i.^/ IN MEMORIAM FLORIAN CAJORI \^f^JL^^^^ FIRST YEAR MATHEMATICS BY GEORGE W. EVANS HEADMASTER OF THE CHARLESTOWN HIGH SCHOOL, BOSTON AND JOHN A. MARSH MASTER IN MATHEMATICS AT THE ENGLISH HIGH SCHOOL, BOSTON CHARLES E. MERRILL COMPANY NEW YORK AND CHICAGO COPTKIGHT, 1916, By CHARLES E. MERRILL CO. El PREFACE This book is the result of twenty years of patient experi- ment in actual teaching. It is intended to be completed in the first year of the high school. It presents algebraic equations primarily as a device for the solution of problems stated in words, and gives a complete treatment of numerical equations such as are usually included in high-school algebra — one-letter and two-letter equations, integral and frac- tional, including one-letter quadratics and the linear-quad- ratic pair. So much of algebraic manipulation is included as is necessary for th^e treatment of these equations. The arithmetic in the book is presented from a new point of view — that of approximate computation — and is utilized in the evaluation of formulas and in the solution of equations throughout the succeeding pages. Geometrical facts are introduced as the basis of many algebraic and arithmetic problems, and wherever they are not intuitively accepted by the pupils they are accompanied by adequate logical demonstration. Proofs, and parts of proofs, are avoided when they seem to the pupils of an unnecessary and hair-splitting kind. AH problems are carefully graded, for it is by means of problems that each successive algebraic difficulty is in- troduced. A great deal of pains has been taken to present new topics clearly and concretely, often dividing them into sub-topics each of which is separately illustrated and apphed to prac- tice. Definitions are generally prepared for by such advance work as will cause the student to feel the need of them; and where no need exists, they are omitted. To introduce defi- nitions of concepts that are already famifiar to the student, 3 t#^^^^ #^^ 4 PREFACE such as angle and right angle, would increase the formalism of the exposition, and be on the whole repellant. The book has been used in manuscript, and substantially in its present form, for four years in one of the largest high schools in Boston. It has been found workable. It gives the pupil who has but one year for mathematics a substantial knowledge of the meaning and purpose of algebra, and more than a hint of what geometry is about. Neither of these things can be said of the conventional first year mathematics. Pupils who go on with mathematics, after the course covered by this textbook, enter upon their second year with habits of self-reliance and of self-criticism, and complete the usual syllabuses of college preparation without duplication of work or loss of time. Teachers who have begun the text with some distrust have finished it with enthusiasm. Other features of the book are: 1. The method of solution of equations: the logical explanation of each step is noted in writing, with a clear and systematic symbolization, by the pupil himself. Technical terms of riianipulation, such as "transposing" and "clearing of fractions," are replaced by briefer statements of the actual thing done to each of the equal numbers. 2. Approximate computation is made not only a time- saver but a training in common sense. 3. The graphical method is used not merely as a prepara- tion for examination questions but as a means of exhibiting the functional relation of the two letters of the equations used, and as a means of clarifjdng the study of elimination; it is made an integral part of the treatment of that subject. 4. The solution of quadratics is treated by the general methods of algebra, instead of using a method pecuUar to high-school work. "Completing the square" is regarded as a method of factoring a quadratic expression; and the sign =b for a square root is deduced from the solution of the equation x^ = n, instead of the reverse. Problems are chosen, first, to show the answers distinct and of equal sig- PREFACE 6 nificance, then successively to show them alike, apparently different but really alike, and so on. The algebraic negative appears here, with its interpretation. Meaningless answers are found, which are to be discarded; irrational answers are exemplified; imaginary answers are mentioned as an indi- cation of some incongruity in the statement of the problem. 5. Two methods of ehmination are given: "by combina- tion" for linear pairs, and "by substitution" for linear- quadratic pairs. 6. Throughout the book, the pupil is required to check all his work, and so to rely upon himself for certitude of his accuracy. 7. The unusual arrangement of topics (see Table of Contents) and the introduction of material from arithmetic, geometry, and even from trigonometry, have been found to arouse an unusual interest in the pupil and to leave him at the end of the year with a feeling of achievement that pupils never get from the mere manipulation of algebraic expres- sions. For all the foregoing reasons it is believed that this book will disarm many of the attacks now being made on the teaching of algebra in high schools. We are greatly indebted to the wisdom and enthusiasm of Mr. Henry M. Wright and Mr. Bertram C. Richardson, who have given invaluable aid in working out the teaching sys- tem embodied in this book. Many other Boston teachers, notably Mr. Walter F. Downey and Mr. Peter F. Gartland, have generously cooperated in testing the various stages of the manuscript. In offering the book to a larger public, therefore, the authors feel warranted in claiming that it is representative of a considerable group of scholarly and pro- gressive teachers. GEORGE W. EVANS JOHN A. MARSH Boston, June 1, 1916. Digitized by the Internet Archive in 2007 with funding from IVIicrosoft Corporation http://www.archive.org/details/firstyearmathemaOOevanrich \ CONTENTS Chapteb Page I. The First Use of Algebra 11 Equations 13 Ratio 16 Measurement of Length 17 Measurement of Angles 21 Valuation Problems 26 Difiference of the Unknown Numbers Given 30 Angles and Angle Relations 32 Multiplications 36 Shortages 37 The Study of a Simple Equation 41 Sum of the Unknown Numbers Given 44 Circles 46 II. Approximate Computation 50 Multiplication 50 Division 54 Square Root 57 III. Measurement Formulas 63 Rectangles 63 Rhomboids 66 Triangles 70 Trapezoids 75 The Number tt 78 Implicit Formulas 80 IV. Additional Types of Problems 83 Fractional Equations 83 Subtracting Fractions and Parentheses 88 The Construction of Equations 91 The Problem of the Digits 92 The Problem of Two Velocities 94 The Problem of AbiUty and Time 100 7 8 CONTENTS CHAPTEB PAGE V. Transformations 103 Negative Numbers 103 Addition 104 Subtraction 106 Parentheses within Parentheses 108 Multiplication 109 The Distributive Law 112 Distributive Factoring 113 MultipUcation of Polynomials 114 Division 116 Long Division 117 VI. Identities and Theorems; Factoring 119 The Proof of Theorems 119 Quadratic Products 121 Factoring by Cross Multiplication 123 Completing the Square 126 VII. Quadratic Equations 130 Two Answers to One Question 130 Both Answers Alike 134 Answers Apparently Different 135 The Meaning of Negative Answers 136 Answers Suggesting Related Problems 138 Meaningless Answers 140 Equations Involving Irrational Numbers 142 Peculiar Quadratics 144 VEIL Some Theorems of Geometry 147 Some Ratios of Area 148 Similar Figures 155 Similar Right Triangles 159 Pythagorean Theorem 162 Angles of a Polygon 164 IX. The Graphical Method 167 The Algebraic Scale 167 The Algebraic Diagram ... 170 The Locus of an Equation 175 CONTENTS 9 CHAPTER PAGE X. Elimination by Combination. 181 Two-Letter Problems 182 Solving for Reciprocals 194 Where Elimination Fails 197 Non-Algebraic Conditions 198 XI. Elimination by Substitution ... 200 Linear-Quadratic Pairs 200 Algebraic Diagrams 206 The Equation of the Circle 208 Other Quadratic Loci 210 The Standard Parabola 215 XII. Supplementary Problems for Practice and Review 218 FIRST YEAR MATHEMATICS CHAPTER ^ I THE FIRST USE OF ALGEBRA 1. Algebra is a method of abbreviating the explanations of arithmetic problems. This was, historically, its first use. It is also used for abbreviating the statement of rules in arithmetic, and for the discussion of numbers in every case where the numbers are described in words instead of being expressed in figures. 2. This chapter illustrates the first use of algebra: for abbreviating the explanation of problems in arithmetic. In the problems which are here given as illustrations, the answers are sometimes evident at a glance. The student is not expected to shorten the arithmetical work, for he can- not do that; he is to put down the explanation briefly and systematically. When that is learned, he will find that he can do complicated problems with greater ease, because he can put down in black and white, as he goes along, the suc- cessive steps of his reasoning. 3. Model A. — A father is 6 times as old as his son, and their united ages are 42 years. Find the age of each. Explanation (1) The father's age + the son's age = 42 years. (2) The father's age = 6 times the son's age. (3) 6 times the son's age + the son's age = 42 years. (4) 7 times the son's age = 42 years. (5) The son's age = 6 years. (6) The father's age = 36 years. 11 12 THE FIRST USE OF ALGEBRA Abbreviated Explanation Let s stand for the number of years in the son's age. Then 6 X s will stand for the number of years in the father's age. © 6xs + s = 42 @ 7 X s = 42 ® s = 6 6 X s = 36 Ans. The father is 36 years old; the son is 6 years old. EXERCISE 1 1. John is twice as old as Henry, and the sum of their ages is 21 years. Find the age of each. 2. John and Will wish to divide 36 cents so that John shall have twice as much as Will. How much money will each boy have? 3. I paid $220 for a horse and saddle, and the horse cost ten times as much as the saddle. What was the cost of each? 4. A sidewalk laid with a flagstone and a curbstone is 6 feet wide; the flagstone is 11 times as wide as the curbstone. What is the width of the flagstone and of the curb? 5. The electoral vote of the state of New York is 9 times that of the state of South Dakota. Both states together have 50 electoral votes. How many votes does each state have? 6. There are 7 times as many sheep as lambs in a pasture, and in all there are 96. Find the number of each. 7. A pole 9 feet long is partly peeled of bark; the bark part is twice as long as the bare part. Find the length of each part. 8. It takes a man four tunes as long to run the first three- quarters of a measured mile as to run the last quarter. He goes the whole mile in six minutes. How long does it take him to run the last quarter? EQUATIONS 13 9. Mercury weighs 13 times as much as water; a quart of mercury and a quart of water together weigh 28 pounds. What is the weight of each? Equations 4. An equation is a statement that two numbers are equal. 5. The algebraic abbreviation for any number, or for any combination of numbers, is called an algebraic ex- pression. 6. It takes two numbers to form an equation. One or both of the numbers may be an algebraic expression. The two numbers which are said to be equal are called the first (left-hand) member and the second (right-hand) member of the equation. 7. The parts of an algebraic expression separated by the + and — signs are called the terms of the expression. The statements numbered ©, 0, ®, in Model A (page 12) are equations, because each states that two numbers are equal. In equation 0, the number obtained by adding 6 X s to s is said to be equal to the number 42. There are three terms in this equation: the first is 6 X s, and the second is s. These terms are separated by the sign +. Either of these terms separately, as well as the sum of both, is an algebraic expression. The second member contains only one term, a number without any letter. 8. In order to fix the mind on the nature of the processes by which his equations are obtained, it is well for the student to indicate beside each equation how it arose from the pre- ceding equations. With very little trouble, he is thus able to complete his explanation. 9. Model B. — In a certain collection of insects, there are twice as many butterflies as moths; of both there are 39. Find the number of each. 14 THE FIRST USE OF ALGEBRA Let X — the number of moths. . Then 2X X = the number of butterflies. © 2XX + X = = 39 ® SXx = = 39 © same values ® X = = 13 ® -^3 ® 2XX-- = 26 Ans ® X 2 . 13 moths; 26 butterflies. In Model B, the numbers in equation © are obtained from ® without any change of value. The numbers in @ are obtamed from the numbers in © by dividing each of them by 3, this fact being indicated on the right of © by the history of the equation, © -r- 3. The numbers in © are obtained from those in © by multiplying each of them by 2, the history being, © X 2. 10. When two number symbols are written together without any sign between them, multiplication is indicated. For representing numbers of which values are not stated, sometimes initial letters are used as in Model A; more often we use a small letter from the latter part of the alpha- bet, such as X, y, or z. These are matters of accepted custom (convention) among writers and students of algebra. 11. The answer in Model B is readily verified because we see at once that the statements of the problem are true for the numbers obtained as answers. That is, 26 = 2 X 13 and 26 + 13 = 39. This verification is called checking. Answers should always be checked by referring to the words of the problem itself, not to the first equation. exercise; 2 1. One man is twice as heavy as another, and both weigh 339 pounds. Find the weight of each. 2. The number of cows and sheep in a certain farmyard is 75, and there are four times as many sheep as cows. Find the number of each. EQUATIONS 15 3. The doctor has twice as many books as the minister, and they both have 2100. How many books has each man? 4. In a dog show there are 36 St. Bernards, twice as many long-haired as short-haired. How many of each kind are there? 5. A locomotive weighs 6 times as much as a car, and both weigh 14 tons. Find the weight of each. 6. A man is twice as heavy as a boy, and both weigh 200 pounds. Find the weight of each. •7. A's house cost 2J times as much as B^s house, and both cost $7000. Find the cost of each. 8. I paid $100 for two lots of tiles; one lot cost 4 times as much as the other. What was the cost of each lot? 9. I walked 4 times as far in the afternoon as I did in the forenoon; I walked 15 miles altogether. How far did I walk in the forenoon? 10. It costs 4 times as much to go from New York to Chicago as from Boston to New York; from Boston to Chicago, by way of New York, it costs $25. How much does it cost to go from Boston to New York? 11. The president of a stock company owns twice as many shares in it as his brother, and both own 165 shares. How many shares does each man own? 12. A man had a big bill in his pocket, and asked the price of a house and barn; he found that the house cost 7 times the value of the bill, and the barn 3 times the value of the bill, both coming to $5000. Find the cost of house and barn. 13. There are three times as many barrels in one cellar as in another; in the store above, there are half as many barrels as in both cellars; altogether there are 216 barrels. How many are there in each part of the building? 16 -THE FIRST USE OF ALGEBRA 14. A fortress is garrisoned by 5200 men; and there j 9 times as many infantry, and 3 times as many artillery, cavalry. How many are there of each? Ratio 12. When one quantity can be obtained from another multiplying, the multiplier is called their ratio. Thus, in Ex. 14 above, the ratio of infantry to cavalry is that of cavalry to infantry is ^. This is onty another way saying that the number of infantry is 9 times the numl of cavalry; and that the number of cavalry is ^ the numl of infantry. 13. Model C. — A line 530 centimeters long is divic into two portions, and the ratio of these is 4. Find the lenj of each. Let X = the number of centimeters in one portion. Then 4 x = the number of centimeters in the other. © a; + 4 X = 530 5 X = 530 ® same values (?) x = 106 ® -H 5 © 4 a; = 424 0X4 Ans. One portion is 106 cm.; the other 424 cm Check: 106 X 4 = 424 + 106 530 EXERCISE 3 1. Divide 78 pounds of cheese into two portions whi will have the ratio 2. 2. Divide a farm of 810 acres into two portions with ratio 8. 3. Two parts of a line 90 feet long have a ratio |. H< long are the parts? MEASUREMENT OF LENGTH 17 4. Of a 200-acre farm, part is sown to one crop and the rest to another. The ratio of the two parts is 4. How much land is there in each part? 5. The ratio of an average man's weight to an average ten-year-old boy's weight is 2. Such a man and boy would together weigh 210 pounds. What is the weight of each? 6. Divide 28 into two parts having a ratio 6. 7. The ratio of the weight of an average St. Bernard dog to that of an average Blenheim spaniel is 15. In the dog show there were 10 St. Bernards and 5 of these spaniels, and altogether they weighed 1085 pounds. What is the average weight of each dog? 8. The ratio of non-voters to voters in a town is 4. If the population is 13,800, how many are voters? 9. The ratio of coin money to bills in a conductor's pocket is ^; he has $14.40 in all. How much of his money is in coin? 10. In a certain transaction the gross receipts were $720. The ratio of the amount invested to the amount received as profit was 8. How much was invested? Measurement of Length 14. A straight line is measured by marking off on it a certain standard straight Hne called a unit of length. This unit is in modern times defined carefully by national govern- ments. The units in commonest use are the foot and the meter, but multiples and subdivisions of these units are also frequently used. When the last measurement mark comes exactly at the end of the straight line we are measuring, the length-number is a whole number, or an integer; when it does not come exactly at the end of the line, the measurement is continued with subdivisions of the unit. The system of subdivision in 18 THE FIRST USE OF ALGEBRA ordinary use for the foot is cumbrous, and is not much used in careful work, such as engineering. Here decimal sub- division is abnost universal. 15. The precision of measurement is indicated by the number of figures required to express the result, rather than •by the number of decimal places. Thus a length of 88.2 inches can be expressed, with equal accuracy, as 7.35 feet, or 2.24 meters, or even as .00139 miles, depending upon the unit used, and not upon the care or skill with which the work is done. These four numbers, representing the same length, are equally accurate measurements, although in one of them only one decimal place is used, and in another five places. Measurements which are expressed, like these, in three figures, are said to be of three-figure accuracy; they may be made with ordinary instruments and skill. Five-figure accuracy requires good instruments and great care, as where a city building lot is measured to thousandths of a foot; six- figure accuracy is possible to experts; and eight-figure accuracy is beyond human skill at present. In solving examples with decimal data, it is impossible for the result to have any greater accuracy than the given quantity. If the measured quantity is expressed in three figures, it is useless to express the result in more than three figures; hence, any answer obtained from decimal data must be cut down to the number of figures of the given data. For example, 1.39 f1^. = 1.39 X 12 in. = 16.68 in. = 16.7 in. EXERCISE 4 1. A flagpole is made in two parts, a heavy base and a lighter upper portion; the entire height is 75 feet. The height of the base bears to the height of the upper portion the ratio 2. How long is each portion? MEASUREMENT OF LENGTH 19 2. A large factory has length and width m the ratio 3. A belt course of ornamental terra cotta around the building measures 640 feet. What is the length of the building? 3. A life-saving crew had two coils of rope whose lengths were in the ratio 2, and neither one alone would reach a wreck offshore. By uniting them end to end, the entire, length of 480 yards carried over the ship. How much rope was there in each coil? 4. Spur tracks were run from the main line of a railroad to two factories, and required the building of 1800 feet of track. The lengths of the two lines of track were in the ratio 5. How long was each? 6. A piece of ground 350 feet long was divided in the ratio 6 for a factory and an adjoining power house. How long was the power house? 6. A line 87.32 feet long is divided into two parts that have a ratio 7. How long is each part? 7. The length of a building lot is twice its breadth, and it takes 248.4 feet of iron fence to inclose it. What are the dimensions of the lot? 8. The widths of two adjoining estates have a ratio 13, and they occupy a total frontage of 493.5 ft. How wide is each estate? 9. The width of the wagonway in a certain street bears to the width occupied by both sidewalks the ratio 2. The total width from wall to wall is 97.29 feet. How wide is the wagonway? 10. Two sides of a triangle have the ratio 3, and the third side is 15.8 inches long. The total perimeter (the distance around the edge) is 36.7 inches. How long is each side? 11. A metal scale .389 ft. long is divided into two parts with the ratio 2. Find the length of each part in inches. 20 THE FIRST USE OF ALGEBRA 12. A meter is 39.37 inches. A meter stick is divided into two parts whose ratio is 8. Find the length of each part in feet. 13. A Hter of dry air at standard pressure and temper- ature weighs 1.293 grams. What is the weight of the two portions of this air when it is divided in the ratio 6? 14. A short piece of railroad is laid with heavy rails weighing 90 lb. to the foot, and lighter rails weighing 75 lb. to the foot. The length of the road is 5.79 miles, and the distance laid with heavy rails bears to the distance laid with the lighter rails the ratio 3. How many rods of each grade of rail are laid? How many tons of each grade of rail are used? (Two rails to the track.) 16. The chemical elements in anthracite coal are mainly carbon, hydrogen, and oxygen, with the weight of carbon bearing a ratio of 38 to each of the other two. In | of a ton of coal, how many pounds of each of these constituents would be found? 16. A kilogram (1000 grams) is equivalent to 2.20 lb. avoirdupois. Eleven kilograms of copper is divided into two masses with a ratio 6. How many ounces are there in each portion? 17. Two distances have a ratio 5 and their sum is .139 miles. How many rods are there in each distance? 18. I wish to divide 3.37 lb. of sulphur into two portions having a ratio 3. How many ounces will there be in each portion? 19. A piece of ribbon 2.46 yards long was divided into two portions whose ratio was 4. How many inches were there in each portion? 20. A book consisting of two parts has in Part II three times as many pages as in Part I. The book has 1436 pages, and the thickness between covers is 2.59 in. How many pages are there in each part? How thick is one page? MEASUREMENT OF ANGLES 21 EXERCISE 6 Oral Work Find the lengths of the pairs of lines described in the fol- lowing examples: 1. Sum, 12 inches; ratio, 3. 2. Sum, 40 centimeters; ratio, 4. 3. Sum, 75 meters; ratio, 2. 4. Sum, 90 inches; ratio, 8. 5. Sum, 144 feet; ratio, 11. 6. Sum, 150 yards; ratio, 5. 7. Sum, 55 miles; ratio, 10. 8. Sum, 3 feet; ratio, 5. G. Sum, 2 inches; ratio, 7. 10. Sum, 4 yards; ratio, 11. 11. Sum, 300 feet; ratio, 5. 12. Sum, 360 yards; ratio, 8. 13. Sum, 400 feet; ratio, 9. 14. Sum, 1200 miles; ratio, 3. 15. Sum, 880 yards; ratio, 7. 16. Sum, 960 kilometers; ratio, 2. 17. Sum, 540 feet; ratio, 8. 18. Sum, 1400 miles; ratio, 13. 19. Sum, 484 yards; ratio, 3. 20. Sum, 1111 feet; ratio, 10. Measurement of Angles 16. An angle is measured by marking off on it a standard angle called a degree. 17. If on a flat surface you revolve a straight line about one of its ends as a pivot, you get one complete rotation when it returns to its starting position, and you have passed 22 THE FIRST USE OF ALGEBRA the line through 360 of the standard angles or degrees. A clock hand, or a watch hand, makes such a movement, and the length of the hand has nothing to do with the size of the angle through which it rotates. In a quarter of an hour every minute hand describes, or passes over, an angle of 90°, and every hour hand describes an angle of 7J°, whether the hands are on a lady's watch or on the town clock. The size of the angle has nothing to do with the length of its sides. 90 260 27 80 mi '"%,. %" 280 EmtractoE 18. A degree is divided into sixtieths, called minutes; and then subdivided into sixtieths of a minute, called seconds. Subdivisions of a degree are expressed also in the form of decimals; that method will be generally used in this book. MEASUREMENT OF ANGLES 23 19. The point from which the two sides of an angle are drawn is called the vertex. The most convenient way of measuring an angle is to place it with its vertex at the center of a circle of which the edge is marked off in degrees. Such a marked circle is called a protractor. On a large circle the degree marks are farther apart than on a small circle, just as a large clockface has longer minute spaces (6° each) than a small one; but the central angle of one degree will, of course, be the same, whatever the size of the protractor. 20. Two angles are called complements when their sum is 90°. Complements intercept a quadrant on a protractor. Two angles are called supplements when their sum is 180°. Supplements intercept a semicircle on a protractor. — Semicircle =iz 21. For representing in algebra the length-number of a straight line, it is advisable to use one of the small letters of the alphabet. For the number of degrees in an angle, use a capital letter. 24 THE FIRST USE OF ALGEBRA EXERCISE 6 1. Two supplementary angles are in the ratio 8; find the angles. a. An angle of 169° is divided into two parts with the ratio 12. How many degrees are there in each part? 3. Two angles are in the ratio 7, and their sum is 152°. What is the size of each angle? 4. One angle is twice the size of another angle, and their sum is the supplement of three times the larger one. Find the two angles. 5. An angle is the complement of 5 times itself; find the angle. 6. The ratio of two complementary angles is 3; what are the angles? •7. An angle of 104° is divided into two parts, having a ratio 7. How many degrees are there in each part? 8. Two supplementary angles have a ratio 17; what are the angles? 9. An angle bears to its complement the ratio 9; how many degrees are there in the angle? 10. A mark on a clockface is so placed that at 4 o'clock it divides the angle between the hands into two angles that have the ratio 4. How many degrees are there in each of these two angles? EXERCISE 7 Oral Work 1. Find two complementary angles whose ratio is 1; 2; 3; 4; 5; 8; 9; 11; 14; 19; 29; 44; f ; f; f ; f ; ^; /^. 2. Find two supplementary angles whose ratio is 2; 4; 5; 8; 9; 11; 14; 17; 19; 29; 44; 59; 89; 179; |; f ; MEASUREMENT OF ANGLES 25 Find the pairs of angles described as follows: 3. Sum, 77°; ratio, 6. 9. Difference, 10°; ratio, 6. 4. Sum, 63°; ratio, 8. lo. Difference, 36°; ratio, 10. 6. Sum, 105°; ratio, 20. ii. Difference, 80°; ratio, 9. 6. Sum, 105°; ratio, 6. 12. Difference, 80°; ratio, 17. 7. Sum, 105°; ratio, 14. 13. Difference, 30°; ratio, 4. 8. Sum, 112°; ratio, 27. 14. Difference, 78°; ratio, 14. 22. Where the number of degrees in an angle is represented by an algebraic expression, it is sometimes necessary to get another algebraic expression to represent the number of degrees in the supplement or the complement. Let us take numbers first, as an illustration, and find how many degrees there are in the supplement of an angle of 133°. Since 47^* + 133° = 180°, 47° is the supplement of 133° (that is, of 180° - 47°). Likewise, 133° is the supplement of 47° (that is, of 180° - 133°). Hence, the number of degrees in the supplement of 7° is 180 — 7; of (c + d) degrees is (180 — c — d). The number of degrees in the complement of (a + 4) degrees is (90 - a - 4), that is, (86 - a). Each of the following algebraic expressions represents the number of degrees in an angle; state in each case the number of degrees in its supplement. 15. M 18. ah 21. c + 8 24. 2a + 3 16. K 19. r + s 22. c + 24 25. 6 R 17. Q 20. Z+F 23. /+11 26. 2/1 + 15 Each of the following algebraic expressions represents the number of degrees in an angle; state in each case the number of degrees in its complement. 27. Z 30. ef 33. ikf + 1 36. 6 a + 7 28. S 31. h-\-k 34. c + 9 37. 3cd 29. g 32. MN 35. 11 + It; 38. m + 5a + 4 26 THE FIRST USE OF ALGEBRA Valuation Problems 23. Model D. — I settled an account of $48 with $2 and $5 bills, using twice as many 5's as 2's. How many bills of each denomination did I give in payment? Let X = the number of $2 bills. Then 2 a; = the number of $5 bills. 2x = the number of dollars in $2 bills. 10 X = the number of dollars in $5 bills. ® 10 X + 2 a; = 48 @ 12 a; = 48 ® same values ® a; = 4 @ ^ 12 2a; = 8 ® X 2 Ans. Four $2 bills; eight $5 bills. Check: 4 $2 bills = $ 8 2 X 4 or 8 $5 bills = 40 $48 EXERCISE 8 1. The coal supply for a school was delivered in two kinds of wagons, the larger holding 5i tons and the smaller 3 tons; there were twice as many of the large loads as of the small ones, and the entire amount deUvered was 182 tons. How many loads were delivered by each kind of wagon? 2. A milk wagon carries certified milk in bottles at 14 cents a quart and 5 times as much milk in cans at 9 cents a quart. The entire load is worth $16.52. How many quarts of each kind of milk are carried by the wagon? 3. A stock farm sold out its horses and cows for $4360. There were three times as many cows as horses. The cows were sold at a uniform price of $110, and the horses at a uniform price of $215. How many of each were sold? 4. Two kinds of wagons, one kind holding twice as much as the other, were used in delivering 220 tons of brick. The smaller wagons delivered 16 loads, the larger 31. How many tons were delivered in each kind of wagon? VALUATION PROBLEMS 27 6. A shoe dealer sold one day a line of shoes at $6 a pair, and four times as many of another line at $3.50 a pair. The receipts on these two lines of shoes that day were $460. How many pairs of each kind were sold? 6. In paying an account of $68, I used only $5 and $2 bills; and the number of $5 bills was three times the number of $2 bills. Find the number of bills of each denomination. How many bills of each kind must I use to settle the fol- lowing accounts in the manner described for each account? 7. To pay $88, I use $5 and $2 bills only, 3 times as many 2's as 5's. 8. To pay $72, I use 5's and 2's only, twice as many 5's as 2's. 9. To pay $42, I use lO's and 2's only, twice as many 2's as lO's. 10. To pay $80, I use lO's and 5's only, twice as many 5's as lO's. 11. To pay $26, I use lO's and I's only, 3 times as many I's as lO's. 12. To pay $45, I use 2's and I's only, 7 times as many 2's as I's. is. To pay $78, I use 5's and I's only, 5 times as many 5's as I's. 14. To pay $84, I use lO's and 2's only, 4 times as many lO's as 2's. 15. To pay $75, I use lO's and 5's only, 7 times as many lO's as 5's. 16. To pay $183, I use lO's and I's only, 6 times as many lO's as I's. 17. Five horses and four donkeys weigh altogether 9600 pounds. Each horse weighs 4 times as much as a donkey. Find the weight of each animal. 28 THE FIRST USE OF ALGEBRA 18. Apples were bought for 39 cents; twice as many red ones at 5 cents apiece as green ones at 3 cents apiece. How many of each kind of apples were bought? 19. Engineers examined a bridge that broke down under the weight of a crowd of people, and decided that the break- ing load was 36 tons. Supposing that there were four times as many men as women, the average weight being 150 pounds for a man and 120 pounds for a woman, what was the num- bei* of each in that crowd? 20. I bought a certain quantity of tea of first quality at 60 cents a pound, and 3 times as much tea of second quality at 45 cents a pound. The tea cost me $13.65. How many pounds of each quality did I buy? 21. Mules at $40 apiece and 7 times as many horses at $125 apiece cost $2745. Find the number of animals in the drove. 22. Lead weighs 4 times as much as marble; 3 lead globes and 2 marble globes, all of the same size, weigh 56 pounds. What is the weight of each globe? 23. A team is made up of oxen and mules, one ox and one mule in each pair; supposing that an ox can pull 4 times as much as a mule, and that the whole team is used to pull 30 tons, how much of this load do the oxen pull? 24. I paid $91 with $5 bills and $2 bills, using the same number of each. How many of each did I use? 26. There are three times as many 5's as 2's in a roll of bills; in all $51. Find the number of bills of each kind. 26. I paid a bill of $1.80 with quarters, dimes, and nickels, using twice as many dimes, and 3 times as many nickels, as quarters. Find the number of each that I used. 27. I bought twice as much coffee at 33 cents a pound as tea at 54 cents, and I paid in all $2.40. How many pounds of each did I buy? VALUATION PROBLEMS 29 28. I bought horses at $120, sheep at $14, and chickens at 50 cents; three times as many sheep as horses, seven times as many chickens as sheep. All cost $517.50. How many of each did I buy? 29. The crew of a towboat consists of an engineer, 2 firemen, 3 deckhands, and a cook, whose wages are, respec- tively, $2, $1.50, $1, and $1.20 per day. For a certain voyage the pay-roll was $73.60. How many days did the voyage last? 30. A certain mill employs men at an average wage of $2.34 a day, 5 times as many women at an average wage of $1.26 a day, and twice as many children (as men) at an average wage of 62 cents a day. The weekly pay-roll of this mill is $592.80. How many men, women, and children are employed in it? 31. I have three times as many nickels as half-dollars; in all $4.55. Find the number of each. 32. There are in a purse three times as many nickels as dimes; in all $1.50. How many nickels are there? 33. Four cows, three calves, and ten sheep cost $168; a cow costs five times as much as a calf, and a calf costs twice as much as a sheep. Find the cost of each. 34. The ratio of two angles is 6. Double the smaller one is the complement of the larger. Find the angles. 35. The difference of two lengths is 32 feet and their ratio is 5. Find the lengths. 36. The ratio of two angles is 6, and the double of the smaller is the supplement of three times the larger. Find the angles. 37. Five wooden posts and two iron ones of exactly the same size and shape weigh altogether 810 pounds. The ratio of the weight of iron to the weight of this wood is 11. Find the weight of each post. 30 THE FIRST USE OF ALGEBRA Difference of the Unknown Numbers Given 24. Model E. — A father is 50 pounds heavier than his son, and both weigh 248 pounds. What is the weight of each? Let X = the number of pounds the son weighs. Then x + 50 = the number of pounds the father weighs. © X + X + 50 = 248 ® 2X + 50 = 248 © same values ® 2x = 198 © -50 ® X = 99 © ^ 2 © x + 50 = 149 © + 50 Ans. Father, 149 pounds; son, 99 pounds, 99 + 50 = 149 + 99 248 EXERCISE 9 Check; 1. Two cannon balls, one being 6 pounds heavier than the other, together weighed 50 pounds. What was the weight of each cannon ball? 2. I have ten cents more than my uncle, and together we have $2.90. How much has each? 3. A marketman sold 11 sheepskins, and then lost $2 of the money he received; he had $7.90 left. For how much did one sheepskin sell? 4. A man owes $181; he pays part of the account with equal numbers of $10, $5, and $2 bills, and gives a check for $28 for the remainder. How many bills of each kind does he give? 5. Find four consecutive numbers whose sum is 94. 6. Two mules and a horse were bought for $400, and the horse cost $40 more than a mule. What was the cost of each? 7. Two bucketfuls of water fill a 15-gallon tub all but 5 quarts. How much does the bucket hold? DIFFERENCE OF THE UNKNOWN NUMBERS GIVEN 31 8. A grocer mixes tea that costs him 20 cents a pound with 3 times as much tea that costs him 38 cents a pound, and sells the mixture for $10, clearing a profit of 62 cents. How many pounds of each kind of tea does he use? 9. My farm is three times as large as the one next to it, and both together, with a 10-acre lot across the road, just make a quarter section (160 acres). Find the size of each farm. 10. There are quarters in one pile, twice as many dimes in another, and $1.13 besides; all this money amounts to $6.08. How many quarters and dimes are there? /^ii. Thirty horses and 40 mules weigh 54 tons; on an average, each horse weighs 100 pounds more than a mule. Find the average weight of each kind of animal. 12. I bought peas at 59 cents a peck, and 1 peck more of beans at 30 cents a peck; also 25 cents worth of lettuce. I paid in all $5. How many pecks of peas and of beans did I buy? 13. I bought sugar at 5 cents a pound, 3 times as much coffee at 50 cents a pound, and paid $1.83 for a ham. The whole bill was $12.68. How much coffee did I buy? 14. I sold a house; then I sold 5 more houses at double the price; then sold 35 tons of coal at $5 per ton. I re- ceived in all $64,448.* What was the price of the first house sold? 15. A man has three times as many dimes as dollars, 4 times as many cents as dimes, 5 times as many nickels as cents. He has in all $17.68. Find the number of nickels. 16. Suppose that each of the 36 boys in a class has the same sum of money, and that the teacher has 15 cents more than all the boys together. All the money amounts to $79.35. How much has the teacher? 32 THE FIRST USE OF ALGEBRA 17. Three times as many $10 bills as $2 bills, and $7.35 besides, make a total of $167.35. Find the number of $2 bills. 18. Seven times as many $1 bills as $5 bills, and $1.45 besides, make $37.45. Find the number of $1 bills. 19. Three times as many $2 bills as $5 bills, 4 times as many $10 bills as $2 bills, and $11.11 besides, make $666.11. Find the number of $2 bills Angles and Angle Relations 25. Two lines are called parallel when they are on the same flat surface and do not meet, no matter how far they may be prolonged in either direction. The figure made by two parallel lines is called a stripe. A straight Hne that crosses two or more straight lines is called a transversal. Fig. 2. 26. When a transversal crosses a stripe, it makes two sets of four angles each. If we number each set with suffixes, beginning each time with the upper angle on the right hand side, it will be seen that All the odd angles are equal. All the even angles are equal. The odd angles are the supplements of the even angles. The angle Ai is read "A one." ANGLES AND ANGLE RELATIONS 33 27. In the triangle ABC (Fig. 2), the sum of the angles at the base is equal to the supplement of the angle at the vertex. That is, A + C = 180 - B. Prolong CB, and draw through B a Une parallel to CA. We now have one stripe with two transversals. On the transversal AB, A and X are even angles and are equal. On the transversal CB, C and Y are odd angles and are equal. That is, © A =X ® C = Y But X + F = 180 - B Hence, ® A+C = 180-B Substituting ® and ® in ® Since this equation may be written ® A+5 + C = 180 + 5 we have the following theorem:* The sum of the angles of a triangle is 180°. 28. Model F. — In a certain triangle, the two base angles have a ratio 2, and the ratio of the vertical angle to the larger base angle is 3. Find the angles of the triangle. Let C = the number of degrees in the smaller base angle. Then 2 C = the number of degrees in the larger base angle. and 6 C = the number of degrees in the vertical angle. C + 2C + 6C = 180 9 C = 180 same values C = 20 -- 9 2C=40 0X2 6C = 12O0X3 Am. 20°, 40°, 120°. Check: 20 2 X 20 = 40 3 X 40 = 120 180 * A theorem is any general statement for which we may expect to find a proof. 34 THE FIRST USE OF ALGEBRA EXERCISE 10 1. In the figure made by a stripe and its transversal, the odd and even angles have a ratio 3. How many degrees are there in each angle? 2. In the same kind of figure, each obtuse angle is 24° greater than any acute angle. What are the interior angles on one side of the transversal? 3. In the stripe shown in Fig. 1, the angle A is 21° greater than - double the angle B. How many degrees are there in each? 4. In the same kind of figure, ^ angle A is 5° more than three ^ times angle B. Find the two ^«- 1- ' angles. 5. In the same kind of figure, angle A is 13.62° more than twice angle B. How many degrees are there in each angle? 6. In the same kind of figure, angle A is 31.13° more than four times angle B. How many degrees in each angle? 7. Three of the odd angles in a stripe make a total 40° less than one of the even angles. How many degrees are there in each angle? 8. One angle of a triangle is 20°, and the other two angles have a ratio 7. What is the largest angle? 9. The two base angles of a triangle have a ratio 3, and the smaller is 15° less than the vertical angle. What are the angles? 10. The vertical angle of a triangle is 10° greater than one. of the base angles, and the other base angle is equal to the sum of the two. Find the vertical angle. 11. The base angles of a triangle are equal, and their sum is 18° less than the vertical angle. What are the base angles? ANGLES AND ANGLE RELATIONS 35 12. In the triangle ABC (Fig. 2), the exterior angle at A is equal to 108°, and the ratio of 5 to C is 2.6. Find the three angles of the triangle. Fig. 2. 13. In the same kind of figure, angle A is 5° more than angle B and 7° more than angle C. How many degrees are there in each angle? 14. In the same kind of figure, angle J. is 3 times angle B, and angle C is 30° more than angle B. How many de- grees are there in A, B, and C? In the same kind of figure, angle A is 15.3° more than 15. B, and B is 5 times angle C. in each? How many degrees are there Fig. 3. 16. In Fig. 3, angle B is 16.38° more than angle A, angles A and B, 17. In Fig. 4, angle C is 20.34° more than angle D, many degrees are there in the angles C and D? Find How 36 THE FIRST USE OF ALGEBRA 18. In the triangle MNO, angle M is 30.5° more than angle 0, and angle N is 11.3° more than angle 0. Find the number of degrees in each angle. 19. In triangle ABC, angle A + angle B = angle C, and A is 32.77° more than B. Find the number of degrees in the angles A, B, and C 20. An angle A + an angle B = 45.38°, and angle A = 2.23 times angle B, Find angles A and B. Let X = number of degrees in angle A; 2.23 x = number of degrees in angle B. 21. In a triangle EFG, E = F = 1.34 G, Find angles E, F, and G. Multiplications 29. Consider the problem: A is twice as old as B; 22 years ago he was three times as old as B. What are their ages now? Let X *= the number of years in B's age now. 22 years ago A was 2 x — 22 years old, and B was x — 22 years old. So the equation is 2 x - 22 = 3 (x - 22) Before solving this equation, we must understand ex- pressions like 3 (a; — 22). Suppose a farmer contracts to deliver 3 bushels of oats and 2 bushels of barley every day. To get the number of bushels of grain delivered in several days, he multiplies not only the quantity of oats but also the quantity of barley by the number of days. Suppose a farmer receiv.es daily x bushels of grain and delivers daily 3 bushels. The increase of grain in his store for, say, 7 days is found thus; X bushels daily for 7 days received 7 x; 3 bushels daily for 7 days delivered 21. Subtracting the amount delivered from the amount received (7 x — 21), we find what the daily increase of x — 3 amounts to in 7 days. SHORTAGES 37 30. Whenever an expression of two or more terms is mul- tiplied, each term of that expression must he multiplied separately. EXERCISE 11 1. 2 (a; - 7) 5. 75(2X + 40) 9. 125 (8 a: + 40) 2. 11 (3 a; - 1) 6. 50 (8 a + 160) 10. 13(180-7 5) 3. 18(3x + 5) 7. 20 (5 A + 70) 11. 30 (5 -3m) 4. 25(?4x+28) 8. 40 (25 x+ 125) 12. 23 (23 - x) Shortages 31. The equation 2 x - 22 = 3 (x - 22) in Section 29 may now be changed to 2 x — 22 = 3 x — 66, which is a change of form but not of value. This equation differs from those we have been using in that it has negative terms on each side of the equation, that is, terms to be subtracted. The left side is not equal to 2 x, but just 22 short of it. In the same way, the right side is 66 short of 3 x. If 22 is added to each side of the equation, the shortage on the left side will disappear, but the right side will still be 44 short of 3 x. That is, the equation will be 2 X = 3 X - 44. This shortage disappears when 44 is added to each side, giving 2 X + 44 = 3 x, an equation Uke many we have been solving. The values on each side have been changed, but the resulting numbers are equal. But if 66 is added at first, we have 2x- 22 + 66 = 3x 2x + 44=3x The entire shortage on the right is made up; and on the left only 22 out of the 66 is needed to make up that shortage, leaving 44 to be added to the complete value of 2 x. 38 THE FIRST USE OF ALGEBRA The shortest way, then, when there are two similar short- ages, is to add the larger. The numbers that were stated as equal in the equation 2a; — 22 =3x — 66 have each be- come greater; they are now different in value from the original numbers, but are still equal to each other. 32. Model G. — A is twice as old as B; 22 years ago he was three times as old as B. What are their ages now? Let X = the number of years in B's age now. Then 2 x = the number of years in A's age now. a; — 22 = the number of years in B's age 22 years ago. 2 X — 22 = the number of years in A's age 22 years ago. © 2 x - 22 = 3 (x - 22) @ 2x-22 = 3a;-66 © same values ® 2x + 44 = 3x ® + 66 ® U = X ®-2aj ® 88 = 2x ®X2 Ans. A is 88 years old; B, 44 years old. Check: B's age: now, 44; 22 years ago, 22. A's age: now, 88; 22 years ago, 66. 66 = 3 X 22. EXERCISE 12 Solve for the value of the letter : 1. 5(A -7) =3A -9 6. 8(X-3) =5X-3 2. 7(a;-3)=5x-3 7. 5(x + 4)=7a:-4 3. 4 ( X + 5) = 6 a; - 20 8. 11 (x + 1) = 13 a; + 1 4. 3 ( X + 2) = 4 X - 5 9. 13 (x - 10) = 3 X 6. 6 (B - 3) = 5 (B - 2) lo. 9 (x + 6) = 10 (2 x + 1) 11. A is now 5 times as old as B, and 5 years hence he will be only 3 times as old as B. What are their ages now? 12. A is 3 times as old as B; 15 years ago he was 5 times as old as B. What are their ages now? 13. A is 3 times as old as B; in 8 years he will be only 2i times as old as B. Find their present ages. SHORTAGES 39 14. A is 7 times as old as B; in 25 years the ratio of their ages will be 2. What are their ages now? 15. The difference between a father's age and his son's age is 28 years. How old was the father when the son was i of the father's age? , How old was the son when the father was 9 times as old? 16. The ratio of two angles is 2, and their complements differ by 21.33°. Find the angles. 17. Of the eight angles formed by the transversal of a stripe, the four odd ones have a sum just 12° less than two of the even ones. Find the angles. 18. Two angles of a triangle are equal, and each is 77.14° less than the supplement of the third angle. What are the angles? 19. There are four steel girders to be used end to end; if they were each 6 feet shorter, they would just do, and if they were each 6 feet longer, 3 of them would do. How long are they? 20. There are four iron pipes to be placed end to end; if they were each 12.4 feet longer, they would be just long enough, and if they were each 12.4 feet shorter, it would take 5 pipes to fill the space. How long are they? 21. I paid 39 cents with nickels and cents, using nine more cents than nickels. Find the number of coins of each denomination. 22. I paid 85 cents in nickels and dimes, using 2 more nickels than dimes. Find the number of coins of each denomination. 23. I paid $2.05 in quarters and nickels, using 1 fewer nickels than quarters. Find the number of coins of each denomination. 24. I paid $4.75 in quarters and half-dollars, using 5 fewer quarters than half-dollars. Find the number of coins of each denomination. 40 THE FIRST USE OF ALGEBRA 25. I paid $1.45 in quarters and dimes, using 4 more dimes than quarters. Find the number of coins of each denomination. 26. I borrowed a certain number of quarters, and paid back the debt with 4 fewer half-dollars. I repaid 25 cents too much. How many coins did I borrow? 27. I borrowed some coins, and paid back 5 fewer coins of another sort; the coins borrowed were quarters, those repaid were halves. I still owe 50 cents. How many coins did I borrow? 28. Rice costs 2 cents a pound more than sugar; 3 pounds of sugar and 10 pounds of rice come to $1.24. Find the cost of each per pound. 29. I walk 3 miles more in the forenoon than in the afternoon; and between Monday noon and Friday night I walk 75 miles. How far do I walk in one afternoon? 30. A street 680 feet long has on one side 10 house lots; 3 are narrow lots, all of the same width; the other lots are each 20 feet wider than the first lots. What is the width of each of the lots? 31. A wall 13i feet high is built with 7 courses of founda- tion stone and 41 courses of brick; each course of stone is 9 inches thicker than one of brick. What is the thickness of the foundation stones? 32. A wall is laid with bricks of two thicknesses, 3 inches and 5 inches; there are 3 more courses of the thicker kind than of the thinner, and the wall is 9 feet, 3 inches high. Find the number of courses of each kind of brick. 33. A hallway is laid with rows of tiles of two widths, 2§ inches and 3§ inches. There are 4 more rows of the wider kind of tile; and the hallway is 5 feet, 2 inches wide. Find the number of rows of each kind of tile. THE STUDY OF A SIMPLE EQUATION 41 The Study of a Simple Equation 33. In either member of an equation, or in both, we may find terms with letters only, terms with figures only, or terms with both. Terms that have the same letters are called similar terms; the terms here that have no letters are also called similar terms. Thus, in the expression 72 x + 45 + 5 X — 15, the terms 72 x and +5 x are similar; and the terms + 45 and — 15 are similar. Similar terms can be united. Thus 72 x + 5 x is the same value as 77 x. On the other hand, terms that are not similar cannot be united. 72 x + 45 is not the same as 117 x; it is not the same as 117; it cannot be simplified. 34. In any case where a number may be separated into two factors, either of these factors is called the coefficient of the other. Thus the term 34 x may be separated into two factors in several ways; for example, it is the same as 17 X 2 X, and consequently 17 is the coefficient of 2 x in this term, and 2 x is the coefficient of 17. In any term, the factor that is expressed in figures is of especial importance; we call it the coefficient of the term. Thus the coefficient of the term 34 x is 34. 35. An axiom is a general statement not requiring proof. The following axioms are used as authority for any changes in value that we make in the members of any simple equation. 36. // equal numbers are increased or decreased by the same Cv^ amount, the resulting numbers are equal. "^--^^ 37. If equal numbers are multiplied or divided by the same amount, the resulting numbers are equal. 38. These axioms specify four different things, — ad- dition, subtraction, multiplication, and division, — that can be done to the numbers on each side of an equation. 42 THE FIRST USE OF ALGEBRA There are other things, such as uniting similar terms, and performing indicated multipHcations, that do not change the value of the equal numbers at all, though they do change the form of the algebraic expressions. In order to avoid confusion, the student is advised to apply these changes systematically in the order indicated by the following Rule (1) Perform any indicated multiplications. (2) Unite similar terms in each member. (3) If there are any terms with minus signs before them, add to each member enough to make up these shortages. (4) Subtract from each member the smaller unknown term. (5) Subtract from each member any known term that stands beside an unknown term. (6) Divide each member by the coefficient of the unknown term. This rule has interesting associations with the history of our subject. There were, in ancient times, two great text- books of algebra: the first, written in Greek about 1600 years ago, gave the substance of the rule; the second, written in Arabic more than 1000 years ago, had for its title the name by which this rule was then known. The first word of that title is our modern word algebra; it means ''restoration" and signifies the operation called for in paragraph (3) of the rule. EXERCISE 13 Unite similar terms: 1. 3x- 5 + 4X + 2 - 5a: + 7-a; 2. 7-7x + 3-x-ll + 9a: + a;-l 3. 3 + 5x-8a;-2 + 7a;-10 + x+15 4. 13x-2x+17-5x-ll-a; + 7x-23 6. H-a; + 6-3a;- 10-|-7a;-5a; + 4 \ THE STUDY OF A SIMPLE EQUATION 43 Perform indicated multiplications and unite similar terms: 6. 3 (4 - x) + 7 (x - 3) + 5x - 4 7. 5 (1 - a:) + 13 (x - 3) + 11 a: - 19 8. x + 5 + 3(2x- 1) + 7 + 2(1 -5x) + 10a; 9. 3 + 4(a;-5)+3x + 7+13(4-3x) + ll 10. 4 (x - 5) + 9 (3x - 2) + X + 4 (1 - 5a;) Solve the following equations for x: li. 3 (x - 7) = 2x - 11 16. 7 (x - 2) = 4 (x + 4) 12. 2 (x + 5) = lOx - 6 17. X + 6 = 7 (10 - x) 13. 2x + 5=3x 18. 8x-l=5a: + 41 14. 5 x - 7 = 3 (x+1) - 4 19. 7 (x + 2) = X + 32 15. lOx + 13 = 3 (x + 9) 20. 5 (1 - x) = 1 - 3a; 21. 11 (x + 3) = 10 (x + 1) + 1 22. 8 (x + 10) = 12 (x + 7) - 8 23. 100 (x - 5) = 60 (x + 3) 24. 4 (x + 5) = 2 (3 X - 10) 25. 3 (x - 10) + 2 (x + 4) = 6x - 22 26.* 8 (10 - x) = 5 (x + 3) 27. 3 X + 4 (x - 2) = 1 + 3 (2 X - 3) 28. 19 (x - 2) = 6 (3 X - 5) 29. 3 (3 X - 5) = 4 (x + 5) 30. 15(5x-3) = 12(4x + 3) 31. 5(x + 4) = 14x + 2 32. 3 (x + 3) = 3 (3 X + 1) - 12 33. 5 (x + 1) = 6 (3 - x) - 2 34. 2 X = 5 (x - 1) - 2 X 35. 17 (x - 2) = 2 (2x + 1) + aj 36. 13 (5 - x) = 3 X + 1 37. 5 X - 8 = 6 (x - 3) 44 THE FIRST USE OF ALGEBRA 38. 3 (2 a; - 5) = 4 a; + 1 39. 25 - 4 a; = 2 (x - 4) - 3 40. 7 - X = 3 (4 - x) + 1 41. 2 (a; + 5) = 5 (5 - x) + a; - 3 42. 4 + X 4- 4 (5 - x) = 3a; + 2 (4 - a;) 43. 2x + 5 =5(x-2)+2x- 10 44. 1 + X + 3 (2 - x) = 7 (3 - x) 45. 13 (8 - x) + 2 (x -- 6) = 2 X + 1 46. 9 (x + 1) + 5 = 5 X -F 6 (x - 1) 47. 5 (x - 6) + 21 = 3 (x + 3) 48. 3 (x - 4) = 4 (x - 3) - 17 49. 11 (x + 3) - 100 = 9 (x + 5) 50. 15(7x-3) =8(13x + 7) -20 Sum of the Unknown Numbers Given 39. Model H. — A merchant has grain worth 9 cents per peck, and other grain worth 13 cents per peck. In what proportion must he mix 40 bushels of grain so that the mixture may be worth 40 cents per bushel? Let X = the number of bushels at 36 cents per bushel. Then 40 — x = the number of bushels at 52 cents per bushel. 36 x = the number of cents value of one grade. 62 (40 — x) = the number of cents value of the other grade. ® 36 X + 52 (40 - x) = 1600 @ 36 a; + 2080 - 52 x = 1600 ® same values ® 2080 - 16 a; = 1600 @ same values 2080 = 1600 + 16 x (3) + lQx © 480 = 16 a; - 1600 30 = a; -M6 10 = 40-a; 40-0 Ans. 30 bu. at 36ff per bushel; 10 bu. at 52jf per bushel Check: 30 x .36 = 10.80 10 X .52 = 5.20 16.00, SUM OF THE UNKNOWN NUMBERS GIVEN 45 EXERCISE 14 1. The total rent of two houses is $75 per month. One of these was idle for three months; the aggregate rental received from both houses for that year was $801. What was the rent of each house? 2. Twenty coins, quarters and half-dollars, amount to $5.75. Find the number of each kind. 3. 108 coins, dimes and cents, amount to $4.32. Find the number of each kind. 4. Fourteen coins, nickels and quarters, amount to $1.70. Find the number of each kind. 5. 100 coins, quarters and dimes, amount to $20.20. Find the number of each. 6. A bridge broke down under a pressure of 28,500 pounds, caused by a crowd of 200 people. If the average weight of a man is 150 lb. and that of a woman is 120 lb., how many men were there in this crowd? 7. I bought 30 pounds of sugar of two grades for $2.28; the better grade of sugar cost 10 cents per pound, and the poorer 7 cents. How many pounds of each grade did I buy? 8. I bought 15 apples for 59 cents; red ones at 5 cents, green ones at 3 cents. How many of each kind did I buy? 9. A grocer was offered $15 for 50 pounds of tea; he furnished 25-cent tea with 35-cent tea so as to make $1 on the transaction. How many pounds of each grade of tea did he sell? 10. A merchant has one grade of grain worth 11 cents per peck, and another grade of grain worth 15 cents per peck. In selling 20-bushel lots, how many bushels of each grade must he furnish in order that each lot may average 48 cents per bushel? 46 THE FIRST USE OF ALGEBRA 11. A storekeeper has $197 in bills of $2, $5, and $10; 3 times as many twos as fives. He has 40 bills in all. Find the number of each. 12. The complement of an angle added to five times the angle is equal to its supplement. How large is the angle? 13. Three times the complement of an angle added to the supplement of the angle gives twice the angle itself. How many degrees are there in the angle? 14. A line 5 feet long is divided into two unequal parts. Seven lines equal to the shorter part and 2 lines equal to the longer part would cover a distance of 18 ft. 9 in. How many inches are there in each part of the line? 15. A rectangular plot of land fronting on a street is divided into 6 house lots by lines running back from the street. Each house lot has the same frontage, and has a perimeter of 241 feet. The whole plot has a perimeter of 745 feet. Find the dimensions of each lot. 16. A storekeeper has $4.04 in dollars, dimes, and cents; 44 coins in all. He had 9 times as many dimes as dollars. Find the number of coins of each denomination. Circles 40. A circle is a figure on a flat surface made by a curved line every point of which is at the same distance from a point within called the center. The curved line of the circle is sometimes called the circumference. Any circle may be supposed to be marked off into degrees like a protractor. A central angle will intercept a portion of the circumference containing the same number of de- grees as the angle, whatever the size of the circle; but in a large circle this portion of the circumference will, of course, be longer than in a small circle. CIRCLES 47 41. The distance from the center to the circumference of a circle is called the radius. Any line drawn between two points of a circumference and passing through the center is a diameter. Any portion of the circumference is called an arc. Two radii that make an angle of 30° intercept an arc of 30°. In a circle having a circumference of 30 inches, an arc of 30° would be 2^ inches long (y\% or xV of the circumference); while on a circle of 100 inches circumference, the same angle would intercept an arc 8| inches long. 42. Two arcs are complementary when their sum is 90°, just as in the case of angles. Two arcs are supplementary when their sum is 180°. EXERCISE 15 1. If each degree of a circle is 3^ in. long, how long is an arc of 4^° in that circle? How many feet are there in the circumference of the circle? 2. Two arcs of a certain circle differ by 13°; one of them is 6| in. longer than the other. What is the length of an arc of 90° in that circle? 3. Two arcs of a certain circle are complementary, and each is 11 ft. 3 in. long. How long is a degree of that circle? How long is the entire circumference? 4. Two arcs of a certain circle are supplementary; one is 7 ft. long, the other 28 ft. How many degrees are there in each arc? How long is the circumference of that circle? 5. One base angle of a triangle is three times the comple- ment of the vertical angle, and the other base angle is 20° less than the vertical angle. What are the angles? 6. One side of a railroad track, laid in a circular arc of 33°, requires 13.8 tons of rails, some of them weighing 40 lb. to the foot, the rest 48 lb. Each rail is 30 ft. long and 48 THE FIRST USE OF ALGEBRA extends over li°. How many rails of each kind are needed for this side of the track? 7. In the figure formed by a stripe and a transversal, three of the even angles have a sum 15° greater than two of the odd angles. Find the angles and draw the diagram. 8. In Fig. 1, XY is parallel to CA, the angle Q is 40° greater than A, and is twice as large as C. Find the angle B. Fig. 1. Fig. 2. 9. In Fig. 2, PQ is parallel to ZX, A = 125°, and X = Y. Find the angle Z. 10. An arc of 30° is 17 in. long. What is the length of the circumference? 11. What is the central angle of an arc 3 ft. long on a circle whose circumference is 45 ft. long? 12. With a circumference of 36 ft., what is the supple- ment of an arc of 17 ft.? What is the complement of an arc of 7 ft.? 13. An arc of 42° has a length of 84 ft. Find the length of a quadrant, and the central angle of an arc of 100 ft. 14. If the quadrant of a circle is 7 ft. long, how long is an arc of 3°? 15. If an arc of 17° is 2 ft. 10 in. long, how long is the circumference? CIRCLES 49 16. If an arc of 9° is f in. long, how long is its comple- ment? Its supplement? 17. If a central angle of 72° intercepts an arc 21 in. long, how long is a quadrant of that circle? 18. If a central angle of 22J° intercepts an arc 8 in. long, how long is the complement of that arc? 19. On a circumference of 25,000 miles, how long is an arcof9°? Of 1.8°? 20. If an arc of 3f ° is 1 ft. long, how long is the semi- circle? 21. If an arc of 1|° is 88 ft. long, how long is a quadrant? 22. If a central angle of 9° intercepts an arc of ^V i^-i how long is the whole circumference? 23. On the outer edge of a protractor, marked to half- degrees, the successive marks are .0125 in. apart. How long is the semicircumference? 24. A railroad track is laid so that the line halfway be- tween the rails is a circular arc of 20°, each degree being 30 ft. long. A total of 382 ties are used on this curve, the first part of them being spaced 15 in. on centers, the rest 3 ft. How many degrees of the arc are in the part where the ties are close-spaced? CHAPTER II APPROXIMATE COMPUTATION 43. Certain methods in arithmetic are of especial value in working with numbers that are obtained by measurement. Such numbers, as has been shown in Section 15, depend for their accuracy upon the precision of the instruments used and the skill of the person using them. Computations based upon such numbers may involve a great deal of tedious and useless labor. The methods described in this chapter will prevent that waste. Multiplication 44. Model A. — Multiply .7854 by 31.07. We begin to multiply from the left side of the 7854 multiplier, instead of from the right. 3J Qy To obtain the first partial product, we are really Qo ceo multiplying by 30, not by 3. The multiplicand is ■ rropj^ approximately equal to .8. Now 30 X .8 would be 054978 10x2-4 or 24. Using common sense, then, instead of a mechani- 24.402378 ^^1 rule, we fix the decimal point after 23 in the first partial product. The other partial products succeed each other as in the other method of multiplying, except that each successive product is displaced one column to the right, instead of to the left; because the value of each place in the multiplier decreases by the ratio xVj instead of increasing by the ratio 10. Thus the 4 of the second partial product is displaced one column to the right of the 2 in the first partial product; and the 8 in the third partial product is displaced two columns to the right, once for the and once for the 7 in the multiplier. One reason why this left-hand or natural order of partial products is preferable is, that the first partial product so 50 MULTIPLICATION 51 obtained is always the largest, and furnishes a sort of trial product, to which the other partial products are added as successive corrections. The student should always use the first partial product, therefore, as a rough check upon his work, and especially should fix the' decimal 'point in it before writing the second partial product; and he should make sure that the number thus obtained as a trial product is what might reasonably be expected. 45. The greatest advantage in using the natural order of partial products is in the saving of labor by discarding un- necessary figures. In Model A, for instance, at least the last four figures are doubtful. If by more careful measurements the multiplier could be carried one place farther, we might find it to be 31.066 or 31.074, and this would change the last four columns. Also the decimal .7854 might have any value between .78536 and 7.8544. The effect of this doubtfulness at the end of each factor is illustrated here by using a star instead of the possible addi- tional figure. 7854* '^^^ ^^^^ ^^^^ there is a star, representing 31.07* ^^ unknown figure, in the fourth decimal 23.562* column shows that everything beyond the 7854* first five figures in the product is value- 54978* less, and should be rejected. It cannot be assumed that the doubtful figures would • ^"^^o i o increase the product, for if the measurement had been carried one place farther, the value might have been less; .78536 instead of .7854, or 31.066 instead of 31.07. Moreover, as the quantities are expressed in but four figures, the results should be cut to four figures, that is, to 24.40. 52 APPROXIMATE COMPUTATION 46. In order to shorten the work of multiplication, much of this rejection can be done in each partial product. We must, however, be careful to avoid errors of rejection. To do this, we add 1 to the last figure we write down, whenever the figure rejected amounts to more than half of that 1. 47. The work of multiplication as thus amended is now given step by step: .7854 One partial product : 31 . 07 23.562 All figures here are obtained from those in the data; there may be a little doubt about the last figure, but no figures appear in the doubtful column. .7854 Two partial products : 31 . 07 23.562 785 The last figure of the multiplicand (4) is marked off to indicate that numbers derived from that figure in this partial product would appear in the doubtful columns. Note that in this case the figure which we reject is less than 5. Consequently we may reject it without adding 1 to the figure in the next place to the left. /// .7854 31.07 Three partial products : 23 . 562 785 65 24.402 or 24.40 The next two figures are now marked off, as there is a zero in the multiplier. Also, in multiplying we say 7X9, since .7854 is nearer .79 than .78. 7 X 9 = 63; hence, carry 6. Then 49 + 6 = 55. Since one figure of the multiplier is zero, we get only three partial products. MULTIPLICATION 53 The computation is now complete; it is fair to assume that our result is accurate to four figures, as the given numbers are expressed to four figures. All figures of the multiplier should he used in the computation. The nearest value of the rejected figure should be used for // / multiplication; 9 for 854; 5 for 54, etc. The successive stages of the work in this example are separated and repeated here for the purposes of study only. Of course no one, after finding the first partial product, would begin over again and put down the first two partial products and so on. All of the actual work is shown in the last stage of the model problem. Check this multiplication by using round numbers (30 X .8 = 24.) and if desirable by interchanging multiplier and mul- tipHcand as follows: 31.07 .7854 21.749 2 486 155 12 24.402 or 24.40 48. Model B. — How much house coal, weighing 45.6 lb. to the cubic foot, can be stored in a space 32 ft. 2 in. long, 21 ft. 5 in. wide, and 12 ft. 10 in. high? In reducing these measurements to feet and tenths of a foot, it is convenient to make a table of equivalents, as follows: Inches: 123456789 10 11 Tenths: 12 2* 34567 8* 8 9 The starred numbers occur where the measurement in inches is just half way between two measurements in tenths. Here it is customary to choose the even number. Since there are just as many even num- bers as odd numbers, we shall be gaining in the long run as much as we are losing. This principle is to be adopted in all our computation where the figure rejected is exactly 5. The working out of this problem is left to the student. 54 APPROXIMATE COMPUTATION EXERCISE 16 Multiply: 1. 3.142 by 144.0 6. 301.7 by 180.6 2. 92.5 by 3.14 7. 496.3 by 13.57 3. 703. by 21.8 8. .0188 by .00379 4. .785 by 1.21 by 288. 9. 1.192 by .3489 6. 47.38 by 71.17 lo. 1.97 by 32.8 by .00523 11. What is the weight of 328 steel beams, each 17.81 ft. long, weighing 178.3 lb. per linear foot? 12. A steel beam is 20.83 ft. long and weighs 42.7 lb. per foot. Find its weight. 13. A foot of wire weighs .109 lb. How much will a mile of this wire weigh? 14. A meter is 39.37 in. How many square inches are there in the surface of a window one meter square? 15. A city lot measures 44.35 ft. front by 77.23 ft. depth. Find its value at $1.63 per square foot. Division 49. In long division, the numbers subtracted are partial products obtained in multiplying the divisor by the succes- sive figures of the quotient, beginning at the left. These successive partial products may be curtailed, as in multipli- cation, when they project into the doubtful columns. 60. Model C. — Divide 63.3 by 27.9. _2 27 9)63 3 One partial product : * cc'o 27.9)63.3 55 8 7 5 5_6 1 9 Two partial products: DIVISION 55 Here we have marked off the 9 in the divisor, since, when multiplied by 2, it would come into the doubtful column to the right of the last given figure of the dividend. But there is a. 2 to carry from the doubtful columns (since 2x9 = 18, which is nearer 20 than 10 in value), making the second partial product 56 instead of 54. ./ 2.26 27.9)63.3 Three partial products: -y-^ 5 6 1 9 1 7 Two figures are marked off; 5 is carried from the doubtful columns. (Why?) xr/ 2.267 Four partial products : 27 . 9) 63 . 3 55 8 Ans. 2.27 I ^ 5 6 Three figures are marked off. Only what is ■'■ " carried from the doubtful columns can now ap- ^ ' pear in the partial product. The next figure in 2 the quotient must be chosen so as to enable us 2 to carry 2. 7 X 279 = 1953, and 8 X 279 = ' ' 2232; evidently 7 is more nearly correct than 8. Since 2.267 is nearer 2.270 than 2.260 in value, we must increase by 1 the last of the three figures which it is allowable for us to retain in the quotient. The answer, then, is 2.27. Check this division by the use of round numbers (60 -i- 30 = 2) . For a more thorough check, multiply the quotient by the divisor, not the divisor by the quotient. Multiplying the divisor by the quotient is not quite so good, because the partial products thus obtained will be the same as the subtrahends used in the process of division. 2.267 27.9)63.3 7 5 1 9 2 561 APPROXIMATE COMPUTATION 51. In the so-called Italian method of division, the sub- trahends are not written down; the computer performs the multiplication and the subtraction at the same time, as fol- lows: Two nines, and five are twenty-three; two sevens and two are sixteen, and seven are twenty-three; two twos and two are six, and zero are six; and so on. The "shop" method of subtraction, which is used here, requires the computer to emphasize the number that has to be added to the subtrahend to give the minuend. This is also called the Italian method of subtraction. Students who are ambitious enough to acquire this trick will find it economical when there is much dividing to do. EXERCISE 17 1. A meter is defined by United States law as 39.370 inches. A carefully surveyed line in France is 283.72 meters long; how many feet long is this line? 2. An arc of 39.3° is 2.73 feet long. How long is one de- gree on this arc? Divide : 3. 308.5 by .5632 7. 15.02 by .1177 ii. 110.4 by .3561 4. 708. by 3.14 8. .350 by 7.83 12. 17.65 by 1.509 5. 256.0 by .7854 9. .309 by 8.47 13. .707 by 88.1 6. 3.142 by 1.303 10. 77.07 by 3.043 14. 3.32 by 885. Perform the operations indicated : 5. 2.303 -^ 359.6 16. 3.835 X 100.2 -^ 3.142 17. 39.37 X 39.37 X 3.142 18. 2.236 X 1.414 X 3.162 19. 4.63 X 4.63 X 4.63 -f- 2.00 20. 729. X .533 X 729. X .533 SQUARE ROOT 57 Square Root 52. The product obtained by multiplying a number by itself is called the square of that number; and the number which is multiplied by itself is called the square root of the product. If a number is divided by its square root, the quotient is the same as the divisor. If either the quotient or the divisor is smaller than the square root, the other must be larger. The square root, then, must be somewhere between the quotient and the divisor. 53. Model D. — Find the square root of 55.90. Since this number is between 49 and 64, its square root wiU be be- tween 7 and 8; let us guess 7.5 and try it. 7.453 7.500)55.90 52 50 3 40 3 00 40 38 2 7.478 7.476)55.90 52 33 3 57 2 99 58 52 6 The square root of 55.90 must be between 7.500 and 7.453. Let us try a number halfway between, say 7.476. To four figures, then, the square root of 55.90 is 7.477, a result which can be verified by multipUcation (or by division, as before). EXERCISE 18 Find the square roots of 1. 57.76 4. 1383.84 7. 3.142 lo. 32.1 2. 94864. 6. 10.7584 8. 125.9 ii. 537.3 3. 0.001444 6. 32.1489 9. 277.7 12. .123 58 APPROXIMATE COMPUTATION Find, to four figures, the square roots of 13. 2 16. 6 19. 8 22. 250 14. 3 17. 7 20. 12 23. 245 15. 5 18. 10 21. 45 24. 252 Compare the results of Exs. 13 and 19. Also compare the results of Exs. 14 and 20. Exs. 15 and 21. Exs. 18 and 22. Exs. 15 and 23. Exs. 17 and 24 54. It sometimes happens that the data of a problem can be stated exactly in figures. For example, if there are just a dozen men, whose aggregate weight is 1954 pounds, the number of men is exact, though the total weight is an ap- proximate number. In the same way, the number of degrees in the three angles of a triangle is exactly 180. Numbers that are thus theoretically exact are expressed with as many figures (zeros) as may be necessary to match the precision of the other data. The weight of the men might vary an ounce or so from the amount stated, without the fact being detected by the most careful weigher; consequently this number is properly regarded as approximate. The number of men, of course, and the number of degrees in all the angles of a triangle, are numbers that are not obtained by measurement, and they cannot admit of fractional error. 55. It sometimes happens, again, that of several data in a problem, one may be much less accurate than the rest. For example: On a carefully surveyed right of way, 3728.7 feet in length, railroad tracks are laid with rails weighing 25.3 pounds per foot. How much steel is required for these rails? Here if we multiply 25.3 by 3728.7, all the columns after the third will be doubtful. And if we multiply 3728.7 by 25.3, all partial prod- ucts after the third will be missing, leaving the same columns in doubt as before. SQUARE ROOT 59 25.3** 3728.7 3728.7 25,3** 759** 74574 177 18644 6 1119 2 ** 94300. 94300. 56. The considerations that apply in addition are slightly different. Thus, if three weights are to be added together, such as 783.4 lb., 2.874 lb., and .005632 lb., the three num- 783.4 bers will have figures in nine different col- 2.874 umns. The last two numbers will have • 005632 figures in columns that are doubtful on ac- 786.3 count of the limited precision of the first number. Consequently the last weight will not affect the result at all, and the second weight might just as well have been reported as 2.9 lb. The sum of the three is 786.3 lb. 57. This principle, then, is to be constantly borne in mind: The accuracy of a result is no greater than the accuracy of its least accurate datum. \ EXERCISE 19 1. A line 738.5 feet long is divided into two parts which have the ratio 1.137; how long is each part? 2. Two supplementary angles have a ratio of 2.003. How many degrees are there in each angle? 3. Two supplementary arcs differ by 3.10°, and their sum is 303.3 feet long. How many feet are there in each arc? 4. A meter is 39.370 inches long; express that length in feet to the same number of figures. 5. A mile is 5280 feet long. How many feet are there in 1.29 miles? 6. The bedplate of an engine lathe is 3.73 feet long. How many inches is that? ^0 APPROXIMATE COMPUTATION 7. A gallon is 3.785 liters, and a liter of dry air weighs 1.293 grams. What is the weight in grams of a gallon of dry air? 8. A gallon is 3.785 liters, and a liter of dry carbon dioxide weighs 1.977 grams. What is the weight in grams of 2.513 gallons of carbon dioxide? 9. One acre is equivalent to 0.4047 hectares. How many hectares are there in 9.538 acres? 10. One pound is equivalent to 0.4536 kilograms. How many kilograms are there in 263.2 pounds? 11. One radian is equivalent to 57.296°. How many de- grees are there in 3.172 radians? 12. One cubic meter is equivalent to 1.3079 cubic yards. An excavation of 35.4 cubic meters is paid for at $2.23 per cubic yard. How much does it cost? 13. One hectare is equivalent to 2.471 acres. How much would a French farm of 11.3 hectares cost at $72 per acre? 14. The two base angles of a triangle have a ratio 1.77, and the vertical angle is 43.4° greater than the difference of the base angles. Find the three angles. 15. Two of the angles made by a stripe and its transversal have a ratio 1.7. How many degrees are there in each acute angle? 16. In the figure made by a stripe and its transversal, if each obtuse angle were 3.83° greater than each acute angle, how many degrees would there be in each of the two interior angles on the same side of the transversal? 17. The angle at the vertex of a triangle bears to the angles at the base the ratios 1.33 and 2.17. Find the three angles of the triangle. 18. Thp angles at the base of a triangle have the ratio .625, and the angle at the vertex is 4° more than three times their difference. Find each angle of the triangle. SQUARE ROOT 61 19. A certain railroad track consists of a straight track 275 ft. long and a curve on which each degree is 16.67 ft. long; it requires 800 ft. of rails. How many degrees are there in the curve? The curve of a railroad track is a line halfway between the two rails. The outer rails make an arc which is really longer than that of the inner rails, but the total length is computed as if both rails were laid on the center line, midway between the actual positions of the rails. 20. In the basket of a balloon are two men, weighing as they stand 160 lb. and 210 lb.; provisions weighing 8 lb.; maps weighing 11 oz.; and instruments weighing 7.23 lb. What is the total load in the basket? 21. A miscellaneous bill of goods purchased from a mail- order house contained a washing machine weighing 125 lb. ; 3 bolts of cloth weighing 9 lb. each; 3 spools of thread weighing 1 oz. each; 2 stoves weighing 360 lb. and 158 lb.; and a packing case weighing 40 lb. What was the freight at 55 cents per hundred pounds? 22. A man's real estate holdings consist of two farms of 230 acres and 157 acres; two village plots of 3 acres and 2 J acres; a ranch of 1700 acres; and a city lot 50 by 85 ft. What is the total area of his holdings? 23. A large corporation owned a mill employing 3500 men, another mill employing 2800 men, a machine shop employing 125 men, an office force of 56 men, and an experimental plant of 12 technical experts. How many men were there on the pay-roll? 24. On a vacation trip of five days, a traveler used an automobile the first day and covered 290 miles, and a bicycle the second day for 32 miles. He tramped up and down a mountain the third day, 6.5 miles. The fourth day he spent at the hotel and walked only 150 yards. The fifth day he took a train home and traveled 368 miles. What was the total distance covered on his vacation? 63 APPROXIMATE COMPUTATION 58. Model E. — Solve 17.2 {x + 3.77) = 3.86 + 43.1 x, © 17.2 (a; + 3.77) = 3.86 + 43.1 x @ 17.2 x + 64.9 = 3.86 + 43. la; © same values ® 61.0 = 25.9a; ©-17.2x-3.86 © 2.35 = x ©-^25.9 Check: 17.2(2.35 + 3.77) = 3.86 + 43.1 (2.35) 17.2(6.12) = 3.86 + 101. 105. =105. Reject no figures until the actual work of computation begins. In checking examples solved in this way, the values found for the two members of the equation may not be exactly equal, but ordinarily they should not differ by more than one per cent if three figures are used. Solve the following equations : 25. 57.3 X + 2.87 = 283.6 26. 2.13 (x + 17.2) = 109.7 27. 8.7 {x - 5.6) = 27.8 28. 1.14 {x + 1.07) = 7.23 + X 29. 21.3 {x - 13.5) = 513 + a; 30. 7.2 {x - 3.6) = 56 - a; 31. .36 (x + .23) = 7.6 - 2.4 a; 32. 2.8 (x - 3.8) = X - 3.3 33. 5.7 X + 2.3 = 3.3 (8.7, - x) 34. 6.4 {x - .77) = .56 (x + 83) 35. 4.38 (2.16 X - 7.55) = 28.9 36. 5.88 X - 3.11 = 2.97 (x + 6.33) 37. 3.34 (x - 1.11) = X + 80.9 38. X - 29.7 = 2.34 (3.45 x - 74.0) 39. .37 (x - .38) = .49 (x + .57) 40. 4.39 X = 32.4 (.127 x + .00525) 41. 11.9 =32.7 (.115 + .0438 x) 42. 6.62 - 5.47 X = 11.1 (x + 10.3) 43. 5.32 - 2.37 (x - 5.89) = 44. 73.7 - 4.90 X = .145 (x + 73.2) CHAPTER III MEASUREMENT FORMULAS 59. Beside being used for abbreviating the explanation of problems in arithmetic, algebra is used also for abbreviating the statement of arithmetical rules. Rectangles 60. The area-number of a rectangle is the product of the length-numbers of two adjacent sides. In the preceding statement three numbers are mentioned; each can be represented by a letter as follows: S = the area-number of the rectangle. a = the length-number of one side. b = the length-number of the other side. Then the rule can be stated by the formula: S = ah, 61. Suppose we find, by measuring, that a rectangular building lot is 93.57 feet long and 64.25 feet wide. The area will be the number of square feet, and decimal subdivisions of a square foot, that could be marked off on the surface of the lot. We know that this area is found by multiplying the two length-numbers together, but it is interesting to prove it in the following way: A strip exactly a foot wide along the short side of the rec- tangle could have a square foot marked off on it for every whole foot in the length of the side (64 of those) ; and a tenth of a square foot for every additional tenth of a foot in length (2 of those) ; and a hundredth for every additional hundredth 63 64 MEASUREMENT FORMULAS (5 of those). A tenth of a square foot here is made by mark- ing one square foot crosswise into ten equal strips, 1 ft. long and iV f^- wide. A hundredth of a square foot is made by splitting each of these narrow little strips into ten very narrow strips, each measuring 1 ft. by y^^ ft. Notice that the area-number of this unit-wide strip is the same as its length-number. The foot-wide strip along the shorter side of the lot, then, has an area of 64.25 square feet; and one of these strips can be marked off on the rectangular lot for every whole foot in the length (93 of them). If we split one of these foot-wide strips into ten equal strips, one of these narrow strips can be marked off on the surface of the lot for every tenth of a foot in its length; and a hun- dredth of a strip, in the same way, for every hundredth of a foot in its length. So that there will be in the whole lot 93.57 strips, each having an area of 64.25 square feet. Since the same reasoning may be applied to any rectangle, it is evident that the area-number of a rectangle is the prod- uct of the length-numbers of two adjacent sides. EXERCISE 20 1. Find the area of a rectangular brass plate 3.28 in. wide and 5.37 in. long. 2. A rectangular city lot has 40.27 ft. frontage, and is inclosed by just 217 ft. of fencing. What is the value of the lot at $2.25 per square foot? 3. A strip of land running back from a city street the whole depth of the lot is in dispute between adjoining owners. The width of the strip is 8.723 in., and the area in dispute is 57.21 sq. ft. How deep is the lot? 4. The front of a lot is 38.0 ft., and the area-number ex- ceeds the other length-number by 2156. Find the depth of the lot. RECTANGLES 65 6. The perimeter of a square is 738.4 ft. What is the area? 62. A surveyor, as will be shown hereafter, can determine the ratios between different lines with- out actually measuring the lines them- selves. For example, in the accompanying figure, the angle X could be measured in degrees, and then by referring to tables the surveyor could find the ratio of 6 to a. EXERCISE 21 1. The ratio of two sides of a rectangle is 1.14 and the long side is 39.3"; what is the area? On drawings, 5 ft. 4 in. is commonly written 5' 4". 2. The ratio of two sides of a rectangle is 3.08 and the short side is 7.46". What is the area? 3. The ratio of two sides of a building lot is 0.385, and the long side is 103.4 ft. What is the area? 4. The ratio of two sides of a building lot is .472, and the short side is 43.7 ft. What is the area? 5. The ratio of two sides of a floor is 1.34, and the area is 301 sq. ft. What is the perimeter? We shall have in this example an equation consisting of x^ = some number. These equations will be studied later in our work, but for the present it is quite correct for the student to find the square root of the number for the value of x. 6. The ratio of two sides of a rectangle is 2.07, and the perimeter is 137 in. What is the area? 7. The ratio of two sides of a rectangle is 1.57, and the two sides differ by just 83 ft. Find the area. 8. The ratio of two sides of a rectangle is .785 and the area is 1025 sq. in. What is the short side? 66 MEASUREMENT FORMULAS 9. The ratio of two sides of a rectangle is .0243, and the perimeter is 500 ft. Find the area. 10. The ratio of two sides of a rectangle is .335, and the two sides differ by 43.7 in. Find the area. Rhomboids 63. The proof given in Section 61 is valid for the areas of rectangles, because the unit squares and their subdivisions fit exactly at the corners of the figure. This is due to the fact that a rectangle is formed by two stripes perpen- dicular to each other. The proof does not, however, ap- ply to a figure formed by two stripes inclined to each other. Such a figure is called a rhomboid. Rectangle 64. In the rhomboid, represent by a the length-number of each of the two sides that lie along one of the stripes; and by pi the breadth of the stripe. Draw two lines straight across the stripe from the ends of one of the lines called a. This makes three spaces within the stripe; two of them, which we call X and Y, are triangles, and the other is an irregular figure that we represent by Q. These letters are understood to represent the area-numbers of the different spaces. RHOMBOIDS 67 Q + X isa, rectangle, of which the sides are a and pi. Consequently ® Q + X = api A tracing of triangle X will exactly cover the triangle Y. Conse- quently © X = Y ® Q -{- Y = api @ subst. in ® But Q + F is our original rhomboid; so we have a convenient ex- pression for the area: ® 5 = api. 65. To state this briefly in words, we need new tech- nical terms : A figure is said to be inscribed in a stripe when all its vertices are in the sides of that stripe. When a figure is inscribed in a stripe, a side of the figure that lies in one side of the stripe is called a base of the figure, and the breadth of the stripe is then the altitude on that base. We have, then, the rule: The area-number of a rhomboid is the product of the length- numbers of a base and the altitude on it. 66. Since the sides of a rectangle are perpendicular, the base in one stripe is the alti- tude in the other; so that whichever side we take as a base, we multiply the same numbers /' for the area. But this is not the case /^ with the rhomboid. Represent now by b the length-number of one of the bases of the rhomboid in the other stripe; and by 252 the breadth of that stripe. Draw two lines straight across the stripe from the ends of b. It happens here that we have four spaces within the stripe, instead of three 68 MEASUREMENT FORMULAS as in Section 64. Two of them are triangles, which we call R and Q; and the other two we call X and Y. Q + Z is a rectangle of which the sides are h and p2. Consequently © Q-\-X = hp2 A tracing of the triangle X + R will just cover the triangle Y ■}- R. Hence, X + R = Y + R X = Y ■ ® -R Q + F = 6p2 subst. in But Q + F in this figure is our original rhomboid; so we have for the area here S = bp2 67. For every rhomboid there are two formulas for the area; two pairs of numbers such that the product of either pair will give the area of the rhomboid. S = api = hpi 68. A parallelogram is any figure formed by two stripes crossing each other, whether the stripes are perpendicular or inclined. Rectangles and rhomboids are both classed as parallelograms. EXERCISE 22 69. Model A. — The area of a rhomboid is 1.23 sq. ft., and the lengths of the two altitudes are 1.02 ft. and 2.74 ft. What are the bases? Since the figure is a rhomboid, we shall use the rhomboid formula. © S = api © S = bp2 Formula (Section 67) S = 1.23 ;S = 1.23 Given pi = 1.02 p2 = 2.74 Given 1. 23 = 1. 02a 1.23 = 2.746 and subst. in © 1.21= a .449 = 6 -M.02, -5- 2.74 Hence, with pi = 1.02, o = 1.21. With p2 = 2.74, 6 = .449. Find a and h for the rhomboids having the following data: 1. >S = 28, pi =2, p2 = 3.5. 2. *S = 360, pi = 12, p2 = 15. 3. S = 2560, pi = 32, p2 = 25.6. RHOMBOIDS 69 4. ,S = 32.7, 5. S = 55.14, e. S = .389, pi = 8.59, pi = 10.16, Pi = .107, P2 = 6.21. P2 = 12.38. P2 = .839. 70. Model B. — The altitudes of a rhomboid differ by 7.23 ft; the bases are 29.7 ft. and 43.2 ft. Find the altitudes. Let pi = the number of feet in the smaller altitude. Then pi + 7 . 23 = the number of feet in the larger altitude. Since the area is the same with either side taken as the base, the shorter altitude must be the one on the longer base. (J) api = hpi @ a = 43.2 ® b = 29.7 ® 43.2 pi = 29.7 (pi + 7.23) ® 43.2 pi = 29.7 pi + 215 (6) 13.5pi = 215 ® pi = 15.9 ® pi + 7.23 = 23.1 With a - 43.2, pi = 15.9; with b = 29.7, pz = 23. 1. Check by areas. Formula Given Given @ and ® subst. in ® ® same values ® -29.7pi ® -J- 13.5 ® + 7.23 Find pu p2, and S for rhomboids as follows: 7. a = 60, h = 72, difference of altitudes = 10. 8. a = 8.1, h = 5.4, 9. a = 30.6, h = 28.3, 10. a = 3.46, h = 5.19, 11. a = .588, h = .961, = 1.2. = 1.47. = 1.17. = .102. ft 71. Model C. — Two sides of a rhomboid differ by 11.2 ; the altitudes are 7.08 ft. and 23.7 ft. Find the perimeter. Let a = 6 + 11.2; and the larger altitude must be taken with the smaller side. ® api = bpi ® pi = 7.08 ® P2 = 23.7 ® (6 + 11.2)7.08 = 23.7 6 Solving, b = 4.78 and 6+11.2 = 16.0; then the perimeter, 2a + 26 = 32.0 + 9.56 = 41.6. Formula Given Given ® and ® subst. in ® 70 MEASUREMENT FORMULAS Find a, h, S, and perimeter for rhomboids as follows: 12. Pi = 36, P2 = 48, difference of sides = 20. 13. Pi = 4, P2 = 5, (I tC 11 = 1.5. 14. Vi = 3.83, P2 = 4.27, (C 11 l( = 1.53. 15. Pi = 11.2, P2 = 16.5, <( it tc = 10.1. 16. Pi = .49, P2 = .73, t( ti a = .26. 72. Model D. — The longer base and the longer altitude of a rhomboid are 26.3 ft. and 18.2 ft., respectively; the perim- eter is 79.2 ft. Find the area. Since the perimeter = 2a + 2 6, a + 6 = half perimeter. Then | (79.2) = 39.6 = 26.3 + 6, from which h = 13.3. © S = hv2 Formula ® P2 = 18.2 Given (longer alt. with shorter side) ® 6 = 13.3 Given >S = (13.3) (18.2) and subst. in ® S = 242. same values To find the other altitude: (6) S = avi Formula 242 = 26. 3 pi subst. in 9.20 = pi ^26.3 Find areas, and missing sides and altitudes of the following rhomboids : 17. Longer base =7.5, longer alt. =5, perimeter = 27. 18. Shorter base =56, shorter alt. =49, perimeter = 288. 19. Shorter base = 15.7, shorter alt. =23.4, perimeter =66.2. 20. Longer base =38.2, longer alt. =32.4, perimeter = 138. 21. Longer base = .728, longer alt. =.418, perimeter = 2.52. Triangles 73. As the following figures show, a triangle can be in- scribed in three different stripes. There will be three bases, and, corresponding to them, three altitudes. TRIANGLES 71 If we take any two of the three stripes belonging to a tri- angle (for example, the two having bases a and h), we find the space common to the two stripes is a parallelogram. / ^v h (y s ^ / y- This parallelogram is composed of two triangles, and we shall find that a tracing of one of them, being rotated on the figure without overturning, can be made to cover the other exactly. But one of the triangles is the one we started with; if we represent its area by S, the area of the parallelogram will be 2 S, Let us represent the altitudes on the bases a, h, and c, by pi, pi, and Pi, respectively. Then © 2S = api © S = ^api ' © -^ 2 In the same way S = hbp2', and if we take the stripe along c instead of that along a, it is evident that S = ^ cps. 74. We have, then, as the rule for the area of a triangle : The area-number of a triangle is half the product of the length- numbers of a base and the altitude on it. S = I api = I bp2 = I cps 72 MEASUREMENT FORMULAS 75. It is evident that, if two parallelograms are inscribed in the same stripe and have equal bases thereon, they will have the same area. If two triangles are inscribed in the same stripe and have equal bases thereon, the two triangles are equal. If a triangle and a parallelogram are inscribed in the same stripe and have equal bases thereon, the triangle will be equal to half the parallelogram. Thus, in the above figure, X = Y, but X figure below, Q = R and Q = i P. The sign 5^ means "is not equal to." Z; and in the EXERCISE 23 1. A triangle and a rectangle are inscribed in the same stripe, and have equal bases thereon; their areas differ by 783 sq. ft. What is the area of each? 76. Model E. — A triangle has two bases that differ by 5.55 ft., and the corresponding altitudes are 15.00 ft. and 10.59 ft. Find the area. Let a = the number of feet in the shorter base. Then a + 5.55 = the number of feet in the other base. Also the longer altitude must be taken with the shorter base. TRIANGLES © S = hapi = hbp2 = h cPi Formula ® api = bp2 ® X2 ® Vi = 15.00 Given V2 = 10.59 Given ® 15.00 a = 10.59 (a + 5.55) @ and ® subst. in @ ® 15.00a = 10.59a + 58.78 same values ® 4.41a = 58.78 ® - 10.59a ® a = 13.33 ©-^4.41 ® a + 5.55 = 18.88 (D + 5.55 73 Then the two bases are 13.33 and 18.88; the area = ^^ (13.33) (15.00) = 100 sq. ft. Check by product, | (18.88) (10.59) = 100. Find two sides and the area of the following triangles : 2. pi = 15 in., p2 = 7.2 in., difference of a and 6 = 13 in. 3. pi =4.8 ft., p2=20ft., 4. pi =4.7 in., p2 = 7.3in., " 6. pi=39.2 cm., p2=52.4 cm., " 6. pi = .0487 in., p2 = .0392 in., " 77. Model F. — The base of a triangle bears to the altitude thereon a ratio of 1.53; and the area of the triangle is 24.5 sq. ft. Find base and altitude. Let pi = number of feet in altitude. Then 1.53 pi = number of feet in base. a IC h = 19 ft. a (( b = 3.2 in. a (( h = 14.8 cm. a (( h = .0152 in. © S = ^api ® 5 = 24.5 ® 24.5 = H1.53pi)(pi) © 24.5 = .765(pi)2 ® 32.0 = (pi)2 ® 5.66 = pi © 8.66 = 1.53 pi Check by area formula, S Formula Given @ subst. in © ® same values © -^ .765 Square root of ® ® X 1.53 Ans. Altitude, 5.66 in.; base, 8.66 in. ^ api. 74 MEASUREMENT FORMULAS Find base and altitude for the following triangles: 7. Area = 24 sq. in., ratio of base to altitude = 1.5. 8. '' = 140 sq. ft., 9. " = 24.0 sq. cm., 10. " = 8.09 sq. in., 11. " = .260 sq. ft., 78. Model G. — Two altitudes of a triangle are 23.7 in. and 15.3 in., and the corresponding bases differ by 3.28 inches. Find the two bases. C( li u 11 = .7. il 11 It u = .83. (( 11 (I cc = 1.29. C( il 11 11 = .873. Let 6i = number of inches in one base; bi + 3.28 = number of inches in other base. © S = lapi = ^hpi = ^ cpz Formula (D api = bp2 © X 2 ® pi = 15.3 Given ® P2 = 23.7 Given ® (&1 + 3.28) 15.3 = 23.7 6i ® and © subst. in ® (D 15.3 6i + 50.2 = 23.7 &i ® same values 50.2 = 8. 46i ® - 15.3 6i (D 5.97 = 6i © ^8.4 ® 9.25 = &i + 3.28 ® + 3.28 Ans. Bases are 5.97 : in. and 9.25 in., respectively. Check by area formula. Find two bases for each of the following triangles: 12. Altitudes, 15 in. and 16.8 in.; difference of corre- sponding bases, 3 in. 13. Altitudes, 5.3 ft. and 4.7 ft.; difference of correspond- ing bases, .80 ft. 14. Altitudes, .255 in. and .431 in.; difference of corre- sponding bases, .287 in. Find two altitudes of the following triangles: 15. Bases, 16 ft. and 25 ft.; difference of corresponding altitudes, 5.4 ft. TRAPEZOIDS . 75 16. Bases, .77 cm. and .61 cm.; difference of corresponding altitudes, .11 cm. 17. Bases, 32.7 ft. and 19.4 ft.; difference of corresponding altitudes, 4.11 ft. Trapezoids 79. A trapezoid has only one pair of sides parallel. It can be inscribed, therefore, in only one stripe; and the bases in that stripe are of different lengths. There is but one altitude. Represent the length-numbers of the two bases by 6i and 62, and that of the altitude by p. The trapezoid can be divided into two triangles by drawing a line from corner to corner (a diagonal). Represent the areas of the two triangles by X and Y. They have the same altitude p, and bases, 61 and 62, respectively. Their areas are ® X = i 61P © F = I 62P Hence, © X + F = | 6ip + I 62P © + © © X + F = § p (61 + 62) ® same values Thus, for the area of the trapezoid, we have the rule: The area-number of a trapezoid is half the product of the length- numbers of the altitude and the sum of the bases. S = ip{b, + h2) EXERCISE 24 80. Model H. — The bases of a trapezoid are 2.386 in. and 2.834 in., and the altitude is 3.862 in. Find the area. 76 MEASUREMENT FORMULAS ® S = hp(J>i + b2) Formula @ fei = 2.386 Given ® 62 = 2.834 Given p = 3.862 Given S = i (3.862)(2.386 + 2.834) @, 0, and subst. in ,S = (1 . 931) (5 . 220) © same values 5 = 10.08 same values Ans. The area is 10.08 sq. in. Find the areas of the following trapezoids : 1. p =32 ft., 6i = 48 ft., 62 = 77 ft. 2. p = 5 yds., 61 = 1.2 yds.,. 62 = 4.8 yds. 3. p = 6.27 m., 61 = 5.83 m., 62 = 9.21 m. 4. p = 83.8 in., 61 = 55.7 in., 62 = 38.2 in. 6. p = .278 cm., 61 = .553 cm., 62 = .839 cm. 81. Model I. — In a stripe 2.38 in. wide, a trapezoid is inscribed whose bases have a ratio 834, and whose area is 9.88 sq. in. Find the bases. Let 61 = number of inches in one base; and .834 61 = number of inches in other base. S = hpihi + h) Formula S = 9.88 Given p = 2.38 Given 9.88 = I (2.38) (61 + .834 61) and subst. in Solving, &i = 4.53, and .834 61 = 3.78. Check by area formula. Find the bases of the trapezoids with the following data: 6. Width of stripe = 12 ft., ratio of bases = 2, area = 288 sq. ft. 7. Width of stripe = 1.8 in., ratio of bases = 1.5, area = 9 sq. in. 8. Width of stripe = 3.27 in., ratio of bases = 3.29, area = 77.8 sq. in. TRAPEZOIDS 77 9. Width of stripe = 75.6 ft., ratio of bases = 1.82, area = 93200 sq. ft. 10. Width of stripe = .428 cm., ratio of bases = 5.07, area =^ .409 sq. cm. 82. Model J. — In a stripe 700 feet wide, a trapezoid is inscribed whose bases differ by 31.8 feet, and have a ratio 7.83. Find the area. Let bi = number of feet in one base; 7.83 6i = number of feet in other ® 7.83 6i -bi = 31.8 6.83 6i = 31 . 8 ® same values (D 6: =4.65 0-5-6.83 7.83 6i = 36.4 0X7.83 S = h P Q>i + h) Formula S = h (700)(4.65 + 36.4) and subst. in /S = 350 X 41 . 1 same values aS = 14400. same values Ans. The area is 14,400 sq. ft. Find the areas of the following trapezoids: r 11. Width of stripe, 30 in.; sum of bases, 120 in.; ratio of bases, 2. 12. Width of stripe, 6 ft.; difference of bases, 6 ft.; ratio I of bases, 1.75. 13. Width of stripe, 5.3 in.; difference of bases, 4.3 in.; ^ ratio of bases, 2.1. [. 14. Width of stripe, 5.32 cm.; sum of bases, 54.0 cm.; I ratio of bases, 1.28. 15. Width of stripe, .294 ft.; difference of bases, .0611 ft.; I ratio of bases, .838. 78 MEASUREMENT FORMULAS The Number tt 83. The length of the circumference of a circle can always be found by multiplying its diameter by a certain famous number, whose exact value is represented by the Greek letter tt (pronounced pi). This exact value cannot be expressed in ordinary figures, any more than V2 can; but it can be expressed approxi- mately as 3.1, 3.14, 3^, 3.142, 3.1416, 3.14159, etc., according to the requirements of the problem in which it is to be used. 84. The area of a circle can always be found by multiply- ing the area of the square whose side is equal to the radius by this same number tt. I i If c = the length of the circumference of the circle, r = radius, 2 r = diameter, S = area of the circle, and r^ = area of square whose side is r, then we have two formulas: c = 2 irr The two theorems represented by these formulas have been known for many centuries, and will be proved in next year's work. EXERCISE 26 1. The radius ofa circle is 21.3 ft. Find the circumference and the area. 2. The circumference of a circle is 283 in. Find the radius. 3. The area of a circle is 1.83 sq. in. Find the diameter. THE NUMBER ir 79 4. The area of a circle is 288 sq. ft. Find the circumference. 5. What is the length of an arc of 21.3° on a circle whose radius is 11.9 ft.? 6. On a circle whose radius is 39.37 in., how many degrees are there in an arc 20.38 in. long? 7. What is the radius of a circle in which a quadrant is 5.8 in. longer than the radius? 8. The radius of a circle is .707 ft. Find the circumference and the area. 9. An arc of 38° is 16 in. long. What is the length of the radius? 10. An arc of 13.7° is 25.8 ft. long. What is the length of the radius? 11. One base of a trapezoid is 1 ft. less than double the other, and the altitude is 4 ft. The two triangles into which the diagonal divides the trapezoid differ by 12 sq. ft. Find the bases of the trapezoid. 12. One base of a trapezoid is 2.46 in. less than 3 times the other, and the altitude is 2.16 in. The two triangles into which the diagonal divides the trapezoid differ by 21. 7 sq. in. Find the bases. 13. The bases of a trapezoid differ by 4 ft., and the altitude is 3.128 ft. The two triangles into which the trapezoid is divided by the diagonal have a ratio 1.5. Find the bases and the area. 14. The bases of a trapezoid differ by 1.053 ft., and the alti- tude is 11.22 in. The diagonal divides the trapezoid into two triangles that have a ratio 1.771. Find the bases and the area. 15. The two bases of a rhomboid are 24 ft. and 15 ft.; the altitudes differ by 3 ft. Find the area. 16. The two bases of a rhomboid are 28.3 ft. and 17.2 ft.; the altitudes differ by 11.5 ft. Find the area. 80 MEASUREMENT FORMULAS 17. The two altitudes of a rhomboid are 7.37 ft. and 5.08 ft., and the sides differ by 5.84 ft. Find the sides and the area. 18. The two altitudes of a rhomboid are .0394 in. and .0288 in., and the sides differ by .0113 in. Find the sides and the area. 19. The altitudes of a triangle are 8", 10", and 12"; the longest and shortest sides differ by 5". Find the three sides and the area. 20. The altitudes of a triangle are 1.24", 1.89", and 2.81"; the longest and shortest sides differ by 3.15". Find the three sides and the area. Implicit Formulas 85. A formula which does not express directly the value of the letter sought, but gives an equation from which that value can be determined, is called an implicit formula. A formula which gives the value of the required letter directly upon substitution is an explicit formula. Thus, the formula is an explicit formula for *S, and an implicit formula for p, 6i, and 62. 86. Model K. — In the formula S = ^ p (61 + 62), find p when S = 29.4, 61 = 2.36, h = 3.52. © 29.4 = I p (2.36 + 3.52) @ 29 . 4 = ^ p (5 . 88) ® same values ® 29.4 = p (2.94) same values 10 = 2? ®^2.94 EXERCISE 26 1. In the formula S = i p (61 + 62), find 61 when S = 690, p = 15, 62 = 55. 2. In the formula S = irr^ find r when S = 600. IMPLICIT FORMULAS 81 3. In the formula ^ = J p (61 + 62), find p when S = 23.8, 61 = 3.57, 62 = 4.83. 4. In the formula c = 2 ttt, find r when c = 1000. 6. The area of a trapezoid is 1.132 sq. ft.; one of the bases is 7.321 in. and the altitude is 19.36 in. Find the other base. Ushig the values given, evaluate the following formulas: 6. n(4:L-\-d) d = 5; L = 30; n = 255. 7. i^^ L = 6; a = 10; tt = 3.1416. 8. Tr(R + r){R-r) R = 25; r = 20; tt = 3.14. 2pP P + P 9. ^^ p = 10; P = 12. Rr {R -^ r) {m — 1) ' -, 11. 2 dnm d = 17326; n = 1440; m = 12.6. 12. / - ^ X ^ / = .372; d = 14.4; /i = 29. cR' 13. ^ Z;^, c = 3; R' = 100; i2 = 21. 14. j4- Q = 110; r = 5; X = 3.1416. ^= 15. '— ^ a = 10; 6 = 17; c = 21. 16. \{-^— + -^) a = .002; 6 = .015; n = 6. 2\n — a n— 5/ ' 17. -(^ - 1 j 7^ = 8; i2 = .75; t = 10. 18. ^^ + ^'^ + ^'^ R = 100; r = 120; R' = 1000. 19. j-^ a = 15; 6 = 28; c = 41; K = 126. 82 MEASUREMENT FORMULAS fi (W' — w') TF' = 1390. W{h-2ti + t) 21- wr/.-9/,4-/^ P = 1244; TT = 2; ^ = 14.5; ^1 = 11.5; t = 10. 22. J(!r-g)-s(r-0-5(5-r) s = 5; S = 1; W =500; w; =240; ^ = 0; T = 44; ^ = 6; r = 5.5. 23- 1 . ^^ c = .002; ^ = 20; H = 60; /i = 3; d = 13.5; D = 1.5. 24. ^(gly) ^ = 243; r = 16; ^ = 12; Q = 96; Tf = 125. =». 2,,. „..(._ '_+..) F = 1000; w = 25.6; T = 100; g = 20; < = 4; s = 5. CHAPTER IV ADDITIONAL TYPES OF PROBLEMS 87. An algebraic expression that forms one entire member of an equation may have its value changed, provided the ex- pression that forms the other member of the equation has its value changed in the same way. For example, if any two numbers are equal, we can in- crease them both by 5, 10, or 17, or any other number; or we can multiply them both by any number, or in any other way apply the axioms of Sections 36, 37 (page 41). The results will be different from the original numbers in value, but the equations obtained will be true, because the new numbers will be equal. In short, to reduce an equation, we may change the value of the two equal numbers, if only we keep them equal; but in reducing an expression which is not a member of an equa- tion, we are not permitted to change its value. Fractional Equations 88. When equations arise that contain fractions, it is necessary to multiply both sides of the equation by numbers which will change the fractions to whole numbers, as in the following examples. 89. Model A. — Solve the equation 3 = 2. ® 1 = 2 X = 6 © X 3 83 84 ADDITIONAL TYPES OF PROBLEMS 90. Model B. — Solve the equation ^ = ?• ^35 © 2y-l © X3 © 10 2/ = G © X 5 © 2/=. 9 © -MO Check: f X T% = f One line of the work in Model B could have been saved if we had multiplied both sides of © by 15 at once, instead of first by 3 and then by 5. T* O* T* 91. Model C. — Solve the equation o ~ o = 7 " 1 2* © X X X 1 2 3 " 4 2 © 2x X © X 2 © 3x-2x=^-3 © X 3 © .-H-. © same values © 2x = 3a;-6 © X 2 © 2 re + 6 = 3x © + 6 © 6 = x © -2a; Here if we had multiplied © by 12, the denominators, we should have had the least common multiple of © X X X \ 2 3 ~ 4 2 © 6x - 4a; = 3a; - 6 © X 12 © 2a; = 3a; - 6 © same values © 2a; + 6 = 3a; © + 6 © 6 = a; © -2a; Check: f - 3- 2 = |-i FRACTIONAL EQUATIONS 85 EXERCISE 27 Find the value of x in the following equations : X . X , ^ . 7a;— 6 , x Zx—b , ^ x ^ x ^ \ ^ 3a; — 7 Zx x — 5 ^•2 + 3 + 2 = ^-^ '°- —r~-ii=^- ^•§--7=6 + ^ 11. ^-g=2a.-10 X , X 1 X -\-2 xH-l,a; + 3 ^•4 + 6-3=^T. la. ^-+^ = x-2 xaja;^ x + 3,a; + 5,^-, "•2-5=2-3 13. -^ + ^-+5=20; 6.^-^+17=0 ,4.2^-^ = 2-. _2 ic_5a; — 8 a; + 5 a: — 5_3x "^^ 3 "^ 6 ~ 36 ^^' ~S~ "^ ~2~ ~ T 2a;— 1 1 -.a; ,3a;, 5a; — 2 _ , . 8.-2 3 = ^ + 6 16. « + ^ + -^-=2x + 4 17. 3x+^ = l^ + 2a; x-l,3x-l,l 4x-3,x-7 19. 3a; + 5 + ^— = — ^ — + 5x ^ .^ 3a;- 1 , 15a;- 3 , „ 20. 10x= — ~ 1 h7a;-l ^ , ^ x + 2 ^ 5a; -2 21. 3a; + 5 jr— =5x — ,^ 2 -3a; 9 - 15a; , ^ , ^ 22. 10a;= — ^ \-5-\-7x 86 ADDITIONAL TYPES OF PROBLEMS 92. Model D. — I bought a certain number of picture postcards at 2 for a nickel, and the same number at 3 for a nickeL I sold them all at a uniform price of 2 cents apiece, and by doing so, lost 25 cents. How many did I buy? Let X = the number of cards of each kind bought. Then 2 x = the number of cards of both kinds sold. -^ = the number of cents paid for cards at 3 for 5 cents. 5 X — = the number of cents paid for cards at 2 for 5 cents. 4 a; = the number of cents received for all. © ^ + ^ = 4x + 25 © 10 X + 15 X = 24 X + 150 © X 6 © 25 X = 24 X + 150 @ same values © X = 150 © - 24x Ans. 150 cards at each price. 150 at 2 for 5 $3.75 300 at 2 cents $6.00 150 at 3 for 5 2.50 Loss $ .25 $6.25 Check: EXERCISE 28 1. A merchant invested $3000 more in one shop than in another. In the first he lost | of the capital invested, and in the second the profits were 40 per cent of the capital in- vested. On both shops he made a net gain of $2080. How much did he invest in each shop? 2. Two lines have a ratio 7, and J of the larger line ex- ceeds double the smaller line by 12 inches. Find the lengths of the lines. 3. Two sides of a rectangle have a ratio f , and the perim- eter is 237.0 meters. Find the area. 4. Two angles have a ratio 2, and the smaller angle is three times the complement of their sum. Find the angles. 6. Two angles have a ratio 3, and double the smaller angle is the complement of } of the larger. Find the angles. 6. A, B, and C shared a sum of money so that A had $30 FRACTIONAL EQUATIONS 87 more than half, B had $33 more than a fourth, and C had the remainder, which was $16. Find the shares of A and B. 7. A had $107 and B had $45. B gave A a certain sum and then had only J as much as A. How much did B give A? 8. From a cask of wine, ^\ had leaked out, f had been drawn out, and there were 11 gallons left. How much wine was there at first? 9. A dealer bought pineapples at the rate of $4 for 7 dozen, and sold them at the rate of $2 for 3 dozen, gaining $4. How many pineapples did he buy? 10. I had a herd of sheep and sold a half interest in the herd, and a half interest in one sheep besides. Then I found that I owned 17 sheep more than J of the herd. How many sheep did I own at first? ' 11. A woman had a basket of eggs, and sold J of them, with i of an egg besides; what she had left were three more than half of what she had at first. How many eggs had she at first? 12. One angle exceeds half of another by 10°, and ex- ceeds the complement of their sum by 9°. Find the angles. 13. In a stripe If inches wide, two triangles are inscribed; their bases differ by 2.58 feet, and their areas have a ratio |. Find the areas. 14. Two men started together on a long-distance bicycle ride. At noon one of them had gone ^ of the distance, while the other had gone f of it; and they were then 30 miles apart. How far had they planned to ride? 15. A and B started eastward from Palmer. At noon A had gone f of the distance to Boston, B had gone ^ of it, and they were 2 miles apart. How far from Palmer to Boston? 88 ADDITIONAL TYPES OF PROBLEMS Subtracting Fractions and Parentheses 93. Model E. — In solving the equation 10 a; = — \-1x- 1 notice carefully that when we multiply by 14, the fraction 15 a; — 3 = becomes a whole number, which must be subtracted from the preceding term of the equation. 140a; - (30a: - 6) = 21 x - 7 + 98x - 14. Here we have 30 x — 6 to subtract from 140 x. If we had to take 30 a; from 140 a;, the remainder would be 110 a;; now, however, since we are taking 6 less than 30 a;, our remainder will be 6 more, or 110 x + 6. The equation is therefore solved as follows: Q 10. l^-^-^ = ^--l + 7. 1 © 140x - (30x - 6) = 21x -7 + 98. - 14© X 14 ® 110x + 6 = 119x - 21 @ same values llOx + 27 = 119 a; ® + 21 27 = 9 a; 0-llOa; 3 = a; ® -9 j5 /p _j_ 3 If the first member of the equation were 10 x = i we should apply similar reasoning. When we multiply by 14, this expression reduces to 140x- (30x + 6). To take away 30 x from 140 x gives a remainder of 110 x. But we are to take away 6 more than 30 x, so that the remainder will be 110 x — 6. Our conclusions may be expressed in the general state- ment: SUBTRACTING FRACTIONS AND PARENTHESES 89 94. In removing a parenthesis with the minus sign, all terms within the parenthesis are to be changed in sign. When we speak of "changing the signs," we mean chang- ing all the plus signs to minus, and all the minus signs to plus. When no sign is expressed before a term, the plus sign is understood. Terms containing the same radical factor and differing only by a numerical coefficient, such as 2 V3 and 5 V3, are similar terms. EXERCISE 29 Remove the parentheses and simplify: 1. 5x + 7- (2a: + 3) 7. 9 x - V^ - {VS - x) 2. 3 a; - 5 - (a: + 2) a. ^S - 9 x - (x - V3) 3. 17x + 13 - (3x - 5) 9. A -35- (90 - A) 4. 11 X - 3 - (2x - 1) 10. 198 - (§ C + 32) 5. 7x-2-(3x-4) 11. l-(x-l) 6. 9a; - 13 - (4 - 2x) 12. 180 - B - (90 - B) 13. 180 -{A+B)- (90 - A) 14. A 4 5 - 180 - (A- 90) 15. 180 - (A + B) - (90 -A) - (90 - C) 16. 360 - (A + 5 + C) - (180 - A - B) 17. V2 - \/3 - (2 VS - V2) 18. VIO - 3 V3 - (3 VIO - 5) 19. 3 - V2 - (5 V2 - V3) 20. a; - V2 - (\/2 - y) 23. 1.73 - (5 - a;) 21. 3 V2 - (1 - \/2) 24. 75 - (V2 - 11) 22. 3 V2 - (1 + V2) 25. 56 + 3 V2 - (13 - V3) 90 ADDITIONAL TYPES OF PROBLEMS EXERCISE 30 Find the value of x in each of the following equations: , x-1 x+S 2 3. 5,_^7^8U 10 2 ' 5 a. x+-^ = 3 g- X re — 5 a; — 8 g; — 4 X — 5 a; + 5 2 8 ~~^ 8. 6 8 4 x + 7 3a; + l _ 7x-ll . ®" 15 25 ~ 45 10. 5 X + 20 - 2 (a; + 2) = 40 - 4 (x - 15) 11. 10 (a^ + 1) - (3 a: + 5) = 30 - lOx + 2 (x + 35) 12. 30 (x + 6) - 10 (a; + 3) = 311 - 3 re ^13. 2 re + 3 (re + 9) = 200 - 4 (90 + 3 re) 14. 5 (re - 6) + 17 = 110 - 2 (re - 5) 15. 13 (re - 13) + 10 = 200 - 2 (75 - re) 16. 200 - 7 (60 - 5 re) = 3 (2 x - 11) - 100 IV. 50 re - 10 (re + 10) = 40 + re - 8 (2 - re) 18. 274 (2 X - 10) - 163 (8 + re) - 48 re = 19. 7re - 103 (re - 17) + 17 (3rc - 103) + 135 = THE CONSTRUCTION OF EQUATIONS 20. |-|(4x-25)=i(5x-61) 21. 2x-7-^-^ = 17-3^;3^ 22. ^, a;-M7 7a;- 2 , 8a; -13 ^^ 2 - 3 ' 9 23. 4a;-7 _ 3a;- 11 a;+17 5 4 6 24. '".""-(-^V)— '» 25. llx+19 Qx-5 6a;+l 6 3 ~ 6 26. 8X-3 4x-7^5^ ^3 27. 7a;- 11 6a; + 5 9a;-5 4 5 22 28. 2x + 7 3a; + 4 . 3 5 -^ ^ 29. a; + 3 2a;- 13 x ^ 5 11 ~ 3 91 The Construction of Equation 95. In constructing an equation for a problem stated in words, the first thing to do is to make a careful list of all the numbers that are referred to in the problem and are not actually given in figures. The next step is to pick out one of these numbers to repre- sent by X, and from that as a starting point to construct abbreviations for the other numbers in the list. When this has been done, it will be found either that (1) one statement in the problem has not been used in making the abbreviations; or that (2) there are two abbreviations for the same number. 92 ADDITIONAL TYPES OF PROBLEMS In the first case, the unused statement may be expressed as an equation. In the second case, the two abbreviations for the same number will form an equation. 96. For the purpose of illustrating the construction of equations, certain types of problems have been used in schoolbooks for many years. They are interesting as a sort of puzzle, but they have real advantages also for the mathematical student; not the least of these advantages being the practice they furnish in picking out separately the material facts in a narrative paragraph. In Exercises 31-34 the student is expected to construct the equations, but not necessarily to solve them. The Problem of the Digits 97. The problem of the digits is based on our notation for numbers. A number of two figures (or digits) is equal to ten times the first digit, plus the second. Thus the num- ber 57 (read it "five-seven," as in a telephone call) means (5 X 10) + 7, not 5 + 7, or 5 X 7. In the same way, if x were the first digit and 5 the second, the number would be 10 x + 5. The same digits inter- changed would give the number 50 + x. Interchanging the digits of a number may be brought about by adding another number; for example, 57 + 18 = 75. 98. Model F. — In a certain number of two digits, the first digit is double the second; and if 27 is subtracted from the number, the order of the digits is reversed. Find the number. Let X = the units figure. Then 2 re = the tens figure. The value of the number is 10 (2 x) + x. The value of the number with digits interchanged is 10 (x) + 2 x. THE CONSTRUCTION OF EQUATIONS 93 ® 20x + x-27 = 10x + 2x ® 21 X - 27 = 12 X © same values ® 21 X = 12 X + 27 + 27 ® 9x = 27 - 12x X = 3 0-9 Ans. 63. Check: 6 = 36 = 2X3 63-27 EXERCISE 31 1. In a certain number of two digits, the first digit is three times the second; and if 36 is subtracted from the number, the digits are interchanged. Find the number. 2. In a certain number of two digits, the second digit is three times the first; adding 54 interchanges the digits. Find the number. 3. In a certain number of two digits, the first digit is four times the second; subtracting 54 interchanges the digits. Find the number. 4. In a certain number of two digits, the first digit is twice the second; subtracting 36 interchanges the digits. Find the number. 5. The sum of the two digits of a certain number is 11, and if 27 is added to the number, the digits are interchanged. Find the number. 6. The first digit of a certain number less than 100 is 1 less than double the second digit; and if 18 is taken from the number, the digits are interchanged. Find the number. 7. The sum of the two digits of a number is 9; and if 45 is added to the number, the digits are interchanged. Find the number. 8. The sum of the two digits of a number is 11; and sub- tracting 45 interchanges the digits. Find the number. 94 ADDITIONAL TYPES OF PROBLEMS 9. The difference of the two digits of a number is one less than the unit's figure; and if 18 is taken from the number, the digits are interchanged. Find the number. The fact that subtraction interchanges the digits shows which digit is the less. 10. The smaller of the two digits of a number is 5 less than 3 times the larger digit; and if 9 is added to the number, the digits are interchanged. Find the number. 11. The difference of the two digits of a number is one less than 3 times the smaller digit; and if 45 is added to the number, the digits are interchanged. Find the number. 12. The sum of the digits of a number is 11 less than 3 times the larger digit; and if 27 is added to the number, the digits are interchanged. Find the number. 13. In a certain odd number which is a multiple of 5, the first of the three digits is 3 less than 4 times the second digit; and if 16 times the sum of the digits is subtracted from the number, the remainder is 3. Find the number. 14. In a number between 300 and 400, the last two digits differ by 4; and if the first two digits are interchanged, the number is increased by 90. Find the number. 15. A number between 100 and 200 -has the middle digit 1 greater than the last digit; if the number is divided by the sum of the digits, the quotient is 11. Find the number. 16. With two digits that differ by two, two numbers can be formed which have a ratio f . Find the numbers. The Problem of Two Velocities 99. The problem of two velocities is based upon the fact that a body may possess two motions at the same time. For example, a ball has two motions if it is rolled along a car floor while the train is in steady motion, for it has its THE CONSTRUCTION OF EQUATIONS 95 own rolling motion in addition to the motion of the train. The ball may go faster than the train, if it is rolled forward; or, if rolled toward the rear of the train, it may still be moving forward, in the direction opposite to the thrower's aim, but more slowly than the train moves. Thus if the train is running 44 feet per second westward, and the ball is rolled with a velocity of 5 feet per second towards the rear of the car, the ball is actually moving through the country westward, 39 feet per second; if it is rolled forward, its total westward speed is 49 feet per second. The problem of two velocities is a very simple illustration of an important principle of physics — the Composition of Motion. 100. The construction of equations under this problem depends upon the formula s = vt where s represents the number of miles traveled in t hours by a body moving at the rate of v miles per hour. Before beginning the following exercise, the student should invent numerical illustrations of this formula. For example : If I walk 3 miles an hour for 7J hours, I travel 22| miles. If a train goes 400 miles in 16 hours, its average speed is 25 miles per hour. If a steamboat keeps up its standard speed of 18 miles per hour, a trip of 192 miles will take it 10 hours and 40 minutes. 101. The model on page 96 will serve for further illustra- tion of the relation between velocity and distance covered. 96 ADDITIONAL TYPES OF PROBLEMS Model G. — The fore wheel of a carriage has a circum- ference of 8 feet, and the rear wheel of 12 feet. When the fore wheel has made 25 turns more than the rear wheel, how far has the carriage gone? Let X = number of turns by the rear wheel. Then a; -j- 25 = number of turns by fore wheel. ® 8 (x + 25) = 12 a; 8 X + 200 = 12 a; same values 200 = 4a; 0-8a; 50 = X 0^4 75 = X + 25 + 25 Since 12 feet is the circumference of the wheel which made 50 turns, the distance is 12 X 50 = 600 feet. Aws. 600 ft. Check: 12 X 50 = 8 X 75 = 600 EXERCISE 32 1. A man can walk 3 miles an hour, or ride 8 miles an hour. How long will it take him to go 132 miles, if he walks half the time and rides half the time? 2. A man can walk 4 miles an hour, or ride 7 miles an hour. How long will it take him to go 66 miles, if he rides half the time and walks half the time? 3. A man can walk 3.43 miles an hour on an average, and he can ride a bicycle at the rate of 11.1 miles an hour. How long will it take him to go 40 miles, if he walks half the time and rides half the time? 4. A street car travels 8.7 miles an hour, and a cab travels 6.2 miles an hour. A man goes 30 miles with time equally divided between these two modes of traveling. How long does it take him to make this trip? 6. The fore wheel of a carriage has a circumference of 5 feet, and the hind wheel of 8 feet. When the fore wheel has THE CONSTRUCTION OF EQUATIONS 97 made 20 turns more than the hind wheel, how far has the carriage gone? 6. The fore wheel of a carriage has a circumference of 6 feet, and the hind wheel of 10 feet. When the fore wheel has made 25 turns more than the hind wheel, how far has the carriage gone? 7. The fore wheel of a carriage makes 29 turns more than the hind wheel in going a certain distance. The circumfer- ence of the fore wheel is 10.8 feet, and that of the hind wheel 17.1 feet. Find the distance. 8. Two wheels are belted together, one being 9.72 feet in circumference, and the other 14.3 feet in circumference. How far has the belt traveled when one wheel has made 10 turns more than the other? 9. A naphtha launch, which averages 9 miles an hour when in running order, broke down when running out to sea. It had to be rowed back at the rate of | of a mile an hour, and reached the pier just 32 hours after starting. How far out to sea had the launch gone when it broke down? 10. A motor boat went downstream at the rate of 9.7 miles an hour, and up the same stream at 5.9 miles an hour. Ten hours were taken for the round trip. How far did the boat go up and down the stream? 11. A toboggan goes downhill at the rate of 20 miles an hour, and is hauled back at 2| miles an hour. If half an hour is taken for the trip down and back, how long is the coast? 12. A boy has an hour for exercise; how far may he ride with his father, at the rate of 10 miles an hour, before getting out to walk back? He can walk 3J miles an hour. 98 ADDITIONAL TYPES OF PROBLEMS 102. Model H. — A man who can row 5 miles per hour in still water finds that it takes him 5 hours to row up- stream to a point from which he can return in 4 hours. How fast does the current flow? Let X = the number of miles per hour the current flows. Then 5 + x = the number of miles per hour the man can row down- stream. 5 — X = the number of miles per hour the man can row upstream. 4 (5 + x) = number of miles in 4 hours downstream. 5 (5 — x) = number of miles in 5 hours upstream. © 5 (5 - a:) = 4 (5 + x) 25 — 5a; = 20 + 4a; © same values © 25 = 20 + 9a; ® + 5a; © 5 = 9 x © - 20 © ^=x © -9 Ans. River flows f mile an hour. Check: The man goes upstream (5 — |) miles per hour, in 5 hours 22| miles; downstream (5 + f) miles per hour, in 4 hours 22| miles. EXERCISE 33 1. A certain river flows 2 miles per hour. A fisherman finds that he can row upstream a few miles in 6 hours, but it takes him only 3 hours to come back. How fast does the fisherman row in still water? 2. A launch goes downstream for an hour, and then takes 3 hours to get back. If the stream flows 4 miles an hour, how fast can the launch go in still water? 3. The tide flows 6 miles an hour in a certain channel. A steamer makes the passage between two lighthouses in this channel in 4 hours against the full tide. With the tide, the steamer makes the same distance in 2 hours. How fast does the steamer go when the tide is not running? 4. A boy can row 4 miles an hour; it takes him 4 hours to row downstream and back on an errand, but only a quarter of the time was spent going down. How fast is the current? THE CONSTRUCTION OF EQUATIONS 99 5. It takes a fishing dory two hours to go down the harbor, with the tide in its favor; coming back, with the tide adverse, the dory took four hours. If the tide is not running, the dory's speed is 6 miles an hour. How fast does the tide run? 6. A man who can row 5 miles per hour finds that it takes him 3 times as long to go up a river as to go down. Find the speed of the current. 7. On a river which runs 1| miles an hour, it takes me twice as long to row upstream as down. How fast can I row in still water? 8. On a river which runs 2 miles per hour, a launch can go only § as fast upstream as down. How fast can it go on a still lake? 9. A man can row 4 miles per hour in still water; going upstream he goes only f as fast as he comes down. Find the rate of the current. 10. The speed of a ferryboat is observed going upstream, and again going downstream; the ratio of these two speeds is .7, and their difference is 6 miles an hour. What is the speed of the current? What is the speed of the boat in still water? 11. A man who can row 6 miles per hour in still water goes 1 mile farther in 3 hours coming downstream than he does in 4 hours going up. Find the speed of the current. 12. A man can row 3 miles farther in 2 hours downstream than he can in 4 hours upstream. The stream flows 2 miles per hour. How fast can the man row in still water? 13. A man wishes to row down the river to the steamboat landing and back in 4 hours. It takes him 2 hours to get down there; he starts back immediately, and at the end of the 4 hours he is 4 miles short of his starting-place. The man can row 3 miles per hour in still water. How fast does the river flow? How far off is the steamboat landing? J 100 ADDITIONAL TYPES OF PROBLEMS The Problem of Ability and Time 103. The problem of ability and time depends upon the fact that the greater the power employed for a task, the less time is required for it. If I can do a certain piece of work in 3 days, and you can do the same work in 2 days, both of us working together ought to be able to finish it in less time than either of us alone; it will not take 5 days then, but less than 2 days. The central facts here are that I can do J of the work in 1 day, and you can do J of it in 1 day; so that both of us are able to do | of it in 1 day. Hence, it is easy to see that the Work can be done in 1^ days by both of us together. 104. Model I. — A can do in 4 days a piece of work for which B would take 12 days. How long would it take them to do the work together? Let X = the number of days in which they can do the work together. Then in 1 day both can do - of the work and and U 11 <( A " '< 1 u 4 (( (( : 1 u 12 ti i( ® i + ^ = l © Xa; ® 3x + x = 12 ® 4 a; = 12 ® x = 3 4 + 12 = 1 X 12 d) same values ® -h4 Ans. Three days for both. Check: EXERCISE 34 1. A can do a piece of work in 15 days; B can do the same work in 18 days. How long will it take them to do it together? THE CONSTRUCTION OF EQUATIONS 101 2. A tank can be emptied by three taps; by the first alone in 80 minutes, by the second alone in 200 minutes, and by the third alone in 5 hours. How long will it take to empty the tank if all three taps are open? 3. A bathtub is filled in 40 minutes and emptied by the wastepipe in one hour. How long will it take to fill the tub with the wastepipe open? 4. A cistern can be filled in 12 minutes by one pipe alone, or in 8 minutes if the second pipe is also turned on. How long will it take to fill the cistern with the second pipe alone? 5. A can do in 2 J hours a job which B can do in If hours and C can do in 3i hours. How long will it take all of them together to complete the work? 6. A does f of a piece of work in 10 days; then he calls in B to help him, and they finish the work in 3 days. How long would B take to do the work by himself? 7. A can do a job by himself in 6 days, B can do it in 10 days, and both together with C to help them can do it in 2 f days. How long would it take C to do it alone? 8. A man can walk the length of a street in 9 minutes, and a boy can run that distance in 6 minutes. If they start at opposite ends of the street, how long will it be before they meet? 9. Two brothers wished, to get a certain sum of money, and agreed that each would earn half of it. One boy earned his half in 10 days, the other in 15 days. If, however, the boys had put their earnings together into one fund, in how many days would they have obtained the sum of money, assuming that each earned at the same rate as before? 10. A merchant can walk to his place of business in 40 minutes, or he can ride to it in 10 minutes. How long will it take him if he walks half the time and rides half the time? 102 ADDITIONAL TYPES OF PROBLEMS 11. Starting at noon, I can get to town on a wagon at 1:48 o'clock; if I walked, I could get there at 2: 15 o'clock. If I agree to walk half the time, at what time must I get off the wagon? 12. A train runs from Portland to Boston in 4| hours, and another train runs from Boston to Portland in 4J hours. If the two trains start at the same time, how long will it be before they pass each other? 13. In Ex. 12, if the trains leave Boston and Portland at 7:45 A.M. and 8:00 a.m., respectively, at what time do they CHAPTER V TRANSFORMATIONS 105. It is often necessary to change the form of a formula or other algebraic expression, or to perform some algebraic operation upon it. In order to be able to do such things intelligently, we must investigate the laws and the methods of addition, subtraction, multiplication, and division, in algebra, to find how they differ from the similar laws and methods of ordinary arithmetic, if at all. Negative Numbers 106. The expression a — h is the algebraic symbol for the result of subtracting h from a. But if h is greater than a, the subtraction is impossible. The question is, in that case, what does a — h represent? 107. To take a particular case: suppose I send to my grocer an order for 10 quarts of berries. If the grocer has a quarts in stock he will have a — 10 quarts left after filling my order. Let us suppose that his original stock was 7 quarts. He would send me those, and would still have an unfilled order for 3 quarts. The situation would not be the same as if he never had had any berries, or any orders for berries; for he now has an unfilled order for 3 quarts. To put these circumstances into arithmetical shape: It is required to subtract 10 from 7. The operation is impossible. But 7 can be taken from 7, leaving 0, and there is still 3 to be taken from any number that mxiy hereafter be 103 " 104 TRANSFORMATIONS added to the expression; just as the grocer is expected to fill his order from any goods that may hereafter come in. We express this condition by writing a minus sign before the figure that remains to be subtracted: 7 - 10 = -3 This means that 3 is to be subtracted frcnr pny number that may hereafter be added to the expression. If we add 5 to the expression in 0, we have ©7-10 + 5 3 + 5, or ® -3 + 5 = 2 If we add 3 to the expression in ©, we have ® 7 - 10 + 3 = -3 + 3, or (D -3 + 3 = 108. The expression —3 differs from all expressions known to arithmetic in this particular, namely, that you can get zero by adding to it. 109. When two numbers added together give zero, one is called the negative of the other; but generally by a negative number we mean a number preceded by the minus sign. That is, 3 is the negative of —3, just as much as —3 is the negative of 3; but if we are asked which of the two is nega- tive, we say —3. 110. Terms preceded by minus signs are called minus or negative terms ; and terms not preceded by minus signs are called plus or positive terms. It may be said that any positive term is the negative of the corresponding minus term; the apparent contradiction is explained when we remember the definition. Addition 111. When several expressions are to be added, we con- sider them as one expression and unite similar terms. In algebraic addition, it is not generally advisable to take the trouble to arrange similar terms under each other. ADDITION 105 EXERCISE 35 Add the following expressions: 1. 3 X + 2 + 4 X + 7; 3 X - 12; 1 - 10 X 2. 5x + 7x-\-ldx - 5x; 3 - 17 x-\-2 - Sx; X- 5; — X 3. S-]-2x + S + 9x-7-x-4:-3x;x + d-Sx; 2 X -\- 1; — x; — X 4. 3 + 5x+13x + 17-8a; + 5-10x; 4x + 2- 3x + 7; 34 - X 5. 171 X - 243 + 318 X - 411 X - 111 + 150; 38 x + 117; 301 - 100 x; 102* x - 299 6. 3(x-5); 5(x-5); 8 (5 - x) 7. 17 (2 X - 3); 11 (4 - 3 x); (x - 1) 3 -^ 8. 7 a + 341 + 2 a - 100 - 200 a - 200; 10 (a - 3); 3 (3 - a) 9. 3 V3 + 7; 2 - V3; 7 V3 - 5; 1 - V3 10. 4^3-6,17 -2 ^y^; -3- ^3; 5 ^3 + 1 11. 5 VIO + 3; 10 + 2 VIO; 5 - VlO; -3 VIO - 10 12. 2 (^3 - 7); 13 (5 - 2 v'3); 7 (5 V3 - 30) 13. 13 X — 4:y-\-z;4y — 10 X — 5 z;4:Z — 4:X + 4: y—4: 14. X + V2; X - 2; X + 3 - 4 V2; 3 - 2 X - 3 V2 15. 2 2/ + X - V3; 3 y-5 x + 2 V3; 3-5 V3-2 x-y 16. V3 - ^3; 2 V3 + 3; 3 ^3 - 5 - V3 17. 5 -v/2-2+2 V5; 3 V5-2 V2; 17 V2+17-2 V5 18. 4 V3 - 5 ^2; - 3 + 4 ^2 - 3 V3; 7; 20 - V2 19. X - V2; X - 2 + 3 \/2; 5 - a^2 20. - 131 X - 100 y - z;- (100 z - 100 x); 50 x+150 y 112. If we make no mistake in reducing the examples of Exercise 35, we get sums that are correct, whatever the values 106 TRANSFORMATIONS of the letters. 2x -{- Sx = 5x whether x stands for 1, for 1000, or for .001. If, now, we substitute 1 for each letter, each of the given expressions will reduce to a number. We may also substitute 1 in the same way in our answer to the example. This result should be equal to the sum of the others. This process enables us to check the correctness of our work. Thus, adding 300 a; + 17 2/ + 2 and 7 a;. + 10 i/ - 9, we obtain 307 a; + 27 2/ - 7. By substituting a; = l,2/ = l, we get 319 + 8 = 327. This result goes to show that the algebraic addition is correct. The symbols Vs and VlO stand for exact values, just as letters would. Consequently, for the purpose of checking our work, we may substitute 1 for Vs or any other root. 113. This process is called checking by coefficients, be- cause in substituting 1 for a symbol used as a factor in a term, we practically leave the symbol out of the term and find the sum of the coefficients in each expression. Checking by coefficients is not an absolute check, but it enables us to detect some possible mistakes. It is easy to apply, and the pupil should form the habit of using it. Subtraction 114. A minus sign preceding a parenthesis indicates that the quantity within the parenthesis is to be subtracted. We have already seen that we simplify an expression containing such a parenthesis by removing the parenthesis with its sign, changing the signs of the terms within the parenthesis, and then uniting similar terms in the entire expression. In subtraction the minuend is the number from which the subtrahend is taken to leave the remainder. From the foregoing it will be seen that the rule for sub- traction is : SUBTRACTION 107 Change all the signs in the subtrahend and add the resulting terms to the minuend. To save time in calculation, the signs of the subtrahend should be changed in your mind, and the terms should be united without writing the expressions again. EXERCISE 36 Perform the following subtractions: 1. From a;3 - 7 a;2 + 16 X - 12 take 3^ - dx" -\- 2x - 4S. 2. From a — h take a — h — c. 3. From X — y take y. 4. From h take — s. 5. From x3 + 9x2 + 2x-48 take a^ - 4:X^ - Sx -\- S. 6. Take a^-5x2-2x + 24 from s^ + 2x^ + 4:X + 3. 7. Subtract 24 -\- U x - 29 x'' + Q j^ from 2t' + 3x^ - nx- 12. 8. Subtract 2 - V3 from 6 + 5 V3. 9. Take a from a — b. 10. From 9p — 5g + 4r take 5q — 2p-jr2r. 11. From 3 - \/2 + 2 V3 take 1 + 3 \/2 - V3. 12. From V3 take' 5 - 2 V3. 13. Subtract \/3 — \/5 from V5 — \/3. 14. Subtract x^ + V3 from 3 \/3 - x. 15. Take 3 V^^ - a; V2 from 2 Vx + S V^- 16. 2 a - 2 6 + 3.C - d - (5 a - 3 6 + 4 c - 7 d) Note that the sign of the first term in a parenthesis is not the sign of the parenthesis. Thus, in Ex. 16, the sign of the parenthesis is — , but the sign of 5 a is +. When the parenthesis with its — sign is removed, the sign of 5 a is — . 17. x^ + 4:x^-2x^ + 7x-l- (x^ +2x^-2 x^ + Qx -1) 18. - (a2 - ax + x2) + 3 a2 - 2 aa; + a;2 108 TRANSFORMATIONS + 2X + 8) 20. 10 a^h-j-S ah^-S a^h^-V- (5 a^b-G aW-1 a^¥) 21. 6 x'y-Z xy^+7 y^-\-a^ -(8 x^y-3 xy^+9 y^+U x^) 22. a; + 1 + (5 - a:) - (3 + 14 x) 23. 2 a - 3 6 + (3 a - 2 6) - (2 a - 3 6) 24. a — 6 + c — (a + ?) — c) 26. 14 a + 27 6 - 13 - (7 a - 110 & - 17) 26. a; - V3 + (\/3 - \/2) - (3 V3 - 2 x) - 27. A - 90 + (180 -SA) - (A - B) 28. X V2 - 3 + (3 a; V2 + 5) - ( V2 - \/3) 29. 13 ^3 - (5 - 10 ^y^) - (13 V3 - \/S) 30. 3.286 - 1.873 x - (11.93 - 28.45 x) 31. 3 - V2 - (\/2 - \/3) + (5 - 3 V3) 32. 180 - 3/ - (f c + 32) +/ - c 33. I ci + 32 - (I C2 + 32) 34. 9 c + 100 - (5/ - 160) + (/ - c) 36. lSO-\- {A +B + C) - {C - A - B) Parentheses within Parentheses 115. Model A. —Simplify 3 x - 2V3 - [2 x - (SVS + X - 10)]. Sx -2V^ -[2x - (5V^ + x - 10)1 = 3 a: - 2 V3 - [2 X - 5 V3 - X + rO] = 3x-2V3-2x + 5V3 4-a;-10 = 2 X + 3 v'3 - 10 Ans. 116. To simplify an expression in which parentheses occur within parentheses, remove them one at a time, beginning with the inmost parenthesis. The vinculum placed over an expression, as 2 x + 1, means the same the parenthesis. The brace { j and the bracket [ ] are also used. MULTIPLICATION 109 EXERCISE 37 1. X - (y - z) + X {y - z)-\r y - (z + x) 2. [k-is-t)]Ms-{t-k)]-lt-{k-s)]-{k+s+t) 3. 2x- {Qy+ \4.z-2x\) - (Qx- \y + 2zl) 4. -iQx- {12y-4x)\ - {Qy- {4.x-7y)l 5. 3 - 2 V3 - [5 (2 - V3) - V3] 6. 10 + 5 VIO + [{x - 10) - 3(V10 - 5)] 7. V2 (3 - V3) - 2 S3 - V3 (2 - V2) - V6! 8. 3.28 (1.403-7.86 x)-5.63 (2.87 x-[3.52x- 1.943]) 9. x{5- V3) - {5- (x + xV^)l ''''l{l-^)-^\l-{l'l)\ Where the answers to the following equations are not whole numbers, express them in decimal fractions. 11. x + 3 - (2x - 17) =4 ' 12. 2 X - 7 - (x + 2) = 2 X - (2 X + 9) ^13. 3x + 2-(2x-5) + (9-7x) =x+(2-5a;) ~14. 13 - (5 - 2 x) - (1 - 7 x) = 25 - (2 + 7 x)' 15. 300 - 2 X - (115 - 3 x) = 1000 - (3 x - 61) . • 16. 100 X - (39 + 8 x) - 10 = 12 X + 345 17. X - (3 X - 2x+ 1) = 8 - (3 x + 7) ^ 18. 12 X - 3 (x - 2) = 20 - (5 + 9 - 7 x) 19. 100 - (2 - 15 X - 23) = 98 - (20 X - 24 - 5 x) 20. 25 X - 10 (1 + x) = 3 - (x - 5x - 7) + 8 x . Multiplication 117. The multiplier and the multiplicand are called the factors of the product. In continued multiplication there are more than two factors. For example, 3 X 5 = 15; 15 X 7 = 105; 105 X 11 = 1155. 3 and 5 are the factors of 15; 15 and 7, or 3 and 5 and 7 are the fac- tors of 105; 3, 5, 7, and 11 are the factors of 1155. 110 TRANSFORMATIONS The factors of any product may be written in any order. For example, 3 X 2 X 4 = 2 X 3 X 4 and 5 a6 = 5 6a. 118. When two or more of the factors of a number are aUke, it is sometimes convenient to indicate the fact as fol- lows: 2X3X3X3=2X33=54;2 aaahbhh = 2 a%\ The small figure denoting the number of equal factors is called an index or exponent. 119. The product of equal factors is called a power. One of the equal factors of a power is called a root. For instance, 27 is the third power of 3. 2 is the fifth root of 32. 64 is a sixth power, 2 being the root, and 6 the index; 64 is also a third power, 4 being the root, and 3 the index; and 64 is a second power, 8 being the root, and 2 the index. 120. The second power is generally called the square, and the second root is called the square root. 7 is the square root of 49 (7 = V49). The third power is called the cube, and the third root is called the cube root. 5 is the cube root of 125 (5 = vT^). The expressions a^, ¥, c*, d^ are read respectively "a square," "h cube" (or third), "c fourth," "d fifth." p^ is read "p kth" or ''p to the kth power." Note the difference in meaning between a^, a?, a^, and 02, 03, ai. The latter symbols are read "a two, a three, a four." 121. The degree of a term is the number of letters that are factors of it. Thus 7 x^y^ is a term of the fifth degree. The degree of an expression is the degree of its highest term. 122. Model B. — Multiply 5 a^y by 3 xY^^. ■ The factors of the product are 5 xxxy X 3 xxxxxyyyyzz. These can be rearranged so as to bring the numerical factors together, and bring all the like letters together: 5x3 xxxxxxxx yyyyy zz which can be simplified thus: 15 ci^i^z'^. MULTIPLICATION 111 123. It is evident that there may be terms that use the same letters as factors, such as 5 x^y and 5 xy^j and yet are not similar; but notice that these terms 5 xxy and 5 xyy have different sets of letters. In order to be perfectly clear, let us define similar terms as terms which have the same sets of letters. EXERCISE 38 Multiply: 1. ZxX^y 6. 'Sa^b^ X4a362 f2. dxy X7xy 7. 8 a^c X 5 a^hc^ 3. 3 ahc X ac &. 12 ab'^d X 15 a%c^ 4. a^ X a2 9. 7 a^c^ X 4 a%(^ 5. a^^ X a lo. a^ X 3 a^ Multiply together the following expressions: 11. x^^y;x^y^ 14. 17 Vx; 2 y 17. 101 a;^; 2 a;ioi; 5 a: 12. rcioy^S'rrV^ 15. x^•2/^•X2/ is- na^h;Sa¥;5a^b^ 13. 2Sa^y;5x^y^ le. Soi^;5y^;oi^;y^ 19. llp";7g^;2pg 20. 7 a;7; 2 x^; 2 ^/^ 124. Checking by coefficients is of little use in multiplica- tion examples such as these in the preceding exercise; the' fact that in each example the degree of the result should be the sum of the degrees of the several factors may be used as a partial check. 125. When an expression is separated into two parts by a plus or a minus sign, it is called a binomial. When an expression is separated into three parts by plus or minus signs, it is called a trinomiaL When an expression is separated into two or more parts by plus or minus signs, it is called a polynomial. When an expression is not separated into parts by plus or minus signs, it is called a monomiaL 1 12 TRANSFORMATIONS The expressions multiplied in the preceding Exercise are all monomials; the multiphcation of binomials and of other polynomials depends upon the multiplication of monomials. 126. Model C. — Multiply 5 a^ by (3 a; - 2 a). 5 a2 (3 re - 2 a) = 5 a2 X 3 X - 5 a2 X 2 a = 15 a^x - 10 a' EXERCISE 39 Multiply: 1. 7 X (a; - 1) 6. 2 (x - 5) 2. 3 a2 (a - x) 7. ih (4 k'^W - 2 h'^j') 3. 10 h^k (2/1-5 /b2) 8. 101 h^k"- (11 h''k'-21 h^k'') 4. 11 p'q (3 p^q - 3 p^) 9. 38 hij (23 hji + 12 ijh) 6. 7 X2/2 (2 X2/7 + 4 x^i/) 10. 10 ax (100 6a; + 1000 ex) Multiply together the following quantities: 11. 3; X - 5; 2 13. 10x^;2y^;ixy^ + i xhj 12. 3 a (a - 6); 2 a;; 3 ax 14. 3 a^h; a^ -b^; 2ab 15. 15 pq^; 9 qr^; ^\ ph — ^V V^ 16. 7 ap2. 4 ct2pj 15 5gr2. ^ly ap — ^ly ft^f The Distributive Law 127. The pupil has noticed that where a polynomial is multiplied by any number, each term is multiplied by that number; that is, 3 (a; — 5) = 3 x — 15. 128. This principle may be illustrated again as follows: A shopkeeper sends every week x dollars to the bank; his messenger uses every week $1 for necessary expenses; his wife draws out y dollars for her personal use; his son remits every week z dollars, which is added to the shopkeeper's account. The increase of the shopkeeper's account each week, then, amounts to x — 1 — y -\- z. In 5 weeks the in- crease would be 5 (a; — 1 — 2/ + 2!). But in 5 weeks the MULTIPLICATION 113 shopkeeper would send 5 x, the messenger would use up 5, the wife would draw out 5 y, and the son would remit 5 z. Therefore b {x — 1 — y -\- z) = bx — ^ — by -\- bz 129. This Distributive Law, as it is called, is one of the important and fundamental principles of algebra. It applies also to minus signs; e.g., — {x — z) = —x + z; and — (x — y-\-z — a — h-\-c) = — x-\-y — z-\-a + h — c. It may be stated as follows: The product of a polynomial by a single term is found hy multiplying each term of the polynomial successively. The negative of a polynomial is found by taking the negative of each term successively. Distributive Factoring 130. A glance will show when a monomial is a factor of every term of an expression; such an expression can there- fore readily be separated into factors. Thus in the expression 5 a2 - 15 a6 + 20 a3 - 5 a 5 a is a factor of every term, and the whole expression is the product of 5 a (a - 3 6 + 4 a2 - 1) EXERCISE 40 Find the factors of the following you cannot: 1. ax -{• ay 5. a^ — ab expressions, or tell why 9. 13 x + 91 2/2 2. ax — ay e. x^ + x^^y lo. x'^y — xy^ 3. 5x+10i/ 7. a; + rr2 n. 7x+49x2+343x3 4. 3 a; - 15 ?/ ^. 5 x + 25 a^ 12. 3 a:+6 xy-\-2 yz 13. 13 x3 _^ 65 3.2^ 4. 117 3-2^2 14. 343 7?y + 98 xV + 28 xf + 8 2/^ 114 TRANSFORMATIONS 15. I5x^ + 9xy + 25 y"- 16. 38 x3 + 57 x^y + 19 xy"^ 17. x^o + 10 x^y + 45 a;V + 120 xY + 210 xY 18. 45 a^^i/s - 120 xy + 210 xV - 270 x^ 19. 71^ — n + n^ — n^ 20. rs — 2 r^s + rsP' + r^s^ EXERCISE 41 Substitute : 1. y = X -\- 2 in the expression 3 x + 4 ?/ — 25 2. 2 = 2 ?/ — 3 in the expression 3 x + 3 2; — 21 3. z = S — y in the expression z -\- 10 — 5y 4. ^=2x + 3in the expression x^ + 2 xy + ^/^ 5. X = y — 1 in the expression 3 x + 2 x?/ + 2 x^ Multiplication of Polynomials 131. In multiplying a polynomial by a polynomial, the multiplicand, considered as one quantity, is multiplied by each term of the multiplier; then these partial products are added. The work is arranged as follows: Model D. — Multiply 2a-75by5aH-3&. 2a -7h 5a + 3b 10 a2 - 35 ab + 6 ab - 21 b^ 10 a2 - 29 a& - 21 62 132 Model E. — Multiply 2x-5by3x-4. 2x -5 3 a; - 4 6 x2 - 15 x -Sx +20 6 a;2 - 23 x + 20 The second term of the multiplier gives a product which is to be sub- tracted, and we change the signs of the terms as we write them down; that is, for -4 (2 X - 5) we write -8 x + 20. MULTIPLICATION 115 A very good check for multiplication is to multiply the multiplier by the multiplicand; this gives a different set of partial products. 133. Model E illustrates all the combinations of signs that are possible in algebraic multiplication. It is worth while to notice, therefore, that when the multiplicand-term and the multiplier-term have. like signs, the product-term is plus; when the multiplicand-term and the multiplier-term have unlike signs, the product-term is minus. These rules, are nothing more than our old ways of dealing with positive and negative parentheses; they are a more convenient way of saying that when the multiplier-term is plus we do not change the signs of the multiplicand, and when it is minus we change them all. Both rules are generally condensed into the following: In multiplication, like signs give plus, and unlike dgns minus. The same rule holds for division, as will hereafter be shown. 134. For further illustration of this important rule, we may represent the multiplication of {x — 3) (x — 2) by a diagram. Here x^ is represented by the entire square; (x — 3)(x — 2) by the rectangle X; 2 x by the -j- rectangle A + Q; 3 x by the rectangle B + Q; \ and 6 by the rectangle Q. i To find the rectangle X, which represents T {x - 3) (x - 2), from x^ we subtract (A + Q), T which represents 2 x, and then (B + Q), which f represents 3 x. J^ But we have thus had to subtract Q twice; and to do this, we must have added Q after one of the subtractions. We may express the work thus: (x-'3)(x-2)=x^-(A+Q)+Q-iB + Q) a;2-2x + 6-3x X 116 TRANSFORMATIONS EXERCISE 42 Multiply: 1. (x + 3)(x + 2) 9. (11 - 5 a) (5 - 2 a) 2. (x + 3) (x - 2) -^10. (101 - 17 x) (11 + 3x) 3. (x + 5) (x - 3) 11. (x + 1.32) (x - 5) 4. (x - 5) (x - 3) 12. (p - 7.36) (p + 1.11) 5. (2 X + 7) (x - 3) 13. (A - 105) (5 - A) 6. (2 X - 7) (3 X - 1) 14. iX-{-nx){X -2x) 7. (1 - x) (2 - X) 15. (A - 5 a) (A + a) 8. (5 - a) (2 + a) 16. (10 + 5 x) (11 + 2 x) 17. (2.75 X- ■f 1.22) (5.08 x + 1.55) 18. (10.33 X + 10.12 y) (1.073 X - 2.224 y) 19. (7.86 X + 8.44) (8.44 x + 7.86) 20. (8 - x) (5 - 2/) 25. (28 a + 5 6) (28 a - 5 6) 21. (x + 1.732) (x - 1.732) 26. (70 p - 1) (70 p - 1) 22. (a - 72) (a - 101) 27. (1 - .037 x) (1 - .037 x) 23. {a - 72) (a - 72) 28. (7 - 12 xY 24. (a - 101) (a - 101) 29. (108 a + 102 hy , 30. (1.11 X - 2.03 yy Division 135. In division, we are given one factor of a number to find the other. The given factor is the divisor, the required factor is the quotient, and their product is the dividend. The signs of the separate terms of the quotient must be such that, when those terms are multiplied by the terms of the divisor, they produce the signs given in the dividend. Express the four possible combinations of signs for sepa- rate terms in tabular form, and show that from the table one may derive the rule: In division^ like signs give plus, and unlike signs minus. jm- EXERCISE 43 Divide: 1. 14x2by2aj 9. -1.032x2 by - 1.110 X 2. 24a;5by4a;2 10. -783x2 by 524 X 3. 24 x^ by 4x2 11. 7.77 X by -1.23 X 4. 9x9 by 3x3 12. 2.346 by -3.132 5. 1056 x2 by 11 X 13. 4.836 X by 3.142 6. 28x2 by -28 X 14. -2.718x2 by 3.142x2 7. 115 by -5 15. 5.43 A by -3.45 8. -708 a by 2 a 16. 7.08x3 by -3.22 X 17. .67 x^ by .37 X Long Division 136. Model F. — Divide 6x2 + lo - 19a; by 3x - 2. The correspondence of multiplication and division may be seen as follows. The arrangement here shown for division is recommended as the most compact. ; Divisor 3x -2 Multiplicand 3a: -2 ' Quotient 2x -5 Multiplier 2x -5 Dividend 6a:2 -19 a: + 10 1st partial prod. 6a:2 - 4a: 1st subtrahend 6x2 -4 a: 2d partial prod. -15x + 10 -15 a: + 10 Entire product 6 x^ -19 a: + 10 2d subtrahend -15 a: + 10 Check: (-3) -(1) = -3 A very good check for division is to multiply the quotient by the divisor; this gives a set of partial products different from the subtrahends obtained in dividing. EXERCISE 44 Divide: 1. x2 - 17x + 30by X - 2 2. 22 + 9/i-/i2by 11 -/i 1 18 TRANSFORMATIONS 3. 9 2/2 _ 49 by 3 2/ + 7 4. 30 s2 + 65 s + 10 by 6 s + 1 6. 60 s2 + 92st-\-S5t^hyl0s + 7t 6. x^ - 1000 X + 999 by a; - 1 7. 240 A2 + 214 A - 63 by 8 A - 9 8. 144 Q2 _ 295 QX + 50 X2 by 9 Q - 10 X 9. 504 m2 - 180 r2 - 1049 mr by 56 m + 9 r 10. 9600 a" - 2541 f + 7436 at by 300 a - 77 t 11. 3 m3 - 11 m2 - 4 m + 30 by m - 3 ^12. 9 2/3 + 3 2/2-22/ + 30by32/ + 5 13. 12 2^ _ 11 ^3 _ 4 ^2 _|_ 25 - 14 by 3 2; - 2 . 14. x'-6x^-\-5x''-x-12hyx''-2x + S >i&. 10 x4 + 29 x3 - 22 a;2 + 17 X - 2 by 5 a;2 - 3 X + 2 16. 7 a:^ + 9 a^ - 37 a;2 + 41 a; - 40 by 7 ^2 - 5 a; + 8 CHAPTER VI IDENTITIES AND THEOREMS; FACTORING 137. Compare the following equations: I. dx-\- 5 ==2x + 7. II. 3 (x + 5) = 3 X + 15. Notice that while the first equation is true only for the particular value x = 2, the second is true for any value that may be chosen. Again, notice that the expression 3 a; + 5 cannot, by any means we know of, be transformed into 2 x + 7; while 3 (x + 5) is in a simple way transformed into Sx -\- 15. 138. These two kinds of equations have distinct names. An equation which is true only on condition that the let- ters in it have particular values is called an equation of condition; while an equation which is true for any values whatever of the letters in it is called an identical equation, an equation of identity, or simply an identity. Equations of condition are the equations ordinarily met with in solving problems. Identical equations may be recognized by the fact that one member can be transformed to the identical form of the other member. The two members are said to be identically equal, and sometimes the sign = is used instead of = be- tween the members of the equation. The Proof of Theorems 139. A theorem is a general statement requiring proof. In algebra, a theorem is often proved by changing the statement to an equation, and showing that one member can be trans- 119 120 IDENTITIES AND THEOREMS; FACTORING formed so as to become exactly like the other member; that is, by showing the equation to be identical. 140. Theorem A. // the sum of any two numbers is multi- plied by their difference, the product is the difference of their squares. Model A. — Proof. Let a and b represent any two numbers. Then the theorem is expressed by the following identity; (a + 6) (a - 6) = a2 - b^ The first member is transformed into the second by a simple multi- plication. 141. Prove the following theorems i 1. The square of the sum of any two numbers is equal to the square of the first number, plus twice the product of the two, plus the square of the second. 2. The square of the difference of any two numbers is equal to the square of the first number, minus twice the product of the two, plus the square of the second. EXERCISE 46 By means of these theorems write the following products at sight, without performing the multiplications: 1. (a + 3)2 3. (a + 132.1)2 5. (a - 7)^ 2. (a + 7)2 4. (a - 3)2 e. (a - 132.1)2 7. (a + 1) (a - 1) 10. (a + 38.7) (a - 38.7) 8. {a + 3) (a - 3) ii. {a + xY 9. (a + 7) (a - 7) 12. (a - xf By means of the same theorems write quotients for the following, without performing the divisions: 13. (a2 - 16) -T- (a - 4) 14. (a2 - 121) -^ (a + 11) QUADRATIC PRODUCTS 121 15. (a2~ 10 a + 25) -^ (a - 5) 16. (a2 + 10 a + 25) -r- (a + 5) 17. (a2 + 100 a + 2500) ^ (a + 50) 18. (a2 - 3600) ^ (a - 60) By means of the same theorems write the factors of the following products: 19. x2 - a2 25. A2 - 81 J52 20. 49 - x2 • 26. 49 a2 - 4 A^ 21. 4 - 4x + a;2 27. B2 - 62 22. a;2 - 20 x-\- 100 28. p2 _ 28 p + 196 23. x' + SOx^ 1600 29. 256 + 32 n + n^ 24. a;2 - 120 a; + 3600 so. 1 - 2 a + a^ By Theorem A find the difference of the squares of each of the following pairs of numbers: 31. 3 and 73 34. 339 and 319 37. 8133 and 8131 32. 9 and 109 35. 1723 and 277 38. 2731 and 269 33. 575 and 425 36. 121 and 120 39. 101 and 99 40. 10001 and 1 Quadratic Products 142. The study of the product of two factors of the first degree, like (x — 5) (a? + 2), leads to certain methods of factoring that are of the greatest use in the solution of an im- portant kind of algebraic equation. An expression in which there is a term of the second de- gree, but no higher term, is called a quadratic expression. Consequently products like {x — 5) (x + 2) are called quadratic products. Equations in which they occur are called quadratic equations. 122 IDENTITIES AND THEOREMS; FACTORING - \ 10 143. Model B. — The diagrams on this page are recom- mended for quadratic products, because by their use one gets a better idea of the way in which such products are formed. The products of terms directly under each other are called straight products. The products of terms diagonally opposite each other are called cross products. The straight products are not similar, and so cannot he united; the cross products are similar, and can he united. In this example, the straight products are x^ and — 10. The cross products are — 5x and + 2 X, and their sum is — S X. The entire product is x2 - 3 X - 10. 144. The same method can be applied to the multiplica- tion of numbers of two fig- ures. For example, 73 may be expressed as 7 i + 3 and 78 as 7 i + 8, where t stands for ten. Then the straight products would be 49 f and 24, and the cross products 21 t and 56 t, with the sum 77 t. Since t = 10, the straight products give 4924, and the cross products 770, the sum being 5694. EXERCISE 46 Multiply the following numbers by the method of Section 144: 1. 34X37 3. 45X25 5. 36X45 7. 83X87 9. 28X73 2. 58X52 4. 38X58 6. 35X65 8. 34X51 lo. 39X96 QUADRATIC PRQDUCTS 123 In the following examples, name the straight products, the cross products, the sum of the cross products, and the entire •odu 11. ct: (x -\-2){x- 7) 26. (11 -h){2 + h) i 12. (a + 5) (a + 9) 27. (4 6 - 1) (2 6 - 3) 13. (a + 3) (a - 10) 28. (2x + S){2x + 3) 14. (b + 7) (6 - 6) 29. (31/ -7) (3 2/ + 7) 15. {h-\-S)(h + S) 30. (6 s + 1) (5 s + 10) 16. (h - 11) {h + 2) 31. (x + 2y){x-7y) 17. {k -l){k- 4) 32. (a + 5 6) (a + 9 &) 18. (x + 10) (x + 10) 33. (3a + 26) (5a + 26) 19. (y - 3) (2/ + 3) 34. (c + x) (3 c + 2 a;) 20. (s + 1) (s + 99) 35. (3 /b - 4 ^) (5 /b + sf) 21. (2 a; + 1) (7 a; - 1) 36. (lOx-2/i) (3x+5/i) 22. (5 a + 1) (9 a + 1) 37. (46-3/b)(36-2fc) 23. (3a + 2) (5a -2) 38. (3:c + 52/)(3x + 5i/) 24. (c + 1) (2 c + 3) 39. (3 2/-ll2)(3 2/+ll^) 25. (/b - 7) (5 fc + 1) 40. (6 s + 5 (10 s - 7 Factoring by Cross Multiplication 145. Model C. — In the product x^ - Q x -\- U, the straight products are x^ and 14, and the sum of the cross products is —9x. The terms that give *^X^ the straight products must have like signs, ^^ and therefore the cross products must have like signs; both cross products, then, are minus. To get the first straight product, we have xXx; to get the second, 2X7; then the signs must be chosen so that both cross products are minus; that gives us the factors (x — 2) (x — 7). >«^^^ ^^.>^ 124 IDENTITIES AND THEOREJ^S; FACTORING 146. or 4 and 6. be negative. Model D. — In the product x^ — 5 x — 24, the straight products are x^ and — 24; — 5 x is the sum of the cross products. The terms giving x^ for a product had hke signs and the terms giving —24 for a product had unUke signs; then the two cross products had unlike signs, and the minus cross product was the greater. To get x^ for a straight product, we should have to multiply x and x. To get 24, we could multiply 1 and 24, 2 and 12, 3 and 8, In each case, the larger number would have to -24 Trjring each in succession, we find that the pair of factors {x + 3) (x — 8) gives the right cross products, so we need not try farther. EXERCISE 47 Name the straight products and the sum of the cross prod- ucts in the following expressions, and find their factors: 1. x2 + 5x + 6 6. x^ + 5x + 4 2. x2 - 2 X + 1 7. x2 + 7 X + 12 3. x2 - 3 X + 2 8. x2 - 7 X + 6 4. x2 - 4 X -f 3 9. x2 + 7 X + 10 5. x2 + 4 a; + 4 lo. a?' - 8 x + 15 QUADRATIC PRODUCTS 125 11. x^ -Sx-{-7 41. a;2 - 4 a; - 60 12. x2 + 8 a; + 15 42. a;2 + 5 X - 84 13. a;2 - 20 a; + 19 43. x^ - 5 x - 150 14. x2 - 20 X + 64 44. a;2 + 6 X - 40 15. a;2 + 20 a; + 36 45. a;^ - 6 x - 91 16. a;2 + 20 a; + 51 46. x^ - 7 x - 30 17. a;2 - 20 X + 75 47. a;2 - 6 X - 55 18. a;2 + 20 x + 84 48. a;^ + 7 a; - 78 19. a;2 - 14 a; + 49 49. a;^ + 7 x - 170 20. x2 - 14 X + 13 50. a;2 - 8 x --- 20 21. a;2 - 14 X + 33 6i. x^ - 9 x - 10 22. x2 - 14 X + 24 62. x2 + 9 X - 22 63. x2 - 10 X - 200 64. x2 + 10 X - 75 65. x2 + 10 X - 96 56. X2 - X - 110 67. x2 + 2 X - 120 68. x2 - 3 X - 130 59. x2 + X - 132 60. x2 - 4 X - 165 61. 2x2 - X - 1 62. 3 x2 - 2 X - 1 63. 3 x2 + 4 X + 1 64. 2x2 + 7x + 6 65. 2 x2 - 5 X + 3 66. 2 x2 - X - 3 67. 3 x2 - X - 10 38. x2 - 4 X - 32 68. 2 x2 + 3 X - 9 39. x2 + 3 X - 70 69. 5 x2 - 9 X - 2 40. x2 _|_ 4 a; _ gg 70. 6 x2 + X - 15 23. x2 + 14 X + 45 24. x2 + 14 X + 48 25. x2 - 8 X + 16 26. x2 + 17 X + 60 27. x2 - 19 X + 90 28. x2 - 18 X + 32 29. x2 + 18x.+ 45 30. x2 + 18 X + 56 31. 'X2 - X - 2 32. x2 - X - 6 33. x2 - 3 X - 4 34. x2 + X - 6 35. x2 + 2 X - 15 36. x2 + 3x-28 37. x2 - 4 X - 21 126 IDENTITIES AND THEOREMS; FACTORING 71. 10x2 + 7a;+ 1 81. 9x2+17x-2 72. 10 a;2 - X - 3 82. 6 x2 - X - 12 73. lOx^ - ISx - 3 83. 6 a;2 + 21 X - 12, . 74. 10 x2 - 17 X + 3 84. 7 x2 - 9 X - 10 75. 3 x2 - 4 X - 7 85. 11 x2 + 27 X + 10 76. 3 x2 + 22 X -f 7 86. 15 x2 - 7 X - 2 77. 5 x2 + 8 X - 4 87. 14 x2 + 11 X - 15 78. 5 x2 - 11 X - 12 88. 21 X2 + X - 10 79. 5 x2 + 28 X + 15 89. 10 x2 - 29 X - 21 80. 10 x2 - 9 X - 9 90. 6 x2 + 17 X - 14 Factor the following expressions: 91. 8x2 - 2 gg 1728x2 - 12 92. 18 x2 - 50 99. 36 xY - 25 26 93. 75 a - 27 a^ 100. 100 h"" - 36 k^ 94. 44 x3 - 275 xY 101. 1210 a^¥ - 10 ah 95. 8 x2 - 162 102. 243 xyh"^ - 12 a^yz^ 96. 18 x2 - 32 103. 1 - 100 «2 97. 27 x2 - 147 104. 75 x^o - 48 a^ 105. 9 a^¥c^ - 9 x^e Completing the Square 147. Although the expression (x — 5)^ — 9 could be sim- plified and factored as follows: (x - 5)2 - 9 = x2 - lOx + 25 - 9 = x2 - lOx + 16 = (x - 8) (x - 2) there is an advantage in having it in its first form, because by Theorem A we can at once factor it into (x - 5) - 3 and (x - 5) + 3 Then simplifying the factors, we get (x — 8) (x — 2). The advantage is greater if the number is large, as in the expression (x — 42)2 — 3^^ ^ COMPLETING THE SQUARE 127 EXERCISE 48 Factor: 1. {x - 18)2 _ 49 ^^ (^ _|. 90)2 _ 36 2. (x - 20)2 - 121 8. (n + 101)2 _ 625 3. (a - 32)2 _ 100 9. (25 - xY - 25 4. (2 a; - 5)2 - 144 10. (x - 38)2 _ 225 5. (x - 35)2 _ 81 11. (3 a; + 51)2 _ 144 6. (p + 28)2 - 900 12. (m - 59)2 - 400 148. In factoring an expression like x^ -\- 5Q x -\- 768, an easier method than finding all pairs of factors of 768 and selecting the right pair is to reduce the expression to the form of the difference of two squares. We may then apply Theorem A to it, as in Exercise 48. Taking the first two terms, and remembering Theorem 1 of Section 141, we can say that x^ + 56 x is a part of a perfect square, the square of some binomial. Since x^ is the square of the first term of the binomial, then the first term of the binomial must be x. Again, 56 x is twice the product of both terms, so 28 x is that product; and since x is the first term, 28 must be the second term. Consequently, a;^ + 56 a; is a part of (x + 28)^ = x^ + 5Qx + 784. 149. Model E. — Factor x^ + 5Q x + 768. x^ + 6Qx + 7Q8 = x^ + 5Qx+ (28)^ - (28)2 + 768 = x2 + 56 re + 784 - 784 + 768 = (x + 28)2 - 16 = (x -\- 28)2 - (4)2 = (a: + 28 + 4) (a; + 28 - 4) = (a; + 32) (x + 24) 150. The process by which we get a perfect square from the first two terms of our given expression is called completing the square. It may perhaps be more clearly 128 IDENTITIES AND THEOREMS; FACTORING understood if we consider the diagram for cross multiplica- tion: Since the middle term is a double cross product, each cross product is half of it. EXERCISE 49 Factor by completing the square: 1. x'^-\-2x - 4S 2. a2 - 28 a + 187 3. x2 - 4 X + 3 4. p2 _ 36 p + 128 6. A2 - 50 A + 264 6. ^2 + 10 a: - 24 7. x2 - 12 X - 45 11. X' 8. n2 - 40 71 - 500 9. x^ - 6x + 5 10. x2 + 10 a; - 96 21. x^ 2 - 62x + 945 12. Z2 + 68X+ 1155 13. s2 - 10 s - 704 14. r2 - 50 r + 616 15. t^ -12t- 1260 16. x^ - 112 X + 3100 17. a2 + 204a + 404 18.^ Q^ + 78Q + 1400 19. h^ -S2h + 1561 20. x2 - 58 X - 759 72 X + 1280 151. The process of completing the square becomes some- what more difficult when the coefficient of the middle term is an odd number. Model F. — Factor x^ + 11 x - 726. a:2 + 11 a; - 726 = a;2 + 11 x + (5.5)2 _ 30.25 - 726 = a;2 + 11 a; + 30.25 - 756.25 = (x 4- 5.5)2 _ (27.5)2 = (x + 5.5 + 27.5) {x + 5.5 - 27.5) = (x + 33) {x - 22) COMPLETING THE SQUARE 129 The method of completing the square should be used when the factors cannot be readily seen by inspection. EXERCISE 50 Factor by completing the square: 1. x' + x-Si 11, x^ + 5x - 594 2. x^ + x-\- t\ 12. x^ -9x- 792 3. x^ + Sx-{- 1.25 13. x2 - 15 X - 1134 4. x^-\- 5x-{-2i 14. x^ - 51x-\- 648 5. a;2 + 7 X + 3 .25 15. x^ + 81 x + 1458 6. a2 + 11 a - 5.75 le. x^ - 75 x + 1386 7. A2 - 13 A + 6.25 17. x2 + 87 X + 1782 8. /b2 - 9 fc + 4.25 18. a;2 + 67 X + 1120 9. X2 + 15 X + 7.25 19. x2 - 79 X + 1350 10. Q^ + 3Q- 897f 2o. a^ + 83 a + 1512 21. x2 - 11 X - 2142 CHAPTER VII QUADRATIC EQUATIONS 152. One of the most important axioms in elementary algebra is the following: The product of two or more factors can never he zero unless at least one of these factors is itself zero. This axiom will hereafter be referred to as Axiom A. 153. In the equation 2x — 7 = 0, x must have the value ^; if we substitute this value for x, the expression 2 a; — 7 becomes zero. 6 a;2 - 5 X - 6 factors into (3 a; + 2) (2 x - 3). To make 3 x + 2 = 0, we must have x = — f . To make 2 x — 3 = 0, we must have x = |. EXERCISE 51 What value must we substitute for x in each factor of the following expressions in order that the factors may severally become equal to zero? 1. x2-3x + 2 6. 14x2 + 5x-l 2. x2-5x + 4 7. 21x2-17x + 2 3. 6x2 -X- 2 8. 3x2-14x + 16 4. 2 x2 - 3 X + 1 9. 12 x2 - 7 X + 1 6. x2 - X - 2 lo. 15 x2 ~ 11 X - 14 Two Answers to One Question 154. Model A. — A square box 7 inches high has 160 square inches more in its lateral surface than in its bottom. What is the size of the bottom? 130 TWO ANSWERS TO ONE QUESTION 131 Let X = the number of inches in one side of the bottom. Then x^ = the number of square inches in the bottom. 7 X = the number of square inches in one side. ® 28 x - a:2 = rBO (D = x2 - 28 X + 160 ® +x2 -28 a; ® = (x-S){x - 20) @ factored ® = x-8 ® by Ax. A ® S = x ® + 8 ® = X - 20 ® by Ax. A ® 20 = X ■ ® + 20 Ans. 8 in. square, or 20 in. square. Check: Area of bottom = 64 sq. in. or 400 sq. in. Area of 1 side = 56 sq. in. or 140 sq. in. Area of 4 sides = 224 sq. in. or 560 sq. in. 224 - 64 = 160 560 - 400 = 160 155. This example has two answers, or, as we sometimes say, there are two values of x which will satisfy the equation. These two numbers are technically called the roots of the equation, using the word root in a sense entirely different from that of Sections 52 and 119. The two answers do not mean that the quantity repre- sented by X can actually have two different values at the same time; that is, the same box cannot, of course, be 8 inches square and also 20 inches square. But a box 8 inches square would have the properties described in the equation,' and so would a box 20 inches square. There are two sizes of square box that could be made so as to be 7 inches high and at the same time so as to have 160 square inches more in the lateral surface than in the bottom. When an equation is so arranged that all its terms are in one member, and the other member is zero — in other words, so that we have an algebraic expression equated to zero — then the factors of that expression are sometimes loosely called the "factors of the equation." 132 QUADRATIC EQUATIONS 156. If we replace by a the number 7 in the problem of Model A, and by h the number 160, we can substitute the following numbers for a and h, and obtain in each case a new problem. (a) 5 3 4 6 10 (&) 36 20 39 EXERCISE 52 44 175 1. In a square storage tank SJ ft. deep, th^e are 32 sq. yd. more of concrete surface in the sides than in the bottom. How big is the tank? 2. At the back of a square building lot, a rectangular piece of ground of the same width as the lot and containing 2000 sq. ft., is added. This makes the lot 90 ft. deep. What is its frontage? 3. A ventilator opening 15 in. high gives 36 sq. in. more space than a square opening of the same width. How wide is it? 4. I bought a rectangular lot of land, containing 3850 sq. ft. There was a billboard fence along the front, but the other three sides required 204 ft. of fencing. What was the depth of the lot? 6. A man contracted to pay 15 cents a foot for gilt picture molding and 20 cents a square yard for straw matting used in furnishing a square room in his house. He was astonished to find that the picture molding cost him $4 more than the matting. What size was the room? 6. Another room was 2 yd. longer than it was wide. For this room, the molding at 15 cents per foot cost $4.20 more than the matting at 20 cents per square yard. What were the dimensions of the floor? TWO ANSWERS TO ONE QUESTION 133 7. There is a cylinder 8 in. in height (h), whose lateral surface (2 irrh) exceeds the area of its base (ttt^) by 88 sq. in. Find its radius (r). (Use tt = 3j.) * 8. A man sold a horse for $102 and found that his loss per cent was one eighth of the number of dollars he had paid for the horse. How much had he paid? 9. In the formula ¥ — {a — xY = & — x^, what value must a have if x = 60 when 6 = 87 and c = 65? 10. In the formula of Ex. 9, if a = 28, 6 = 17, and c = X 4- 5, how much is xl Use the following numbers to make new problems from Exs. 1-9. Example 1. (a = 81; h = 32) (a) (Jb) 6 15 9 35 7h 24 8 61 10 421 Example 2. (a = 2000; h = 90) (a) (&) 1200 70 420 44 360 46 2100 100 2400 110 Example 3. (a = 15; 6 = 36) (a) ib) 22 72 19 60 33 175 20 91 36 180 Example 4. (a = 3850; b = 204) (a) (b) 2240 136 3000 160 3920 196 4140 226 4400 260 Example 6. (a = 15^; b = 20ff; c = $4) (a) 10^ (6) 15^ (c) $1.05 m $2.30 8^ ISi 18?^ 36^ 56^ $1.20 25,i $1 $2.00 Example 6. (a = 2yd.; & = 15^; c = 20jf; d = $4.20) (a) 2 ft (b) 10^ (c) 21^ (d)$1.05 ,. 2 ft 7?i 15^ $1.08 2 yd. 2 yd. 3 yd. 10?^ 15^ 11^ 18^ 27^ IQ^ $2.00 $3.00 $2.78 Example 7. (a = 8; 6 = 88) (a) (&) 25 154 5 66 6 110 9 176 13 418 Example 8. (a = $102; b = i) (a) (6) $144 $147 $105 $192 $105.60 Example 9. (a; = 60; 6 = 87; c = 65) (:^) (&) (c) 37 41 47 95 70 118 66 47 79 48 46 62 10 40 38 134 QUADRATIC EQUATIONS Both Answers Alike 157. Model B. — In the problem of Model A (p. 130), if the number 160 had been 196, both answers would have been alike. The problem would read: A square box 7 inches high has 196 square inches more in its lateral surface than in its bottom. What is the size of the bottom? Let X = the number of inches in one side of the bottom. Then x^ = the number of square inches in the bottom. 7 X = the number of square inches in one side. © 28 X - a:2 = 196 ® = a:2 - 28 a; + 196 © + a;^ - 28 a; ® = {x - U){x - 14) © factored © X - U = © by Ax. A ® X = U © + 14 Ans. Bottom of box, 14 in. square. Check: Area of bottom = 196 sq. in. Area of 1 side = 98 sq. in. Area of 4 sides = 392 sq. in. 196 + 196 = 392 Here, on account of a special selection of figures (such as might accidentally appear in any problem), the two answers are alike. Similar results are obtained if, instead of 7 inches and 196 square inches, we say 9 inches and 324 square inches, or 5 inches and 100 square inches. EXERCISE 53 Solve the problems of Exercise 52, making the substi- tutions indicated for each problem: 1. For the numbers 8 J and 32 substitute 25i and 289, respectively. 2. For 2000 substitute 2025. 3. For 15 and 36 substitute 16 and 64, respectively. 4. For 3850 substitute 5202. 5. For 15 jz^ and $4 substitute lOj?^ and $1.80, respectively. / ANSWERS APPARENTLY DIFFERENT 135 6. For 15j?f and $4.20 substitute 10^ and $2.20, respectively. 7. For 8 and 88 substitute 7 and 154, respectively. 8. For 102 substitute 200. 9. For 87 substitute 25. 10. For 28, 17, and 5 substitute 18, 14, and 4, respectively. Answers Apparently Different 158. If we find two different numbers as the values of x in solving a quadratic equation, we naturally conclude that this indicates two answers for the problem from which the equation was formed. We sometimes find, however, that the two answers thus indicated are actually the same. Model C. — Divide 28 into two parts whose product shall be 75. Let X = one part; then 28 — x = the other part. ® same values @ + a:2 - 28 a; (D factored by Ax. A ® + 25 ® by Ax. A ® + 3 = 3, 28 - X = 25. 25 and 3, or 3 and 25. The first answer means that one part can be 25 and then the other must be 3; and the second, that one part can be 3 and then the other must be 25. The two answers are therefore really the same. Check: 25 X 3 = 75; 25 + 3 = 28. EXERCISE 54 1. A rectangular field containing one acre requires 924 feet of fencing. What are its dimensions? (An acre is 10 square chains and a chain is 66 feet long.) © X (28 - x) = 75 (D 28 X - x2 = 75 ® = x2 - 28 X + 75 ® = (x- 25) (x - 3) = X - 25 (D 25 = x ® = X - 3 ® 3 = x ® When X = 25, 28 - X = 3; when x Ans. 136 QUADRATIC EQUATIONS . 2. Two cubical bins, side by side, extend the whole length of a 16-foot wall, and will exactly contain OJ cords of kindling wood. What is the size of each? (A cord of wood measures 4 by 4 by 8 feet.) 3. The number 30,551 has two factors whose sum is 360. What are the factors? 4. A rectangular room requires 68 feet of picture molding. The same room requires 32 square yards of carpet. What is the size of the room? 5. A wharf projects 3^7 feet into the water, and is made of two platforms; its total area is 377 square feet. What is its greatest width and its least? 6. Forty-six rods of fencing are required for a rectangular field whose diagonal is 17 rods. What is the size of the field? 7. A rectangular corner lot and two square lots adjoining it contain in all 9972 square feet, and have altogether 114 feet frontage on each street. All the lots on a street have the same depth. What are the dimensions of the lots? 8. By one pipe a certain tank is filled, and then by an- other immediately emptied, the whole operation requiring an hour all but ten minutes. Then both pipes are used to fill the tank again, which is done in 12 minutes, How long would it take each pipe separately to fill the tank? The Meaning of Negative Answers 159. Sometimes one answer is negative and still can be interpreted as a reasonable answer. It is generally neces- sary to change the sense of some word, so as to get a meaning exactly opposite to the meaning in the original statement of the problem. Model D. — Of two pipes, one takes 3 hours longer than the other to fill a cistern, and both together would fill it half THE MEANING OF NEGATIVE ANSWERS 137 full in an hour. How bng would it take each pipe alone to fill it? 1 Let X = the number of hours for one pipe to fill it. Then x + S = the number of hours for the other pipe. the fraction of the cistern filled by the first pipe in one hour. the fraction filled by the other pipe in one hour. X + 3 11 1 ® x^ {x + S)~ 2 ® 2x + Q-\-2x = x{x + S) (D Ax + 6 = x^ -\-Sx = x2 - X - © = (a; - 3) (i)X2x{x + S) @ same values @ - 4x - 6 2) ® factored Whence we obtain x = S and x + 3 = 6; that is, one pipe would fill it in 3 hours and the other in 6 hours. But we also obtain x = —2 and x + 3 = 1; that is, the first pipe is emptying the cistern, and by itself could do that in 2 hours, while the second would fill it in one hour. Check: The first pipe fills ^ of the cistern in one hour; the second fills i of it in one hour; both, ^. Check: The first pipe takes out ^ the cistemful in one hour; the second pours in the whole capacity; there is left enough to fill ^ the cistern. EXERCISE 55 1. A field 20 rods by 18 rods is made 352 square rods in area by cutting off a strip of uniform width on the side, and adding a strip of the same width on the end. What is the width of the strips? 2. If a fraction having 70 as its numerator is inverted and multiplied by 6, its new value will exceed its original value by 1. Find the fraction. 3. A boat which is able to go 5 miles an hour in still water goes 21 miles southward on a certain river and then 24 miles northward, the whole time spent in travehng being 11 hours. How fast does the stream flow southward? 138 QUADRATIC EQUATIONS 4. Both pipes of a certain water tank are opened, and the tank, being empty at noon, is filled at 6 o'clock. One of the pipes, if used alone, would take 35 hours longer than the other pipe alone to fill the tank. At what time would the tank be full in each case? 5. In the equation c^ = a^ + b^, if a = 2h + 2 and c = 13, what does h equal? 6. In the equation of Ex. 5, if 6 = a — 2 and c = 58, what does a equal? In Exs. 1-4, the following numbers may be used to form new problems: Example 1. (a) 20 16 25 32 35 (o = 20; 6 = 18; (&) 13 10 22 30 12 c = 352) (c) 230 105 540 880 342 Example 2. (a) 8 72 16 45 80 (o = 70; 6 = 6; (h) 4 36 144 900 16 c = 1) (c) -3 5 7 11 15 Example 3. (a) 3 7 6 13 8 (a = 5; 6 = 21; (&) 12 34 30 62 22 c = 24; d = 11) (c) 10 33 32 63 30 id) 8 10 19 10 8 Example 4. (a) 3 6 4 2 8 (a = 6; 6 = 35) (&) 8 5 15 3 63 Answers Suggesting Related Problems 160. A negative answer to a quadratic equation may not be interpretable as a reasonable answer to the problem which gave rise to the equation, but may suggest a new problem, closely related to the given one, and having the negative and positive answers interchanged. Such a case is the following: Model E. — A farmer bought a herd of cows for $400. If there had been 4 more in the herd, the price would have been $5 less apiece. How many were there? ANSWERS SUGGESTING RELATED PROBLEMS 139 Let X = the number of cows in the herd • Then 400 — = the number of dollars paid for X 400 400 , ^ one cow. © X -x + 4 + .^ ® 400 x + 1600 = 400 X + 5 a:2 + 20 a: © X X (x + 4) ® = 5 x2 + 20 x - 1600 ® - 400 X - 1600 ® = rc2 + 4 a; - 320 © -5 © = (x - 16) (x + 20) ® factored © x = 16; X = -20 © by Ax. A Ans. 16 cows. Check: 16 cows for $400, $25 apiece ' 4 more, or 20 cows, for $400, $20 apiece 25 - 20 = 5 Now we can change the algebraic sense of the problem by changing words like " more " to ''less " and vice versa. The new problem would read: A farmer bought a herd of cows for $400; if there had been 4 less in the herd, they would have cost $5 more apiece. How many were there? Let X = the number of cows in the herd, ^ 400 400 ® = A — 5 ^ X X — 4: Whence x = 20, or —16. Check: 20 cows for $400, $20 apiece 4 less, or 16 cows, for $400, $25 apiece 25 = 20 + 5 EXERCISE 56 In the following problems, the negative answers suggest related problems in the opposite algebraic sense. 1. There are 40 rooms in a house, and 3 more rooms on each floor than there are floors in the house. How many floors are there? 2. Comparing a pail with a 140-quart tub, a man con- cluded that the tub held as many pailfuls as the pail held 140 QUADRATIC EQUATIONS quarts. He found, however, that the tub held 4 pailfuls more. What was the capacity of the pail? 3. In order to save 4 hours on his regular trip of 126 miles, a stage driver finds it necessary to go 2 miles an hour faster than his usual rate. What is his usual rate? 4. What is the price of eggs when 10 more in a dollar's worth lowers the price 4 cents a dozen? 5. A rectangular field contains IJ acres. If its length is increased by 20 feet and its breadth diminished by 18 feet, its area will be diminished 2340 square feet. What are its dimensions? For Exs. 1-5 the following numbers may be used to form new problems: Example 1. (a) 35 50 60 44 . 112 (a = 40; 6 = 3) (&) 2 5 4 7 6 Example 2. (a) 88 176 99 90 180 (a = 140; b -- = 4) (6) 3 5 2 ^ 3 Example 3. (a) 6 12 4 2 5 (a = 4; 6 = : 126; (b) 176 135 120 180 150 c = 2) (c) 3 4 11 1 2i Example 4. (a) 5 2 25 50 20 (a = 10; 6 = 4) (&)' 1 1 8 2 5 Example 5. (a) 2 3 h 4 2h (a = U; 6 = 20; ib) 24 14 12 -60 -45 c = 18; d = = 2340) (c) 20 30 10 -16 -30 id) 3120 7680 780 6240 -3600 Meaningless Answers 161. Jn each of the following problems, an answer will be obtained which satisfies the quadratic equation, but has no reasonable interpretation in the problem which gave rise to that equation; and the answer may not even suggest a related problem. MEANINGLESS ANSWERS 141 EXERCISE 57 1. A square box 7 inches high and without a cover has 288 square inches of inside surface. What is the size of the bottom? 2. Of two large wheels, one makes 48 turns more than the other in roUing a mile. Their tires, straightened out and laid end to end, would reach 21 feet. What are their cir- cumferences? 3. Two sprinters start in opposite directions from the middle post of a straightaway 440-yard track. In one second they are 21 yards apart, and one of them finishes his sprint 2 seconds ahead of the other. What is the speed of each? 4. A tank containing 4 gallons is filled by one pipe and then immediately emptied by another, both operations requiring just 5 hours; one pipe dehvers 3 gallons per hour more than the other. How many gallons can each pipe deliver in an hour? 5. A man rows 5 miles downstream and back in 4 hours. The stream flows 3 miles an hour. How fast can the man row in still water? In Exs. 1-5 the following numbers may be used to form new problems: Example 1. (a = 7; 6 = 288) (a) (b) 8 185 10 176 15 ^44 20 249 3 36i Example 2. (a = 48; b = 21) («) (6) 40 23 10 32i 88 22 96 241 88 27 Example 3. (o = 440; b = 21; c = 2) («) (6) (c) 400 18 5 300 22 21 1000 20i 22^ 900 201 10 500 241 Example 4. (a = 4; 6 = 5; , c = 3) (a) (c) 6 5 . 7 8 7 9 10 ' 3 7 4 3 5 21 13^ 5 142 QUADRATIC EQUATIONS Example 6. (a) 15 24 30 25i 10^ (a = 5; 6 = 4; (&) 9 14 12 10 5 c = 3) (c) 4 5 6 7 2 Equations Involving Irrational Numbers 162. Model F. — What is the length of one side of a square whose area is 3 square feet? Let X = the number of feet in one side of the square. © a;2 = 3 @ x2-3 = ©-3 This last equation could be wTitten ® x2 - (V3)2 = © same values In equation ©, V3 is the symbol for a number that, multiplied by itself, would give 3 for a product. ® (x + V3) {x - V3) = ® factored 163. The exact value of \/3 cannot be written in ordi- nary figures. The square of 1.7 gives 2.9, the square of 1.732 gives 2.999, the square of 1.73205 gives 2.9999, and so on. The squares are nearer and nearer to 3, but never exactly equal to it. Again, the numbers If, If, ly\, 1|^, If T> give squares that are nearer and nearer to 3, but never exactly equal to it. Numbers that cannot be exactly expressed in ordinary figures are called irrational numbers. 164. Whether we use the symbol \/3, which represents the exact value of the number that multiplied by itself will give 3 for a product, or whether we use a more or less ac- curate approximation (such as 1.7, or 1.73, or 1.732), will depend in every case on the use that we intend to make of our results. In general, if the results are to be used for arithmetical computation or for measurement, it is desirable to use the approximate value, the number of figures being chosen ac- cording to the needs of the problem. If, on the other hand, EQUATIONS INVOLVING IRRATIONAL NUMBERS 143 we intend to use our result in further algebraic work, the algebraic symbol V3 may be found more convenient. Sys- tematic methods of dealing with such symbols will be given in later years. 165. The roots of the equation x^ = 3, or of any equation which reduces to x^ = some number, are said to be equal and opposite. That is, the value of x turns out to be either a particular number or its negative. Thus the equation x^ = 16 gives X = 4: and x = —4. Notice that the word root means an entirely different thing in the following sentence: There are two square roots of 16 (or of any number); one is + and the other — . 166. A quadratic equation without any first-degree term is sometimes called a pure quadratic equation. 167. We cannot factor by inspection the equation x2 - 10 X + 20 = By completing the square of which a;^ — 10 x are the first two terms, we change this equation into the form of the difference of two squares, and obtain (a;^ - 10 a; + 25) - 5 = which can be written {x — 5)^ — (V^)^ = and factored thus (x — 5 + ^y5) {x — 5 — \^5) = Using the value V5 = 2.236, these factors become {x - 2.764) {x - 7.236) = . EXERCISE 58 Factor the following expressions and find the approximate numerical values of your factors, using \/2 = 1.414, ^3 = 1.732, V5 = 2.236, V6 = 2.450, VIO = 3.162. 1. (x - 2)2 - 2 5. (a - 10)2 _ 6 9. (A + 10)2 _ iq 2. (x - 5)2 - 3 6. (a +1)2-2 10. (16p - 1)2 - 6 3. (x-7)2-5 7. (5 + a)2-5 ii. (2a; -5)2 -3 4. (2x - dy - 10 8. ip - 3)2 - 3 12. (4m + 7)2 - 5 144 QUADRATIC EQUATIONS Solve the following equations: 13. a;2 + 6 a; - 3 = 20. 2 A;^ - 24 A; + 62 = 14. x2- 10a;+ 15 = 21. 4x2+16^-24=0 15. x2 + 2 X - 5 = 22. 2 p2 + 12 p + 8 = 16. A^ + QA-1 =0 23. x2 - 17.2 = 17. p2 _ 24 p + 141 = 24. p2 + 3 p _ 8 75 = 18. A^ + 20^ + 98 = 25. X2 + 5X = 11 19. ^2 4- 30 ^ + 220 = 26. a" + 17.3 = 17.3 a 27. 344 - a:2 = 52.6 x 168. Peculiar Quadratics Model G. — Solve the equation X' = 3x ® x^ = 3x ® a:2 - - 3a; = (D-^x ® X {x -3) =0 @ factored X = 0, a; = 3 by Ax. A Notice that the value x = satisfies the equation. This equation is said to have one null root. 169. Model H. — Solve the equation x2 =60:- 11 Q a;2 = 6a;-ll @ x2-6x + ll=0 ®-6a: + ll ® (a;2-6a; + 9) + 2 = ® same values Since ® is the sum of two squares, we cannot factor it; the appearance of such an equation shows that some absurdity existed in the statement of the problem that gave rise to it. The equation is technically said to have imaginary roots. Equations of this type will be studied in later years of the mathematics course. PECULIAR QUADRATICS 145 EXERCISE 59 1. Three times the square of a number is equal to 10 times the number itself. What is the number? 2. A room is twice as long as it is wide, and its floor con- tains 144 J square feet. How wide is it? 3. A square room 9 feet high takes twice as much paper to cover the walls as to cover the ceiling. What is the size of the ceiling? 4. In a current of 2 miles an hour, a man takes 8 hours longer t9 row 24 miles upstream than to row the same dis- tance downstream. How fast can he row in still water? 5. A field 20 rods by 18 is made 353 square rods in area by cutting off a strip of uniform width on the longer side and adding a strip of the same width on the end. What is the width of the strips? 6. A square box 9 inches high has 300 square inches more on the sides than on the bottom. What are the dimensions of the box? 7. One man takes 4 hours longer than another to saw a cord of wood, and both working together can saw it in 3 hours. How long would it take each man working alone? 8. Two wheels, the sum of whose circumferences is equal to 8 yards, roll the same distance. The number of turns made by one wheel, added to the number of turns made by the other wheel, gives a sum equal to the number of yards in this distance. Find the circumferences of the wheels. Use a letter for the number of yards traversed; also one other letter. 9. A line 100 centimeters long is divided into two parts, such that the longer part is contained in the whole line as many times as the shorter part is contained in the longer part. What is the length of each part? 146 QUADRATIC EQUATIONS 10. Across the back of a square building lot, a rectangular piece of ground containing 1820 square feet is added. The lot is then 75.3 feet deep. What is its frontage? 170. Model I. — Solve the equation 23.4 a;2 + 508 X =3420 ® 23.4 x' + 508 X = 3420 23.4 x' + 608 X - 3420 = -3420 ® ^ + 21.7 a; - 146 = + 23.4 y? + 21.7 X + (10.8)« - 263 - same values {x + 10.8)« - (16.2)2 „ same values (x 4- 10.8 + 16.2) (x + 10.8 - 16.2) = factore 7. The slant side of a right triangle is 34.2 in., and the perpendicular sides are in the ratio 2.31. Find the perpen- dicular sides. \ 8. The ratio of the slant side of a right triangle to one of the perpendicular sides is 3.55, and the other side is 44.9 cm. Find the sides of the triangle. 9. The frame of a kite is made with two pieces of wood, 30 in. long and 54 in. long. They cross at right angles at the middle point of the shorter piece and 18 in. from one end of the longer. Find the perimeter of the kite, and its area. ^ 10. A ladder 40 ft. long is placed, for safety's sake, with its foot 7 ft. from a building. How far up the side of the building will the ladder reach? Angles of a Polygon 190. The sum of the angles of any polygon is given by the formula W = (n — 2) 180°, where W represents the sum of the angles and n the number of sides of the polygon. Divide the polygon into triangles by drawing all possible diagonals from one vertex. This will be a common vertex for all the triangles. Two of the sides of the polygon pass through this common vertex, but each of the other sides serves as the base for one triangle. There will be, then, just n — 2 triangles. ANGLES OF A POLYGON 165 Letter the angles of each triangle with the same letter that the poly- gon had at that point, but attach suffixes according to the serial number of the triangle; that is, in the first triangle the angles would be Ai, Bxy Ci) in the second A2, C2, Di] in the third A3, D3, Ez', and so on. Then, since the sum of the angles of a triangle is 180°, we have the following series of equations: © Ar + Bi + Ci = 180 ® A2 + C2 + D2 = 180 ® A3 + D3 + ^3 = 180 and so on. There would be n — 2 of these equations, one for each triangle. Adding the equations, we should find Ai+A2 + A3+ . . . =A; Ci + C2 = C; D2 + D3 = D; and so on. In the last two triangles, (^6 + (?6 = G, and H^ = H, just as Bi = B. Then the sum of the left sides of all the equations would be A + -B + C + . . . = TF, and the sum of the right sides would be {n — 2) 180; that is, W = {n - 2) 180°. 191. An equiangular polygon is a polygon of which all the angles are equal. An equilateral polygon is one of which all the sides are equal. Except in the case of a triangle, the sides of a polygon do not have to be equal when the angles are equal. Thus a square and a rectangle are both equiangular; but though the sides of the square are all equal, the sides of the rectangle are not. 192. In the case of an equiangular triangle, if we take a tracing of the triangle and turn it over on the original triangle 166 SOME THEOREMS OF GEOMETRY (making the tracing of one side fit on the side itself), the angles at the end of that side on the tracing will exactly fit on the angles of the triangle, because they are equal; and the other two sides will be exactly covered by their tracings, so that the other vertex-angle will also fit. Hence the theorem: An equiangular triangle is also equilateral. It is also true that an equilateral triangle is equiangular. 193. A regular polygon is one that is both equilateral and equiangular. Model J. — How many sides are there in a regular polygon one angle of which is 175°? Let n = number of sides of the polygon. Then 175=^^?— ^H?5 and n = 72 n EXERCISE 66 1. How many degrees are there in each angle of an equi- lateral triangle? 2. How many degrees are there in each angle of a regular 5-sided polygon? 3. How many degrees in each angle of a regular polygon of 6 sides? Of 10 sides? Of 12 sides? Of 180 sides? Of 100 sides? Of 40 sides? Of n sides? Of 20 sides? Of 23 sides? Of 33 sides? 4. How many sides in a regular polygon one angle of which is 144°? 5. How many sides in a regular polygon one angle of which is 174°? 176°? 172°? 156°? 170°? 150°? 168°? CHAPTER IX THE GRAPHICAL METHOD 194. A foot rule has numbers marked for each inch, and the subdivisions, though not marked, are known to cor- respond to certain definite fractional numbers. In the same way, any straight line may be marked at convenient equal distances, starting from a point which may be called the zero point. Then the successive marks, or their distances from the zero point, represent successive whole numbers; and points between these marks, or their distances from the zero point, represent fractional numbers. 195. Some points on such a line cannot be expressed in figures because they represent irrational numbers. Thus, if we construct a square with one side equal to the (unit) dis- tance between two successive marks, and then lay off on the line from the zero point a distance equal to the diagonal of that square, the point so obtained will represent \/2. The number tt is represented by a point which can be found by roUing a circle with unit diameter from the zero point one complete turn to the right. The Algebraic Scale 196. Points in regular order to the left of the zero point can be taken to represent negative numbers. This device has the advantage that the distance between two points on the line represents their algebraic difference. If the distance runs from left to right, the difference is plus; if from right to left, it is minus. 167 168 THE GRAPHICAL METHOD For example, in Fig. 1, the points marked with dots are, respectively, —6, —2.5, +3.5, and +8. Fig. 1. 197. A Une which is used for representing numbers in this way is called an algebraic scale. The scale of a thermometer is a good illustration of an algebraic scale, running up and down instead of right and left. The rise of the thermometer is the algebraic difference, going up, between two temperatures; and the fall is the difference, going down, — a negative number. EXERCISE 67 1. Construct an algebraic scale, or select one on a piece of squared paper, and on it mark as accurately as possible points to represent -10, -6, -4.5, 0, 5, 8.5, 10.5, \/2, - V2. 2. Draw a rectangle 2 units long and one unit high. What numbers can be represented by marking off a dis- tance, equal to its diagonal, to the right and to the left of the zero point on an algebraic scale? To the right and to the left of the point representing 1? What approximate deci- mal values might serve to locate the points so marked? 3. Draw a rectangle one unit high, of which the base ex- tends from the point to the point V2. What number is represented by marking off a distance equal to its diagonal? 4. On an algebraic scale, locate the point — \/6. 5. The temperature of the body of a human being in good health is 98.4° Fahrenheit; the temperature at which copper melts is 1900° F. These two points cannot be accurately located on the same algebraic scale. Why not? THE ALGEBRAIC SCALE 169 198. When two numbers are related to each other so that to each value of one corresponds one value of the other, a diagram can be made in which each pair of corresponding numbers is represented by one point on the diagram. Ex- amples of such numbers are the distances traveled by a train and the times elapsed since starting; the premium charged for a life insurance policy of $1000 and the age of the person insured at the date of the insurance contract; the diameter of a sphere and its area. 30 10 / > / / y / f/ / ,.<'■ ■f / w ^^/ A f y 7 y / / V J /y / 1 S 8 Fig. 2. In Fig. 2 is shown the record of distance and time for two trains from Boston to Worcester, a total distance of 45 miles. This diagram is constructed with hours and minutes along the bottom, and distances up the side, so that any point on the diagram represents a certain distance from Boston and a certain time of day. Diagrams of this kind are used in train dispatchers' offices. Any point on the line marked /'Express train'' shows by its height above the bottom of the diagram how far the train has gone; and by its distance from the left side of the diagram it shows the time of day. Similarly for every point on the line marked "Local train." 170 THE GRAPHICAL METHOD When the express train overtakes the local, the intersection of the two lines warns the train dispatcher that he must provide, at the place and time indicated by the point of in- tersection, opportunity for the trains to pass. The Algebraic Diagram 199. Algebra, however, must include negative numbers as well as the ordinary numbers of arithmetic. Conse- quently, an algebraic diagram must be made to include representations of negative numbers, whether paired with each other or paired with positive numbers. X -4 -i -2-1 5 6 7 Y Fig. 3. For convenience in discussing the matter, we will use x and y to represent any pair of corresponding numbers. Our diagram is constructed by taking two scale-lines like the one described in Section 196, and placing them perpendicular to each other with the zero points together. Call the horizontal Hne the X-scale, or axis of Xy and the vertical line the Y-scale, or axis of y. Then imagine vertical hues drawn through each point of the axis of x, and horizontal lines drawn through each point of the axis of y. THE ALGEBRAIC DIAGRAM 171 A point which represents x = —2, y = 4, must be the point which Hes opposite —2 in the axis of x, and opposite 4 in the axis of y. X Y n III 1 I I I 1 — _ 5-< _• 4-, I W J i i 1 o VI V • I V VI ■ Y X Fig. 4. The points in Fig. 4 that represent the following values of x and y are marked with the corresponding Roman numerals: I X = 4, II X = -2 III X = 3, IV X = 6, V X = 2i, VI X = -2|, y VII X = -5, ?/ 2/ = 2 ^ = 3 y = h 2/= -3 2/= -2^ -3 -31 The construction of points to represent in this way cor- responding numbers is called plotting. 200. The size of the space between successive whole numbers may be made as large as convenient, and with large spaces a good deal of subdivision is possible. It would seem possible, therefore, to secure great accuracy in a diagram by 172 THE GRAPHICAL METHOD taking a large unit space. When this is done, however, the drawing becomes so unwieldy that new sources of error arise and the results lose much of their presumable accuracy. With small diagrams and ordinary care, it is unhkely that more than two figures can be properly represented; but in large and careful drawings three figures can be used. When data are given more accurately than the diagram can represent, only the available figures are used. For ex- ample, 38.07 and 57.93 would probably be drawn as 38 and 58. 201. With paper already ruled, the intervals, or spaces between rulings, may be made to correspond to units, or to hundredths, or to tenths, or to any other decimal position of the digits. Thus, in plotting temperatures of melting for metals, the interval might well represent 100° Fahrenheit. In plotting the temperatures of a day, the interval might represent 1°; and in plotting the temperatures of a fever patient, tenths of a degree are used. The "scale of the diagram" must be decided upon in each case. EXERCISE 68 V 1. Insurance companies expect a person in good health at the age of 10 years to live 48.7 years longer; at 20 years, 42.2 years longer; at 30 years, 35.3 years longei-; at 40 years, 28.2 years longer; at 50 years, 20.9; at 60 years, 14.1; at 70 years, 8.5; at 80 years, 4.4; at 90 years, 1.4. Plot the expectations according to the ages. 2. At 25 years of age, one must pay for an ordinary life insurance policy $21.49 per thousand; at 30 years, $24.38 per thousand; at 35 years, $28.11; at 40 years, $33.01; at 45 years, $39.55; at 50 years, $48.48; at 55 years, $60.72; at 60 years, $77.69; at 65 years, $101.48. Plot the prices according to the ages. THE ALGEBRAIC DIAGRAM 173 y 8. The train leaving Boston for Portland at 9 a.m. is 12 miles away at 9:21. At 9:32, it is 17 miles away; at 10:05, 40 miles; at 10:35, 58 miles (here the train waits ten minutes) ; at 11:37, 100 miles; at 12, 115 miles (Portland). The train leaving Portland for Boston at 9 a.m. is 100 miles from Boston at 9:26. It is 84 miles away at 9.44; 70 miles at 10:23; 58 miles at 10:45 (here the train waits ten minutes); 40 miles at 11:24; 17 miles at 11:55; 12 miles at 12:06; and reaches Boston at 12:25. Plot, on a single diagram, the distances from Boston corre- sponding to the times for each train. y 4. On a certain Monday the moon rises at 8:35 p.m. On Tuesday it rises at 9:07; Wednesday at 9:34; Thursday at 9:57; Mday at 10:25; Saturday at 10:51; Sunday at 11:21. Plot the times of rising corresponding to the days. 5. The number of marketable feet, board measure, in a spruce tree of the Adirondack forests is 64 if the diameter breast-high is 9 inches, 83 if the diameter is 10 inches, and 101 if 11 inches. Then 127, 156, 189, 225, 263, 303, 349, 396, 447, 503, 564, 630, 702, 776, 858, 948, 1013, 1089, 1260 feet, for diameters increasing an inch at a time up to 30 inches. Compute the value of each size of tree at $15 per thousand feet, and plot their values corresponding to their diameters. 6. The Centigrade and the Fahrenheit scales, for ther- mometers, are used for recording the same temperatures, though with different numbers. Thus 32° F. is the same temperature as 0° C, and 212° F. is the same as 100° C. On a diagram, take one of the axes to represent the Centigrade scale, the other the Fahrenheit, and mark on the diagram the two points that represent the temperatures here men- tioned. Now suppose that all points representing corre- sponding readings of the two scales lie in line with the two 174 THE GRAPHICAL METHOD points just marked; draw the line. By means of the line you have drawn, answer the following questions: What temperature Centigrade corresponds to 68° F.? To 104° F.? To 5° F.? To -40° F.? What temperature Fahrenheit corresponds to 15° C? To 75° C? To-5°C.? To-40°C? 7. The centimeter is used as generally on the continent of Europe as the inch is with us, and much scientific work in the United States is done with the metric system. 100 centimeters is approximately 39.4 inches, and 12 inches is approximately 30.5 centimeters. Take one axis for inches and the other for centimeters; locate the two points de- scribed and through them draw a straight line. By means of the line, answer the following questions: How many centimeters are there in 24 in.? In 17 in.? In 50 in.? In 1 in.? In 44 in.? In 29 in.? How many inches are there in 20 cm.? In 50 cm.? In 80 cm.? In 16 cm.? In 38 cm.? In 53 cm.? 202. Model A. — Plot the corresponding numbers which satisfy the equation 2x — Sy =6. In this equation, for any value of x we can obtain a value of y. Thus, if a; = 5, 2 a: = 10, and we have © 10 - -3y = 6 ® 10 = 3 1/ + 6 (D + ^y (D 4 = 32/ @ -6 n = y ® -3 Again, if a; = 6, ?/ = 2; if x = —3, y = —4; and so on. Thus we obtain the following table of corresponding numbers: X = - ^, y = - ^ X = 1, y = - Ih X = 5, y = H x=-S, y=-4: X = 2, y = -I x = Q, y = 2 X = -2, y = -Sk X = S, y = x = 1, y = 2% X = 0, y = -2 a; = 4, 2/ = | x = 8, y = 3i THE LOCUS OF AN EQUATION 175 These pairs of numbers will be represented by the points in Fig. 5, running up toward the right. Fig. 5. The Locus of an Equation 203. It would be found, on trial, that all the other pairs of corresponding numbers which satisfy this equation lie on the same line, i.e., their points fall into line with those already put down. In fact, the straight line drawn through two points that represent any two answers to the equation is a complete list of all possible pairs of corresponding num- bers that satisfy the equation. In order to be complete, this line and the two axes are supposed to be endless in extent. This Hne is called the locus of such points as satisfy the equation, or for brevity, the locus of the equation. The word locus is the Latin word for place; it means here the place to find all such points, and no others. 204. If for any temperature / on the Fahrenheit scale the reading on the Centigrade scale is c, the two readings are connected by the equation ©/ = |c + 32 This is an explicit formula for /. 176 THE GRAPHICAL METHOD In the same way the equation @c = |(/-32) is an explicit formula for c. And the equation ® 5/-9C = 160 is an implicit formula for either / or c. Show how equations ® or could be derived from ®. Define an explicit formula. Also an implicit formula. Notice that an answer to a two-letter equation is a pair of numbers that will satisfy the equation. We are said to plot the equation when we plot such pairs of numbers. Plot the equations ®, ®, and ®. Why are they all represented by the same Une? 205. With the information we already have about similar triangles, it is easy to prove that every straight line in an algebraic diagram is the locus of an equation of the first degree. In the accompanying figure, where a straight line crosses the axis OX at A, and the axis 07 at B, OA and OB are constant; let us represent them by a and 6, respectively. Then, if P represents any point on the line AB, OM and MP are the variables that would be represented by x and y, re- spectively. In other words, a and h are two numbers repre- sented by the points at, which this line crosses the axes; x and y are the two numbers represented by any point on the line AB; they will vary as we take different points on the line. For any one position of the point P, such as the one shown in the illustration, there are two right triangles, which are similar. AMP ~ AOB. (Under what condition are two right triangles similar? How do these two right triangles fulfill this condition?) In these two triangles, MP {= y) in the triangle AMP corresponds to Y B ^V^ ^ \. "^ ^ X 1 r~ '^ ^\ Fig. 6. THE LOCUS OF AN EQUATION 177 DB {= b) in the triangle OAB; and MA (= a — x) corresponds to OA (= a). Then we have the equation ® ^ — L ^ a a — X If we multiply each side of this equation by a, we get 6= "y a — X Then, multiplying by (a — a;), h (a — x) = ay Or, in one operation ® b {a — x) = ay (i) X a (a — x) This equation can be changed in form thus ® ab — bx = ay same values ab = bx + ay (3) + bx Thus we have established the fact that every point on the straight Une represents two numbers that satisfy a certain equation of the first degree, the coefficients of which can be ascertained in advance by the measurements of the figure. Take a point not on the straight line, and call it Q. Draw QN ± OX, cutting AB as at R. Then R represents a number-pair that satisfies our equation. Does Q? Can you give an algebraic reason for your answer? Draw your own diagram. The sign ± means "perpendicular to." EXERCISE 69 1. Draw the diagram and follow through the proof given in Section 205 for the case where OA = 9 and OB = 4. 2. Proceed as in Ex. 1, for the case where OA = — 3 and OB = 5; only, for the sake of simplicity, mark the point P on the right of OY. 3. In the same way, find the equation of the line AB where a = —4, and 6 = 1. 4. Find also the equation of the line AB where a = 2, b = -5. 5. In the equation Sx — 4y =36, what value of x cor- responds to the value 2/ = 0? Where is the point that rep- 178 THE GRAPHICAL METHOD resents this pair of values? What value of y corresponds to the value a: = 0? Where is the point that represents this pair of values? 206. The points where a line crosses the axes, in an algebraic diagram, represent numbers that are called the intercepts of the equation. Thus, in the demonstration of Section 205, a is the intercept on OX; b is the intercept on OY, The intercepts of an equation, then, may be defined as the points that satisfy the equation and lie on the axes. The intercept on OX will have y = 0. The value of x corresponding to y = can be found from the equation, and will be the value of a. In the same way, the value of b can be found by letting x = in the equation. EXERCISE 70 1. What are the intercepts of the equation 2 x -\- 5 y = 40? Follow through the proof of Section 205 for the equation of a straight hne having these intercepts. 2. What are the intercepts of the equation Sy — 4:X = 60? Follow through the proof, in the same way, for the equation of the straight fine having these intercepts. 3. Proceed in the same way for the equation 3y -\- 9 = 2x. 4. Find the intercepts of the equation S x -{- 7 y =27. What is the tangent of the angle OAB in the algebraic dia- gram for this equation? 5. The point 2, 3 satisfies the equation in Ex. 4. Mark the point N on OX to represent x = 2; draw NQ J. OX, making NQ = 3. What is the tangent of NAQ'! How does this show whether Q is on AB? 207. From the preceding examples it is clear that for any two-letter equation of the first degree we can find a pair of THE LOCUS OF AN EQUATION 179 intercepts and a straight line passing through them. With- out a formal proof, for the present, we may take it for granted that the equation we start with will be the equation of the straight line so obtained. We may conclude, then, that every two-letter equation of the first degree has for its locus a straight line. It requires some reflection to see that this statement differs from that of Section 205; but it is very important to appreciate the difference. It becomes clearer when the statements are expressed in the conditional form, thus: In the case of an equation for which we have plotted the locus on an algebraic diagram, (1) If the locus is a straight line, its equation is a two-letter equation of the, first degree. (2) // the equation is a two-letter equation of the first degree^ its locus is a straight line. Two statements related in this way, so that the con- ditional part of each is the conclusion of the other, are called converse statements. / EXERCISE 71 For each of the following statements, state the converse. Then answer, for each, these two questions: first. Is the original statement true? Second, Is its converse true? 1. If a man is very rich, he can afford to pay his carfare. 2. If Leander has broken his leg, he cannot swim two miles. 3. If it is raining, there are clouds in the sky. 4. If I am your teacher, you are my pupil. 5. If his name is on the voting list, he can vote. 6. If a man plays the cornet badly, he annoys his neighbors. 180 THE GRAPHICAL METHOD 7. If you feel tired, you have walked ten miles to-day. 8. If my straw hat is ruined; a horse has stepped on it. 9. If Caius lived in Caesar's time, Caius has been dead for centuries. 10. If this book were written in Greek, it would be hard to understand. CHAPTER X ELIMINATION BY COMBINATION 208. An equation with two unknown letters imposes a condition upon the values of those letters. For example, the equation 2 x — S y = Q is an impUcit formula for y when the value of x is given, or for x when the value of y is given. Thus it imposes a condition upon the two numbers x and y, such that when a numerical value is assigned to either there is one and only one corresponding value of the other. The nature of the condition imposed by the equation is indicated by drawing its locus. Every point of that locus represents a pair of numbers which will satisfy the equation. Two equations, with two unknown letters in each, may be plotted upon the same algebraic diagram. If the condi- tions expressed by these equations are understood to be imposed upon the same pair of numbers, — in other words; if the abbreviations x and y have the same meaning in both equations, — then the equations are called simultaneous. The conditions that they impose can be satisfied by the points that He upon both loci and by those only. To determine the values of two unknown letters two conditions are necessary. EXERCISE 72 Construct the loci of the following equations, putting each pair on a separate diagram: 1. x-{-2y =5, Sx + 7y = 17 2. 15 X + 3 2/ = 39, 17 a; - 2/ = 31 181 182 ELIMINATION BY COMBINATION 3. 7x + 4?/=41, Saj — 2/=4 4. 2x + 2/ = 11, 3 re + 7 2/ =22 5. 5x + 62/=51, 6a:-52/ = -12 e. 2x- y = -4, 7x + Qy =Q2 It will be found that these pairs of lines intersect in the points given below; the points are numbered to correspond, with the pairs of equations. Each of these points, then, satisfies both equations in the pair to which it corresponds. 1. X = 1, y = 2 3. X = S, y = 5 5, x = S, y = Q 2. X = 2, y = 'S 4. x=5, 2/ = l e. X = 2, y = S Two-letter Problems 209. Model A. — Six horses and 11 cows sell for $810; 13 horses and 5 cows sell at the same rate for $1190. Find the price of each animal. Let X = the number of dollars for one horse; y = the number of dollars for one cow. (I) Qx + ny = SlO (II) 13 a; + 6y = 1190 210. Model B. — Six pounds of tea and 11 pounds of coffee sell for $9.23; 13 pounds of tea and 5 pounds of coffee sell at the same rate for $11.90. What is the price of each? Let X = the number of cents for one pound of tea; y = the number of cents for one pound of coffee. (III) ' Qx + Uy = 923 (IV) 13 a; + 5y = 1190 211. In the four equations that come from these two problems, x and y do not have the same meaning. In (I) and (II), for example, x stands for the number of dollars in the price of a horse; in (III) and (IV), for the number of cents in the price of a pound of tea. When the unknown letters in two equations stand for entirely different quantities, as in TWO-LETTER PROBLEMS 183 (II) and (III), they must be treated as if they were written with different letters; the terms which seem to be similar are really not similar. In (I) and (II), we should find that X = $80 and y = $30; while in (III) and (IV) we should find that x = 75 cents and ?/ = 43 cents. 212. In two simultaneous equations, however, terms hav- ing the same letters are similar terms. Any member of either equation can be added to or subtracted from a member of the other equation, and if the x terms (or the y terms) are the same, a resulting equation may be obtained free from x (or y). Thus for Model A: (I) 6rc + ll2/ = 810 (II) 13a:+ 5y = 1190 ® 30 x + 55 2/ = 4050 (I) X 5 143 X + 55 2/ = 13090 (II) X 11 The y terms are now alike. Subtracting ® from 0, 113 a; = 9040 0-0 a; = 80 -^ 113 To find the value of y, we may return to the original equations, (I) and (II). We make the x terms alike by multiplying (I) by 13 and (II) by 6, and then subtract the resulting equations. Or, since we know now that x = 80, Q X = 480, we may write instead of (I), 480 + 11 2/ = 810 subst. in (I) .0 11 2/ = 330 - 480 2/ = 30 -^ 11 The answer to Model A, then, is $80 for each horse, and $30 for each cow. In the same way we find, for Model B, 75 cents a pound for tea, and 43 cents a pound for coffee. This is the method by which the points of crossing were found for the six pairs of lines in Exercise 72. If the x terms (or the y terms) are of opposite signs in the two equa- tions, the equations must be added, instead of subtracted, to 184 ELIMINATION BY COMBINATION get rid of that letter. The algebraic work really effects the reduction of the two given equations to two new equations in the form of particular values for x and y. Model A. Check: 6 horses at $80 = $480 11 cows at $30 = 330 • $810 13 horses at $80 = $1040 5 cows at $30 = 150 $1190 Model B. Check: 6 pounds at .75 = $4.50 11 pounds at .43 = 4.73 $9.23 13 pounds at .75 = $9.75 5 pounds at . 43 = 2.15 $11.90 EXERCISE 73 Solve the following simultaneous equations for x and y^ choosing for elimination the letter which requires less work : 1. 2x + 2/=5, 3x + 2/=7 2. 2x-\-y = n, x + 2y=7 3. 3x + 22/ = 13, x + 2y =7 4. 3 X + 4 !/ = 13, 4 a; + 2/ = 13 6. x-\-y = 17, x-y = 5 e. x + dy =7, Sx + 5y = 13 7. 2x + 5y = 17, 4:X + Sy = 13 s. x + 2y = 12, x-2y = -S 9. 2x + y = 12, Sx + 2y = 19 10. x-\-10y = IS, x + Qy =9 11. 2 re + 5 2/ = 29, Sx-2y = -4: 12. 3x + 2/ = ll, 10x-Sy=5 13. 5x + 42/=27, 5y — 4:X = —5 14. Qx + 5y =27, 5x-\-y = 1S 15. Qx — dy = 9, 5x — y = 17 le. 2x- y = -2, Sx + y =7 \ TWO-LETTER PROBLEMS 185 17. 18. 19. 20. 21. 22. 23. 24. ^ 25. 26. 27. 28. 29. ^30. 31. 32. 33. 34. 35. 36. 37. 38. 39. <3r40. 41. 42. 43. 44. 45. 2x + y = lh dx-y=9 2x-y =2, dx- y = 5 3x + 4:y=S2, 4x-2/ = ll Sx — 4:y = —l, 4:X — y = lQ 2x-\-Sy =27, 3x-2y =S x + y = 11, x-Sy = -13 Sx- 5y = -27, 5x-Sy =27 4:X + 5y=SS, X — y = —4: 3x + 42/=22, 2x-y = ll x + Sy = 12, Sx-2y = 14: 2x + 52/=36, Sx-4y = -15 3x + 42/=36, 2a:-52/ = -22 6 X - 2/ = 20, Qx + 5y =80 5x+10y = 15, 5x-Sy =2 2x-y =9, Sx+lOy =4:8 2x + y = 16, 3x-\-10y = 58 5x-4y =4Q, 4 a; + 10 ?/ = 50 lOx + Sy =40, 5x-2y = -15 -2y = -18 4x Sy = -22, a; 3x + 52/=35, 5a;-22/=48 5x-Sy =44, 4x- lOy =2\ Qx + 5y =41, 5x- 4y =2Q Qx- 5y =2Q, 5x + 4y =S8 6x — 5y = 17, 5x — 4?/ = 15 Qx — 5y = 17, 5x — 4?/ = 15 5x + 3y = 53, 4x- lOy = -32 5x-32/=5, 4a:+10 2/=66 5 X + 3 2/ = 46, 3 X + 10 2/ = 85 5a;-32/=25, 3x-10 2/ = -26 5 a; + 3 2/ = 38, 3 a; - 10 2/ = 11 ^ 186 ELIMINATION BY COMBINATION 46. 5 rc - 3 2/ = 37, 3 x + 10 2/ = 34 47. 3x - 51/ = -41, 5 a; + 2 2/ =35 48. 3a; - 52/ = -38, 5a;-22/=0 49. 3x — 42/ = 18, 2a; — 52/=5 50. 3 X - 4 2/ = 14, 2 a; + 5 2/ = 40 213. The process of getting rid of an unknown letter is called eliminating that letter. The method used above is called elimination by combination. There is another method, which will be described later. 214. The student is advised to check his numerical work in advance by constructing the loci of the equations he is to work upon and finding the point where they intersect. The most convenient way of constructing the loci is by means of the intercepts. Instead of actually drawing the lines on the algebraic diagram, the student will find it easier to take a sheet of tracing paper on which a straight line has been drawn. Lay the tracing paper over the algebraic diagram so that the line lies on the intercepts of one of the two equations. Lay a ruler with its edge on the other two intercepts; and with a pin mark the point where this edge crosses the tracing-paper line. This point should correspond to the values of x and y in the answer. EXERCISE 74 Solve the following equations: 1. 2a; + 72/=36, 5x + 32/=61 2. 5a; + 82/ =34, 9a; + 22/=24 3. 13 X + 17 2/ = 107, 2 X + 2/ = 10 4. 14x + 92/ = 100, 7x + 22/=30 6. X + 15 2/ = 37, 3 X + 7 2/ = 35 6. 15 X + 19 2/ = 79, 35 X + 17 2/ = 157 TWO-LETTER PROBLEMS 187 7. 6 a; + 4 2/ = 90, 3 x + 15 2/ = 123 8. 39 a; + 27 1/ = 213, 52y + 29x= 318 q9. 18 a; + 98 2/ =664, Q3x + 7y = 112 10. 26 a; + 101 2/ = 511, 103 a; + 39 2/ = 941 "^11. 2 x + 7 ^ = 49, 6 a; - 5 2/ = 17 12. 7 a; - 4 2/ = 2, 25 2/ - 13 X = 49 j^ 13. a; + 2/ = 90, x — 2/ = 13 14. 4 x + 9 2/ = 50, 7 x - 17 2/ = 22 ^ 15. 5 a; - 3 2/ = 17, 12 2/ - 7 a; = 23 r 16. 171 x-21Sy = 642, 114 a; - 326 2/ = 244 17. 9 2/ - 7 a; = 43, 15 X - 7 2/ = 43 18. 12 a; + 7 2/ = 176, 3 2/ - 19 x = 3 19. 43 x + 2 2/ = 266, 12 a: - 17 2/ = 4 020. 5x4-92/ = 188, 13 X - 2 2/ = 57 21. 4 X + 3 2/ = 22, 3 X + 5 2/ = 11 22. 5 X - 4 2/ = 57, 17 X - 3 2/ = 162 23. 3x — 52/ = 51, 5x + 72/=39 24. 7 2/ - 5 X = 143, 3 X + 5 2/ =89 Q)25. 4x+ll2/ = 144, 172/ — 8x=414 26. 2x-72/=8, 32/-9x=21 27. 17 X + 12 2/ = 59, 19 X - 3 2/ = 148 28. 8x + 32/=3, 4x + 32/ = l C29. 59 2/ - 17 X = 123, 2x - 13 2/ = -17 30. 3X + 22/ =42, 13x + 232/=225 EXERCISE 75. 1. For 8 cows I get $20 more than I pay for 40 sheep; and for 16 sheep I pay $21 more than I receive for 3 cows. Find the price of each animal. 188 ELIMINATION BY COMBINATION ' 2. Find two numbers whose difference is 3V of their sum, and is 3 less than ^ of the larger number. 3. In 10 hours A walks 1 mile less than B does in 8 hours; and in 6 hours B walks 1 mile less than A does in 8 hours. How many miles does each walk per hour? 4. If A's money were increased by 36 cents, he would have three times as much as B; if B's money were increased by 5 cents, he would have half as much as A. How much money has each? 6. A pound of tea and 6 pounds of sugar cost 72 cents; if sugar were to rise 50 per cent and tea 10 per cent, the same quantity would cost 84 cents. Find the price of the tea and of the sugar. 6. Find two numbers such that 3 times the greater ex- ceeds twice the less by 29, and twice the greater exceeds 3 times the less by 1. 7. Three men and 16 boys together earn $107.25 in 5§ days, and 4 men with 10 boys earn $192.50 in 11 days. What are the daily wages of each of these men and boys? 8. A farmer bought land for $7500, part of it at $80 per acre and part at $50 per acre. The cheaper land he sold at a loss of 10 per cent, and the more valuable land at a gain of 10 per cent. On the whole, he profited $50 by the trans- action. How much land did he buy? 9. If I divide the smaller of two numbers by the greater, the quotient is .36 and the remainder .64. If the greater of the two numbers is divided by the smaller, the quotient is 2 and the remainder 27. Find the numbers. -i- 10. The sum of the ages of a father and a son will be doubled in 25 years; the difference of their ages is J of what the sum will be in 20 years. How old are they? TWO-LETTER PROBLEMS 189 11. Find two numbers such that 3 times the greater ex- ceeds I the less by 439, and 3 times the less exceeds | the greater by 49. 12. Eight years ago A was 5 times as old as B; 7 years hence he will be only twice as old. How old are they now? -- 13. A merchant offers me 8 pounds of black tea and 24 pounds of green tea for $10, or 14 pounds of the black tea and 17 pounds of the green tea for the same sum. If it makes no difference to him which offer I accept, what are his prices per pound? 14. A and B buy a horse for $550. A can pay for it if B will advance f of the money he has in his pocket; and B can pay for it if A will advance J of the money in his pocket. How much money has each man? 15. A sum of money was divided equally among a certain number of persons. If there had been 10 more persons, each would have received $2 less; if there had been 10 less, each would have received $3 more. How many persons were there and what was the share of each? 16. If a rectangular lot of land were 8 feet wider and 2 feet longer, it would contain 960 square feet more. If it were 2 feet narrower and 8 feet shorter, it would contain 760 square feet less. What is its area? 17. A farmer bought a certain quantity of eggs at 2 for 5 cents, and other eggs at 3 for 8 cents. Wishing to sell them to a relatiye at cost, he named a price of 31 cents per dozen, but found he would lose 5 cents. His relative then sug- gested 5 for 13 cents, and the farmer accepted that price, but made a profit of 10 cents on the transaction. How many dozen were there of each kind? 190 ELIMINATION BY COMBINATION 18. It takes me 16 times as long to walk around the edge of a long rectangular field as to walk directly across it. The next field, which is 30 feet wider and 150 feet shorter, has the same area. What are the dimensions of each field? 19. An electric car 40 feet long, running parallel to a rail- road, passes a train 455 feet long, running in the opposite direction, in 9 seconds. The electric car is later overtaken and passed in 15 seconds by the same train returning at the same speed. How fast was the car going? How fast was the train going? 20. A cruiser 386.8 feet long passes a battleship going in the same direction in 1.12 minutes; returning, she passes the same battleship in 12 seconds. The length of the battle- ship is 352.4 feet. Supposing each ship maintains the same speed throughout, how many miles per hour do they go? (5280 feet = 1 mile.) Find the fractions described as follows: 21. If 4 is added to the numerator, the value of the fraction will become 1. If 3 is added to the denominator, the value of the fraction will become |. 22. If 1 is subtracted from the numerator, the value of the fraction will become J. If 11 is added to the denominator, the value of the fraction will become J. 23. If 8 is added to the numerator, the value of the fraction will become f . If 8 is added to the denominator, the value of the fraction will become J. 24. If 17 is added to numerator and tO denominator, the value of the fraction will become |. If 10 is subtracted from numerator and from denominator, the value of the fraction will become |. 25. If 3 is taken from the numerator and 1 is added to the denominator, the value of the fraction will become |. TWO-LETTER PROBLEMS 191 If 3 is added to the numerator and 5 to the denominator, the value of the fraction will become |. 26. Find two fractions with numerators 4 and 11 re- spectively, such that their sum is 2Jf , and if their denomina- tors are interchanged their sum is 2|J. 27. A fraction is such that, when it is multiplied by f , the sum of its terms is 107. If 8 is added to each of its terms, the value of the fraction becomes |. 28. A proper fraction is such that if the numerator is halved and the denominator increased by 3, its value be- comes i. If the fraction is multiplied by |, the difference of its terms will be 53. 29. Two fractions have denominators 20 and 10, re- spectively. The fraction formed from these two by taking for its respective terms the sums of the corresponding terms of these two fractions is f ; and the fraction similarly formed by taking the differences is f . Fiiid the numbers described as follows. Unless otherwise stated, the number has two digits. 30. A number divided by the sum of its digits gives 7f for a quotient. If 54 is subtracted from the number, the digits are interchanged. 31. If I divide a number by the sum of its digits, the quotient is 6 and the remainder 7. But if I invert the order of the digits and then divide by the sum of the digits, the quotient is 4 and the remainder 6. 32. A number exceeds 5 times the sum of its digits by 8. If the order of the digits were reversed, the number would be 15 less than 7 times the sum of the digits. 33. A number is 4 times the sum of its digits; and if 27 is added to the number, the order of the digits is interchanged. 192 ELIMINATION BY COMBINATION 34. A number is two more than 5 times the difference of its digits. If its digits are interchanged, the resulting num- ber is 8 times the sum of its digits. 35. Two numbers have the same digits in opposite order. The difference of the numbers is 3 times the sum of the digits; and the sum of the digits is 2 more than 2| times the difference of the digits. 36. If 45 is subtracted from a number, the digits are in- terchanged; but the same result might have been obtained by subtracting 7 from the number and then dividing by 2. 37. Two numbers which have the same digits in opposite order differ by 18; and the smaller number is 4 times the sum of the digits. 38. If a certain number is divided by the sum of its digits, the quotient is 4 and the remainder 6. If the digits are interchanged, the resulting number divided by 5 gives 4 more than the sum of the digits. 39. A number consists of 3 digits, the middle digit being zero. If the order of the digits is reversed, the number is increased by 396. If the second and the third digits change places, the number is increased by 3 more than 6 times the sum of the digits. 40. A number consists of three digits, the middle digit being 6. It has the order of its digits reversed if we add to it 22 times the sum of its digits; and the number so formed is 72 less than double the original number. 41. There is a number between 1000 and 10,000, of which the second digit is 6 and the fourth is 3. Reversing the order of the digits increases the number by 909; but if only the first and third digits change places, the number is in- creased by 2970. TWO-LETTER PROBLEMS 193 The following numbers may be used to form new problems for Exs. 1-20. Substitute the numbers here, in the order given, for the numbers in the corresponding problem. That is, the problem formed from Ex. 1 of Set A will read: For 5 cows I get $45 more than I pay for 20 sheep; and for 19 sheep I pay $48 more than I receive for 2 cows. Set a. 1. 5, 45, 20, 19, 48, 2. 11. 5, i 508, 5, |, 404. 2. },d,l 12. 3,3,7,2. 3. 6, 2, 4, 4, 1, 7. 13. 20, 12, 15, 10, 21. 4. 9, 4, 4, I 14. 100, i i 5. 10, 8, 5.88, 10, 25, 6.54. 15. 6, 8, 6, 12. 6. 5, 3, 1.06, 3, 5, 22. 16. 6, 10, 1260, 5, 10, 1000. 7. 5, 3, 80.50, 7, 2, 7, 18.75, 2^ 8. 1860, 75, 60, 5, 20, 57. 17. 4, 3, 3, 4, 12, 4, 9, 10, 44. 9. .28, .76, 3, 3. 18. 12, 12, 48. 10. 60, i, 20. 19. 38, 298, 7, 14. 20. 380, lj\, 13, 335. Set B. 1. 8, 80, 15, 20, 35, 5. 11. 4, i 302, 4, i, 268. 2. A, 2, h 12. 4, 4, 10, 2. 3. 8, 1, 6, 4, 1, 6. 13. 60, 25, 52, 64, 20. 4. 18, 5, 21, i 14. 340, i, I 5. 3, 5, 1.96, 25, 5, 2.37. 15. 8, 2, 8, 3. 6. 7, 4, 1.77, 4, 7, 54. 16. 5, 8, 1600, 5, 2, 1100. 7. 8, 5, 92, 4, 3, 1, 38.75 ,5. 17. 4, 7, 6, 13, 25, 10, 20, 39, 6. 8. 2760, 88, 72, 12^ 15, 18. 18. 7, 42, 60. 9. .47, .03, 2, 3. 19. 30, 245, 5, 11. 10. 33, h 41. 20. 383.5, 1.4, 12, 355.7. Set C. 1. 4, 109, 11, 20, 165, 3. 11. 2, |, 31, 2, i, 36. 2. h 14, i 12. 6, 2, 9, U. 3. 12, 3, 10, 8, -1, 10. 13. 24, 40, 44, 11, 55. 4. 20, 5, 2, i. 14. 430, I |. 5. 2, 10, 1.50, 20, 30, 1.87. 15. 6, 5, 6, 7. 6. 11, 8, 1.84, 8, 11, 25. 16. 7, 2, 776, 3, 8, 824. 7. 2, 12, 204, Sh 5, 7, 76.50, 3. 17. 3, 2, 6, 5, 9, 1, 14, 11, 20. 8. 3100, 49, 38, 25, 20, 107. 18. 15, 50, 200. 9. .33, .66, 2, 32. 19. 37, 243, 7, 20. 10. 36, i 4. 20. 211.2, 1^, 11, 369, 6. 194 ELIMINATION BY COMBINATION Set D. 1. 8, 50, 26, 13, 30, 3. 11. 6, i 282, 6, |, 208. 2. ^Q, 4, }. 12. 8, 7, 3, 3. 3. 15, -5, 10, 8, -4, 9. 13. 12, 20, 9, 8, 15. 4. 21, 6, 14, i 14. 620, i ^V- 5. 2, 8, 1.30, 30, 20, 1.65. 15. 6, 5, 6, 15. 6. 19, 16, 200, 16, 19, -25. 16. 4, 7, 672, 6, 3, 528. 7. 7, 5, 128, 8, 4, 11, 106.75, 7. 17. 6, 7, 3, 5, 17, 3, 2, 3, 68. 8. 3360, 35, 28, 5, 7h, 77. 18. 8, 27, 54. 9. .78, .22, 1, 23. 19. 48, 502, 11, 25. 10. 43, i 26. 20. 391.8, 1.5, 14, 301.2. Set E. 1. 7, 105, 5, 18, 26, 5. 11. 7, }, 483, 7, }, 333. 2. h 3, h 12. 7, 10, 6, 4. 3. 10, -3, 7, 18, 5, 15. 13. 20, 20, 14, 18, 23. 4. 28, 7, 1, i 14. 365, ^\, ^J^. 5. 5, 1, 2.28, 5, 12i, 2.40. 15. 25, 9, 25, 15. 6. 17, 13, 1060, 13, 17, -460. 16. 9, 3, 543, 1, 7, 257. 7. 3, 5, 68.75, 5, 10, 13, 103.125, 21. 8. 10000, 90, 70, 7, 9, 452. 17. 6, 11, 3, 7, 25, 3, 6, 13, 48. 9. .56, .4, 1, 15. 18. 10, 77, 154. 10. 63, }, 63. 19. 37, 443, 8, 20. 20. 398.1, 1.45, 15, 367.5. Solving for Reciprocals 215. In equations like the following it is better to make the X terms (or the y terms) alike without first clearing of fractions. The reciprocal of x is -. X Model C. — Solve the f oUowing pair equations: ^ 3,2 2 ^5,4 11 ® i+-y = 2i ® hH © X2 ® l-ii ©-© © 21 = x © X 21 2; The value of y may be found in a similar way. wb>^ I .. SOLVING FOR RECIPROCALS 195 EXERCISE 76 1. A and B together can do a piece of work in 8f days. If A worked 3 days and B 5 days, only half of the work would be done. How long would it take A alone to do the work? How long for B alone? 2. Five boys and 10 men could do a certain job in 3 days; one man and one boy would take 24 days to do it. How long would it take one man alone to do the work? One boy alone? 3. Twenty-four pails and 20 cans of water will just fill a certain tank; 6 pails and 14 cans will half fill it. How many pailfuls will fill it? How many canfuls? 4. Two pipes fill a cistern in 20 minutes; if the first pipe were twice as large and the other half as large, the cistern would be filled in 15 minutes. How long would it take to fill the cistern, using only the first pipe? The second? (The word large here refers not to the width but to the capacity of the pipe.) 5. Two pipes can fill a cistern in 5 minutes; if one of the pipes is closed half the time, it will take 7| minutes. How long would it take, using each pipe alone? 6. A and B together could dig a well in 12 days; but at the end of the third day B quits, so that the job takes A 6 days longer than it otherwise would. How long would it take each man working alone? 7. After working 6 days on a certain job with B, A says to him, "I can finish this job alone in 10 days." B replies, ''If we work together one more day, I can finish it alone in 5 days." If what they say is true, how long will it take each alone to finish the job? 8. A can row 11 miles downstream in the time it takes him to row 7 miles against the stream. He rows downstream for 196 ELIMINATION BY COMBINATION 3 hours, then rows back, and at the end of 6 hours he is 5 miles from his starting place. How fast does he row? How fast does the stream flow? 9. B rows 9 miles downstream in 45 minutes. He rows back, near the bank where the current is only half strength, in one hour and a half. Find the speed of the boat and of the stream. 10. C rows 6 hours downstream and 15 hours upstream, covering in all 72 miles. His speed upstream is f of his speed in still water. What is the speed of the stream? 11. D rows downstream for an hour and a half, but it takes him 3 hours to row back. In the first three hours he rows 12 miles. What is the speed of the boat and of the stream? 12. A and B run a mile race. A gives B 12 seconds start and beats him by 44 yards. Then A gives B 165 yards start, and is beaten by 10 seconds. What is the speed of each? 13. A traveler started on a journey of 330 miles, having 4 hours and 12 minutes to spare to make connections with another train at the end of that journey. An accident, which occurred when he was 2 hours out, held up the train for 2 hours, and diminished its speed for the rest of the run. Then he missed his connection by just three minutes. If the accident had happened 6 miles farther on, he would have been just in time to make the connection. How fast did the train go before the accident? How fast did it go after the accident? 14. A was sent to a town 147 miles away, and 7 hours later B was sent after him. After traveling 71 miles, B was handed a letter to deliver to a person living 17 miles beyond the town to which A had been sent. He overtook A just as he was entering the town, and handed him the letter; the letter was delivered just 9 hours and 40 minutes after B re- ceived it. At what speed did A travel? B? WHERE ELIMINATION FAILS 197 Where Elimination Fails 216. Model D. — A certain number consisting of two digits is equal to 4 times the sum of its digits. If the digits are interchanged, it is equal to 21 times the difference of its digits. Find the number. Let X = the tens digit; y = the units digit. Then x -\- y — the sum of the digits. X — y = the difference of the digits. I® X -\- y = the number. lOy -{■ X = the number with the digits interchanged. © lOx + y ^4{x + y) (D 10y + x = 21ix-y) ® 10x-\-y = 4iX-\-4Ly © same values Qx-dy=0 ® - 4.x - 4:y © 2x-y = © -^3 ® lOy -\- X = 21x — 21y @ same values ®3l2/-20a; = ®-21x + 21y © 20 a; - 10 y = ® X 10 ® 2ly = ©+© @ y = ® -T- 21 @ 2 a; = @ subst. in ® @ a; = @ ^ 2 This result is absurd, since a number with zero digits is no number at all. Such a result points to some mistake in the statement of the problem, or in our understanding of it. 217. We took for granted, in forming equation ®, that the tens digit was larger than the units. Let us see if that is the source of the absurd result. © 10x + ?/ = 4x + 4i/ @ lOy + x = 21y -2lx ®22a;-ll2/ = ® + 21x-21|/ @ 2x-y = ®-Ml Equation ® is the same as ©. 198 ELIMINATION BY COMBINATION In this case also, then, we cannot determine, from the two equations given, the particular values of x and 2/, because the two conditions are not independent. 218. Model E. — A certain number of two digits, if added to half the sum of the digits, gives a total of 21. The units digit added to 7 times the tens digit gives a total of 26. Find the number. Let X = the tens digit; y = the units digit. Then a; + ?/ = the sum of the digits. 10 .-c + 7/ = the number. ® 10 X + 2/ 4- H^ + 2/) = 21 (D 7 a; + 2/ = 26 ® 20x + 2i/ + a; + 2/ = 42 ® X2 ® 21 X + 3 y = 42 ® same values ® 7 a: + 2/ = 14 0-3 Equations (2) and © cannot possibly be true for the same values of x and y. Such equations are called inconsistent. Draw their loci and find out how this inconsistency is repre- sented on the algebraic diagram. Non-Algebraic Conditions 219. It happens sometimes that a problem is given, like Model D, which implies only one algebraic condition, but still the answer is restricted to a few particular values, or even to only one. Such a restriction is due to a condition, implied in the statement of the problem, such that it cannot be stated algebraically. In Model D, our digits must be from the nine Arabic numerals. We found by trial that our result was absurd unless the second digit were the larger; and our algebraic conditions both reduced to the fact that the second digit NON-ALGEBRAIC CONDITIONS 199 was twice the first. That leaves, for the only possible solutions: X = 1, 2/ = 2; the number 12 X = 2, 2/ = 4; the number 24 x = 3, 2/ = 6; the number 36 X = 4, y = S; the number 48 Four particular answers to the problem. Check: 12 = 4 (1 + 2); 21 = 21 (2 - 1); and similarly for the other three numbers. 220. Equations which do not give a solution from the algebraic conditions alone are called indeterminate equa- tions. The problems which give rise to such equations, although they may contain enough non-algebraic con- ditions to determine the answer, are often called inde- terminate problems. Find three numbers, each of which will satisfy one of the two in- consistent conditions in Model E. CHAPTER XI ELIMINATION BY SUBSTITUTION 221. A two-letter equation of the first degree is called a linear equation. Any equation in which at least one term contains two unknown letters as factors, and no term contains more than two, is called a quadratic equation. These two unknown letters that are thus used as factors of a term may be the same letter, as in the terms 5 x^ and 32 2/^; or they may be different letters, as in the term 3 xy. The equations that follow are examples of two-letter quad- ratic equations: 5x2- 3a;2/ + 32 2/2 =0 ^y-x'' = ll Zx-2y^ = l Linear-Quadratic Pairs 222. Model A. — A certain rectangular field, containing 480 square rods, requires 104 rods of fence to inclose it. What are its dimensions? I. Let X = the number of rods in length. mu 480 Then — = the number of rods in breadth. ® 2. + 2(^^«)=104 ® rc2 + 480 = 52 X ® X X ■ ^ 2 ® :» ;2 - 52 X + 480 = @- 52 X and so on. Or otherwise. 11. Let X = the number of rods in length. Then 52 — X = the number of rods in breadth. © X (52 - x) = 480 ® 52 a; - a;2 = 480 © same values ® = x2 - 52 a; + 480 - 52 X+X^ and so on. 200 LINEAR-QUADRATIC PAIRS 201 Still another way would be to use two unknown letters, as follows: III. Let X = the number of rods in length; y = the number of rods in breadth. © xy = 480 @ 2 X + 2 ?/ = 104 ® x + y = 52 (2) ^2 Equations ® and ® we cannot solve by the method of elimination heretofore tried. We can find the value of x from the third equation, — not the numerical value that satisfies both ® and ®, but an explicit formula for x. We may then substitute in the other equation; that is, we can use in the first equation a formula for x instead of x itself. Then, if the formula for x had no letter x in it, we shall have an equation from which x has been eliminated. Thus: ® X = 52 - y (D - y ® y (52 - y) = 480 subst. in ® (e) 52y — y^ = 480 ® same values = 2/2 - 52 2/ + 480 ® -52 y + y^ and so on. Or we could in the same way have eliminated y. The use of a second unknown letter may often, as in this case, be a way of explaining how to get, with one letter, abbreviations for two unknown numbers. The method just given is the one by which elimination should usually be done in equations of this kind, but the valuie of one letter may be found from the quadratic equation and substituted in the linear; as: ^ 480 ^ ® X =-. — 0-^2/ y 480 ® [• y = 52 subst. in ® ® 480 + 2/2 = 52 2/ ® X ^ 2/2 - 52 2/ + 480 = ® - 52y 223. Model B. — A cistern can be filled by two pipes, running together, in 2 hours, 55 minutes. The larger pipe by itself will fill the cistern 2 hours sooner than the smaller 202 ELIMINATION BY SUBSTITUTION pipe by itself. How long will it take the larger pipe alone to fill the cistern? The smaller pipe? Let X = the number of hours for the smaller pipe. y = the number of hours for the larger pipe. (1) x-y = 2 ® x^ y 35 ® S5y + S5x ^ 12xy (2) xS5xy ® x^y + 2 ® + y ® S5y + S5(y + 2) = 12yiy + 2) ® subst. in ® ® 701/ + 70 = 12y^+2^y same values ® = 12 7/2 - 46 2/ - 70 ® -70 2/ -70 ® = 6 2/2 - 23 1/ - 35 ®-2 Then y = 5, or — |. Arts. 5 hours for the larger pipe, 7 for the smaller. Check: 5 hours = 7 hours — 2 hours. ^ of the tank + | of the tank = ^f of the tank, in 1 hr. 2 hrs. 55 min. = ff hrs.; ff X B = 1- 224. Model C. — A number of two digits is 9 less than the square of the sum of its digits; and if 45 is subtracted from the number, the digits are interchanged. Let x = the tens digit; y = the units digit. Then lOx + y = the number. (a? + yY = the square of the sum of the digits. lOy + X = the number with digits interchanged. ® 10 X + 2/ + 9 = x2 + 2 xi/ + ?/2 @ 10a; + i/-45 = 10 2/ + a; ® 9x-9i/ = 45 ® - IO2/-X + 45 ® X - y = 5 ® ^9 ® x=y+5 ®+l/ (6) 10 (1/ -h 5) + 2/ + 9 = y^+10y-\-25+2y (y+5)+y^ ® subst. in ® ® Uy + 69 = 47/2 + 20^ + 25 ® same values ® = 42/2 + 91/ -34 @ - 11 2/ - 59 Then y ^ 2, y = -V". a; = 2/ + 5 = 7. Am. ' Check: 72 = (9)^ - 9 72 - 45 = 27 72. LINEAR-QUADRATIC PAIRS 203 225. The rule for this method, which is called elimination by substitution, is : Find an explicit formula for x (or for y) from the linear equation^ and substitute it in the quadratic equation. Both in Model B and in Model C we used in the answer only that pair of values which was easily interpreted in the problem. In general, however, when two values of x and y occur, both pairs should be given in the answer, if there is no implied condition in the problem making either of them unreasonable. EXERCISE 77 Solve: 2. xy = 1] Zx — by =2 3. a;-^±l=_l; a;2 + i/2=25 4. 7 a;2 - 8 x?/ = 159; 5 a; + ? 2/ = 7 5. X - 4 2/ = -8; x2 + 2 2/2 = 34 6. 3 a:2 = 336 - 4 2/2; X + 2/ + 14 = 7. x2 - 2 X2/ + 2 2/2 = 58; a; + 2/ = 13 8. a;2 + 5 2/2 = 4 a;2/ + 13; 3 X - 10 2/ = 1 9. 3 2/ - 2 a; = 6; 4 0:2 = 9 (2/2 + 12) ^°- 2 + 1 = 1'^ + ^ = ^ 11. 3x2- 2x2/ = 15; 2 a; + 3 2/ = 12 12. -+^=5; 3X-22/ = 16 2/ X 2' ^ 13. x2 + 24x2/ = (3 2/ + 4)2 + 8; x =32/ + 2 '^•"^ 2 - 2'^ x + 2-^ 15 3x- lOi/ = 1; x2-x2/ =52/2 + 79 204 ELIMINATION BY SUBSTITUTION 16. 11 2/ + 6 a: = 62; x^ - xy + y"^ = 13 17. -3- - -/ = ^ ; a; + 2/ + 3a^2/ = 93 18. ^±^-^:^=Z; x2 + 6x2/ + 2|/2H-4x+32/=29 19. x2 - 2 XT/ - 2/2 = 31; x + 2/ = 13 20. 10-^tl^=^; 3x2-60^ + 52/ =20 21. 2^+A_li = ^; x2 + 4x^ = 32/2 + 20x + 6 x-y-l y 23. Find two numbers whose sum is 4 times their differ- ence, and the difference of whose squares is 196. 24. A path around a rectangular garden is 7 feet wide and 1806 square feet in area. The area of the path is 994 square feet less than the area of the garden. Find the area of the garden. 25. A number is 7 times the sum of its two digits; and if the number is multiplied by its first digit, the product is 672. Find the number. 26. The difference of two numbers is y\ of the greater number, and the difference of their squares is 380. Find the numbers. 27. If a certain rectangular field were 75 rods longer and 20 rods wider, its length would be double its breadth, and its area would be double what it is now. Find its dimensions. 28. In a certain block containing exactly similar houses, there are 300 rooms. There are 5 more houses in the block than there are rooms in one of the houses. How many houses are there in the block? LINEAR-QUADRATIC PAIRS 205 29. The front wheel of a bicycle makes 16 turns less than the hind wheel in going a mile. If the front wheel were 6 inches more in circumference, it would turn 60 times less than the hind wheel in going a mile. Find the circumference of each wheel. 30. When a certain train has traveled 5 hours, it is still 60 miles short of its terminus. One hour can be saved on the whole trip by running 5 miles an hour faster. Find the speed of the train, and the length of the trip. 31. A and B start on a road race together. A is a sure winner; looking back once on the road, he sees B 60 rods behind. A crosses the line 4 minutes after that, and B comes in 6 minutes behind A. When A looked back, he had as far to ride as B had already ridden. Find the speed of each. 32. The fore wheel of a carriage turns 132 times more than the hind wheel in going a mile. 6 turns of the fore wheel cover 2 feet less than 5 turns of the hind wheel. Find the circumference of each wheel. 33. I rowed 24 miles downstream and back again. I took 8 hours longer on the return trip, because the current reduced my speed to J of what it was going downstream. Find the rate of the current. 34. Two trains start at the opposite ends of a double- track railroad 300 miles long. After they pass, one train takes 9 hours, the other 4 hours, to complete the journey. Find the speed of each train. 35. If 1 is added to the denominator of a certain fraction, the value of the fraction becomes J. If 2.1 is added to the fraction, the fraction is inverted. Find the fraction. 36. If 2 is added to the denominator of a certain fraction, the value of the fraction becomes ^. If 4 is added to each term of the fraction, its value is increased by .125. Find the fraction. 206 ELIMINATION BY SUBSTITUTION 37. A rectangular field 238 square rods in area loses 58 square rods because a strip one rod wide is taken off all around the edge for roads. Find the dimensions of the field that remains. 38. Five leaks and a drainpipe empty a cistern in 4 hours. The average time for one of the leaks to empty the cistern would be 24 hours more than half the time the drainpipe requires. How long will it take to empty the cistern if the leaks are stopped? 39. When a certain kind of cloth is wetted, it shrinks J in length and tV in width. A piece of this cloth used for a rectangular awning was found to have shrunk 8f square yards in area. Then the edging bought for it before it shrank (intended to go on one long side and both ends) was found to be 2 yards too long. What were the length and the width of the cloth before shrinking? 40. The area of a rectangular plot of land is 16 square feet less than a square plot of equal perimeter; and its breadth is 1 foot more than | of its length. Find the di- mensions of the rectangular plot. 41. Two boys run in opposite directions around a rec- tangular field, the area of which is one acre. They start at one corner and meet 13 yards from the opposite corner; and one of the boys could go around 6 times while the other was going 5 times. Find the dimensions of the field. An acre = 160 square rods. A square rod is a square 5.5 yards on a side. Algebraic Diagrams 226. The algebraic diagram for a two-letter quadratic equation cannot be a straight line; for, as we learned in Chapter IX, a straight line is always the locus of a two- letter equation of the first degree. To get a diagram for a two-letter quadratic, we must use the equation as an im- ALGEBRAIC DIAGRAMS 207 plicit formula in which we can give a series of values to one letter, in order to find the corresponding value (or values) of the other. 227. The following example furnishes one of the simplest illustrations of this method of plotting a two-letter quad- ratic equation: Model D. — The diagonal of a rectangle is 5 inches long. The perimeter of another rectangle of the same width but twice as long is 22 inches. Find the dimensions of the first rectangle. Let X = number of inches in the length of the rectangle; y = number of inches in the width of the rectangle. Then x^ -{- y^ = the square of the diagonal. 4x + 2y = the perimeter of the other rectangle. x2 + ?/2 = 25 @ 4:x + 2y = 22 ® 2x-\-y = 11 0^2 y = 11 -2x - 2a; © a;2 + (121 - 44 re + 4 x2) = 25 subst. in 5a;2-44x + 96 = 0-25 , © (5 X - 24) (x - 4) = factored Equation © leads to two solutions, one giving a rectangle 4 inches by 3 inches, the other a rectangle 4.8 inches by 1.4 inches. In drawing the two loci that form the algebraic diagram for this example, gives us a straight line with intercepts 5| and 11. We must consider ® an implicit formula, so that we can give a series of values to one letter and find the corresponding values of the other. Thus we find that when x = 1, 1 -{- y^ = 25, and y^ = 24. Whence, y = 4.9 and y = —4.9. We shall have the same values if a; = —1. In that way we can construct a table as follows: X 1 1 2 233445 y 4.9 -4.9 4.6 -4.6 4-4 3-3 When X is greater than 5, y^ becomes negative, and con- sequently we cannot find a corresponding value of y. 208 ELIMINATION BY SUBSTITUTION 228. If we now plot the values in the above table, we shall find that they lie on a semicircle, to the right of the 2/-axis, with its center at the point where the axes cross {the origin). If we plot the values of y corresponding to the negative values of x, we shall find them lying on the other half of the same circle. The locus of equation ©, then, is a circle with radius 5, having its center at the origin. 229. With our knowledge of geometry we can prove that the circle just described is the ^^^- ^' required locus of equation ®; for the equation x^ + y^ = 25 simply means that the slant side OP of the right triangle OMP (where P is the point represented by x and y) must always be equal to 5. Consequently the point P will have a constant distance 5 from the origin. The Equation of the Circle 230. Find the equation of a circle with radius 10, having its center at the point which represents the number-pair 5,7. The "number-pair 5,7" means the pair of values a; = 5 and y = 7. Every point on the circle will represent a pair of numbers. Thus P represents the numbers x and y, of which a; is de- FiG. 2. ALGEBRAIC DIAGRAMS 209 fined by the length of OK, and y by the length of KP, just as the length of OL is 5 units and the length of LC is 7 units. Since P is on the circle, CP must be equal to 10 units; and that gives us the equation {x - 5)2 +{y- 7)2 = 100 (Why?) This will be true whenever P is on the circle. Even when X is less than 5, making x — b negative, the square {x — 5)^ will be positive. In the same way, {y — 7)^ will always be positive. We shall be concerned, then, only with the differ- ence of y and 7, not with its sign. For every point on the circle, therefore, © {x - 5)2 + {y- 7)2 = 100 ® a;2 - 10 X + 25 + 2/2 _ 14 y _}_ 49 = 100 Q same values ® x2 + 2/2 - 10 X - 14 ?/ - 26 = 0-100 EXERCISE 78 1. Find the equation of the circle with center at 3,2 and the radius 5. 2. Find the equation of the circle with center at 5,0 and the radius 3. 3. Find the equation of the circle with the center at 0,3 and the radius 8. Find the equations of the circles having; 4. Center at 3,5 and radius 6. 5. Center at —2,4 and radius 7. 6. Center at 5,-3 and radius 4. 7. Center at —4,-3 and radius 9. 8. Center at 7,— 2 and radius 11. 9. Center at —5,8 and radius 3. 10. Center at 4,— 4 and radius 4. ^' c-^-. 210 ELIMINATION BY SUBSTITUTION 11. Find the equation of the circle with center at the origin, and radius 17. Where is this circle intersected by the straight line that passes through the points 3,20 and 11,12? 12. Find the equation of the circle that has its center at the point 2,5 and its radius equal to 5. Where is this circle intersected by the line that cuts each axis 14 units from the origin? Other Quadratic Loci 231. The equations of circles that we have thus far found have, for terms of the second degree, the sum of the squares of x and y. We have not yet proved, though it is true, that every equation of this type represents a circle. Quadratic two-letter equations of other types, however, are very common. Let us consider first those appearing in the model problems of this chapter. In Model A, we have the equation xy = 480 and we take this as an implicit formula for y, using, only such values of x as make computation easy and give points con- veniently placed in the diagram: x 8 10 12 15 16 20 24 30 32 40 48 60 y 60 48 40 32 30 24 20 16 15 12 10 8 If we changed the sign of each value of x, we should have the negatives of the same values of y. The points thus represented would lie in the lower left-hand quarter of the diagram. Drawing the line represented by the linear equation of Model A, we find that the loci intersect in two points which correspond to the two solutions of the equations. .HiA .-^ ..Ac^r^ir Jo^^^ ALGEBRAIC DIAGRAMS 211 "^~ ■^~" > V -80 "^■~ \ s \ \ I V \ s> in s ^ k^ >!o ^ s50 ■ ' -^ \ V ^ s \ V \ V \ \ \ Fig. 3. 232. In Model B, equation gives for each value of x one value of y, just as in Model A; but in this case the negative values of x give totally different values of y. Thus we have, for the value x = 10, the following equation for y: 35 2/ + 350 = 1201/ from which we obtain. i/ = 4.0. On the other hand, for the value x = — 10, the equation becomes ® 35 2/- 350 = -1202/ @ 155 2/ = 350 from which we obtain y = 2.3. © + 1202/ + 350 212 ELIMINATION BY SUBSTITUTION The table of corresponding values is: a: -10 -7 -5 -2 -1 1 2 3 4 5 y 2.3 2.0 1.8 1.3 .7 -1.5 -6.4 105 10.8 7 We can plot the locus of x — y = 2 and find the two points of intersection of the loci. 7 10 5 4 Fig. 4. 233. In Model C, the equation we have for an implicit formula is 10 X + 2/ + 9 = x2 + 2 a;?/ + 2/2 and when we substitute successive values of x we get in each case a quadratic for y, which would generally have two roots. Thus we get the following table: X Quadratic for y Values of y 2/2-2/-9=0 3.5, -2.5 1 2/2 + y - 18 = 3.8, -4.8 2 2/2 + 32/- 25 =0 3.7, -6.7 3 i/2 + 5i/-30 = 3.5, -8.5 5 2/2 + 9?/- 34 =0 2.9, -11.9 7 2/^ + 13 2/ -30 =0 2,-15 -1 2/' -3 2/ + 2=0 1,2 -2 2/^ - 5 2/ + 15 = none -3 2/^^ - 7 2/ + 30 = none -.5 2/2 - 2 2/ - 3.75 = 3.2, -1.2 ALGEBRAIC DIAGRAMS 213 By plotting the values, we get the curve given in the fol- lowing diagram. This curve is cut by the line x — y = 5 in the point 7,2, and also in the point |,— 4j. Fig. 5. ^ ^ ■^ /^ -— ^ s I ^ 1 ^ Ji^ ^ S £ ; ' : . ) I \ \ / ^ ^ — Fig. 6. 234. The equations of Ex. 5, Exercise 77, give the diagram of Fig. 6. 214 ELIMINATION BY SUBSTITUTION 235. The curves obtained for Models A and B are called h3rperbolas. Each of these curves consists of two branches, and does not inclose a space. The curve obtained for Model C is a curve of one branch ; this also is open, that is, it does not inclose a space. It is called a parabola. The curve shown in Fig. 6 is a closed curve, not all the diameters of which are equal, as in a circle; it could be described as a circle shrunk crosswise. It is called an ellipse. A quadratic locus can never be a single straight line, but it can be a pair of straight lines. Such figures as these form the subject matter of a very interesting study called Analytic Geometry. For our pur- pose here it is enough to notice that the locus of a quad- ratic two-letter equation can generally be cut by a straight line in two points. EXERCISE 79 Plot ihe following loci: r 1. (Circle) x" + y^ =Sx-\-Qy 2. (Circle) x" + y^ - S x = 2S 3. (Pair of straight lines) 24 y^ — 25 x^ = 25 xy 4. (Parabola) 2/' = 4 x - 20 5. (Parabola) S {x - 2yy = 7 x + Q2 6. (Parabola) x^ - Qxy + 9y^ = Ax + 4S 7. (EUipse) x2 + 3 2/2 = 5 X + 18 8. (EUipse) x^ + xy + y^= 28 9. (Hyperbola) 2x^ — xy — Qy^ =28 10. (Hyperbola) 5 x^ — 3 xt/ =65 ALGEBRAIC DIAGRAMS 215 The standard Parabola 236. The locus of the equation y = x^ is called the Standard Parabola. It is plotted as follows (Fig. 7) : \ X : ±1 ±2 ±3 ±4 ±5 ±6 ±7 ±8 \ y 1 4 9 16 25 36 49 64 237. The standard parabola and its equation can be utilized, as will now be shown, for a graphical check on the solution of a one-letter quadratic equation. Consider the equation x^ -\- 5 x = 7. This can be written x^ = — 5 x + 7. This equation will be satisfied by any value of x that will make x^ and — 5 a; + 7 equal; that is, by any value of x that will give the same value of y in the two explicit formulas y = x^ and y = — 5 a; + 7. The first of these formulas is the standard parabola, and the other is a straight line having an intercept 1.2 on OX and 7 on OY. Where these two loci intersect, the same number-pair will satisfy both formulas. Our check, then, is as follows: On a carefully drawn dia- gram of the standard parabola, lay a straight line (a ruler- edge, or a line ruled on tracing paper) through the intercepts of the line y = — 5 x + 7. Either point where this line crosses the parabola will represent a value of x which gives the same value of y in both loci. This value of x is, there- fore, a root of the equation x^ + 5 x =7. 238. In the same way, the equation 5 x^ — 14x + 2 =0 can be written x^ = 2.8 x — .4. Its roots will be indicated by the intersections of the standard parabola with the straight line y = 2.8 x — A, or 5y = 14 x — 2. It is easy to see that the straight line will sometimes touch the standard parabola in only one point. What may 216 ELIMINATION BY SUBSTITUTION 50 \ OK 1 "n u / \ 4u\ 20 -1 5 -1 S 0' ■ ) 1 1 5 2( Fig. 7. ALGEBRAIC DIAGRAMS 217 we then conclude about the roots of the equation? Also, the line might miss the parabola altogether. Then we know that the roots of the equation are imaginary. EXERCISE 80 By using the diagram of the standard parabola, Fig. 7, find roughly the roots of the following equations, and verify them by the algebraic solutions. 1. a;2=x + 30 6. 2a;2-16=3x 2. x2 - 15 = 2 X 7. 7 x2 = 32 3. 2 x2 + X = 15 8. 2 x2 - 9 a; + 14 = 4. x2 = 16 a; - 20 9. 2 x^ - 19 x + 14 = 5. 4a;2-28a; =49 lo. lla;^ + 13a; + 3 = CHAPTER XII SUPPLEMENTARY PROBLEMS FOR PRACTICE AND REVIEW Invent algebraic expressions for the following sets of numbers. 1. Any two numbers; any three numbers. 2. The sum of any two numbers; the difference of any I wo numbers. 3. The product of any two numbers; the product of any three numbers. 4. Any two consecutive numbers; any three. 5. The product of any two consecutive numbers. 6. The product of any three consecutive numbers. 7". The sum of any two consecutive numbers. 8. The sum of any three consecutive numbers. 9. Any odd number; any even number. 10. Any two odd numbers; any two even numbers. 11. Any two consecutive odd numbers; their sum; their product. 12. Any two consecutive even numbers; their sum; their product. 13. Any three consecutive odd numbers; their sum; their product. 14. Any three consecutive even numbers; their sum; their product. 218 PROBLEMS FOR PRACTICE AND REVIEW 219 15. The square of any number; the squares of any two numbers. 16. The sum of the squares of any two numbers. 17. The difference of the squares of any two numbers. 18. The square of the sum of any two numbers. 19. The square of the difference of any two numbers. 20. The difference of the squares of two consecutive odd numbers. 21. The difference of the squares of the two different numbers formed by the same two digits; the square of their difference. II 1. What is the supplement of X degrees? 2. What is the complement of 3 X degrees? 3. What is the complement of Y degrees? : 4. What is the supplement of 5 F degrees? 6. Two angles have a ratio 3. If the number of degrees in the smaller is represented by A, how many degrees are there in their sum? In the complement of their sum? 6. Two angles have a ratio 11. If the number of degrees in the smaller is represented by B, how many degrees are there in the supplement of their sum? 7. Two angles have a ratio 4. Represent the number of degrees in the smaller by some letter, and obtain a way of expressing, first, their difference; and then, the comple- ment of their difference. 8. What is the complement of C degrees? Of C + 1 degrees? Of C + 2 degrees? What would these three values be if the value of C were 20? 9. What is the supplement of D degrees? Of D — 1 degrees? Of D — 2 degrees? What would these three values be if the value of D were 68? 220 PROBLEMS FOR PRACTICE AND REVIEW 10. What is the complement of A + 20 degrees? Of A — 33 degrees? 11. What is the supplement of X + 30 degrees? The complement of X — 25 degrees? 12. What is the complement of 2 X — 3 degrees? The complement of 3 X — 2 degrees? The supplements of the same angles? 13. If two angles have a ratio 5, how would you repre- sent them? If the angles were supplements, how could you state that fact by means of an equation? 14. In the stripe in Fig. 1, the angle A is 8° less than X; A and Y have the ratio 3. How many degrees are there in X? Fig. 1. Fig. 2. 15. In the stripe in Fig. 2, the angle X is 12° less than three times the angle 7. How many degrees are there in X? 16. In Fig. 3, A = 38.3°, C = 53.6°. How large is B? Fig. 4. Fig. 3. 17. In Fig. 4, Z = 56.4°, X and Y have the ratio 2. How large is X? PROBLEMS FOR PRACTICE AND REVIEW 221 18. In Fig. 5, the ratio of Q to P is 1.6, and R is 24° less than their sum. How large is each angle? 19. In Fig. 6, A = C — 5, and the ratio of A to 5 is ^. How large is each angle? Fig. 5. Fig. 6. Fig. 7. 20. In Fig. 7, X is 35° less than 7, and Z is 2° greater than Y. How large is each angle? 21. In each of the following stripe diagrams, how many degrees does A represent? ^ '' U+30 Fig. 8. 22. In each of the following diagrams, how many degrees are there in the angle X? Fig. 9. 222 PROBLEMS FOR PRACTICE AND REVIEW 23. The ratio of the angles at the base of a triangle is If; and the ratio of the angle at the vertex to the larger base angle is 1.4. Find the angles of the triangle. Let one angle equal 180° minus the sum of the other two angles. 24. The ratio of the two smaller angles of a triangle is 1.33; and the ratio of the largest angle to the smallest is 2.17. Find the angles of the triangle. Ill Convert into Fahrenheit the following temperatures Centigrade : 1. -80° C. 6. 20° C. 2. -40° C. 6. 31.32° C. 3. 0°C. 7. 98.16°C. . 4. 4°C. 8. 458. 1°C. 9. 33.44° C. Convert into Centigrade the following temperatures Fahrenheit : 10. 0°F. 14. -40° F. 11. 16° F. 15. -80° F. 12. -15°F. 16. 68° F. 13. 32° F. 17. 91.56° F. 18. What temperature Fahrenheit is just double the cor- responding temperature Centigrade? What temperature Centigrade is just double the corresponding temperature Fahrenheit? 19. What temperature has the same number of degrees in both Centigrade and Fahrenheit scales? 20. A certain temperature, observed with Centigrade and Fahrenheit thermometers, is recorded by writing numbers that differ by 60. What are the two thermometer readings? PROBLEMS FOR PRACTICE AND REVIEW 223 IV Solve: 1. 7 (19 X - 21) = 11 (12 X - 3) - 30 2. 9 (25 x + 11) = 19 (12 a; - 5) + 2 3. 51 X - 1135 -f 13 (103 -x) = 11 (2 X + 20) 4. 203 - 17 X = 3 (x - 505) - 2 6. 1170 + (x - 4) = 11 (x - 3) + 22 6. 9 (103 X - 820) = 206 (2 - x) + 139 7. 800 {x - 10) + 2 (11 X - 328) - 1 - 35 x = 8. 3000 (x - 5) = 10 (100 - 27 x) H- 355 -x 9. 10 (2 X - 3) = 2 (x + 15) + 6 10. 3(3x- ll) + 2(5x + 2) =5 + 9(2x-5) 11. 5 (7 X - 2) + 2 (5 X - 13) = 3 (11 x + 2) + 26 12. 17 (3x- l)+3(5x- 17) = ll(5x-3) + x + 5 13. 13 (x - 11) + 17 (2 X - 27) - 1 = 3 (x - 29) 14. 101 (5 X + 17) = 203 (3 X + 23) + 3 X - 3059 15. 4 (x - 13) = 13 (x - 4) - 81 16. 117 (x - 25) + 25 (x + 117) = 3x + 50 + 3 (7 - x) 17. 8 (x - 1) + 4 (9 - 4x) = 4 + 17 (3 - x) 18. 17 (2x + 4) + 13 (2x - 5) = 3 (x + 17) 19. 51 X - 1137 + 13 (103 - x) = 11 (2 X + 20) - 2 20. I (18x - 252) = § (15 - 3x) + 1 21. i (24x - 800) = f (45 - 5 x) - 5 22. I (60x - 875) = i (35 + lOx) - f X 23. f (1185 X - 5925) = A (11 X - 55) • 24. i (93x - 138) = f (lOx - 25) 25. t (56 X + 42) = 5 (2 X + 16) 26. f (42 X - 48) - X = 3 (x -f 17) 27. I (60 - 22 x) + T% (30 - 50 x) =0 28. i(5x + 25) + f (4X-28) =f (3x + 9) 224 PROBLEMS FOR PRACTICE AND REVIEW 29. f (21 a: - 57) = 2 (6 X + 17) 30. ISx + 3 + t (18 - 12a:) = 3 (7 - a;) + 11 - a; 31. 31 (x - 22) + f (21 x + 18) = 2a: + 883 ^^' 4 - 10 +20 + ^1 5 ; 5 a; + 1 _ 21 - 3 a; » /2 a; + 3\ 6- 1. A is 12 years older than B; in 2 years he will be 4 times as old as B. What are their ages now? 2. A is 6 times as old as B; in 4 years he will be only 4 times as old as B. What are their ages now? 3. A is 18 years older than B; in 5 years he will be 3 times as old as B. Find their ages now. 4. A is 9 times as old as B; in 6 years he will be only 6 times as old as B. Ages now? 5. A is 7 years younger than B; in 4 years he will be half as old as B. Ages now? 6. In 3 years A will be i as old as B; now he is 20 years younger than B. Ages now? 7. A's age at present is J of B's; 10 years ago it was } of B's age. Ages now? 8. A was 8 times as old as B one year ago; he is now only 5 times as old as B. Ages now? 0. A is 30 years older than B; 5 years hence his age will be 6 times B's. Ages now? 10. A is 7 times as old as B; in 10 years he will be only twice as old as B. Ages now? 11. The difference between a father's age and his son's is 28 years. How old was the father when the son was i of his PROBLEMS FOR PRACTICE AND REVIEW 225 father's age? How old was the son when the father was 9 times as old as the son? 12. I paid $68 with $10 bills and $2 bills, using 10 more twos than tens. How many bills of each kind did I use? 13. I paid $2.35 with quarters and nickels, using 5 more nickels than quarters. Find the number of coins of each denomination. 14. I paid $139 with fives and twos, using 3 more twos than fives. Find the number of bills of each denomination. 15. I paid $7.55 with dimes and quarters, using 12 more quarters than dimes. Find the number of coins of each denomination. 16. I paid $6.40 with nickels and half-dollars, using 4 more halves than nickels. Find the number of coins of each denomination. 17. Three times as many dimes as nickels, 5 times as many quarters as dimes; in all $28.70. How many of each? 18. A man borrowed some coins and paid back 33 more coins of another sort. The coins borrowed were half-dollars, those repaid were nickels; he still owed $1.50. Find the number of coins borrowed. 19. I sent for deposit two rolls of bills; a roll of twos and a roll of fives. The latter roll contained 50 bills fewer, but was worth $26 more, than the first roll. How many bills were there in each roll? 20. In a packet of $10 bills there are 3 fewer bills than in a packet of $2 bills which lies near by. The first packet is worth $74 more than the second. How many bills are there in each packet? 21. A shopkeeper, counting his cash, finds 26 more nickels than quarters. The nickels amount to 10 cents less than the quarters. How many coins of each kind has he? 226 PROBLEMS FOR PRACTICE AND REVIEW 22. A man paid $2.20 in quarters and nickels, using two more nickels than quarters. Find the number of each kind of coin. 23. A carpenter worked on a job, and at the end of the day sent in his time as 10 hours, although he had worked but 8 hours. He had worked overtime, and for every hour overtime he counted IJ hours. How many hours did he work overtime? 24. A carpenter worked from 2 : 30 to 8 : 30 p.m. He sent in his time as 10 hours, counting every hour overtime as 2 hours. How many hours did he work overtime? 25. A plumber worked from 11:45 a.m. to 8: 15 p.m. and sent in his time as 10 hours, counting every hour overtime as IJ hours. How many hours did he work overtime? 26. A gas fitter worked from 11:30 a.m. to 9:30 p.m. and sent in his time as 12J hours, counting every hour overtime as IJ hours. How many hours did he work overtime? 27. A steam fitter worked 7| hours and sent in his time as 11 hours, charging overtime double. How many hours did he work overtime? 28. A plumber worked 9 hours and sent in his time as 10§ hours, charging overtime as time and a half. How many hours did he work overtime? 29. A gas fitter spent on a job 12| hours, and sent in his time as 15 hours, charging overtime as time and a third. How many hours did he work overtime? 30. The common stock of a gas company pays 4 per cent and the preferred stock 7 per cent. 45 shares, part of which is preferred stock, pay $240 per year. How many of these shares are common stock? How many are preferred stock? 31. An insurance agent proposes to insure a man for $9000 by two limited-payment policies, one at $40 per thousand, the other at $35.60. The aggregate annual PROBLEMS FOR PRACTICE AND REVIEW 227 premium on the two policies is $338. How many thousand dollars are there in each poUcy? 32. An insurance agent proposes to insure a man for $11,000 by two policies, one at $38.53 per thousand, the other at $43.15 per thousand, the aggregate premium be- ing $442.31. How many thousand dollars are there in each policy? 33. A college student borrowed $1000 and 20 years after- ward he repaid it with $1080 interest. The interest ran at 6 per cent for some years and was afterward reduced to 5 per cent. How long was the higher rate of interest in force? 34. An investor bought 24 shares of stock, from which he received annually $100 in dividends. Some of the stock paid 5 J per cent dividends; four times as many shares paid 4| per cent, and the rest paid 3 per cent dividends. How many shares of each kind did the investor buy? 35. A town issued 40 $1000 bonds, some bearing interest at 3| per cent, and the rest at 5 per cent. The town treasurer had to make annual payments of $1730 interest on these bonds. How many bonds of each kind were issued? 36. A man took out a life-insurance policy when he was 21 years old, paying $19.62 per thousand. When he was 30 years old, he took out more, paying $24.38 per thousand. At 40, he took out enough to double his insurance, paying for the additional pohcy $33.01 per thousand. He was then carrying $16,000 insurance, and paying in premiums $449.60 per year. What was the amount of his first policy? 37. The ratio of births to deaths in a certain country is annually about 1.3; the ratio of immigration to emigration is about 17; and the ratio of deaths to emigration is 20. The net annual increase of population is 132,000. Find the annual immigration. 228 PROBLEMS FOR PRACTICE AND REVIEW 38. I drove westward in a carriage for ten hours, and then rode in a train 5 times as fast for two days and nights. The total journey was 600 miles. How fast did the carriage go? 39. Two travelers started toward each other from op- posite ends of a straight road 132 miles long; one went 4 miles an hour, the other 7 miles. How long was it before they met? 40. Two trains start in opposite directions from the same station at the same time, and in 5 hours are 337| miles apart. If the ratio of their speeds is 2, what is the speed of each? 41. A man starts on a journey of 163 miles. He walks 13 hours, then rides twice as fast for 20 hours, and finally has to stop 4 miles short of his journey's end. How fast did he walk? 42. Two trains start at the same moment from opposite ends of a 33-mile track, one going. 20 miles an hour, the other 24 miles an hour. They stop when two miles apart. How long were the trains running? 43. A and B start at 8 a.m. from opposite ends of a 10- mile road and walk toward each other, A walking one mile an hour faster than B. They meet at 9:15 a.m. How fast does each walk? 44. A frontiersman escaped at midnight from a besieged post and started at 4 miles per hour for a fort 64 miles away. He was met by a cavalry troop which had started from the fort at 3 A.M., going 9 miles an hour. At what time did they meet? 45. A trolley car started at noon from Boston for Wor- cester, a 44-mile run. At 12 : 30 a carriage, which was 8 miles an hour slower than the car, started from Worcester for Boston. They passed at 2 p.m. How fast did the carriage go? PROBLEMS FOR PRACTICE AND REVIEW 229 46. I rode 6 miles an hour, then walked 7 times as long at the rate of 4 miles an hour, then went 23 miles by train. I traveled 91 miles in all. How long did I walk? 47. A man starts, on a wager, to walk 1000 miles in 50 days. After traveling for 30 days, he finds that he can go a distance of 5 miles less per day and still have time to go 20 miles extra before the end of the 50 days. How many miles per day did he walk at the start? 48. A village block 680 feet long has on one side 10 house lots. Three are narrow lots, all of the same width; the other lots are 20 feet wider. How wide are the lots? 49. A wall 20 feet high is built with 8 courses of granite blocks and 16 courses of brick. Each granite block is 22J inches higher than the thickness of a brick. How thick is each brick? 50. A platform is laid with boards that are all either 6 inches or 8 inches wide. There are 5 more of the 8-inch boards than of the others, and the platform is 15 feet wide. How many boards were used? 51. A platform 8| feet wide is laid with 43 hardwood strips, of which 16 are each an inch wider than each of the 27 others. What are the widths of the strips? 52. A boy starts from his room in college at noon one day to walk to his home, 41 miles away. At two o'clock of the same day, the father starts on horseback for the college, riding 3 miles an hour faster than the son can walk. They meet at 5 p.m. How fast can the boy walk? 53. A pile of 27 cannon balls in three sizes weighs 254 pounds. There are 13 of the smallest size, and 10 that are twice as heavy. The others are each by 2 pounds the heaviest in the pile. Find the weight of each size. 54. A tank holding 5940 gallons is filled in three hours by three pipes. The first pipe carries twice as much as tiie 230 PROBLEMS FOR PRACTICE AND REVIEW third, and the second pipe carries 3 gallons less, per minute, than the third. Find the number of gallons per minute passing through each pipe. 55. A reservoir holding 10,080,000 gallons is filled in one day by three pumping stations. The first station delivers 4 times as much as the third, and the second station delivers 200 gallons less, per minute, than the third. How much does each deliver? 56. A tank holding 8820 gallons is drained in 2 hours by 4 pipes, marked A, B, C, and D. The pipe marked A car- ries one gallon per minute more than B, and 2 gallons per minute less than C. D carries half as much as all the other three. What is the capacity of each pipe? 57. A cistern can be filled from the water main in 12 hours. If also an extra pump is rigged from a neighboring cistern, it can be filled in 8 hours; and with all the fire engines in town to help, by pumping from a lake near by, it could be filled in 6 hours. How long would it take to fill the cistern if the main were shut off, and the filling depended entirely on the extra pump? How long if the filling depended entirely on the fire engines? 58. If 50 is subtracted from \^ of a number, the result is the same as if 1 were subtracted from j\ of it. Find the number. 59. Subtracting half a number from 50 leaves as much as subtracting a third of it from 35. Find the number. 60. Seven ninths of a certain number is greater than 25 by as much as tV of it is less than 6. Find the number. 61. Two thirds of a certain number exceeds 99 by as much as ^ of it is less than 31. Find the number. 62. Five sixths of a certain number is less than 260 by as much as f of it exceeds 45. What is the number? PROBLEMS FOR PRACTICE AND REVIEW 231 63. Five sixths of a number exceeds 27 by as much as tV of it is less than 6. What is the number? 64. If a number is added successively to 40 and to 22, and ^ of the first sum is added to f of the second .sum, the last sum will be 28. Find the number added to 40 and 22. 65. If a certain number is subtracted successively from 40 and from 55, and then ^ of the first remainder is taken from f of the second, the last remainder will be 10. Find the number first subtracted. 66. If a certain number is subtracted successively from 53 and from 62, and then f of the first remainder is taken from y of the second, the last remainder will be 10. Find the number first subtracted. VI 239. Model A. — At what time between 4 and 5 o'clock are the hands of a clock 9 minutes apart? Let X = the number of minutes past 4. Then —^ = the number of minute spaces traversed by the hour hand in X minutes. And 20 + ^2 = the number of minute spaces from 12 o'clock to the hour hand. If the hour hand is 9 minutes ahead of the minute hand, the equation becomes ® 240 + X = 12 X + 108 ®' X 12 ® 132 = 11 a; @ _ a; _ 108 12 = a; ® ^ 11 Ans. 12 minutes past 4. Check: At 4:12 o'clock, the hour hand is ^ of the way from 4 to 5, that is, at the 21st minute mark; the minute hand is at the 12th minute mark, or 9 minute spaces behind. © 20+^ + 9=: @ 29+^2 = . ® 348 + a; = 12 a; 348 = 11 a; 31tV = X 232 PROBLEMS FOR PRACTICE AND REVIEW If the minute hand is ahead of the hour hand, the equation becomes ® same values ® X12 ® -a; ® - 11 Ans. 28^ minutes of 5. Of these two answers, the first only can be realized on an ordinary clock, because the mechanism of the clock is such that 28x\ minutes of 5 is never indicated by the position of the hands. They move by jerks, one for each tick of the pendulum, and pendulums that tick elevenths of a second would have no reason for existence other than the need of illustrating this problem. Instead of the numbers 4, 5, and 9, in the statement of Model B, use the following sets of numbers, and solve the resulting problems, for an actual clock: 1. 4,5,2 6. 5, 6, 14 11. 12, 1, 27 16. 12, 1, 22 2. 3,4,4 7. 6, 7, 14 12. 9, 10, 26 17. 4, 5, 24 3. 6,7,3 8. 8,9,7 13. 1, 2, 21 18. 7, 8, 24 4. 7,8,2 9. 8,9,4 14. 2, 3, 26 19. 9, 10, 23 5. 3,4,7 10. 5,6,3 15. 10, 11, 21 20. 1, 2, 17 21. A student of music always practiced at the. piano be- tween 5 P.M. and 7 p.m. One day he noticed as he began to practice that the hands of his watch were exactly 3 minutes apart. He practiced until they were again 3 minutes apart, and then stopped. How long did he practice? 22. Find when the hands of an actual clock are exactly together. 23. Find when the hands of an actual clock are at right angles. 24. When are the hands of an actual clock pointing oppo- site ways? PROBLEMS FOR PRACTICE AND REVIEW 233 25. When are the hands of an actual clock making with each other an angle of 30°? 26. Find when, on an actual clock, the hands are exactly one minute space apart. VII 1. One rhomboid has a base measuring 128 ft. and an altitude of 56 ft. Another has a base of 64 ft. and an altitude of 112 ft. What is the ratio of their areas? 2. One triangle has an altitude of 1.732 in. and a base 2 in. long. Another has a base 2 in. long and an altitude of 3.464 in. What is the ratio of their areas? 3. One rectangle has an area of 50 sq.'ft. and its base is 5 ft. long. Another has an area of 125 sq. ft. and its base, also, is 5 ft. What is the ratio of the altitudes? 4. One triangle has an area of 128 sq. ft. and an altitude of 16 ft. Another has the same area and an altitude of 40 ft. What is the ratio of the bases? 5. One rectangle has a base of 12.5 ft. and an altitude of 10 ft. Another rectangle has a base of 100 ft. and an altitude of 10 ft. What is the ratio of their areas? 6. One triangle has a base of 12 ft. and an altitude of 2i ft. Another triangle with the same base has an altitude of 36 ft. What is the ratio of their areas? 7. One circle has a radius of 68 ft. and another a radius of 17 ft. What is the ratio of their circumferences? 8. One circle has a radius of 10 ft. and another a diameter of 125 ft. What is the ratio of their circumferences? 9. Two trapezoids have altitudes of 125 ft. and 1000 ft. One trapezoid has bases 31 ft. and 19 ft. long; the other has bases 24 ft. and 36 ft. long. What is the ratio of the areas? 234 PROBLEMS FOR PRACTICE AND REVIEW 10. A parallelogram has a base of 110 ft. and an altitude of 44 ft. Another parallelogram has a base of 44 ft. and an altitude of 110 ft. What is the ratio of the areas? 11. One rectangle has an altitude of 15.7 ft. and a base of 52.9 ft. Another rectangle has a base of 75.6 ft. and an altitude of 15.7 ft. What is the ratio of their areas? 12. One triangle has a base 1.732 in. long and an alti- tude 3.713 in. long. Another has a base of 1.732 in. and an altitude of 2.902 in. What is the ratio of their areas? 13. One rhomboid has a base 196 ft. long and an altitude 973 ft. long. Another has a base of 392 ft. and an altitude of 973 ft. What is the ratio of their areas? 14. One rhomboid has a base of 4967 ft. and an altitude of 8961 ft. Another has a base of 59,670 ft. and an altitude of 8961 ft. What is the ratio of their areas? 15. One trapezoid has bases 2.12 in. and 3.56 in. long, and an altitude of 4.56 in. Another has bases 2.72 in. and 2.96 in. long, and an altitude of 7.42 in. What is the ratio of their areas? 16. One trapezoid has bases of 391.1 ft. and of 892.2 ft. and an altitude of 10,910 ft. Another has bases 409.6 ft. and 873.7 ft. long and an altitude of 1910 ft. What is the ratio of their areas? 17. A circle has a radius of 6 in., and another circle has a radius of 1§ ft. What is the ratio of their areas? 18. Two circles have radii of 2.76 ft. and 5.02 ft., respec- tively. What is the ratio of their areas? 19. Two squares have perimeters of 1.682 in. and 3.084 in., respectively. What is the ratio of their areas? PROBLEMS FOR PRACTICE AND REVIEW 235 240. Model B. — The perimeter of a rhomboid is 2.00 in. and the altitudes are 1.03 in. and .387 in. Find the area. S S Since S = api = bp2, a =— and 6 = — and a + b = half-perimeter. Pi P2 Whence,^3 + -4 = 1.00. Solving this equation, we find S = .281. Ans. To CHECK, get 2 a -\- 2 b after the sides have been found. Find Sj a, and b, for the rhomboids with the following data: 20. pi = 20, p2 = 15, perimeter = 140 21. pi = 12, p2 = 9, perimeter = 56 22. pi = 1.1, p2 = 1-7, perimeter = 5.0 23. pi = 15.9, p2 = 18.1, perimeter = 157.6 24. pi = .371, p2 = .586, perimeter = 4.00 25. The ratio of the sides of a rectangle is 1.23, and the perimeter is 78.5 in. What is the area? 26. One base of a rhomboid exceeds the altitude thereon by twice as much as the other altitude exceeds the other base. The longer base is 30.0 ft. long and the shorter base is 18.3 ft. Find the area. 27. The altitudes of a rhomboid differ by 3.50 in. and the bases by 4.30 in. The perimeter is just 2 ft. Find the area. 28. The altitudes of a rhomboid have a ratio 1.05, and they differ by 6 in. If the shorter base is 100 ft. long, what is the area? 29. The altitudes of a rhomboid have a ratio 2.13, and the perimeter is 23.8 ft. Find the area. 30. The ratio of the altitudes in a rhomboid is 2.30, and the ratio of one altitude to the base it stands on is 1.09. The area is 2.87 sq. ft. Find the perimeter. 31. The ratio of two sides of a triangle is 2.04, and the third side is 2.48 in. less than their sum. The perimeter is 1.86 in. Find the sides. 236 PROBLEMS FOR PRACTICE AND REVIEW 32. Two triangles are inscribed in the same stripe 2.46 ft. wide. Their bases differ by 7.62 ft., and their areas have a sum of 39.37 sq. ft. Find the bases. 33. A triangle and a rhomboid have the same base, 12.37 ft. long; the altitudes on this base differ by 8.23 ft., and the sum of the areas is 215.6 sq. ft. Find the two altitudes. 34. Two triangles inscribed in a stripe 3.572 in. wide have bases that differ by 2.873 in., and the ratio of their areas is 11.57. Find the bases and the areas of the triangles. 35. A triangle and a rhomboid are inscribed in the same stripe. The base of the triangle is 35.8 in. long and that of the rhomboid is 13.7 in. The triangle is 32.1 sq. in. larger than the rhomboid. What is the width of the stripe? 36. Two bases of a triangle are respectively 65.4 ft. and 38.8 ft. long, and the corresponding altitudes differ by 8.22 ft. Find the two altitudes. 37. In the same stripe a triangle and a rhomboid are inscribed, with the base of the rhomboid bearing to the base of the triangle the ratio 3.79. The rhomboid is 635 sq. in. larger than the triangle. The stripe is 20.6 in. wide. Find the bases of the triangle and the rhomboid. 38. A triangle and a rectangle have bases measuring 41.2 cm. and 53.7 cm., respectively. The triangle bears to the rectangle the ratio .327. The altitude of the triangle is 17.8 cm. What is the altitude of the rectangle? 39. Two triangles inscribed in the same stripe have bases 15.7 in. and 19.3 in., respectively, and the difference of their areas is 50.8 sq. in. Find the width of the stripe. 40. The altitude of a trapezoid is 3.74 ft. The bases differ by .0364 ft. The area-number exceeds the sum of the length-numbers of the bases by .0140. Find the bases. PROBLEMS FOR PRACTICE AND REVIEW 237 41. The bases of a trapezoid are 1.032 in. and 1.182 in. long. The area-number exceeds the length-number of the altitude by 1.09. Find the altitude. 42. The number of degrees in a certain arc is equal to the number of inches in its complement. If each degree of arc is 2 in. long, how many degrees are there in the arc first mentioned? 43. The number of degrees in an arc exceeds by 6 the num- ber of inches in its complement; and the entire circumference is 75 ft. long. How many degrees are there in the arc first mentioned? 44. Two supplementary arcs have a united length of 90 ft. The number of inches in one, added to the number of degrees in the other, gives a total of 775. How many degrees are there in each arc? 45. The number of degrees in an arc is equal to the number of inches in the whole circumference, and the rest of the cir- cumference measures 50 in. What central angle intercepts this arc? 46. On a circle 308 in. in circumference, the number of inches in a certain arc is equal to the number of degrees in its complement. How many degrees are there in the arcs? 47. Of the two angles at the base of a triangle, one is 3° less than double the other. The angle at the vertex bears to its own supplement the ratio If. Find the angles of the triangle. 48. The number of inches in an arc is J the number of de- grees in its complement. The circumference is 72 in. Find the length of the arc and the number of degrees in it. 49. The altitude of a parallelogram is 3 in. Its base is 7 in. longer than the base of a triangle inscribed in the same stripe. The areas of the two figures have a ratio 3.4. Find the area of the triangle. 238 PROBLEMS FOR PRACTICE AND REVIEW 50. In a stripe 5.02 in. wide, a triangle and a parallelogram are inscribed. The base of the parallelogram is the shorter by 3.37 in., and the ratio of its area to the area of the triangle is 1.73. Find the bases. 51. A triangle and a rhomboid have bases respectively 3.00 and 5.00 ft. long. The altitude of the triangle is the longer by 14.52 in., and its area bears to the area of the rhom- boid the ratio .500. Find the two altitudes and the area. A sector is a figure bounded by an arc of a circle and by the sides of the central angle that intercepts the arc. If the central angle is A degrees, the area of the sector will be ^^7: of the area of the entire circle. 52. Find the area of the sector of a circle of radius 20 in., if the central angle is 75°. 53. Find the area of a circle, if a sector of 38° has an area of 43.8 sq. in. 64. Find the area of the sector of a circle of radius 32.7 cm., if the central angle is 83.7°. 55. In a circle whose radius is 3.37 in., how wide a central angle will contain a sector of 3.37 sq. in.? 56. In a circle of 28.7 ft. radius, a sector has a total per- imeter of 83.3 ft. How many degrees are there in the sector? 57. In a circle of radius 30.8 in., a sector contains 283 sq. in. How many degrees are there in it? 58. Two trapezoids inscribed in the same stripe have one base the same; the other bases are f of an inch and | of an inch respectively. The ratio of their areas is f . Find the comimon base. PROBLEMS FOR PRACTICE AND REVIEW 239 59. The areas of two circles differ by 92.8 sq. in. The area of a sector of 40° on one circle exceeds the area of a sec- tor of 60° on the other by 8.35 sq. in. Find the areas of the circles. 60. The base angles of a triangle differ by one degree. The sum of these angles exceeds the angle at the vertex by as much as half the larger base angle. Find the angles of the triangle. 61. In two circles, one circumference is double the other. An arc of the smaller circle exceeds its complementary arc on the larger circle by tV of the smaller circumference. How many degrees are there in each arc? 62. The point C cuts off from the quadrant AB an arc, AC, of 38°. How far shall the point C be moved in order that the ratio of AC to CB shall be GJ? 63. The bases of two rhomboids differ by 2f ft. Their altitudes are 17.3 ft. for the one with the shorter base, and 23.8 ft. for the other. The ratio of their areas is |. Find the areas. 64. A certain circle has twice the area of a certain rectangle. If the rectangle were 125 sq. ft. smaller, the circle would be 3 times as large as the rectangle. What are the areas? 65. A rectangle is 380.1 sq. ft. larger in area than a tri- angle. If the triangle were 9.52 times as large, the two areas would be the same. What are the areas? 66. One triangle is 4 times as large as another, and the sum of fhe areas of both is one half the area of a certain circle. If the larger triangle were 5 times as large, and the smaller triangle were 100 sq. ft. less, the area of the circle would be •just equal to the area of the two triangles. What are the three areas? 6V. What is the complement of an angle of 47.15°? 240 PROBLEMS FOR PRACTICE AND REVIEW 68. What is the supplement of an angle of 47.15°? What is the supplement of the complementary angle? 69. Whatis the complement of an angle of 89.175°? What is the supplement of the same angle? What is the supple- ment of the complementary angle? 70. What is the supplement of an angle of 126.52°? What is the complement of the supplementary angle? 71. What is the complement of an arc of 62.33°? What is the supplement of the same arc? 72. Two complementary arcs have a ratio 4. What are they? 73. Two supplementary arcs have a ratio 6J. What are they? 74. Of two supplementary arcs, one is 6.89 times the other. What are the arcs? 75. The ratio of two complementary arcs is 1.232. What are the arcs? 76. Two angles of a triangle are 56.32° and 72.49°. What is the third angle? 77. The sum of tw^o angles of a triangle is 171.2°. What is the third angle? 78. One angle of a triangle is 36.2°. Of the other two angles, one is twice the other. What are the other two angles? 79. One angle of a triangle equals half the largest and twice the smallest angle of the triangle. What are the angles of the triangle? 80. One angle of a triangle is 91.3°. The other two angles are equal. What are they? 81. In a triangle with two equal angles, the ratio of the third angle to the sum of the other two is 1.672. What are the angles of the triangle? PROBLEMS FOR PRACTICE AND REVIEW 241 82. In a triangle with two equal angles, the ratio of each of the equal angles to the third angle is 2.179. What are the angles of the triangle? 83. In a triangle, the ratio of the largest angle to the smallest is 5.234, and the ratio of the third angle to the smallest is 1.212. What are the three angles? 84. One of two supplementary arcs of a circle of 16.3 in. radius is 2.7 in. longer than the other. Find the number of degrees in each arc. VIII 241. In multiplying and dividing algebraic polynomials, it is of great importance to arrange not only the given expressions but all the expressions obtained in the course of the work according to the powers of some one letter. For example, the expression Qa'h + Q ah^ + a« + 6' + 15 a^b^ should be rearranged in one of the two following ways: a6 + 6 w^h + 15 a^62 + 6 ab^ + h^ or 6^ + 6 afes + 15 a'W + 6 a^b + a« Rearrange the following expressions, if necessary, before multiplying: 1. (x + 9) (x2 + 4a; -45) 2. {x + 9) (x2 - 36 - 5 a;) 3. (9 + a;) (27 + 6 re - x") 4. (x + 9) (72 - x^ + x) 5. (3 a: + a;2 - 10) {x - 5) 6. {p'-q' + 7vq)(7p-q) 7. {x^ - 14 - 5 x) (2 - a;) 8. (x2 - 14 - 5 a;) (x + 7) 242 PROBLEMS FOR PRACTICE AND REVIEW 0. (4 A; + h) (/i2 - 4 /c2 - 3 hk) 10. {d' -{-g'- dg) id' + g' + dg) 11. (10 2/2 - 3 X2/ + x') (x + 3 2/) ^^12. (10 2/2 + a;2 - 3 xy) (x + 5 2/) 13. (10 2/2 + a;2 - 3 xy) {x-\-2y) 14. (c2 + 6 cd - 5 d") (6 d - c) 15. (c2 + 6 c(^ + 5 ^2) {bd- c) 16. (14 y' -]- 5xy — x') (x -\- 7 y) 17. (14 2/2-^2 + 5 0:2/) (x -2y) 18. (5 a2 + 5 a;2 _|_ 2 aa;) (a^ + x' - 5 ax) 19. (a;2 + 4 + 2 x) (a;2 + 4 - 2 x) 20. (x2 - 15 + 2 x) (2 X + 15 - x2) 21. (r^ — r's + ^s^ — s^) (r^ + rs + s^) 22. (2 x2 + 9 2/2 + 12 X2/) (12 X2/ - 2 x2 - 9 y') 23. (x2 + 6 X + 18) (18 + x2 - 6 x) 24. (10 p5 + 2 p2 + 25 g2) (10 pg - 2 p2 _ 25 g^) 25. (3 /i2 + 4 s2 + 2 hs) (2hs-Sh'-4: s^) 26. (x2 + 3 X2/ + 2 2/2) (x2 - 6 X2/ + 4 2/^) 27. (3 a2 + 3 52 - 10 ah) (Sa' -Sh' -S ah) 28. (4 x2 - 9 2/2) (3 2/2 - 5 X2/ - 2 x2) 29. {p^ + 2pq + Sq){Sp + 2q) 4 30. (5 a2 - 3 ax + 2 a^) {a' + 2 ax) 31. (a3 + 53 + 3 ^25 _^ 3 ^52) (3 ^5 _ ^2 _ 2 62) 32. (/i^ + /b^ + 6 /i2/c2 - 4 /i/b [/i2 + /c2]) (4 /i/c + /i2 + h^) 33. (5 2/2 - 2 X2/ + 3 x2) (2 X2/ + 5 2/2 + 3 x2) 34. (2 X2/ - 5 2/2 - 3 x2) (4 X2/ + 10 2/2 + 6 x2) 35. (3 x2 + 3 X2/ + 2 2/2) (2 2/2 - 3 X2/ + 3 x2) 36. [5-p3 + p(2_3p)][l-3p(p-3)] 37. (x2 + fx + i)(f -ix + x2) PROBLEMS FOR PRACTICE AND REVIEW 243 38. (3 a2 + ^ - 2 a) (5 a2 - i - i a) 39. (x2 + xy + i y^) (f y^ - xy + x^) 40. (ix^-ix-\-^){ix + ^x^ -i) 41.- (a^ + ¥) {a - h) (a^ + a6 + ¥) 42. (a2 + 2 a6 + 62) (a^ - 2 ah + ¥) 43. (a^ + a% + a62 + 6^) (a - b) 44. (a2 + 62) (a2 - 62) - (a - 6)^ 45. (a3+3 a26+3 a62+63) (a+6) - (6-a) (a3+a26+a62+63) Rearrange the following expressions, if necessary, before dividing: 46. (x3 - 9 x2 + 8 a; - 12) -^ (x2 + 7 x - 6) 47. (7 fc3 + 9 /c2 + 14 A; + 12) -^ (7 fc2 + 2 A; + 12) 48. (7 x3 - 12 x2 - 8) -^ (7 x2 + 2 a; + 4) 49. (2 a^ - a62 + 6^) -- (2 a2 - 2 a6 + 62) 50. (61/3 + 6 X2/2-34 X22/-4 x^) -t- - (6 2/2+18 x?/ + 2 x^) 51. (x3- 2x + 1) ^ (x- 1) 52. (2 a3 + a26 - 13 a62 + 28 6^) -v- (2 a + 7 6) ''^. (12 x3 + 8 ^22/ - 21 x?/2 + 14 2/3) -- (3 X + 2 ?/) 54. (2 xV - 11 0^22/2 - 57xy + 31) -^ {2 xy - 1) 55. (14 d3 + 15 d2g, - 58 dg^ + 21 g^) -^ {7 d - 3 g) 56. (a' + 4) -^ (a2 + 2 a + 2) 57. (4 2/^ + 81) ^ (2 2/2 + 6 2/ + 9) 58. (d^ + 324) ^ (^2 + 6 c^ + 18) 59. (64 p' + 81) -^ (8 p2 _ 12 p + 9) 60. (6^ + 962 + 81) -^ (62 + 36 + 9) 61. (n' + n2/b* + k^) -^ (^2 + nk^ + fc^) 62. (p4 - 27 p2g2 _|. q4) _^ (p2 _ 5 p^ _ ^2) 63. (81 xy + 8 xY + 16) -^ (9 x22/2 - 8 X2/ + 4) 64. (a«6« + 14 a^64 + 625) -^ (a^ft* + 6 a%^ + 25) 244 PROBLEMS FOR PRACTICE AND REVIEW 65. (a2 - ¥ - 2hc - c^) -^ {a - h - c) 66. (x^ + x^y + xY + xY + xy* + y^) -i- (x^ + y^) 67. (8 c^-22 c3a;+43 c^o^^-SS 0x^+24 x^) -^ (2 c^-S cx+4 a;2) 68. (a^ + 63 + c^ - 3 a6c) -f- (a + 6 + c) 69. (s^ - 2 s3 + 1) ^ (§2 - 2 s + 1) Vo. (x^ - oi^a^ - x2 + a2) -^ - (x2 - xa - X + a) 71. (r® — r^s^ — r^ + s^) -^ (r^ + y*^ — ^'^s + r — rs — s) 72. (/i^o + /iV' + g'') - (/i' + /ig' + Sf=^) 73. (la:3 + 2)-^(ix+l) 74. (|a3-ia62-^63) ^(3^ + 5) 75. (I a^ + I a63 - I 64) -- (a + 1 6) 76. (i rc3 _|_ 2 ^5^2 _ ^252^ _ I ^353) ^ (1 a, + ^6) 77. aa^ + ia262+|64)^Q62 + ict6 + ^a2) Wherever there is a remainder in the following divisions, write the remainder as the numerator and the divisor as the denominator of a fraction which is an additional term of the complete quotient: 78. (x3 + 2/3) -^ (x- y) 79. (x^ _ 6 ^2 _^ 12 X - 6) -f- (x - 2) 80. (a^ + 65) -^ (a2 + 4 a + 8) 81. (a + a^ - 34 a2 + 226) -^ (a^ - 25) 82. (200 + 3 a - 34 a2 + a^) -^ (a^ - 4) 83. (300 + a^ - 2 a - 34 a2) -^ (a2 _ 2 a - 15) 84. (150 + a*-34a2) -^ (a3 + 20-a(5a + 4)) 85. (a3 + a26+^a62 + |)^(|-a) 86. (xio + xy + 2/'") -^ (x^ + 2/') 87. [{x^ - y^y + x!/ (x2 + 2/2 + 3 xi/)] -^ (x? + xy + y^) PROBLEMS FOR PRACTICE AND REVIEW 245 IX Plot the loci of the following equations: 1. 2 x2 - 5 X + 3 2/ = 171 11. xy = 36 2. 3 x2 + 3 X - 13 2/ = 239 12. x"" -\- 2Q = 3xy + 2y 3. y^-4:xy-\-Sy = lQx-7-4:X^ 13. 5x^-{-Qxy = 144 4. x2 + 2 X2/ - 24 14. x2 + 1/2 + 6 X - 4 2/ = 12 5. x2 + x?/ + 2 2/2 = 74 15. X2/ + 3 X - 4 2/ = 36 6. 24 2/2 - 25 X2/ = 25 x^ le. 9x^+42/^+145 =54x -402/ 7. x2 + 2/2=5 17. 9x2-55 =42/2+54x+402/ 8. 2 x2 + X2/ + 2/2 = 8 18. 2/^ - 8 X + 28 = 4 2/ 9. x2 - X2/ + 2/2 + 3 X = 34 19. x2 = 8 2/ 10. x2 + 2/2 = 53 20. x2 + 25 = 6 X + 8 2/ 21. Find where each of the loci of the equations in Exs. 1-20 intersects the line x + 2 2/ = 5. X Solve the following pairs of simultaneous equations: I. x-22/=9 e. x-2y = 6.42 X2/ + X = 2/ + 9.98 7. 2 X + 2/ = 8.45 a;2 _ 2/2 = 6.18 8. 2 X + 2/ = 9.45 x2 + 22/=2/' + 2x + 6.18 9. 140 x2 - 30 X2/ = 23 5 X + 2/ = 2.7 10. 7 x2 - 6 X2/ = 40.6 5 X + 2 2/ = 9.85 II. In two triangles, the vertical angles are equal, and the ratio of the right sides exceeds by 1 the ratio of the left sides. The ratio of the areas of the triangles is \^. What are the ratios of the sides? xy = 68 2. X - 2 2/ = 472 xy = 29952 3. X - 2 2/ = 3.42 xy = 8.98 4. 2 X + 2/ = 13 x2 - 2/2 = 16 5. 2 X + 2/ = 755 a;2 - 2/2 = 6487 246 PROBLEMS FOR PRACTICE AND REVIEW 12. In two triangles, the vertical angles are equal, and the ratio of the right sides is f the ratio of the left sides. The ratio of the areas is |. What are the ratios of the sides? 13. Two triangles have vertical angles equal, and the ratio 9f the left sides is 90.21 per cent of the ratio of the right sides. The ratio of the areas is 1.216. What are the ratios of the sides? 14. Two sides of a triangle are 28 cm. and 27 cm. Half the area is cut off by a line that cuts the longer side 3 cm. higher up than the point where it cuts the shorter side. Where does the line cut the shorter side? 15. The ratio of the areas of two similar triangles exceeds the ratio of the bases by 42. If the area of the smaller tri- angle is 40 sq. in., what is the area of the larger? 16. In a trapezoid, the upper and lower bases are 56 cm. and 70 cm., and the right side is 15 cm. If the sides are extended until they meet, how far up will the left side cut the right? 17. The ratio of the areas of two similar triangles is 10 less than .763 times the ratio of the bases. If the area of the larger triangle is 82.31 sq. ft., what is the area of the smaller? 18. The ratio of the areas of two similar triangles is 12 less than 7 times the ratio of the bases. If the area of one triangle is 5 sq. mi., what is the area of the other? (Four answers.) 19. The ratio of two rectangles is If greater than the ratio of their bases, and 3 less than the ratio of their alti- tudes. What is the ratio of the rectangles? 20. In two rectangles which have equal areas, the ratio of the bases exceeds by | the ratio of the altitudes. What is the ratio of the rectangles? 21. The ratio of the bases of two rectangles exceeds by 1 the ratio of the altitudes; and the sum of these two ratios is PROBLEMS FOR PRACTICE AND REVIEW 247 1 less than the ratio of the rectangles. What is the ratio of the rectangles? 22. The altitude of a triangle is 25 per cent greater than the base. Its area is equal to a rectangle having an altitude of 9 cm. and a base 2 cm. shorter than the base of the triangle. Find the altitude and the base of the triangle. 23. One side of a triangle is 11 cm. longer than the other. A line cutting these sides, at distances of 6 cm. and 7 cm., respectively, from the vertex, cuts off a quadrilateral equal to f the original triangle. Find the lengths of the sides. 24. The two sides of a right triangle differ by | the hypot- enuse, and the hypotenuse is f their sum. Find the three sides. The slant side of a right triangle is called the hypotenuse. 25. The two sides of a right triangle differ by 2 cm., and the hypotenuse is 8 cm. long. Find the length of the two sides. 26. One side of a right triangle is 12 cm. more than double the other, and the hypotenuse is 78 cm. long. Find the lengths of the legs. 27. One side of a right triangle is 1 cxa. less than 4 times another, and the hypotenuse is 14 cm. less than the sum of these two sides. Find the three sides. 28. The hypotenuse of a right triangle is 11 cm. greater, and the larger leg is 2 cm. greater, than double the shortest side. Find the three sides. 29. In a triangle with two equal sides, the area is 120 sq. cm. One of the equal sides is 17 cm. Find the base and the altitude. 30. The length of a rectangle is J centimeter less than 5 times the width, and the diagonal is 44 millimeters. Find the area. I meter = 100 centimeters 1 centimeter = 10 millimeters 248 PROBLEMS FOR PRACTICE AND REVIEW 31. One side of a right triangle is | meter less than the hypotenuse, and the whole perimeter is 66 centimeters less than 3 times the hypotenuse. Find the sides. 32. Of the numbers representing the three angles of a triangle, two are formed by the same digits interchanged, and the other is the square of the sum of those, digits. The first two angles would be complements if one were diminished by 9°. What are the angles? 33. The area of a rectangle is 480 square centimeters, and the diagonal is 34 centimeters. Find the perimeter. 34. The length of a rectangle exceeds the width by 49 miUimeters, and the perimeter exceeds the sum of the diago- nals by 2 centimeters. Find the sides. 6 35. The diagonal of a rectangle exceeds 5 times the width by 2 centimeters, and the perimeter is 1Q4 millimeters. Find the sides. INDEX Abbreviating explanations of prob- Axiom, 41 . lems, 11. Ability and time, problem of, 100. Accm-acy, 18. limited by data, 59. of diagrams, 171. Acre = 160 square rods, 206. = .4047 hectares, 60. 2.471 acres in a hectare, 60. Addition, approximate, 59. Age problems, 38. Air, weight of, 20, 60. Algebra, 11. ancient textbooks, 42. derivation of the word, 42. Algebraic expression, 13. Algebraic negative, 136. Alphabet, use of, in algebra, 23. Altitude, 67. Analjdiic geometry, 214. Angle, sides of, 22. Angles, measurement of, 21. of a polygon, 164. in a stripe, 32. Answers, complete Ust of, 175. meaningless, 140. Approximate addition, 59. computation, 50. division, 54. Arc, 47. Area, ratios of, 148. of circle, 78. of trapezoid, 75. of triangle, 71. Axiom A, 130. Axis of X, 170. of y, 170. Base, 67. Binomial, 111. Brace, 108. Bracket, 108. Center, 46. Centigrade and Fahrenheit, 175. Change of form, 42. Changes of value in an equation, 41, 83. Check, by constructing loci, 186. for approximate division, 55. for approximate multipUcation, 53. for long division, 117. Checking, 14. by coefficients, 106, 111. quadratics with the standard parabola, 215. Circle, 46, 78. equation of, 208. Circumference, 46, 78. length of the, 78. Clock problems, 231. Coal, components of, 20. weight of, 53. Coefficient, 41. of a term, 41. Coefficients, check by, 106. Complementary arcs, 47. 249 250 INDEX Complements, 23. Complete quotient, 244. Completing the square, 127. Composition of motion, 95. Computation, approximate, 50. Condition, equation of, 119. Conditions, to determine two un- knowns, 181. Continued multiplication, 109. Convention, 14. Converse statements, 179. Cord (of wood) =4X4X8 feet, 136. Correspondence of angles and sides, 155. of numbers, 169. Cross products, 122. Cube of a number, 110. Cube root, 110. Cubic meter = 1.3079 cubic yards, 60. Cubic yard (1.3079 cubic yards in a cubic meter), 60. Current problems, 98. Curve of a railroad track, 61. Cylinder (lateral surface), 133. Data, accuracy of, 18, 58. Decimal data, 18. point, placing of, 50, 51. subdivision, 18. Degree, 21. (57.296° in a radian), 60. of a term, 110. of an expression, 110. Diagram, algebraic, 170. Diameter, 47. Difference of squares, 120. of unknown numbers given, 30. Digits, problem of the, 92. Discarding figures, 51. Distributive factoring, 113. law, 112. Division, approximate, 54. Doubtful columns, 52, 54. figures, 51. Egyptians, 147. Eight-figure accuracy, 18. Elimination by combination, 186. by substitution, 203. Ellipse, 214. Equation, 13. linear, 200. of condition, 119. quadratic, 121, 130, 200. Equations, construction of, 91. inconsistent, 198. independent, 198. indeterminate, 199. simultaneous, 181. Equiangular polygon, 165. triangle is equilateral, 166. Equilateral polygon, 165. triangle is equiangular, 166. Equivalents of inches in tenths of a foot, 53. Errors of rejection, 52, 53. Explanations of problems abbre- viated, 11. Explicit formulas, 80. Exponent, 110. Factoring, by cross multipUcation, 123. distributive, 113. Factors, 109. of an equation, 131. Fahrenheit and Centigrade, 175. Figures inscribed in a stripe, 67. Five-figure accuracy, 18. Fixing the decimal point, 50, 51. INDEX 251 Foot (5280 feet in a mile), 190. Formula, 63. S = ab, 63. S = api = hp2, 68. S = hapi = hbp2 = h cps, 71. 5 = 1 p (6, + 62), 75, 80. c = 2irr, 5 =7rr2, 78. s = vt, 95. TF = (n - 2) 180^ 164. Fractional equations, 83. Gallon = 3.785 liters, 60. Geometry, 147. Greeks, 147. Hectare = 2.471 acres, 60. .4047 hectares in an acre, 60. History of equations, 14. Hyperbola, 214. Identical equation, 119. Identities, 119. Imaginary roots, 144. Implicit formulas, 80. Inches reduced to tenths of a foot, 53. Inconsistent equations, 198. Independent equations, 198. Indeterminate equations, 199. Index, ilO. Indirect measurement, 65. InequaUty, sign for, 72. Inscribed, figures, in a stripe, 67. Integer, 17. Intercepts, 178. Irrational numbers, 142. Italian method of division, 56. of subtraction, 56. Kilogram = 2.20 pounds, 20. .4536 kilograms in a pound, 60. Law of signs, 115, 116. Left-hand order of partial prod- ucts, 50. Length, measurement of, 17. of the circumference, 78. Limits of accuracy, 18. Linear equation, 200. Linear-quadratic pairs, 200. Liter (3.785 Uters in a gallon), 60. Locus of an equation, 175. of a two-letter quadratic, 214. Long division, 117. Measurement, indirect, 65. of angles, 21. of length, 17. Measurement-numbers, 50. Member of an equation, 13. Meter, 17, = 39.370 inches, 20, 56. Mile = 5280 feet, 190. Minus terms, 104. Minute (60 minutes in a degree), 22. Monomial, 111. Motion, composition of, 95. Multiplication, continued, 109. (distributive law), 36, 37. how indicated, 14. of polynomials, 114. reversed, 50. Natural order of partial products, 50. Negative answers, meaning of, 136, 138. Negative numbers (subtrahends), 103. Negative terms (subtrahends), 37, 104. Notation in geometry, 147. NuU root, 144. 252 INDEX Number-pair, 208. Number TT, 78. Numbers, irrational, 142. Odd and even angles in a stripe, 32. Pair of straight lines, 214. Parabola, 214. Parallel lines, 32. Parallelogram, 68. Parentheses, removal of, 89, 106. within parentheses, 108. Partial products, as successive corrections, 51. in division, 54. order of, 50. Perimeter, 235. TT, 78. Plotting, of points, 171. a quadratic, 207. Plus terms, 104. Polygon, angles of a, 164. equiangular, equilateral, 165. regular, 166. Polynomial, 111. Polynomials, multiplication of, 114. rearrangement of, 241. Positive terms, 104. Pound = .4536 kilograms, 60. Power, 110. Precision of measurement, 18. of instruments, 50. of data, 58. Problem of ability and time, 100. of the clock, 231. of the digits, 92. of two velocities, 94. Protractor, 23. Pure quadratic, 143. Pythagorean theorem, 162. Quadrant (quarter of a circle), 23. Quadratic equation, 121, 130, 200. Quadratic expression, 121. Quadratic loci, 210. Quadratic products, 121. Quadratics, peculiar, 144. Radian = 57.296^ 60. Radius, 47. Railroad curve, 61. Ratio, 16. Ratios obtained by measuring an- gles, 65. Ratios of area, 148. Reciprocals, solving for, 194. Rectangle, area of, 63. Reducing an equation, 83. an expression, 83. Regular polygon, 166. Rejecting figures, 51, 62. Rejection, errors of, 52. Rejection of, 5, 53. Removal of parentheses, 89, 106. Rhomboid, area of, 67. Right triangles, similar, 159. Rod = 16.5 feet, (320 rods in a mile). Root, 110. Roots, equal and opposite, 143. of an equation, 131, 143. Rough check, 51. Rule for solving simple equations, 42. for subtraction, 107. Scale, algebraic, 168. of a diagram, 172. INDEX 253 Second (60 seconds in a minute), 22. Sector, 238. Shop method of subtraction, 56. Shortages, 37. Sign of identity, 119. of perpendicularity, 177. of similarity, 163. Signs, law of, 115, 116. Similar figures, 155. ^ polygons, 155. right triangles, 159. terms, 41, 111. triangles, 156. Simple equation, study of, 41. Simultaneous equations, 181. Six-figure accuracy, 18. Square of difference, 120. of a number, 57, 110. of sum, 120. Square rod = 30j square yards. (160 square rods in an acre), 206. Square root, 57, 110. in equations, 65. Standard parabola, 215. Straight-line locus, 176. Straight lines, pair of, 214. Straight products, 122. Strip, area of, 64. Stripe, 32. figures inscribed in a, 67. Study of a simple equation, 41. Subtraction, rule for, 107. of fractions, 88. of parentheses, 88. Subtrahends in division, 55. Suffixes, 32. Sum of the angles of a polygon, 164. of the angles of a triangle, 33. Sum of the base angles of a tri- angle, 33. of the unknown numbers given, 44. Summation, 104. Supplementary arcs, 47. Supplements, 23. Tangent of an angle, 160. Tangents, table of, 160. Terms in algebraic expressions, 13. Theorem, 33, 119. Theorem A, 120. Theorem of Pythagoras, 162. Three-figure accuracy, 18. checking, 62. Tracing paper, use of, 71, 186, 215. Train dispatcher's diagram, 169. Transversal, 32. Trapezoids, 75. Trial product, 51. Triangles, 70. having one angle the same, 151. inscribed in a stripe, 148 similar if equiangular, 156. Trinomial, 111. Unit of length, 17. Unit-wide strip, area of, 64. Valuation problems, 26. Velocities, problem of the two, 94. Vertex of an angle, 23. Vinculum, 108. Weight of air, 20, 60. of carbon dioxide, 60. of coal, 53. Zero obtained by adding, 104. 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