0% Chicken 21.5% 2.5% 0 % 1.1% 74.8% Wheat, Flour 13.8% 1.9% 71.9% 1.0% 11.4% Eggs 14.8% 10.5% 0 % 1.0% 73.7% Milk 3.4% 4.0% 5.0% .6% 87.0% Butter 1.0% 85.0% 0 % 3.0% 11.0% 4. Make a table of values and a straight line graph for each of the following. (a) y = a -2 (d) y = a +1 (g) y = }~ (b) y = 2a: (e) y = #2 +4 (h) y = 32 — 4 (c) y = 4–2 (f) y = 1 +a; (i) y = 6–32 — 242 — SUPPLEMENTARY AND REVIEW WORK FOR. UNIT 5 1. Add. 3a; 9ab –39 –2a:”y —bc - Cl3: 5t 7s 9y –8w Q2. 2ab —y —32°y 6bc 3aac 4t –2s 4y 4v. 4a: ab –4y –2a:”y —bc 2aac —3t 4s –3y — M. 24. 3ab —54) —52°y 3bc — 5aac —2t –3s –67) — 5u. 2. Add. - 2a – 4 y 2a –2y+ 2 3uv-H2u — 50 7m2–6mm +5m2 3rs +2st +5t 62 – 29 ac-H7 y – 2 –2^uv-H 6v — v —3rm?--47mm — m2 — rS — St. — i. 2a –H39 — 5a: –42 7uv —4tu +4v … 5m2 +5mºn — 4n? —5ts +3st +3t —52 – 4ty 10a: —3/ - 2u +59 mºn +2n? 4rs —2st —2t 3. Copy with like terms in the same column and add. (a) 22 –3 y +2, —3a –y-H62, — 53:--31) —62, and – a –y–H32 (b) 2w-H2v — 5 w, 7u —6v-H3tw, -3v —6v, 3v-H2u, and 3 w —2v (c) 9k –m —n, 3rm +4k —5m, m —n, k + m, 4n —k, and m +n —2k 4. Subtract. 2a: 57m. —52 —5b 5b —2^w –2k 2s 17; 7t 52 2m. 22 —2b —2b 5w) —5k — 53 —7t 17t 5. Subtract. 122 – 20ay +6ly 4v. —6v --5 —5m — m. 2p-H.34 – r 2s – 5t-H 19 9ac – 2a:y-H4y 3u + v — 3 –2m ––2m. —p — q +2r –2s — 9 6. Copy with like terms in the same column and subtract. (a) 4ay —y from —24:y-Hy (c) 5 m + n from n –3m. (b) wy –v —3% from u +v (d) r—s — 5t from 2r–H5t 7. Remove parentheses and combine like terms. (a) (ac – y) — (a +y —2–3) (d) — (r—s) + (t–3r—s) (b) (rt —ts) + (2ts +2rs) — 5rs (e) m — (3m — m) — (3m — m) (c) — (p +2g) + (2g —p) – 5p (f) 8a —[2b — (3c-H2b) —a) 8. Multiply. 5a 72 —3t —6s 777.70, —9st —a:”y 7nn? —ty 8efg _2 –32 4s —2t 2mn. — 10t 5ty? — ??, 3rt 2fgh 9. Multiply. a —3b 2u —3v-H2w 3m —2n + k, 4r-H2s +5t 2a –3y +52 20. – 57) m — 3k 3r —2t a —23) –32 10. Multiply. - (a) (a —2b) by 3b - (c) (32–4y) by 42 (e) (2m – m —k) by (2n+k) (b) (8r-H.3s) by 2: (d) (2u +3v) by 4w (f) (a —2b-i-3c) by (2b –c) 11. Divide. 5a) 10a –23)162 7b) —21b –3y) – 18y 2m)10mm. – 12rs)72rst — 243 — 12. 13. 14. 15. Divide. ay)a?acy – 2day” 3m2n)12m2n3 +15mºnº –5a)15ac”y – 20acy2 —7tu) 14twv-H7tw” Divide. a-H2b)a? — ab–6b° 2a – 3 y) 10a:” – 11acy – 6y” 5u +39) 10 w” +wy – 30°–H5uv” +3v3 Divide. Check by multiplying the divisor by the quotient and adding the remainder. (a) arº-H.32°y +15acy?--10y? by a -2\, (d) acº —2a:”y?--y4 by a -y (b) a “ —3ry +y3 by a -y (e) a”--2a:y-Hy?--2a2+2y2+z” by a +y+2 (c) 28-H 10a:”y – 10ay” —yº by a +y (f) ac”--2a:y--y?–222 –2g2+2” by a +y –z Check each division and each multiplication in exercise 14 above by letting a = 2, y =3, and z = 4. 16. Finish the magic squares, including the sums of rows, columns, and diagonals. (a). sº Sums 2a –H4y +62 3a –H59 – 72 6a-H89 + 102 J 4a ––6/+82 Sums (b) f Sums a +3b +5c 6a +8b+10c 5a +7b ––9c Sums 6a+12b +186 17. Remove parentheses and combine like terms. (a) (a –2b +3c) — ( — c-H2b) — (–2a – b +c) —8c (c) (3k — m + n) — (k — m —n) + (3k +2m –3m) — 5m (b) (a +y –52) + (2 –z) — (a +y –32) — (–5y-Hz) (d) — (–r-Hs —t) + (−r —s +t) — (5r-H2s) + (5s –2t) 18. Find the product of- (a) a +3, and 22 – 5 y (e) 92+7y —52 and 32-H22 (b) 52 +y and 22 —y (f) 11a; +y —72 and 5a –y +32 (c) 7a: +y and 82 – 5 y (g) —82+3y –2 and 3a –2y+62 (d) 52 – y and 3y —z (h) ac – 29 —52 and 52 – 29 – 2 19. Divide. (a) acº —yº by a -y (f) a 3-H2a:”y +2a:y?--yº by a +y (b) acº — y4 by a -y (g) 23:4–7a:8–182% — a +3 by 22--1 (c) a 9-Hyº by a +y (h) 10a:3+33** –52a:--9 by 2a2+72 –9 (d) acº — y” by ac” + y” (i) 324-H.3a*-i-2a:” – a – 1 by 32% – 1 (e) a 3–2a:”y —y” by a -y (j) 52%–8a.”y +8ay” —30/3 by 52 –3y 20. Copy with like terms in the same column and perform the indicated operations. (a) From the sum of a -39 +2 and – 53:-H62 –y, subtract 22–2–y. (b) From 32 —y –122, subtract the sum of 89–2, 22--2, and 3y-ac. (c) From the sum of 72 – 29 +z and 82 –y, subtract the sum of 82-y and 3y-H22. — 244 — 5. SUPPLEMENTARY AND REVIEW WORK FOR UNIT 6 . Write the products. (a) (ac-H3) (a —4) (d) (y–7) (y-º-6) (g) (.52 +3) (.32 +5) (j) (acł-Hy?)? (b) (y –5) (y –4) (e) (1 —ac) (1+z) (h) (ab – c.) (ab-i-c) (k) (22–3) (22–H3) (c) (a +3) (a +6) (f) (32 – 1) (22 +1) (i) (23 —y?)? (1) (acº —2) (acº +2) . Factor completely. (a) a”--4a–H4 (d) #b2+-bc}-H cº (g) aºy” — 10a:yz+252% (b) a 9–6ay-H9y? (e) 81a2+36ab +4b* (h) 42°2+12a:yz+9y”z (c) c”—c-i-.25 (f) .25c” —2cd+4d? (i) ; c3–3c2d +9Cd? . Write the special products. (a) (a +b)? (d) (22 —y)* (g) (c’--#d)? (j) (#2 + y)? (b) (a —b)? (e) (.52 –29)% (h) a (32 —y)” (k) (23–4y")? (c) (a +29)% (f) (#c +5d)? (i) (cy —z)? (1) 5c(.5d —e)? . Factor completely and check by multiplying the factors together. (a) a.”—52y+6/? (g) 10a2+9:cy —9/? (m) 182°–2.1ay +.05y” (b) 32°-Ha.2 — 22? (h) 3a2+22ab +70? (n) 23:” – 11a2+52? (c) ac”z —acz — 422 (i) a 2–7ay —60y? *= (o) ++a” —4:#ab +30 (d) 50:8–20.cy? (j) a” — 16b+ (p) a 24 —y?a (e) 1–922+ 3 ºzº (k) as —4ab? (q) ac” +5.cºyº ––6/24 (f) aºy?–9y? (1) 3a?–10ab +7b2 (r) a 3–8wy? The formula for the area of the border around a square plot is A = S^-s”, where S and s are the sides of the larger and the smaller square, respectively. Compute A for the given values of S and s. Use (S-s) (S+s) = S2 — sº = A. 1. 2 3 4 5 6 7 8 º . . S 8.5 6# 5 2.7 2% 3# .85 1.75 º S 1.5 || 2% 3 1.3 1} 1% . 25 1.25 º S—s * º S+s º % A S Factor each of the exercises 6-50 and check by finding the product of the factors. 6. 5a-2 — 15a, 21. ac” + 17a: +60 36. ac” – aa — bac-Hab 7. a 2–H 11a; +24 22. a”--9ab +20b3 37. acº — 7a;+10 8. a7b2+18qb +32 23. ac” —8a; +12 38. 22 — 572 ––56 9. a.2 – 7a; +6 24. r2 – 5ar — 50a2 39. m2 + 13m — 140 10. a 2–H62 – 7 25. uż +13 w — 300 40. ac”—H·34a: +289 11. 22-i-7a; +6 26. y” +13) +12 41. acº -- 11a; – 12 12. a 2+5ac—84 27. a 2 – 9a: — 36 42. a 2–H3a; +2 13. a 2-H21a; +110 28. w? --23v —-102 43. a 2–292-1-190 14. y?--35y +300 29. y” +12y — 45 44. uż — 13u – 14 15. 1/2 +7 y–60 30. p?--25p–150 45. acº — 232 + 132 16. ac2–3a – 28 31. b2 – 30b --200 - 46. a 2 — 15a, +50 17. y?–79 – 18 32. a 2–20a: +91 - 47. ac” – 20a: +100 18. a7+9aac +82° 33. y” +19ay +48a.” 48. w? --29wy -- 1000? 19. p?–11p–60 34. ac” – 23a ––120 49. w” — 15v — 100 20. 22+13az+36a” 35. acº — 43a;+460 50. mac – mim --aac -i-an. — 245 — SUPPLEMENTARY AND REVIEW WORK FOR. UNIT 7 1. Solve for a and check. (a) a -3 = a (g) A = B+a; *(m) #2+c=d •º ºt-º (b) ac – b = c (h) C+a: = D *(n) #2 – m = p *(t) **** == Q. (c) m = n – a (i) 3a ––2 = 11 *(o) #–E–b *(u) . 7a: — a = .4a: (d) k — a =l (j) 52 – 3 = 12 ‘(p) +-a=4 “(y) ºrt-º- 2 6 . 5 3 (e) a +4 = m. (k) a = b +2a: -(q) tºº-ºº: “(w) ºf *-*. 2 3 Cl, 2 2a – 1 a.a. 1.2a ––1 a. f = P 1) A = B — 2 × [ ]" = –- × Sk — = — (f) a +q (1) ac () ==== (x) →-- 2. Remove parentheses and combine like terms. (a) (3a – a) — (2a – a) + (a — ac) (e) (52 – 29) — (2y–3a) +4 (b) 2ay — (ay +b)+(b–2ay) (f) (u-H3v) — (–2u –30) – 10u (c) (abc-Hbe) — abo — (2bc-H.3abc) (g) a(3a–b) —2b (b — a) +3b* (d) — (– a – ar) — (3a – 2d) + (2d —2) (h) ac(y – 22) + ( – y–32) +33% 3. Solve for a and check. (a) (32–2) (a —6) = a (32–8) (c) a (22 — a) = 2(22 +b) (b) (63 – 1) (a — 5) — (3a – 1) (22 – 1) = 0 (d) 10a: — (4a: +3)=3(82–2) 4. How old am I? (Set up appropriate equations and solve.) (a) Add seven years to my age and multiply the sum by 5, and the product is 100. (b) Add 3 to my age, multiply the sum by 6, subtract 2, and the result is 100. (c) Subtract 2.5 from my age, multiply the difference by 8, and the result is 100. (d) Subtract a from my age, multiply the difference by b, and the result is c. 5. How much money do I have? (Set up appropriate equations and solve.) (a) Add eleven cents to what I have and multiply the sum by 5. The result is $4.85. (b) Subtract what I have from $1, multiply the difference by 10, and add 30 cents. The result is $6. (c) Subtract what I have from $1, multiply the difference by 4, and add $1. The result is $5. (d) Take three cents from what I have, divide the remainder by 4, and only one dime remains. 6. Find two consecutive integers such that— (a) The smaller plus twice the larger equals 101. (b) One half the smaller plus twice the larger equals 47. (c) Three times the larger minus the smaller equals 65. (d) Twice the larger minus five times the smaller equals – 46. 7. Find the acute angles of a right triangle when— (a) The larger is 4 times the smaller. (b) The larger is 16° greater than the smaller. (c) Twice the smaller is 8° more than the larger. (d) The larger minus 8° equals the smaller plus 6°. **8. A messenger leaves P and travels toward Q at the rate of r miles an hour; h hours later a second messenger leaves P and travels after the first at the rate of S miles an hour. In how many hours will the second messenger overtake the first? **9. At what time between three and four o’clock does the minute hand overtake the hour hand? **10. At what time between nine and ten o’clock do the hour hand and the minute hand point in exactly opposite directions? — 246 — SUPPLEMENTARY AND REVIEW WORK FOR. UNIT 8 1. By properly changing the signs of the numerator, the denominator, and the fraction, write each of the following in three equivalent forms. (a) º (b) -º @ H. (d) -º 2. Reduce the following fractions to their lowest terms. (a) º (c) #. (e) #, (g) #. (b) #. (d) #. (f) º:# (h) *## () tº G) ºr * = 3. Perform the indicated multiplications and divisions. Express results in the simplest form. ac” – acy ac” – acy — 2ny 3–2 9–62 +zº 32–15 a 2+32–40 (a) a – 2 y acº — y” (c) 4+2a 6+52-1-22 (e) 22+8 a 9-1-92-i-8 (b) *-i-ry” cº-42y+4y” (d) 2°--22–8 cº–2–12 (f) *-32-H2. ac” + 1 a:3—24:2y ac” +y? a 2-1-102 –24 cº--52–84 ac? — 1 a. – 2 (g) *-i-yº.2%-H2:ry-Hy” 2-y (h) *—62–27 tº +10++24 22–32–28 g a +y a;4 — y” 3a; wº-i-92--18 24-72–18 cº-i-32 +2 4. Find the L. C. M. of the following. (a) ac”— 5a, +6, a 2–6a -i-8, a 2–72–H12 (c) 23:? – 32 +1, 4a:” – 1, 1+2a: (b) acº — 4, a 2-H42 +4, ac”—3a; +2 (d) a 9–y", a -y, a 2–y? 5. Find the L. C. D. of the fractions. 1 3 2 5ac -i-2 2a – 1 3 (a) y y (c) y y a;2 – a – 6 ac2 – 9 ac? --52 --6 52% + 10a: 23:2 – 8 acº —H4+ +4 2 . 3a – 1 2a –H3 I 1 1 (b) (d) g-Hº-T2. ' 2–73 IET2 . ac2 — 16 2–gy-Hyº g-Hy’ acº — y? 6. Find the L.C.D. of each pair of fractions; then reduce each fraction to an equivalent fraction having the L.C.D. as denominator. 5 2 y 3a; 5 * $/ 1 (a) y (c) y *(e) → y →- 4 — y 4 +y ac2 — 22 —8 ac” — 4 y” —60) +9 y” +4y —21 3a; 2 4 2 3 2 (b) = ** – # (d) → *-*— *(i) = *—, ± 3a –H2y 3a. 4a:” –y” 23:y—y” 6 y” —y – 1 3y” +7 y +12 _* 7. Perform the indicated additions and subtractions. Express results in the simplest form. 3 2 2a: ac” –-1 6 ac – 6 acº — 26 (a) z-3"3-2 (d) 23:? – 32 + 1 T1-22 (g) 2–4 Tri-3 **E=12 Ol, ac – y 3 a – 7 2 1 *-i-º-º: tº h I *-*. (b) *ET*2E. (e) acº — y” a +y (h) ac” +3.c — 28 "24-7 4 – 2 3 2 1 — a 2 — as ac” + acy +y? 1 3 | - >k:k / ; am sºme smºmºmº-mºm. --- (c) * – 4 'aº –32-H2 (f) a – 2 1 — ac (i) acº — iſ 3 ac – y ax *8. Simplify the complex fractions. 1+* +". 1 –– +2 a a” †II $/ $/ (a) –F– (b) —- (c) 2 sºm-º: sº mass mºs 1 —— 1 —— a tº b -- 1 y -- Q/ — 247 — SUPPLEMENTARY AND REVIEW WORK FOR UNIT 9 1. Solve for w and check. (C 3 - a; +7 12 Ol. b C — — = 1 d — — = 1 — — -: (a) a — 1 a. (d) a – 5 a. – 5 (g) a 'a –1 ac(a – 1) Q. (C 6 2 8 — 26 Ol. b C b - - - - -: - - - (b) a — 3 a +3 ac” – 9 (e) a +4 a -4 3(a +4) (h) QC T.I. Q. 3a – 1 3a; +4 1 1 — 1 (l, b C = f - — = † - - (*) #TTT5...i ° 3. LäT.II-II (i) ; +z=T-z- 2. Find two numbers differing by 12, whose ratio is 5/7. 3. Find two numbers differing by 40, whose ratio is 5/7. 4. Twenty years hence I shall be twice as old as I was ten years ago. How old am I now? 5. John is twice as old as Morris. If John were four years younger and Morris were four years older, their ages would be in the ratio of 4 to 3. How old is each? 6. What number may be added to both the numerator and the denominator of the fraction 3/5 to obtain the frac- tion 5/9? 7. A certain number is added to both numerator and denominator of the fraction 3/5; the same number is sub- tracted from both numerator and denominator of the fraction 4/7. The two resulting fractions are equal. Find the number added and subtracted. - 8. The denominator of a certain fraction is 13 greater than the numerator. If 2 be added to the numerator, and 3 be subtracted from the denominator, the fraction equals 4/5. Find the original fraction. © 9. The rate of one automobile is twelve miles an hour less than that of another. The faster goes 231 miles while the slower goes 165 miles. Find the rate of each. 10. How many pounds of salt need be added to forty-six pounds of water to make a brine containing eight per cent salt? 11. How much cream containing forty per cent butterfat need be added to twenty pounds of milk containing four per cent butterfat to make a mixture containing twenty per cent butterfat? 12. An auto radiator contains ten gallons of a twenty-five per cent mixture of alcohol. How much needs to be drained off and replaced with pure alcohol to make a fifty-five per cent mixture? 13. A dairy company uses milk averaging 3.9% fat and other milk averaging 5.1% fat to make a mixture averaging 4.2% fat. How much milk of each kind need it use to make 100 pounds of the mixture? *14. It is fourteen miles by automobile road from A to B, and eight miles by footpath. Travel by automobile is thirty- one miles an hour faster than walking, and it takes 1/5 as long to go by automobile as to walk. Find the rate of each. **15. The formula V = Th(R*—r”) gives the cubical contents of the material needed in making a hollow cylinder (tile) having outer radius R, inner radius r, and length h. (a) Solve for h. Find the value of h when V = 440, R = 8.5, and r = 1.5. (b) Solve for R2. Find R2 when W = 3960, r = 4, and h = 15. **16. C = r Solve for E; for R; for r; for n. 70, **17. A man and a boy together can do a job in six days less time than the man working alone. If the man works three times as fast as the boy, how long would it take each to do the job working alone? **18. One pipe will fill a tank in T' hours, and another pipe will empty it in t hours. If the tank is full and both pipes are opened, in how long a time will it be completely emptied? In order that the tank be emptied when both pipes are opened, it is necessary that T be greater than t. Explain. Check by substituting definite numbers for T and t. — 248 — SUPPLEMENTARY AND REVIEW WORK FOR UNIT 10 1. Use the method of substitution to solve for a and y. Check. (a) 3a; +2y =0 (b) 42-y-1 (c) 22 — y = 2 (d) a + y = –1 5a, +39 = 1 22+y=5 3a – y = 4 a +2y = 2 (e) 52 –39 = 14 (f) 64 – 29 = 11 (g) 20a: –7 y = 20 (h) 9a, +12y = –12 2a:--9y =26 4a –H2y =9 5a, +149 = 5 4a –39 = – 17 2. Solve graphically. - (a) 2a: +3y = 12 (b) 2a: --y =8 - (c) 52 + 2y = 10 (d) 32-H2y = 10 2a –39 = 0 4a – y = 4 4a ––3y = 15 52 +y=8. 5 (e) 2a –3y = —24 (f) ac –30) = 15 (g) — 42 + y = 15 (h) 4a: +8y = 18 a +5 y = 27 2a – y = 20 2a – y = —7 6ac – y = 1 3. Solve exercises 1 and 2 above by the method of addition and subtraction. (Multiply the coefficients and the constant term of one or both equations by suitable numbers when necessary.) 4. The sum of the digits of a given two-digit number is 12. The number formed by reversing the digits is 36 greater than the given number. Find the given number. 5. In a given two-digit number, the first digit exceeds the second by 1. The given number plus the number obtained by reversing the digits equals 99. Find the given number. 6. The current in Red River is known to be three miles an hour. A man rows down the river for two hours; it takes him five hours to return. Find the distance he rows, and his rate in still water. 7. A rectangular garden plot is ten feet longer than it is wide. If its width were increased by three feet and its length were increased by eight feet, its area would be increased by 296 square feet. Find the dimensions of the plot as given. 8. Mrs. Harrison invested $8,000, part at 5% and part at 53%. Her income was $415 a year. How much did she invest at each rate? - 9. A man and a boy working together can do a piece of work in five days. If the man were to help the boy only one day, it would take the latter ten additional days to finish the work. How long would it take each to do the work alone? 10. Separate 80 into two parts such that the quotient of the larger divided by the smaller is 6 with a remainder of 3. 11. A rectangular pool is fifteen feet longer than it is wide. A walk three feet wide, surrounding the pool, has an area of 426 square feet. Find the dimensions of the pool. 12. The sum of A's age and B's age is 98 years. Four years ago, A was exactly twice as old as B. Find the age of each now. 13. Ten years ago, the quotient of A's age divided by B's age was 3. Ten years hence A will lack one year of being twice as old as B. Find the age of each now. - 14. The sum of two numbers is k, and their difference is m. Find the numbers in terms of k and m. 15. A coal dealer contracts to deliver twenty-four tons of coal. If he could have used both his large and his small truck, he could have made the delivery in four trips. But after the first trip the small truck broke down, necessitating one additional trip with the large truck. Find the capacity of each. 16. Five times one number decreased by 4 times a second number equals 3, while twice the first number plus the second number is equal to 80. Find the numbers. 17. The sum of two numbers is 30. If three times the larger number is subtracted from four times the smaller number, the remainder is 1. Find the numbers. 18. The sum of the digits of a given two-place number is 11. The number formed by interchanging the digits is 27 more than the given number. Find the given number. — 249 — SUPPLEMENTARY AND REVIEW WORK FOR. UNIT 11 1. Solve for ac. (a) ar:3 = 5:7.5 (c) 4:a: = 3:10 - (e) ac:# = 4:3 (b) 6:4 = 2:10 (d) 8:3 = 12:a: (f) .2:a: = 5:20 2. To the left of the double line below is a recipe for twelve biscuits. For a larger family, for company, or for a church dinner, more biscuits are needed. Find the number that should be in each blank of the table to show the amounts of the various ingredients needed in making the number of biscuits written at the top of the column. Number of biscuits 12 18 24 30 54 60 84 120 300 Flour (cups) 2 -*- *-*. - *- - - *- e-m-mºse Baking powder (teaspoons) 2 - *- -º- - - *-me *- e-me Salt (teaspoons) # -*. - *- - - *º- -e *-mºs Shortening (tablespoons) 4 --- -- - *- - *m- -e -*. Milk (cups) # -º-e - *-*. *- - --- *- *- 3. In each part below, the three sides of one triangle and one side of a similar triangle are given. Find the other two sides of the second triangle. (a) 3, 4, 5 6, y — (c) 6, 8, 9 3, −, - (e) 5, 6, 8 , 3, (b) 3, 4, 5 , 6, (d) 6, 8, 9 — — 3 (f) 5, 6, 8 — – 12 4. The rainfall from the roof twenty-four feet by thirty-two feet drains into a cistern eight feet square. Write a for- mula expressing the relationship between h, the rise in inches of the water in the cistern, and r, the rainfall in inches on the roof. This is an example of variation. In each part below, either h or r is given. Find the other. Fº .4" .3" . 17 # # r § h 4" 6” 2” 4.5" 5. A large cog wheel has 480 teeth. The teeth of a smaller wheel fit into those of the large one so that they turn together. Let n be the number of revolutions of the smaller wheel to one revolution of the larger, and let t be the number of teeth in the smaller wheel. Then nt = 480. This is an example of variation. Complete the table below. 70, 30 32 48 5 15 t 12 24 36 - 60 20 6. (a) In direct variation, the of the variables is a constant. (b) In inverse variation, the of the variables is a constant. SUPPLEMENTARY AND REVIEW WORK FOR. UNIT 12 1. Raise to the powers indicated. (a) (–32)? (d) (2ab?)? (g) (22–2/)” (j) (a +2b)* (b) (22:2y)? (e) (a —b)? (h) (23 +y})? (k) (2a3b – c.)? (c) (ałb)? (f) (a +3b)? (i) (52 –39)% (1) (ałb} – }c)? 2. Find the indicated roots. 3 / -5. (a) – V4 (d) Jää (g) V(a-b)? (j) ; (b) V-8 (e) — VFT255 (h) važ(a-2)? *(k) ; 3/8 GT * ...as ſº (c) v/27 (f) W; (i) Viaºbº () W; — 250 — 3. Find the third side of a right triangle when two sides are given. Let a and b be the legs, and c the hypotenuse of the right triangle. - (a) a = 5, b = 6 (c) c = 1.3, a = .5 (e) a = 8, b = 4 (b) c = 15, b = 12 (d) a = 17, c = 19 (f) a = 7, b =5 4. Use the table on page 191 to find the following square roots. Remove from beneath the radical sign the largest possible square before referring to the table. - (g) a = 3, b = 2 (h) c = 10, a = 6 (a) V162 (c) V704 (e) V1728 (g) V275 (b) V252 (d) V1331 (f) v.203 (h) v272 5. Simplify the fractional radicals. 1 Ta 3 / GT ...) iſ 8 (a) V3, © W b © V 52 (g) V a. 2 T 4a ºf (b) V3 @ V a —b (i) Wii do V 1 — h 6. Change into like radicals and combine. (c) V3+ V3 (d) V3 – V12 (g) VIS-VHS 8 — *— (h) V4–Vršs (e) W54+VIö (f) Vi-Jä (a) V18– V72 (b) V108+ V27 7. Solve the radical equations. Check. (a) V2-2=2 © v7=-2V. (e) V2-3 = } V2-F9 **(g) Vºr-- V62 =8 (b) V32-ET=4 (d) VIT-2 – V2-F2T=0 (f) 2 V15–2 = V2+5 **(h) V22 – Vºz = 2 8. Perform the indicated operations. Express results in the simplest form. (a) V12+ V3 (d) vſ4a2–b2+ V2a–b (g) (al/2 +b)alſº (j) va. Vå (b) V128. V2 (e) al/2, al/3 (h) ab --a8/?bl/? (k) acy ** (c) Va-b; Va-Hö (f) a 2 +a;2/3 (i) Vic H-3, V2-37 (l) vºm-- Wm. SUPPLEMENTARY AND REVIEW WORK FOR. UNIT 13 1. Solve for ac and check. (a) acº = 49 (e) a 2 = .25 (i) a 2+32 = 0 (m) acº — .09a = 0 (b) a 2–64 = 0 (f) a 2–3 =0 (j) a.2 = 5.c (n) ac2 = 1.21 (c) a 2–25 =0 (g) a.” = 1.44 (k) acº — .5a = 0 (o) ac2 = .3a; (d) a 2 = 121 (h) acº = 1.69 (l) ac” = }a: (p) a 2 – 32 =0 2. Solve by factoring and check. (a) a 2–7a: +12 = 0 (d) a 2–7a:--6 =0 (g) a 2–22 – 15 = 0 (b) a 2–7a: —8 = 0 (e) acº — 14a – 15 = 0 (h) ac” +14a – 15 = 0 (k) acº +4+2 –3 = 0 (c) acº —7a: —60 = 0 (f) a 2+2a – 15 =0 (i) 23:2 – 7a: +6 = 0 (l) acº — .22 — . 15 = 0 3. Solve by completing the square, and check. Use the table of square roots on page 191 when you need it. (j) 622 — 13a; +6 = 0 (a) a 9–6a --8 = 0 (d) acº —84. H-12 = 0 (g) ac” – a – 6 = 0 (b) a 2+4a – 12 = 0 (e) acº +3a;+1 =0 (h) ac” +a; – 12 = 0 (c) a 2–10a: —24 = 0 (f) a 2–53 +6 =0 (i) ac” – a – 1 = 0 4. Solve the exercises in 3, above, by use of the formula. . Simplify the fractional equations and solve. Check. (j) 2a:” – 32 + 1 = 0 (k) 5a-2 +2a – 3 = 0 (l) 33:2 —62 – 1 =0 (C 1 3 4 ac — 4 1 .., 2a: +1 1 (a) a 2"3. (d) a — 1 3 2a: (g) a; a - 1 3 (j) 2. 2a – 4 3. 1 (C 3. a — 1 a +1 5 8a; +1 — = 0 - = 1 h = —- k) — —-2a: = 2 (b) ºr it. (*) ITT.I., (h) Ti-H-I-5 (k) Hi-42, 3 2 Q. 1 3a; Q: 16 .5ac — 2 55 tº- -: - 2 = — e = —- l = - (*) iT.I., (i) = +2+2-4 () , Istri-fi () +...+z=y. . Solve for a, and y. Check. (a) a +y=5, a " +y} = 13 (b) a +2y = 1, a2 —y? = 21 (c) a +4y =0, y = a 4–15 (d) y =3a, y = 2a:” –2 (e) at – y = 7, 2% +y^ = 169 (f) 3a – y = 15, a 2+y” =25 . A table top is fourteen inches longer than it is wide. Its area is 12# square feet. Find its dimensions. — 251 — 8. The base of a triangular lot exceeds its altitude by four rods. The area is 126 square rods. Find the base and the altitude. 9. A man traveled 105 miles. If he had gone seven miles an hour faster, he would have made the journey in thirty minutes less time. How fast did he travel? 10. A motorboat goes sixteen miles upstream in two hours, and returns in a half hour. Find the rate of the current and the rate of the motorboat in still water. 11. A motorboat can go a certain distance against a current of nine miles an hour in forty minutes, and can go the same distance with the current in five minutes. Find the distance, and find the rate of the motorboat in still water. 12. Mr. Jones makes an automobile trip of eighty miles in two and a half hours. His rate inside the city limits (one fourth of the entire trip) is twenty miles an hour less than the rate on the open road. Find his rate inside the city, and on the open road. 13. The sum of two integers is 9, and the sum of their reciprocals is #. Find the integers. 14. Mr. Jones left city A and walked in the direction of city B. Two hours later Mr. Smith, in an automobile, started after Mr. Jones and overtook him at the end of fifteen minutes. If Mr. Smith traveled thirty-two miles an hour faster than Mr. Jones, find the rate of each. SUPPLEMENTARY AND REVIEW WORK FOR. UNIT 14 Solve the right triangles for the parts required. 1. Given A =58°. (a) Find a when b = 17.2. (c) Find c when b = 24.4. (e) Find a when c = 40. (b) Find c when a = 20. (d) Find b when c = 32. (f) Find b when a = 27.2. 2. Given A = 43°. Solve all the parts of exercise 1, above. 3. Given a = 10. (a) Find B when b = 5.1. (b) Find B when c = 20. (c) Find B when b = 10. 4. Given B = 32°. ! (a) Find a when b = 18. (c) Find c when b = 8.6. (e) Find a when c = 4.2. (b) Find c when a = 24.2. (d) Find b when c = 3.6. (f) Find b when a = 1.8. 5. Given B = 43°. Solve all the parts of exercise 4, above. 6. Given c = 25. (a) Find A when a =20.5. (b) Find B when b = 6. (c) Fºnd A when b = 15. 7. Solve for the remaining parts when the following parts are given. (a) a = 3, A = 32° (c) c = 18, A = 47° (e) b = 40, B = 10° (g) a = 16, A =81° (b) a = 15, B = 14° (d) c = 24, B =61° (f) b = 14, B = 56° (h) b = 32, A = 64° 8. A road goes up a hill whose average slope is ten degrees. How many feet does the road rise in going a horizontal distance of one mile? 9. A taut kite string 300 feet long makes an angle of fifty-six degrees with the (horizontal) ground. How high is the kite above the ground? 10. A roof has a slope of thirty-four degrees. What rafter length is needed for a vertical rise of eight feet? - 11. In determining the distance across a river from A to C (points on opposite banks), a surveyor measures line CB equal to 250 feet along one bank, making CB perpendicular to -— A.C. With his transit he measures angle CBA to be sixty-two degrees. Find A.C. %iver 12. Two buildings stand exactly sixty feet apart. From a point midway between them, a boy finds that the angle of elevation to the top of one is thirty-eight degrees, and the angle of elevation to the top of the other is forty-seven degrees. Find the height of each building. 13. From the top of a lighthouse, 320 feet above the water level, the keeper observes two ships in line with the light- house. The angle of depression to the nearer ship is five degrees, and to the more distant ship is three degrees. Find the distance between the ships. 14. From a platform thirty-two feet distant from a telephone pole, a boy finds that the angle of elevation to the top of the pole is forty-eight degrees and the angle of depression to the bottom of the pole is thirty-eight degrees. Find the height of the telephone pole. 15. Find the area of triangle A BD when AB = 100 yards, angle B =38°, and angle D = 32°. — 252 — UNIT TEST.S. to accompany FIRST - YEAR ALG E BRA A TEXT-W ORIS BOOK BARTOO AND OSBORN WEBSTER PUBLISHING COMPANY 1808 WASHINGTON AVENUE ST. LOUIS, MISSOURI Name TEST ON UNIT 1. FORMULAS AND LITERAL NUM Form A 1. Add. 2. Subtract. 2d 15d 23:2 3aac 5m. 3a d 1022 9aac 2m &=ºmºmº- * Write the answers at the right. 3. form: 9. is 2c. 10. 2–H8+4+6, 2–7 = ? 4. If a = 2 and y =3, then a 4-Hy”–2ay = ? 5. 6 . Write an algebraic expression which represents: 4–H3(4–H3)(4–3) = ? (a) The sum of 23, and 3y. (b) Their difference. (c) Their product. (d) Their quotient. . Combine like terms: (a) 3a–22-H42-H 62 –3a; (b) 3a–H4b –2b–H4a–6a Perform the indicated operations and leave each result in its simplest (a) a +a;+a; (b) a a a (c) 3a 6b (d) 4a: + 2y Find the perimeter of a rectangle whose length is 3a and whose width The area of a given triangle is K. What is the area of a second triangle whose base is the same as, and whose altitude is twice that of the given tri- angle? 11. Write the term whose literal part is m, whose exponent is 3, and whose coefficient is 4. 12. Write the algebraic expression which represents the sum of a and b multiplied by their difference. 13. Write as algebraic expressions: (a) 2n decreased by 4a. (b) The product of 4a and 59. (c) The remainder when 3m is subtracted from 2b. (d) The sum of 2d and 7c multiplied by their difference. 14. The sum of two numbers is 36. One number is t. What is the other number? 15. 16. lationship between P and s as given in the table. -º-º-º-º-º-º-º- ºmºm-m- --- A dozen apples costs y cents. What is the cost of one apple? Write a formula to express the re- P| 0 || 3 || 6 || 9 || 12 || 15 S | 1 || 2 || 3 || 4 || 5 || 6 Copyright, 1937, by Webster Publishing Co. 4y : 10. 11. 12. 13. 14. 15. 16. P == 6. 27. .4r Answers Perfect score, 32 My score, #b #b (a) (b) (c) (d) (a) (b) (a) (b) (c) (d) (a) (b) (c) (d) Name Perfect score, 33 My score, TEST ON UNIT 1. FORMULAS AND LITERAL NUMBERS Form B 1. Add. 2. Subtract. 3a 3C 2d2 67mm. 7p 8a; 4.5 y #a 50 17a: 862 97.07, 4p Q. .8y #a Write the answers at the right. Answers 3. 6––4+2+ 6. 3 – 5 = ? 3. 4. If y = 3 and w = 4, then y?--2w”—yw = ? 4. 5. 7+2(3+5)(7–2) = ? 5. 6. Write an algebraic expression which represents: 6. (a) The product of 4a and 2e. (a) (b) Their difference. (b) (c) Their quotient. (c) -- - - - - (d) Their sum. (d) 7. Combine like terms: 7. (a) 4m – m+6m – 7m-H.3m (a) (b) 6a+2y—y-H2a –3a. (b) 8. Perform the indicated operations and leave each result in its simplest 8. form: (a) 32.4y 4. (a) (b) m + m + m, (b) (c) b. b. b (c) (d) 3m + 6m (d) 9. Find the perimeter of a rectangle whose width is 3a and whose length 9. is 72. 10. The area of a given triangle is G. What is the area of a second triangle 10. whose altitude is the same as, and whose base is one half that of the given tri- angle? 11. Write the term whose literal part is y, whose exponent is 2, and whose 11. coefficient is 5. 12. Write the algebraic expression which represents the sum of 2a, and y 12. multiplied by their difference. 13. Write as algebraic expressions: 13. (a) 6a increased by 7 y. (a) (b) The remainder when 4c is subtracted from 3m. (b) (c) The product of 7g and 8m. (c) (d) The sum of 2b and 7c multiplied by their difference. (d)— —--— 14. John and Fred together have thirty-one marbles. If John has m mar- 14. bles, how many has Fred? 15. If six pencils cost c cents, how much will one pencil cost? 15. 16. Write the formula to express the 16. C = relationship between C and n as given in C | 3 || 5 || 7 || 9 || 11 13 the table. m | 1 || 2 || 3 || 4 || 5 || 6 Copyright, 1937, by Webster Publishing Co. Name - Perfect score, 20 My score, TEST ON UNIT 2. THE EQUATION Form. A Write as equations. Let n represent the unknown number. Answers 1. Five increased by ten times the number equals thirty-five. 1. 2. Three times the number less seven equals the number. 2. 3. The number plus five times itself equals twenty-seven. 3. 4. One-third the number is three greater than six. 4. 5. Five years ago, Ruth was seventeen years old. 5. 6. Solve mentally. 6. (a) ac-3 = 7 (a) (b) #g = 4 (b) (c) 2+} =3; (c) (d) 3 — w = 2 (d) Solve. 7. 3 = #m–F1 7. 8. 3a;+3 = %a;+28 8. 9. 3y–7=y-1 9. 10. 85–4 =2t — 1 10. 11. #s-H3 = s— 1 11. 12. John is three times as old as Henry. The 12. Henry's age = yr. sum of their ages is fifty-six years. Find the age of each. John's age = — yr. 13. A board seventy-two inches long is cut 13. Shorter part = in. into two parts, one of which is eight inches longer than the other. Find the length of each part. Longer part = — in. 14. A rectangle is twice as long as it is wide, 14. Width - and its perimeter is ninety-six inches. Find its width and its length. Length - Copyright, 1937, by Webster Publishing Co. Name - Perfect score, 20 My score, TEST ON UNIT 2. THE EQUATION Form B Write as equations. Let y represent the unknown number. Answers 1. Ten years hence, Leroy will be twenty-five years old. 1. 2. Half the number is six greater than fifteen. 2. 3. Six times the number is ten greater than the number. 3. 4. James gave his sister one third of his money and had three dollars left. 4. 5. Twice the number less six equals fifteen. 5. 6. Solve mentally. 6. (a) a +3 = 10 (a) (b) y--#=7# (b) (c) #w = 2 (c) (d) 4–g = 3 (d) Solve. 7. 7 =#m-i-2 7. 8. 3a;+ 4 = a +9 8. 9. 2\w-H4 =#w-H 19 9. 10. 89–7 =2g–2 10. 11. 4k — 1 = k–H 6 - 11. 12. The sum of two numbers is 360, and one 12. Larger number = . of the numbers is three times as large as the other. Find the numbers. Smaller number = 13. In a class of thirty-two pupils, there are 13. Boys -: six more boys than girls. How many boys are there? How many girls? Girls – 14. The perimeter of a rectangle is sixty-six 14. Length = feet. Its length is three feet greater than its width. Finds its dimensions. Width = Copyright, 1937, by Webster Publishing Co. Name Perfect score, 60 My score, TEST ON UNIT 3. DIRECTED NUMBERS Form A Add. 1. 2. 3. 4. 5. 6. 7 8. 8 — 7 — 12 — 13 – 14a 1642 7ab 13ay” — 7 4 15 33 — 5a — 16ac? 6ab — 3ay” Subtract 9. 10. 11. 12. 13. 14. 15. 16. 9 — 7 8 6 — 497 m? 0 — 68c –27p? –4 —5 –2 11 — 51 m.” —9 +68c 6p” Multiply. 17. 18. 19. 20. 21. 22. 23. 24. — 7 —8 12 9a — 27t — 13ty 14ab –3a. — 5 +5 9 — 3 — 3 8 — 3 –4y Divide. 25. —21-- —3 = 27. 56-i-7 = 29. —21a;+ — 7 = 31. – 42a:y-:- —3 = 26. --63+ – 21 = 28. — 42+3 = 30. 24a:*-i- –4 = 32. 27b3+ –9 = Add. 33. 34. 35. 36. 37. 38. 39. 40. + 6 6acy 24a — 7 – 17ab m? 67mm. — 17ed — 2 –9ay – 6a 16 — 100b 47m? –4mm. 7cd +14 4a:y — 7a — 9 — ab — 67m.” –7mm. 27c6, Perform the indicated operations. 41. (–16) + (–2) = 46. (–51a:”) + (17) = 51. 3a–H4a–9a = 42. (–4a)(+8) = 47. 4a:-H (–10a) = 52. 3+6+2= 43. (4a)+(–5a) = 48. (–3)(–;) = 53. (–2)(–3)(–4) = 44. (9a) — (–4a) = 49. 3– (–10) = 54. (–2)(–4)(–2) = 45. (632?)--(–9) = 50. —3+(–3) = 55. 6+ (–2)+3 = Evaluate, when a = -1, y = -2, z=3. 56. a 2–2ay-Hy”= 57. 2. – 2 a;+2 - 58. 3a;” — 42% = 59. (a +y)(y—z) = 60. A ship sailed along a meridian from 17° south latitude to 61° South latitude. Through how many degrees did it sail, and in what direction? O Copyright, 1937, by Webster Publishing Co. Name Perfect score, 60 My score, TEST ON UNIT 3. DIRECTED NUMBERS Form B Add. 1. 2. 3. 4 5. 6. 7. 8. — 9 6 — 15 32 — 17a: 13m? 9ba; — 16abº +11 — 10 +17 — 12 — 6a: — 137m2 —23ba. 30 bº Subtract. 9. 10 11 12 13. 14 15 16 +8 —9 — 7 8 –63ry –54g 0 –29 r? —5 —6 +2 15 –79ay 54g — 15 772 Multiply. 17. 18. 19. 20. 21. 22. 23. 24. 13 —6 —9 8c —18s — 16m. 16ay –7a 7 +7 —6 –4 — 4 + 4 — 7 —8b Divide. 25. (–16)+(–2) = 27. 49-3-7 = 29. (–180) + (–3) = 31. (–34mm) + (–2) = 26. (48)+(–24) = 28. —81 +3 = 30. 24a2+ (–6) = 32. — 16a,” +8 = Add. 33. 34. 35. 36. 37. 38. 39. 40. + 9 3ay 360:2 — 7 – 14ca: $/ 9b2 – 15ay — 6 –20y — ac” 17 — 7ca: 69 — 16b” – 4acy +14 6ay — 15a,” — 10 — 5ca: –7y — b” 27acy Perform the indicated operations. 41. (–24) + (–3) = 46. (–369°)+(9) = 51. 7a-6a–4a = 42. (–4a) + (+12) = 47. (9a)+(–15a) = 52. 6–6–3–2 = 43. (7b)+(–9b) = 48. (–;)(—#) = 53. (–3)(–4)(–2)= 44. (9m) — (–6m) = 49. 6-(–14) = * 54. (–2a)(–b)(–c) = 45. (60y”) + (−4) = 50. —7–(+6) = 55. 6-3-(–3)+7= Evaluate, when a =2, b = -3, c = -4. 56. a”—20.b-H b% = 58. 20.2–3c2 = b—c 57. = 59. (a-c)(b+c) = b-Hc 60. A ship sailed along a meridian from 16° north latitude to 29° south latitude. Through how many degrees did it sail, and in what direction? O Copyright, 1937, by Webster Publishing Co. Name TEST ON UNIT 4. GRAPHS Form A 1. The numbers of millions of bushels of apples grown in the United States in the years 1929–1934 are given in the table below. Show these facts in a broken- line graph. Year 1929 1930 1931 1932 1933 1934 Millions of bushels 136 153 202 141 143 120 2. The numbers of millions of bushels of peaches grown in the United States in the years 1929–1934 are shown in the bar graph. Write these facts in the table below. Year 1929 1930 1931 1932 1933 1934 Millions of bushels 3. The percentages of income of a certain state derived from classified sources are given in the table below. Show these facts in a circle graph. In the table, write the number of degrees of the circle graph given to each of the four items. (Use the space at the bottom of this page for computation if necessary.) Agriculture Manufacturing Trade Other 30% 20% 163% 33.9% 4. Given the equation y=22–5, write six pairs of values of 2 and y in the table, and plot these pairs of values on the graph. Copyright, 1937, by Webster Publishing Co. | 1 4680 0000 | 2 0 § # Perfect score, 26 My score, § § § # Name Perfect score, 26 My score, TEST ON UNIT 4. GRAPHS 17 Form B 16 1. The numbers of millions of bales of cotton grown in the United States in 15 the years 1929–1934 are shown in the broken-line graph. Write the facts in the table below. 1 3 Year 1929 1930 1931 1932 1933 1934 ! 2 | Millions of bales 1 | ! 0 2. The value (in millions of dollars) of the tobacco grown in the United States in the years 1929–1934 is shown in the table below. Show these facts in a bar graph. § # § § # § Year 1929 1930 1931 1932 1933 1934 Millions of dollars 286 211 131 108 178 223 | 3. The table of percentages below indicates the way one family spends its income. Show these facts in a circle graph. In the table, write the number of de- grees of the circle graph given to each of the four items. (Use the space at the bottom of this page for computation if necessary.) Food Shelter Clothing Other 25% 35% 20% 20% O O O O 4. Given the equation y = 3–2, write six pairs of values of a; and y in the table, and plot these pairs of values on the graph. Copyright, 1937, by Webster Publishing Co. Name Perfect score, 24 My score, TEST ON UNIT 5. THE FUNDAMENTAL OPERATIONS Form A Add. 1. 2. 3. 3a;+4y—62 6a–7b-H 16 4a:” – 6ay-H4y” – 62–7 y-H42 4a – b – 1 –9a2+ 7acy – 7 y” 7a: — y-H22 —8a–H.9b — 9 — 10a:y—4y” Subtract. 4. 5. 6. 3m-H 47)—H 6 3a–4b +6c 3r–6s–H 4t –67m — 10n+10 4b – 6c-H7 –7p-H 6s–11t Remove parentheses and combine like terms. 7. (a –y) — (32–4y)+(a –2y) = 8. (20–b-Hc) — (3a–46) — (b–2C) = 9. – (2a–g) — (4a:--3y) — (2a: —6) = Write the products. 10. (2a)(–4a:*) = 12. (–2acy)(4y”) = 14. (–4ab)(–;bac) = 11. (–4ab”)(3a*b) = 13. (–4a)(5a) = 15. (3a.”)(–20") = Multiply. 16. 17. 18. 22–3y 3a – b g°-Hy-H1 ac-H4y tº 2a+3b gy – 1 Write the quotients. 19. 328-3-(–22) = 20. —64a3b*-i- (84b*) = 21. (14b*–24b*)-;-(–2b) = Divide. 22. ac-4)a?–7a;+12 23. 3a–2b)3a*–200b-H 12b^ 24. b –c)b*—cº Copyright, 1937, by Webster Publishing Co. Name Perfect score, 24 My score, TEST ON UNIT 5. THE FUNDAMENTAL OPERATIONS Form B Add. 1. 2. 3a – 4b-H7c — 4a:-H 9y—16 —60-H.7b —9c — 6a–10y-H17 7a —3b —8c 11a: – 4y – 9 Subtract. 4. 5. 16k — 14m-H17 4a – 6y-H72 –21k-H 4m – 14 7y–82+17 Remove parentheses and combine like terms. 7. (2d –36)–(4a–H7c) — (a+6c) = 8. (32–29–2) — (–69-H42)+(62–8y) = 9. — (g-H6h-Hi) — (–g-H6h) — (-h) = Write the products. 10. (–4a)(–6a”) = 11. (6ay)(–4a:”y°) = Multiply. 16. 7a-2b —4a–H3b Write the quotients. 22. a-6)a”-Ha-42 Copyright, 1937, by Webster Publishing Co. 12. (–7&d”)(–2c%d) = 13. (–6a)(7y) = 17. 32–29 6a+ y –81 cºdë 9C2d 23. a -7y)3a*–193:y– 14y” 3. 7a3+ 4ac-H 16c.” —80% — 4c” — 180c-H 5c” 6. 40–H4C – 16b –4a–H4c–34b 14. (6bc)(#ca) = 15. (2a2/8)(–3a118) = 18. a” — a +1 a +1 2.17m,3–637m2n — 37m.” cº-º- * 21 24. m—2)m°–8 — 10 — Name - Perfect score, 40 My score, TEST ON UNIT 6. SPECIAL PRODUCTS AND FACTORING Form A Perform the indicated operations. 1. (–5a)?= 4. (–16bc) + (–2b) = 7. (#bºy)?= 2. V4a2b.4 = 5. (–7a3a)? = 8. V36aºbº = 3. (32°y)? = 6. – V16b?c6 = 9. —3a*(3a*—b”) = Write the products. 10. (32–4y)(3a;+4y) = 16. (6–c)(2+c) = 11. (a-3)?= 17. (4a–3)?= 12. (49–H3e)*= 18. (4a:-H5a)(42–5a) = 13. (y–5)(y-H 10) = 19. 4a2b (a?–5°) = 14. 6aº (a+3y”) = 20. (322–1)?= 15. (22–3y)(32–5g) = 21. (4a –y)*= Factor completely. 22. 4a2+40–H1 = 28. a”—ay” = 23. 3a*b*-ī-6ab3 = 29. g°–49–5 = 24. 16 — a”ac” = 30. acº — 7a:”—H 12a: = 25. 9–6b+b% = 31. mº– m = 26. bac-Hby-Hb = 32. acº—9a2+20a = 27. a 2–5ac-H6 = 33. 27a —3aac’ = Perform the indicated operations. 34. Va”–8a–H 16 = 37. V36–60b-H25b” = ss. * * 38. (2?--2–12) + (2+4) ſº == tº 2 2 — 12 ) + (2 -: ac-H.3a c” – c – 12 4.d3b2 – 80 bº 36. — = 39. — = c—H3 20 b? 40. The base of a triangle is a +6 and its altitude is ac-H4. Find the area. Area = Copyright, 1937, by Webster Publishing Co. — 11 — Name Perform the indicated operations. 1. 2. 3. Perfect score, 40 My score, TEST ON UNIT 6. SPECIAL PRODUCTS AND FACTORING (–32)” = - v/ºff- (4ab3)? = Form B 9b2C6 = 5. (–4em) + (–e) = 6. (–7b2a:3)?= 7. V25.cº- 8 . (#a?m)? = 9. —2aº(a 2–y”) = Write the products. 10. 11. 12. 13. 14. 15. 22. 23. 24. 25. 26. 27. 34. 35. (a +4)*= 16. (2–a)(4+a) = (2a–3b)(2a+3b) = 17. 4.-)- (2a –39)*= 18. (4e +ay) (4e–ay) = (b–10)(b+7) = 19. (2e2+3)? = 3m”(m—ac”) = 20. 6aºy (a —2y) = (2C+7)(3c.—6) == 21. (2a–5c)? = Factor completely. 2a2+6ay = 28. ba’—by” = 92°-H62–H1 = 29. m”—3rm — 4 = 902 — 16b2 = 30. a” — a = ay-Haac-Ha = 31. e8–7e?--10e = 9 — 250222 = 32. Tr”—HTrh = 36–H 12A + A* = 33. b3 — b” —90b = Perform the indicated operations. v/22-H42-E4 = 37. V25–80a–H64a2 = (4a”—b”)+(2d-i-b) = 38. (a 2–2–12) -- (a —4) = 36. *+5++6 39. 4&y’-7′y 40. The area of a triangle is a”-H9a–H20 and its altitude is a +5. Find its base. Base = a;+2 a’y? Copyright, 1937, by Webster Publishing Co. — 12– Name Perfect score, 12 My score, TEST ON UNIT 7. LINEAR EQUATIONS IN ONE UNKNOWN Form A Solve. -* Answers Answers 1. 3(a: —2)+2(2a–H4)=3 1. **— 6. A = p +prt (Solve for t.) 6. t = 2 2 2. ——— = 2 2. **— 7. An angle less its comple- 7. O 6 8 ment equals forty degrees. Find the angle. 2u) & 3. — = a – b (Solve for w.) 3. **— 8. One half of a certain 8. b number is six more than one third of it. Find the number. 4. (y–-2)”— (y-H2)*=24 4. y = *** 9. Find two consecutive odd numbers whose sum is 320. 9. 5. P=3s-H4b (Solve for s.) 5. S = for March was $5.21 more than for February. The bill for both months was $67.15. What was the bill for February? For March? 10. Mr. Smith's grocery bill | 10 Copyright, 1937, by Webster Publishing Co. — 13 — Name Perfect score, 12 My score, TEST ON UNIT 7. LINEAR EQUATIONS IN ONE UNKNOWN Form B Solve. Answers Answers 1. 2(z+4)+3(2–2) = 4 1. * *— 6. A = p +prt (Solve for r.) 6. r = Qſ) ºl) 2. — — — = 5 2. wº— 7. Find two consecutive 2 3 even numbers whose sum is 74. 7 3a; 3. †-a-b (Solve for a:.) 3. **— 8. An angle less its supple- 8. —” ment equals twenty-four de- grees. Find the angle. 4. (y–2)” – (y-H4)?=12 4. 9=— 9. One third of a certain 9. number is six less than one half of it. Find the number. 5. T=4y-H5 (Solve for y.) , 5. y = 10. A rectangular play- W =—— g ground is thirty-five feet longer 10 than it is wide, and its perimeter L =— is 470 feet. What is its width.” Its length? Copyright, 1937, by Webster Publishing Co, * — 14 —- & * Name Perfect score, 15 My score, TEST ON UNIT 8. FRACTIONS Form A Leave each result in its simplest form. Add. 1 a # 5a: 3 1. –––– = 3. = 3a 4 * a — 1 ac-H 1 Ol C 20% a -2 2. * — = 4. + > ba, acy 3a–H6 6 Subtract. 7 2 a-H 1 6 5. — - – = 7. + = = - 2a: ac a”--3a, a 3C 6. –4 = ab be Multiply. a(c.—d) c gº 9 yº–5y--6 2/+4_ cal c – d. * > g”—y–6 3y—6 Divide. 3c2 5c O. b 10. — —H — = 11. —H· = 8a; 2a: a;+y ‘a’—y” 3aw-H6b. e 33; a”--4ab+45° a--25T Simplify the complex fractions. 1 a-H– 0. 13. 0. 6 ac-H5+--— Q. 14. — = 1 3 → Q: Q. Supply the missing numerator or denominator and its correct sign so that each fraction is equivalent to the given fraction. \ Copyright, 1937, by Webster Publishing Co. — 15 — Name Perfect score, 15 My score, TEST ON UNIT 8. FRACTIONS Form B Leave each result in its simplest form. Add. 2 3 - 2a: 3 1. —–H– = 3. -: Oy Q2 a –2 at-H2 a 1 37/? 3a; aa, 2y – 1 2 Subtract. 6 2 a – 1 3 5. ———= 7. - - - 5a, a ac”—H2a: a, a b \ 6. — — — = acy y? Multiply. ac(a;+y) 4y 9 ac” – 2da;+ a” 4a: 3y g-Hyº tº 2a:-H2a. 2—as Divide. a?b ab? ac” – ſº ac 10. —--— = ! 11. tº ºtu c20 col? 24/ a’y” a” — 50 +6 tº 30 – 6 12. —- - a’— a – 6 20-H.4 Simplify the complex fractions. 1 1—— QC 13. - 1 30 — — 3C b2 a —b "a Hº 14. – ab a-Hb Supply the missing numerator or denominator and its correct sign so that each fraction is equivalent to the given fraction. Copyright, 1937, by Webster Publishing Co. Name TEST ON UNIT 9. FRACTIONAL EQUATIONS Form A Solve. Place your results in the answer column. 6 2 1 1. – = 3 1. a = *- 4. — = — (C 2–1 2–3 3 1 1 ac-H7 ac-H 5 2. — — — = — 2. y = 5. E + g 8 4 a;+3 at-H2 3 2 Ol. C 3. — = —-H4 3. w = 6. F QU) QU) b–y d-y 7. Separate 60 into , two 7. parts whose quotient is two- thirds. What is the larger part? Copyright, 1937, by Webster Publishing Co. 8. If the same number is added to both numerator and denominator of the fraction H%, the resulting fraction is equivalent to #. What is the number? Perfect score, 8 My score, • QX = ... y = — 17 — Name TEST ON UNIT 9. FRACTIONAL EQUATIONS Form B Solve. Place your results in the answer column. *- 5 - 1. — = 10 1. a = 4. –––––––* QC a –2 ac-H2 ac”— 4 2 - 2. ——H·8 =9 2. y = s. *-i- -- 3/ 2 2–1 2 1 1 70, 7%, *= * - - - 3. w = 6. – = 3v 12 w n+y n-y 7. The sum of two numbers 7. 8. If the same number is is 18. The larger divided by the smaller gives a quotient of 2. What is the larger number? subtracted from both the numer- ator and denominator of the fraction #, the resulting frac- tion is equivalent to #. What is the number? Perfect score, 8 My score, Copyright, 1937, by Webster Publishing Co. Name Perfect score, 18 My score, TEST ON UNIT 10. LINEAR EQUATIONS IN TWO UNKNOWNS Form A 1. Solve graphically. (Score, 3) a;+2y = 4 2a – y = 3 Q. QC $/ !/ Solve for a, and y. Check. 2. 4a:-Hy = 10 6. The sum of two numbers is forty-six, and their 2a:-Hy = 4 difference is twelve. Find the numbers. 3. 2a:— y =3 3a;+2y =8 7. Three years ago, a father was five times as old as his daughter. Three years hence, he will be only three times as old. How old is each now? 4. 7a;+3y = 4 22–5 y=7 8. A pound of butter and six loaves of bread cost $1.14; two pounds of butter and two loaves of bread cost $1.08. What is the cost of one loaf of bread? Of one pound of butter? 5. The sum of two angles of a triangle is 136 de- grees, and their difference is 36 degrees. Find all the angles of the triangle. Copyright, 1937, by Webster Publishing Co. Name Perfect score, 18 My score, TEST ON UNIT 10. LINEAR EQUATIONS IN TWO UNKNOWNS Form B 1. Solve graphically. (Score, 3) ac-H y = 5 a – 29 = –1 Q. 3C $/ $/ Solve for a and y. Check. 2. 3a –y = 5 6. The sum of two angles of a triangle is 100 de- 2a:-Hy =5 grees and their difference is 22 degrees. Find all the angles of the triangle. 3. 5a +3/=8 2a+ y = 3 7. Ten years ago, a father was four times as old as his daughter. Ten years hence, the father will be only twice as old. How old is each now? 4. 5a —2) = 6 2.c-3y = —2 8. John has ten dollars. He can buy four shirts and six neckties and have thirty cents left, and he lacks forty cents of having enough to buy five shirts and * * ** ſº ies. is th f a shirt? Of ktie? 5. The sum of two numbers is fifty-three and their five neckties. What is the cost of a Shir 2, IA62CKU16 difference is twenty-one. Find the numbers. Copyright 1937, by Webster Publishing Co. — 20 — Name Perfect score, 14 My score, TEST ON UNIT 11. RATIO, PROPORTION, VARIATION Form. A Write these ratios as fractions and reduce to lowest terms. * 3 1 1. 96:64 = 3. — . — = 8 4 2. 12:18 = 4. (a”—b”): (a-b)”= | Solve these proportions for the unknowns indicated. 5. ac:4 = 3:2 (C = 7. a . b = wic QD = 4 – 2 r—H2 3 6. — = — g = 8. = —- ºr = $/ 7 r — 2 4 Form proportions and solve. 9. Divide 72 into two parts whose ratio is 8:1. 10. A man uses seventeen gallons of gasoline in driving 323 miles. At the same rate, how many miles can he go on ten gallons of gasoline? Express as formulas. 11. The area of a circle depends on its radius. A = 12. The perimeter of an equilateral triangle depends on the length of a side. P = 13. The area of a square depends on the length of a side. A = 14. The perimeter of a rectangle depends on its length and its width. P = Copyright, 1937, by Webster Publishing Co. — 21 — Name Perfect score, 14 My score, TEST ON UNIT 11. RATIO, PROPORTION, VARIATION Form B Write these ratios as fractions and reduce to lowest terms. 5 2 1. 32:24 = 3. — ; – = 9 3 2. 6: 16 = 4. (a”--ab): (a+b)”= Solve these proportions for the unknowns indicated. 5. 7: a = 2:8 QC = 7. 2: a = b : c 2 = 5 y w—3 2 3 6 w—H·3 5 Form proportions and solve. 9. Divide 90 into two parts in the ratio of 7:11. 10. A man uses twenty-one gallons of gasoline in driving 357 miles. At the same rate, how many gallons would he use in driving 272 miles? Express as formulas. 11. The circumference of a circle depends on its radius. C = 12. The area of a triangle depends on its base and its altitude. A = 13. The perimeter of a square depends on the length of a side. P = 14. The total surface of a cube depends on the length of an edge. S = Copyright, 1937, by Webster Publishing Co. Name TEST ON UNIT Simplify the radicals. 1. 4 W64 = 2. 5 V/a2b% = Form A 3. Combine terms which are similar or can be made similar. 5. 2 V3+7.V3 = 6. 3 WI6— Vö4 = Multiply or divide as indicated. 9. V4. V5= 10. 2 V3.3 M6= Perform the indicated operations. Find the square root of: 16. 6 7 6 7. 8. 11 12 14. (al/8)?= 17. 3 0 2 5 Use the table to find the missing side of each right triangle. 19. a = 2, b = 5, c = 20. a = 4, c = 6, b = Copyright, 1937, by Webster Publishing Co. 21. a =4, b = 2, c = 22. b = 5, c = 8, a = 12. POWERS AND ROOTS . V45+ v5 = . V7+ V56= 15. 02/8-i- a 1/2 = 18. 6 8.8 9 Perfect score, 23 My Score, 12 20 21 29 39 52 89 v/n 2.646 3.464 4.472 4.583 5.385 6.245 7.211 9.434 Name Perfect score, 22 TEST ON UNIT 12. POWERS AND ROOTS Simplify the radicals. 1. 3.V4 - 2. a V8b5– Form B Combine terms which are similar or can be made similar. 5. 6-V5+7 V5 = 6. 3M/8–VT8 = Multiply or divide as indicated. 9. V8. V2= 10. 2W2.3 V3 = Perform the indicated operations. 13. ac1/2 . acl 16 = Find the square root of: 16. 2 8 9 \ 7 8 11 12 14. (3:28)/2 = 17. 1 1 5 6 Use the table to find the missing side of each right triangle. T * s/. 2\/2 = g+2v 3 /a: 3 / y Q/ Q. . V48+ V3 = . W4-- VIOS= 15. ac1/2 –-acl/8 = 18. 1 8.4 9 My score, 70, 19. a = 2, c = 4, b = 20. a = 4, b = 6, c = 21. b = 2, c = 5, a = 22. a = 5, b = 8, c = 12 20 21 29 39 52 89 v/n 2.646 3.464 4.472 4.583 5. 385 6. 245 7.211 9.434 Copyright, 1937, by Webster Publishing Co. Name Perfect score, 16 My score, TEST ON UNIT 13. QUADRATIC EQUATIONS Form. A 1. Solve by factoring. 2. Solve by completing the square. ac” – 32 — 10 = 0 ac”--8.r-H 7 = 0 Solve by any method you choose. 3. a 2–10a;+9 = 0 5. 2.c4+3a –2 = 0 4. a 2–93 – 10 = 0 6. 3rº — Sar-i-4 = 0 7. A rectangle is seven inches longer than it is wide, 8. The product of two consecutive odd integors is and its area is sixty square inches, Find its dimensions. 323. Find the integers. Copyright, 1937, by Webster Publishing Co. Name Perfect score, 16 My score, TEST ON UNIT 13. QUADRATIC EQUATIONS Form B 1. Solve by factoring. 2. Solve by completing the square. ac”——2a – 24 = 0 w?–53:--6=0 Solve by any method you choose. 3. a 2–H 7a: — 18 = 0 5. 32% — 8a; — 3 = 0 4. a7%–42 — 12 = 0 6. 23:?–53:--3 = 0 7. A rectangle is seven inches longer than it is wide, 8. The product of two consecutive even integers is and its diagonal is seventeen inches. Find its length and 528. Find the integers. its width. Copyright, 1937, by Webster Publishing Co. — 26 — Name Perfect score, 38 My score, TEST ON UNIT 14. NUMERICAL TRIGONOMETRY Form. A Fill the blanks below. Use the table on page 238 of your textbook. No. Angle Sine *COsine Tangent Work space 1. 18° 2. 23° 30' 3. 38° 15' 4. 47° 54/ 5. 64° 10' Write below as ratios the values of the sines, cosines, and tangents of the in- dicated angles of the triangle at the right. 6. sin A = cos A = tan A = 5 7. Sin B == cos B = tan B = A 4 Use the table on page 238 of your textbook to find the values of angles A. 8. Sin A = .2924 A = 11. sin A = .4035 A = 9. COs A = .4027 A = 12. COs A = . 1607 A = 10. tan A = .2886 A = 13. tan A = 2.0660 E Fill the blanks below. Use the table on page 238 of your textbook. 14. 15. 16. Work space A 38° 60° B 45° Ol. 86.6 b 70. 7 C 12 17. A man stands 100 feet from the base of a tower and sights the top. The angle of elevation is seventy- three degrees. How high is the tower? 18. A ladder twenty feet long leans against a house and exactly reaches a window nineteen feet high. What angle does the ladder make with the ground? Copyright, 1937, by Webster Publishing Co., — 27 — Name Perfect score, 88 My score, — TEST ON UNIT 14. NUMERICAL TRIGONOMETRY Form B Fill the blanks below. Use the table on page 238 of your textbook. No. Angle Sine *Cosine Tangent Work space 1. 68° 2. 33° 15' 3. 46° 30' 4. 71° 36' 5. 15° 20' Write below as ratios the values of the sines, cosines, and tangents of the in- dicated angles of the triangle at the right. B 6. sin A = cos A = tan A = 13 § 7. Sin B = cos B = tan B = A 12 C Use the table on page 238 to find the values of angles A. 8. Sin A = .7489 A = . 11. sin A = .7933 A = 9. COs A = .8192 A = 12. cos A = .6053 it: 10. tan A = .8098 A = # 13. tan A = 1. 1573 == Fill the blanks below. Use the table on page 238. 14. 15. 16. Work space A. 46° 30° B 22° O), 86.6 100 b C 15 17. A tall tree stands on a level plain. From a point 125 feet from the base of the tree, the angle of elevation of its top is thirty-nine degrees. How high is the tree? 18. A guy wire 125 feet long reaches from the top of a smokestack to a point on the level ground and 85% feet from the base of the smokestack. What angle does the wire make with the ground? } Copyright, 1937, by Webster Publishing Co. — 28 — Page Abscissa. . . . . . . . . . . . . . . . . . . . . . 62 Addition. . . . . . . . . 6, 9, 42, 43, 44, 68, * * * * * * . . . . . . . . . . 69, 132, 135, 200 Ahmes. . . . . . . . . . . . . . . . . . . . . . . 110 Algebraic expressions. 5, 6, 13–20, 27 Algebraic symbols. . . . . . 6, 16, 21, 50 al-Khowarizmi. . . . . . . . . . . . . . . 39 Angles . . . . . . . . . . . . . . 113, 229, 233 Archimedes. . . . . . . . . . . . . . . . . . . 19 Areas. . . . . . . . . . . . . . . . . . . 3, 4, 8, 12 Arithmetic scale. . . . . . . . . . . . . . . 41 Aryabhata. . . . . . . . . . . . . . . . . . . . 26 Babbage, Charles. . . . . . * * * * * * * * 68 Babylon, ruins of . . . . . . . . . . . . . . 177 Bagdad. . . . . . . . . 's e-- a-- * * * * * * * J. .. 39 Barlow, Peter. . . . . . . . . . . . . . . . . 191 Bhaskara. . . . . . . . . . . . . . . . . . . . ’. 26 Binomials. . . . . . . . . . 7, 79, 82, 92, 97 Brahe, Tycho. . . . . . . . . . . . . . . . . 197 Brahmagupta. . . . . . . . . . . . . . . . . 26 Carroll, Lewis. . . . . . . . . . . . . . . . . 198 Coefficients. . . . . . . . . . . . . . . . . . 6, 42 Combining like terms . . . . . . . . . . 2 Common denominator. . . . . . . . . 133 Complex fractions. . . . . . . . . . . . . 139 Compound interest. . . . . . . . 195, 196 Coordinates, system of . . . . . -. . . . 62 Copernicus, Nicholas. . . . . . . . . ... , 155 Cosine . . . . . . . . - - - - - - - - J - - - - - - 229 Cumulative reviews.', . 18, 39, 55, 67, - • * * * * * * * * * * *‘88, 105, 125, 142, 154, • * g º º ºs . . . . . 176, 187, 206, 225, 237 Degree of an equation... . . . . . . . . .22 Descartes, René. . . . . . . . . . . . 62, 161 Diophantuš. . . . . . . . . . . . . . 163, 201 Discriminant. . . . . . . . . . . . . . . . . . 214 Division. . . . . . . . . . . . . . . . . . . * . . . . . 15, 52, 81, 82, 84, 130, 131, 202 Eads, James Buchanan. . . . . . . . . 42 Egyptian writing. . . . . . . . . . . . . . 16 quations - checking. . . . . . . . . . . . . . . . . . . 27 conditional . . . . . . . . . . . . . . . . . 19 cubic . . . . . . . . * . . . . . . . . . . . . . 22 definition of . . . . . . . . . . . . . . . . 19 equivalent. . . . . . . . . . . . * * * * * * * 158 fractional. . . . . . . . . . . . . . . 143, 173 graphic representation of . . . . . 156 identity. . . . . . . . . . . . . . . . . . . . l laws of . . . . . . . . . . . 25, 27,29, 107 linear . . . . . ‘. . . . . . . . . . 22, 106, 155 literal. . . . . . . . . . . . . . . . . . . . . . 117 members of . . . . . . . . . . . . . . . . . 19 parentheses in . . . . . . . . . . ‘. . . . , 111 quadratic. . . . . . . . .: . . . . . . .22, 207 radical, . . . . . . . . . . . . . . . . . . . . 204 roots of . . . . . . . . . . . . . . . . . . . . 188 solutions of . . . . . . . . . . . . . . . . . . . . . 19, 22, 30, 107, 111, 1:19, 143 translating words into. . . . . . . . 1 * * * * * * * * * * * . ... , 8, 13, 14, 20, 21 Evaluating algebraic expressions. ' ' ' ' ' ' ' - - - - - - - - - - 9, 118, 151, 152 Exponents... . . ........ ". . º, 78, 203 Factor. . . . . * * * * * * * * * * * * * * * 91, 101 Factoring. . . . . . . . . . . . . .Cº - - - - - - . . . . 89,91, 93, 96, 98, 100, 101, 103 Fermat, Pierre de... . . . . . . . . . . . . 97 Formulas. . . 2, 3, 8, 10, 12, 15, 16, 18, . . . . . . . 61, 63, 107, 151, 195, 214 Fractional equations. . . . . . . l43–150 Tà Ctions addition of * * * * * * * * * * * * * * * * * 135 clearing of... . . . * * * * * * * * * * * * 143 Complex. . . . . . . . . . . . . . . . . , , , 139 definition of . . . . . . . . . . . . . . . . 126 ivision of . . . . . . . . . . . . . . . . . . 130 multiplication of . . . . . . . . . . . . 129 ~. | *-----, ~~--- - INDEX age reduction of . . . . . . . . . . . . . . . . 126 Signs. . . . . . . . . . . . . . . . . . . . 128 subtraction of . . . . . . . . . . . . . . . 136 terms. . . . . . . . . . . . . . . . . . . . . . 126 Function . . . . . . . . 4, 16, 182, 183, 184 Fundamental operations, order of * * * * * * * * * * * * * * * * * * * * * * * * * * 9, 68 Galileo. . . . . . . . . . . . . . . . . . . . . . . 155 Goethals, George Washington. ... 1 Golden section. . . . . . . . . . . . . . . . 217 raphs bar. . . . . . . . . . . . . . . . . . . . . . 56, 57 broken line . . . . . . . . . . . . . . . 56, 59 - circle. . . . . . . . . . . . . . . . . . . . 56,60 quadratics, of . . . . . . . . . . . . . . . 222 rectangular . . . . . . . . . . . . . . . . . 56 straight line . . . . . . . . . . 56,63, 156 trigonometric functions. . . . . . . 229 Heraclitus. . . . . . . . . . . . . . . . . . . . 19 Hindu method of solving quad- ratics. . . . . . . . . . . . . . . . . . . . . . 218 Index numbers. . . . . . . . . . . . . . . . 197 Interest, compound. . . . . . . . 195, 196 Interpolation . . . . . . . . . . . . . 192, 232 Inventory tests. . . . 17, 38,54, 66, 87, . . . . . . . . . . . 104, 124, 141, 153, 175, * * * * * * * * * * * * * * * 186, 205, 224, 236 Irrational numbers. . . . . . . . . . . . 197 Jefferson, Thomas . . . . . . . . . . . . . 126 Repler, Johann. . . . . . . . . . . . . . . . 155 Least (lowest) common multiple. 133 Letters in algebra. . . . . . . . . . . . . . 3 Levers. . . . . . . . . . . . . . . . . . . . . . . 185 Like terms. . . . . . . . . . . . . . . . 6, 32, 53 Line graphs. . . . . . . . . . . . . . . . . . . 156 Linear equations. . . . . . . . . . 106, 155 Literal equations. . . . . . . . . . 117, 119 Eliteral numbers. . . . . . . . 2, 14, 15, 43 Lowest terms. . . . . . . . . . . . . . . . . 127 Magic squares. . . . . . . . 73, 75, 79, 87 Mahavira. . . . . . . . . . . . . . . . . . . . 26 Maxwell, James Clerk. . . . . . . . . . 106 Measurement, indirect. . . . . . . . . 226 Members of an equation. . . . . . . . 10 Michelson, Albert A. . . . . . . . . . . 226 Monomials. . . 7, 15, 42, 43, 78, 81, 90 Multiples. . . . . . . . . . . . . . . . . . . . . 133 Multiplication . . . . . . . . . . . . . . . . . . 9, 15, 50, 78, 79, 90, 129, 131, 201 Multiplication signs. . . . . . . . . . . 2, 50 Negative direction. . . . . . . . . . . . . 41 Negative number. . . . . . . . . . . . . . . 41 Newcomb, Simon. . . . . . . . . . . . . . 89 Newton, Sir Isaac . . . . . . . . . . . . . 134 Numbers directed. . . . . . . . . . . . . . . . . . . . 40 irrational. . . . . . . . . . . . . . . . . . . 197 literal. . . . . . . . . . . . . . . . . . . . . . 1 negative. . . . . . . . . . . . . . . . . . . . 41 numerical value of . . . . . . . . . . . 42 positive. . . . . . . . . . . . . . . . . . . . 41 prime. . . . . . . . . . . . . . . . . . . . . . 101 rational . . . . . . . . . . . . . . . . . . . . 197 signed. . . . . . . . . . . . . . . . . . . . . . 41 Number scale. . . . . . . . . . . . . . . 40–44 Numerical coefficient. . . . . . . . . . . 42 Ordinate. . . . . . . . . . . . . . . . . . . . . 62 Pacioli. . . . . . . . . . . . . . . . . . . . . . . 28 Panama Canal . . . . . . . . . . . . . . . 1, 18 Parallelogram, area of . . . 8 Parentheses. . . . . . . . . . . . . 10 inclosing terms within . . . . . . . . 72 in equations. . . . . . . . . . . . . . . . 111 ~ in problems. . . º. - 1. one within another . . . . . . . . . . 72 removing. . . . . . . . . . . . . . . . 71, 110 Pascal, Blaise. . . . . . . . . . . . . . . . . 68 Peirce, Benjamin . . . . . . . . . . . . . . 207 Polynomials . . . . 7, 69, 74, 81, 83, 84 Positive numbers. . . . . . . . . . . . . . 41 Powers. . . . . . . . . . . . . . . . . . . . . . . 188 Prime factor. . . . . . . . . . . . . . . . . . 101 Prime number. . . . . . . . . . . . . . . . 101 Problems angle. . . . . . . . . . . . . . . . . . . . . . 113 compound interest. . . . . . . 195, 196 courier. . . . . . . . . . . . . . . . . . . . 122 geometric. . . . . . . . . . . . . . . 170, 171 golden section. . . . . . . . . . . . . . . 217 €Weſ. . . . . . . . . . . . . . . . . . . . . . . 185 mixture. . . . . . . . . . . . . . . . . . . . 123 number. . . . . . . . . . . . . . . . . . . . . . .34, 35, 147, 148, 168, 169, 221 tank (or work). . . . . . 149, 150, 217 using quadratics. . . . . . . . . 216, 221 using reciprocals. . . . . . . . . . . 174 using right triangles. . . . . . . . . 194 using similar triangles. . . . . . . . 181 using trigonometry . . . . . . . . . * * * * * * * * * * * * * 228, 231, 233, 235 Proportion. . . . . . . . . . . . . . . . 177, 179 Pupin, Michael. . . . . . . . . . . . . . . . 40 Quadratic equations. . . . . . . 22, 207 Quadratic equations, solving by completing square . . . “211, 212. factoring. . . . . . . . . . . 208, 209, 210 formula . . . . . . . . . . . . 213, 214, 215 Hindu method. . . . . . . . . . . . . . 218 Quadratic equations, two un- knowns. . . . . . . . . . . . . . . . . . . 220 Quadratic trinomials. . . . . . . . . . . 98 Radicals. . . . . . . . . . . . . . 189, 197–202 Radicand. . . . . . . . . . . . . . . . . . . . . 197 Ratios. . . . . . . . . . . . . . . . . . . 177, 178 Reciprocals. . . . . . . . . . . . . . . 173, 174 Rectangle . . . . . . . . . . . . . . . . . 3, 4, 10 Recorde, Robert. . . . . . . . . . . . . . . 6 Reviews, cumulative. See Cumu- lative reviews. Roots. . . . . . . . . . . . 189, 190, 191, 197 Scale, number. . . . . . . . . . . . 40–44 Special products. . . . . .89–92, 95, 102 Square root. . . . . . . . . . . . . . . 190, 191 Square, trinomial . . . . . . . . 93, 211 Steinmetz, Charles P. . . . . . . . . . . 19 Stevin, Simon. . . . . . . . . . . . . . . . 114 Stock Exchange. . . . . . . . . . . . . 2 Subtraction . . . 6, 46, 74, 90, 136, 200 Supplementary and review work. e - A * * * ºr * * * * * * * * * * * * * * 239–252 Tables compound interest. . . . . . . . . . . 196 squares and square roots. . . . . 191 trigonometric functions. . . . . . . 238 Tests, inventory. See Inventory teSt.S. Thermometer. . . . . . . . . . . . . . . . . 11 Translating words into symbols. . * * * * * * * * * * * * * * 8, 13, 14, 20, 21, 22 Transposition . . . . . . . - - 33 Trapezoid . • a - a - a sº tº ºn 12 Triangle. . . . . . . . . . . . . 4, 181, 193 Trigonometric functions. . . .229, 238 Trinomial . . . . . . . . . . . . . . . . . . . 7 Trinomial Squares. . . . . . . . . . 93, 211 Variation direct . . . . . . . . . . . . . . . . . . . . . 182 indirect . . . . . . . . . . . . . . . . . . . 1 S3 Vieta, Francois. . . . . . . . . . . . . . . . 21 Widman, John . . . . . . . . . . . . . . . . 16 № = ∞， ∞） = ∞ √≠ ≤ ≥ ± − × . （±√（√3.）: 5 （-） %%ſ'， §. 33 ± &*）'）; ſae...? §§ &, #.*? ș *& } # * ~