1
ALGEBRA
FIRST COURSE
0
IN MEMORIAM FLORIAN CAJORI
Digitized by the Internet Archive
in 2007 with funding from
IVIicrosoft Corporation
http://www.archive.org/details/algebrafirstcourOOIongrich
CORRELATED MATHEMATICS FOR SECONDARY SCHOOLS
ALGEBRA
FIRST COURSE
BT
EDITH LONG
DEPARTMENT OP MATHEMATICS, HIGH SCHOOL, LINCOLN, NEBRASKA
AND
W. C. BRENKE
PROFESSOR OF MATHEMATICS IN THE UNIVERSITY OF NEBRASKA
NEW YORK
THE CENTURY CO.
1913
Copyright, 1913, by THE CENTURY CO.
TABLE OF CONTENTS
CHAPTER I
Paoxs Introduction 1-13
(The numbera refer to articles.)
1. The Operations of Arithmetic. 2. The Whole Numbers, or integers. 3. Addition. 4. Subtraction. 5. MultipUcation. 6. Division. 7. Subject Matter of Algebra. 8. Algebraic Ex- pressions. Their Forms and Values.
CHAPTER II
Measurements — Lengths, Angles, Areas, Volumes 14r-27
9. English Units of Length. 10. Metric Units of Length. 11. Comparison of Enghsh and Metric Units of Length. 12. Angles. — Definitions and Notations. 13. Classification of An- gles. 14. Measurement of Angles. 15. The Number tt. 16. Measurement of Areas. 17. Measurement of Volumes. Sum- mary.
CHAPTER III
Measurements Continued. Temperature, Weight and
Density, Force 28-41
18. Measm-ement of Temperature. 19. English Units of Weight. 20. Metric Units of Weight. 21. The Beam Balance. 22. The Lever. 23. Density or Specific Gravity. 24. Sum- mary. 25. Questions and Problems for Review.
CHAPTER IV Graphic Representation of Quantity 42-48
26. The Amount of any Quantity Represented by the Length of a Line.
CHAPTER V Positive and Negative Numbers 49-54
27. Exercises. 28. Opposite QuaUties. 29. Positive and Negative Numbers. 30. Graphic Representation of Positive and Negative Numbers. 31. Summary,
ill
IV TABLE OF CONTENTS
CHAPTER VI
Pagbs Addition and Subtraction 55-70
32. Notation. 33. Addition of Directed Line Segments. 34. Geometric Addition of Numbers. 35. Exercises and Problems. 36. Subtraction. 37. Summary.
CHAPTER VII
Multiplication and Division 71-80
38. Meaning of Multiplication. 39. Geometric Illustra- tions of Multiplication. 40. Multiplication of General Num- bers. 4L Division. 42. Summary.
CHAPTER VIII
Addition and Subtraction of Polynomials 81-89
43. Definitions. 44. Addition and Subtraction of Poly- nomials. 45. Rules for Adding and Subtracting Polynomials. 46. Removal of Parentheses. 47. Summary.
CHAPTER IX
Multiplication and Division of Polynomials 90-104
48. Multiplication of a Polynomial by a Monomial. 49. Dis- tributive Law. 50. The Commutative Law of Addition. 51. The Commutative Law of Multiplication. 52. Exponents. 53. Multiplication of One Polynomial by Another. 54. Division of Polynomials. 55. Summary.
CHAPTER X
Problems Leading to Simple Equations 105-126
56. Problems. 57. Formation of an Equation. 58. Solu- tion of the Equation. 59. Summary,
CHAPTER XI Simple Areas and Their Algebraic Expressions. Elements
OF Factoring 127-149
60. Rectangles. 61. Geometric Theorems on Factoring. 62. Parallelograms. 63. Theorem. 64. Triangles. Theorem. 65. Regular Polygons. 66. Theorem. 67. Trapezoids. Theorem. 68. Circles. 69. Factoring Continued. 70. Summary. Exer- cises.
TABLE OF CONTENTS V
CHAPTER XII
Pages
Fractions 150-165
71. Reduction of Fractions to Lowest Terms. Cancellation. 72. Multiplication of Fractions. 73. Division of Fractions. 74. Reduction to a Common Denominator. 75. Addition and Subtraction of Fractions. 76. Complex Fractions. 77. Long Division. 78. Equation Involving Fractions.
CHAPTER XIII
Quadratic Equations 166-205
79. Definitions. 80. Some Problems Leading to Quadratic Equations. 81. Square Roots of Numbers. Pythagorean Theorem. 82. Geometric Construction of Square Roots. 83. Rational and Irrational Numbers. 84. Factoring of Quadratic Expressions. 85. Factors of ax^ -\- bx -\- c. 86. Algebraic So- lution of the General Quadratic Equation, 87. Imaginary Roots. 88. To find the Square Root of a Number. 89. Sum- mary and Problems.
CHAPTER XIV
Variables. Constants. Functions of Variables. Graphic
Representation 206-224
90. Variables. 91. Constants. 92. Functions of Variables.
93. Functions of a Single Variable. 94. The Function Notation.
95. Graphic Representation. Linear Functions. 96. Graph of
of Quadratic Functions. 97. Graphs of Other Functions. 98.
Summary.
CHAPTER XV
Uses of the Graph 225-240
99. Graphic Representation of Measurements of a Variable
Quantity. 100. Graphic Solution of Equations. 101. Nature
of the Roots of a Quadratic Equation. 102. Graphic Solution of
Equations of Higher Degree. 103. Summary.
CHAPTER XVI
Loci. Simultaneous Equations 241-262
104. Meaning of the Word Locus. 105. Coordinates. 106. Straight Line Loci in General. 107. Simultaneous Equations. 106. Summary and Problems.
Vi TABLE OF CONTENTS
CHAPTER XVII
Pages
Exponents and Radicals 263-281
109. Definitions. 110. To Multiply a"* by a". 111. To Mul- tiply aT" by 6"». 112. To Divide oT' by a^. 113. The meaning of a". 114. The Meaning of (a*")". 115. Summary of Results. 116. Fractional Exponents. 117. Irrational Numbers or Surds. 118. Examples Involving Surds. 119. Equations with Irrational Terms. 120. Summary and Exercises for Review.
CHAPTER XVIII
Binomial Theorem 282-283
121. The Binomial Theorem.
Protractor Inside of back cover
PREFACE
In the preparation of this text the authors have had in view two primary aims, first to emphasize and vivify the treatment of Algebra by a systematic correlation with Geometry, and secondly to present the subject-matter in a style sufficiently simple to be easily grasped by students of high school age.
The first object has been accomplished by a free intro- duction of constructive exercises from Geometry, including theorems and problems of sufficient range to give to the student a fair working knowledge of the elementary prop- erties of important geometric figures. No attempt has here been made at formal demonstration, that being re- served for the book on Geometry proper; the sole aim has been to bring out the fundamental facts regarding such figures as angles, intersecting lines, polygons, circles, and some simple solids. These facts are brought out by drawing, paper cutting, and super-position, and they are then made the basis of further algebraic work.
In bringing out the laws which govern the four fundamental operations, systematic use is made of graphic representation of these operations on a number scale. Experience has shown that this is a simple and effective way to fix these laws so firmly in the mind of the student that they are not easily forgotten.
A lengthy discussion of the contents of the book would be superfluous here. Its scope, aims, and methods are indicated by the table of contents, and can be more fully appreciated only by reading the pages of the text.
The second object, namely to produce a simple treat- ment, is accomplished largely by the use of a more nar-
vii
Vlll PREFACE
rative style than is usual in mathematical text-books. Without this the book could have been made considerably smaller, but it was felt that the addition to the number of pages would be more than compensated by the gain in clearness and fullness of explanation. The time and effort required to cover the year's work will be less than it would be with a more compact and concentrated form of pres- entation.
The exercises and problems are drawn largely from the student's own experience, and include many that are in the nature of experiments, to be performed by the student individually, or with the cooperation of the class. It is urged that most of these experiments be actually performed, because of the discussion and interest which comes from such concrete applications.
In using the book for the first time there will be perhaps some tendency on the part of the teacher to omit features which at first sight may seem unusual. Such omissions might easily detract materially from the force and usefulness of the text, and it is strongly urged that they be reduced to a minimum.
For those who wish to make a still closer correlation with Geometry, references to that subject (Part II of this series) are given, which outline in a general way a two years' course in Mathematics. For this purpose, if the teacher desires it, the two parts. Algebra and Geometry, may be had bound in one volume. This will be especially advantageous in schools where free text-books are fur- nished, because the work of the first year is then always at hand for reference during the second year.
E. Long,
W. C. Brenke.
Lincoln, Nebraska, July, 1913.
ALGEBRA
FIRST COURSE
CHAPTER I INTRODUCTION
1. The Operations of Arithmetic. In arithmetic we deal with the nmnbers 1, 2, 3, . . ., and with combinations of these numbers by use of the four fundamental operations, namely, addition, subtraction, multiplication, and division. Let us briefly review the meaning of these processes.
2. The Whole Numbers, or Integers. We first recall that the numbers 1, 2, 3, . . ., 10, 11, 12, called whole numbers or integers, are merely used as counters, to indi- cate how many objects are contained in a given group. The same thing could be done in many other ways, as, for example, we might make strokes, one for each object in a group. Thus we have the Roman numerals I, II, III, IV.
3. Addition. If we have two groups of objects, say one containing 3 and the other 4 separate objects, their com- bination into a single group will contain how many objects? Having counted the objects in one group, say 3, we must now count forward 4 more, and so we arrive at the number 7.
We write this result in the form 3 + 4 = 7.
This is read: ''Three plus four equals seven."
This means that when three things of any sort are united into a single group with four others, the result is a group
1
2 ALGEBRA — FIRST COURSE
containing seven distinct objects. The process is called addition.
The numbers 3 and 4 are called addends, and the number 7 their sum.
Evidently we might reverse the order of our counting, and count first the four objects and then the three more to complete the group. This gives us
4 + 3 = 7.
That is, two numbers may be added in either order; the result is the same.
In like manner we interpret the equation
4 + 3 + 5 = 12
as the union of three groups into one. In all cases the order of writing the numbers to be added together is immaterial; the result is the same. For example, we may add a column of figures either up or down; we may first rearrange the num- bers in the column and then add.
Now in applying the operations of arithmetic to actual problems, we usually deal with objects of a definite sort. We say 3 apples, 3 seconds, 3 feet, 3 quarts, and so on. For example, if we put 3 oranges and 4 apples into a basket, the basket contains 7 objects, but neither 7 oranges nor 7 apples. That is, if the number which results from the addition of several others is to represent so many objects of a certain kind, then all the numbers which are added must represent objects of the same kind.
4. Subtraction. The expression 7 — 3 (seven minus three) means that number which when added to 3 will give 7. From arithmetic we know that this is 4, so we have 7 — 3=4. Subtraction means to find the difference of two numbers, and to do this we find what number must be added to the subtrahend to produce the minuend.
Thus at a store when you buy goods whose value amounts
INTRODUCTION 3
to $1.37, and you pay with a five-dollar bill, the salesman hands you the purchased article and then counts on the change. In other words, in order to subtract $1.37 from $5, he determines the amount which must be added to $1.37 to make $5.
5. Multiplication. The number 3 can be considered as 3X1, that is, as a unit taken three times to measure the quantity considered. Thus 3 hours is 3 X 1 hour; 3 feet is 3X1 foot, and so on.
To multiply 3 by 4 means to take 3 units 4 times, giving 12 units; that is, 4 X 3 = 12.
Instead of the cross, X, a dot is often used to indicate multiphcation; thus 4 • 3 = 12.
From arithmetic we know that the product is the same when the order of the numbers to be multiplied together is reversed; that is, 3 • 4 = 12.
6. Division. To divide one number by another means to find a third number which when multiplied by the divisor will give the dividend.
Thus 12 -T- 3 means find a number which when multi- plied by 3 will give 12. Our knowledge of multiplica- tion enables us to guess the right number in many cases. When we cannot easily guess the number, certain rules are used for getting it. We shall later see the reason for these rules, when we study the subject of division further.
The common symbols for division are
io _!. o. 12 12 . 6, 3.
7. Subject Matter of Algebra. In the subject of algebra, we continue the study of these four fundamental operations as applied to nmnbers, and take up other operations. Also we freely use letters to designate numbers and it is chiefly in its free use of symbols that algebra differs from arith- metic.
4 ALGEBRA — FIRST COURSE
This can best be brought out by some examples.
Example 1 (a). If a train travels 5 hours at the rate of 30 miles an hour, what is the distance passed over?
Solution: 5 • 30 miles = 150 miles.
(b) If a train travels t hours at the rate of v miles an hour, what is the distance passed over?
Solution: t • v miles = d miles,
where t stands for the number of hours (time), v for the rate per hour (velocity), and d for the total distance.
Under (b) we have the solution of every possible problem of the sort in (a).
The result is usually written thus
tv = d,
the multiplication sign being omitted.
In algebra it is understood that when two letters are written side by side without any sign between them, the numbers for which these letters stand are to be multiplied together.
Thus: a*h ' c = ahc.
Example 2 (o;. If a rectangle is 3 feet wide and 2 feet high, what is its area?
Soluiion: 3*2 = 6, number of square feet in the area.
(6) If a rectangle is b feet long and h feet high, what is its area? • Solviion: b • h = a, number of square feet in the area.
Here again the equation under (6) contains all possible equations like those under (a).
Example 3 (a). A rectangular tank, whose base is 6 feet by 4 feet, and which is 3 feet high, is filled with water weighing 62^ pounds per cubic foot. What weight of water is contained in the tank?
Soluiion: 6 • 4 • 3 = 72, number of cubic feet in the contents of
the tank. 72 • 62i = 4500, number of pounds in the tank.
(6) A rectangular tank whose base is a rectangle a feet by b feet,
INTRODUCTION 5
and whose height is h feet, is filled with a Uquid which weighs w pounds
per cubic foot. What is the weight of the Uquid in the tank?
Solution: a-b'h = abh, number of cubic feet in the contents of
the tank.
ahh ' w = abhw, number of pounds in the tank.
If we let W stand for the number of pounds in the tank, we have
the^ormula
W = abhw.
Note that if w stands for the number of pounds per cubic inch, abh must be reduced to cubic inches.
Example 4 (a). What is the amount of a principal of 300 dollars placed at simple interest for 6 years at 5%?
Solviion: j^^ = number of dollars interest on $1 in 1 year.
300 • if 7 = 15 = number of dollars interest on $300 in 1 year. 6 • 15 = 90 = number of dollars interest on $300 in 6 years.
300 + 90 = 390 = number of dollars, amount of $300 in 6 years.
(6) What is the amount of p dollars placed at simple interest for n years at r%? .
Solution: ^7^ — number of dollars interest on one dollar in one year.
p • — — = YKF. = number of dollars interest on p dollars in one year. 1 UU J. uu
n • :^ = 7^ = number of dollars interest on p dollars in n years. 100 100
p + -^ = number of dollars in the amount of p dollars in n years.
If we let a stand for the number of dollars in the amount,
, npr
"=P + Io6-
In this formula replace p by 300, n by 6, and r by 5; what do you get for the value of a?
This result is usually written thus:
'^-A^'-m)'
where the right-hand member of the equation is to be understood as follows: Multiply the number standing before the parentheses, namely, p, into each of the numbers in the parentheses, and add the results. Thus: 5(4 + 3-|-2) = 5.4 + 5-3 + 5-2
= 45; a (6 + c + d) = ab -\- ac -[• ad.
6 ALGEBRA — FIRST COURSE
We may first add together the numbers within the parentheses, and then multiply their sum by the number outside. Thus: 5 (4 + 3 + 2) =5.9
= 45.
In this way we may look upon 6 + c + c? as a single num- ber which is to be multiplied by a. So we may look upon
TIT
1 + YqT^ as a single number to be multiplied by p.
Example 5. What principal must be put at simple interest at 6% so that it shall amount to $480 in 10 years? According to Example 4 we always have
Then we have
^(1+^)
P =
1+ — ^100
But a = 480, n = 10, r = 6. Therefore, p = j^-^
^"^"loo"
Multiplying numerator and denominator by 100, we have
48,000 p =
100 + 60 ^ 48,000
160 = $300. We here use the fact that if a = b'C, then r = c.
No matter what numbers are used for a, h, c, the second equation is always true if the first is true.*
We of course use this rule as a check on division in arith- metic. For example, dividing 357 by 17 gives 21. Check the correctness of this answer by showing that 21 times 17 gives 357. That is, since 357 = 21 • 17, then Vt' = 21.
* There is only one exception to this rule, namely, when b is zero. Division becomes meaningless when the divisor is zero.
INTRODUCTION 7
8. Algebraic Expressions — Their Forms and Values. In the preceding examples we have given some indication of the way in which letters are used to stand for numbers and have made use of some of the notation which is common in Algebra. We shall now give a list, although not a com- plete one, of the various forms used in the symbolic arith- metic which we call Algebra. Much of this is repetition of what has been said in the examples above.
In what follows all the letters and their combinations are supposed to stand for ordinary arithmetic numbers.
(1) Addition.
a-{-h means to add the number h to the number a; we read it ''a plus 6" just as in arithmetic.
6 + a means to add a to 6; we read it "6 plus a.'' Then we always have
a + 6 = b -\- a. That is, the sum is the same in whatever order we add.
Likewise, a + h -\- c means the sum of a and b and c. We read it "a plus b plus c."
The sum is the same in whatever way the letters may be written. The same is true of any number of letters.
(2) The expression (a + b) means that we are consider- mg the sum of the numbers a and b and are regarding it as a single number.
For example, the expression .
(a + 6) + (c + d) means *Hhe sum of a and b plus the sum of c and d. But this is exactly the same as adding together the four numbers a, 6, c, d. That is
(a + 6) -f- (c + d) = a + 6 + c + cZ. In the same way {a-{-b -\- c) means that we are to regard the sum of the numbers a, 6, c as a single number, and so on. Then we would have for example
(a-h6 + c) + (d + e)=a + 6 + c + d + e.
8 ALGEBRA — FIRST COURSE
In place of parentheses, ( ), we often use brackets, [ ], or braces, { J , or the vinculum.
Thus (a + b), [a-\-h], {a-\-hl, a + 6; each means that we are to regard the sum of the numbers a and ?> as a single number. These marks are called signs of aggregation.
Exercises. By giving arithmetic values to the letters in the following, show that the equations given are true.
1. (a + 6) + c = a + 6 + c. (Replace a by 5, h by 2, c by 3.)
2. a + {h + c) = a + h + c. (Replace a by J, 6 by J, c by i)
3. (a + c) + 6 = a + 6 + c. (Replace a by 6J, h by 8J, c by 51)
4. c+ (a + h) = a-\-b + c. (Replace a by 3f , h by 5^, c by If.)
5. {a + c) + {h -\- d) = a + h + c -\- d.
6. (a + d) + (c + h) =a + h-{-c + d.
7. {a + h + c) + {d + e-\-f)=a + h + c + d + e+f.
8. (a + 6) + (c + c^) + (e +/) = a + 6 + c + d + e +/.
9. [m + n] + [A; + Z] = m + n + A; + Z.
10. [s -\- 1 -^ u] + (v -\- w) = s -\-t + u-\-v + w.
11. [s + V -\- t] -{- [w + u] = s + t + u + V + w.
12. /i + A; + b + 5 + r]+Sa: + 2/J=/i-hA; + p + 5 + r + a; + 2/.
We may now state our first rule.
Rule I. In forming a sum the signs of aggregation, ( ),
[ ], { J , , may he omitted.
(3) Subtraction.
a — b means the number which must be added to b to give a.
b — a means the number which must be added to a to give 6, as in arithmetic.
We read these as "a minus fe" and "b minus a" respec- tively.
INTRODUCTION 9
Likewise: a — h-\-c means ''start with the number a, subtract the number 6, and to the result add the number c." Evidently this is the same as a + c — 6.
The signs of aggregation are used as before.
Thus: (a + 6) + (c — d) means "to the sum of a plus h add the difference of c minus rf." This is the same as a -\-b -r c — d, that is we may omit the parentheses.
(a + 6) — (c — d) means "from the sum of a and h sub- tract the difference of c minus c?." But we cannot remove the parentheses and say our answer is the same as a + & — c — (i, as a simple example will show.
We have (10 + 7) - (12 - 4) = 17 - 8
= 9. But 10 + 7 - 12 - 4 = 1.
The two results are not the same, so that in subtraction we cannot drop the parentheses as in addition. We shall soon have a rule to cover this case.
Exercises. Remove as many as possible of the signs of aggregation in the following exercises.
1. (a-h) + (c-d) = ? 2. (b-c) + id- a) =?
3. ia-\-b-c)-d=? 4.. d-\-{b-c-d) =?
5. [a-d-{-c]-\-[d-e] = ? 6. [a + b] -{- [d - e - c] = ?
7. \h-kl+[m-n] = ? 8. s -\- t + u - v -{- p - q = ?
9. (f-\-9)-(h-j) + k^'^ 10. (u+v)-(x±y)-(u-z) = l
11. n-{p + q)-{s-t) = ? 12. (a-6)-c + d-[e+/] = ?
Show by inserting arithmetic numbers for letters that the expressions in the last four exercises change their values when the signs of aggregation are omitted.
(4) Multiplication.
aXb means to multiply the number b by the number a. Since this is the same as multiplying a by b, we have
aXb = bXa,
10 ALGEBRA — FIRST COURSE
Using the dot to indicate multiplication,
aXb=a'b=b'a.
Usually we do not use either the cross or the dot; we simply write ah, thus:
aXh = a'h = ah — ha.
Notice that this is contrary to the rule in arithmetic:
2i = 2 + i, not 2 X i
The numbers a and 6 are called factors, and ah is their product.
Similarly we have
aXh X c = a'h * c = ahc,
and the "factors" a, h, c may be taken in any order. Thus:
ahc = ach = bca, and so on.
This applies to the product of any number of factors. We also may use the signs of aggregation:
aX{b-\-c) =a{b + c) = ah + ac. aX[h + c - d] = a\h -\- c - d] == ah + ac — ad.
(5) Division. The symbol j- means the quotient of a
divided by h, this quotient being a number which when multiplied by h will give a.
In place of t we may use a-r-h.
If we call the quotient q, then since the quotient multiplied by the divisor equals the dividend, we have,
a = bq equivalent to ^ = q-
Let the student carefully pote our definition:
I, § 8] INTRODUCTION 1 1
TJie operation of dividing ahy b consists in finding a number q which if multiplied by b gives a. That is, r or a-r-b, each stands for that number q, sv^h that bq = a.
The expression r is called a fraction, a being the numer- ator and b the denominator. The numbers a and b are called the terms of the fraction. Then, as in arithmetic, we have
2 _2j^ = !LL?.
3 2-3 n-S'
so we may have a in place of 2 and b in place of 3, and may write
a _3a _na
b~Sb~nb'
Rule. The two terms of a fraction may be multiplied by the same number without changing the value of the fraction. That is,
.<. a ., na
if T = Qf then -r = q. b ^ nb ^
(6) Cancellation. If both terms of a fraction contain the same factor, this common factor may be cancelled out. Thus,
y^a _ a ^b~b'
(7) Division of Products. When a product of numbers is to be divided by a number, or by a product of numbers, common factors should first be removed by cancellation.
144 » 56 - 18 _ ^'^'^'?'?-?'f'?'2'2'^'S'2 _ 36-12.14 ^.^.^.^.^.^.^.y.^ '^'
12 ALGEBRA — FIRST COURSE tt.$8
Likewise
24 amnr _ 3 • ^ • ^ • ? amyir 8 hn %^%*%h^
3amr
Especial attention is called to the fact that if you are called upon to divide a given number by a certain number and then to multiply by that same number, the result will be the given number. Thus, if you are told to divide 629 by 250 and then to multiply by 250 the result is 629. Or, in arithmetic form,
Aoq 250-250 = 629.
In all cases,
a*- = 0,
no matter what the values of a and h.
The rule that you can multiply the numerator and de- nominator by the same number and not change the value of the fraction, gives us the simplest method of reducing a complex fraction to a simple one. You have but to find the least common denominator of all the fractions found in both numerator and denominator, and multiply both terms of the fraction by it. Thus,
3f given ^^•
The least common denominator of the fractions is 15. Multiply the numerator and denominator by 15:
^^55 2/ir 37;
A little practice, to make the student familiar with this method, is worth while.
I. § 8] INTRODUCTION 13
Exercises. Simplify the following fractions:
11 3 1 _L 1
1. -0-* ^' T ^' 1 I 2
81|' f - iV ■ 824
Exercises for Review. Find the numerical value of each of the following expressions when the letters are replaced by the given numbers.
|
1. a + h + c; |
o = 3, 6 = 4, c = 5. |
|
2. a — h + c; |
a = 10, 6 = 3, c = 4. |
|
3. (a + 6)H-(c-c^); |
a = 21, 6 = 15, c = 40, d = 28. |
|
4. (a-h)-{c + d); |
a = 93, 6 = 22, c = 12, d = 18. |
|
6. p(q + r - s); |
p = 15, g = 65, r = 32, s = 12. |
|
6. (p + g) (p- q); |
P = n,q = M. |
|
^' h-k' |
^=5'^=6- |
|
mn Q « ' m-\-n' |
3 15 |
|
. x-\-(y-z), ^' x-(y + z)' |
3 2 1 ^=2'2/=3''=6- |
10. ,^^^, ; p = 56, g = 112, r = 28.
i+f
11. ; a =4, 6 = 5.
12. 13.
14.
|
'-f |
||
|
[r+(s-t)]- |
(s + t): |
; r = i,s = i,t = h |
|
a + (6 - c) . a-(6 + c)' |
||
|
TO n n m mn ' |
3 2 |
CHAPTER II
MEASUREMENTS — LENGTHS, ANGLES, AREAS, VOLUMES
9. English Units of Length. You are familiar with the yardstick, the foot rule, and the inch. To tell some one how long a certain line is you might say that it measures so many yards; the yard is then your unit of length. To state the length of a shorter line you might tell the number of feet which it contains; the foot is then your unit of length; for a still shorter line, such as you might draw on a sheet of paper, the inch would be a convenient unit. Quite long lines, as the length of fence around a farm, are measured with the rod as the unit of length; for distances between cities we use the mile as the unit.
' So we have the following common English units of length: inch, foot, yard, rod, mile.
Exercise. How many feet in 3 yards? In 5| yards? In n yards? How many inches in each of these? How many rods in 1 mile? In 2j miles? In m miles?
10. Metric (or French) Units of Length. The French have chosen a different set of units to measure lengths of lines. This is based on the decimal system and is com- monly used in scientific work. In this system there is a unit corresponding nearly to the English yard, and called a meter. It is a little longer than the yard and represents one ten-millionth part of the distance from the earth's equator to the pole, measured along a meridian. One tenth of a meter, about four inches, is called a decimeter; one hundredth of a meter is a centimeter, a little less than half
14
II. § 11]
MEASUREMENT OF LENGTHS
15
an inch; one thousandth of a meter is a millimeter. For long lines, where we would use the mile, the French unit is the kilometer, or one thousand meters, a little more than half a mile.
Thus the common metric units of length are: millimeter, centimeter, decimeter, meter, kilometer.
Exercise. How many decimeters in half a meter? In 2.3 meters? In m meters? How many centimeters in each of these?
11. Comparison of English and Metric Units of Length. The figures below show parts of a ruler, one edge of which is divided into inches and tenths of inches, the other into centimeters and millimeters. Such a ruler will be found inside the back cover of this book.
|
MM MM |
Inches and Tenths M M 1 MM M M M M 1 |
M |
1 1 1 |
|
|
1 1 9. |
s h '5 |
6 7] |
8 |
9 |
|
WMMMm |
MMMiniMI |
II III! III! nil nil |
Hill |
II Hill |
Centimeters and Millimeters
Inches and Tenths
I I I M I M II M M M I M M M I M M M M M I I
33 \ 83 \ 8U\
851
36 91]
Centimeters and Millimeters
Exercise 1. From the first figure read off, as exactly as possible, the number of inches in a centimeter; the number of centimeters in an inch.
Exercise 2. Draw a line several inches long. Measure it with the edge of the ruler marked in common units; measure it again in metric units. From these measurements find the number of inches there are in a decimeter. Solve as in the following example.
16 ALGEBRA — FIRST COURSE [ii.§i2
Example. Suppose the line measures 4.5 inches and also 1.14 deci- meters. To find the number of inches in one decimeter:
Let i = the number of inches in one decimeter.
Then 1.14 i = the number of inches in 1.14 decimeters.
But 4.5 = the number of inches in 1.14 decimeters;
therefore, 1.14 i = 4.5, since both stand for the measurement of the same line.
Therefore i = 3.95. (Why?)
So we have 3.95 inches in a decimeter, as nearly as we can tell from the given measurements.
Repeat this experiment, using different lengths of lines. Exercise 3. From the second figure above, how many inches in 9 decimeters? Hence how many inches in one decimeter?
Exercise 4. Draw a line 2.75 inches long. Measure it in centimeters. Now solve as above to find how many centi- meters in an inch. Let c stand for the required number.
Exercise 5. Draw a line 0.75 inch long. Measure it in millimeters. Solve as above, to find the number of milli- meters in an inch. How does this answer compare with the answer to Exercise 4?
Exercise 6. A line measures x inches. How many deci- meters does it contain?
Exercise 7. A line measures y centimeters. How many feet does it contain?
12.* Angles — Definitions and Nota-
/^ tion. Suppose that we partially open
,///^ a fan, or two of the arms of a folding
^/^'' ruler. Suppose also that a line is
/^^ drawn on each arm of the ruler, start-
O-- — -" IJ ij^g ixom the pivot and following the
middle of the arm. Definitions, The figure so formed by two straight lines starting out from the same point is called an angle. The
* At this point Chapter I of Geometry may be taken up if closer correlation is desired.
II. § 13]
MEASUREMENT OF ANGLES
17
point marked A in the figure (the pivot) is called the vertex of the angle. The lines AB and AC are called the arms of the angle.
Notation. We shall often use a single letter, usually a capital, placed near a point in a figure to designate that point. If we designate a certain point by A, and another point by B, the straight line through these two points is called the line AB or the line BA. We would say 'Hhe line AB" when the line is drawn from A to B; we would say "the line BA " when the line is drawn from B to A. Often it makes no difference which we use; in other cases a dis- tinction is necessary.
To designate the angle formed by the lines AB and AC, we say ''the angle BAG/' or, ''the angle CAB.'' The first means that in opening up the angle we regard AB as a fixed arm and AC as revolving; the second means that AC is the fixed arm and that AB is revolving. Often it makes no difference which notation is used. In either case, to desig- nate an angle, first name a point on the fixed arm, then name the vertex, and finally name a point on the movable arm.
In fixed figures, angles may usually be read either way.
The symbol for the word "angle" is Z ; so ABAC means "angle BAC."
Exercise. In each of the figures below read off the points, lines, and angles marked in it.
13. Classification of Angles. Angles are classified accord- ing to the amount of turning done in separating the arms.
18
ALGEBRA — FIRST COURSE
tll, § 14
In the figure on p. 16 suppose AB to be fixed and AC to be revolving; when AC has revolved half way around, so that it lies just opposite to AB and forms one straight line with it, the angle BAC is called a straight angle; half of a straight angle is a right angle; if AC turns through less than a right angle it forms with AB an acute angle; if AC turns through more than a right angle but less than a straight angle, it forms with AB an obtuse angle ; more than a straight angle is called a reflex angle; a complete turn is called a perigon.
Z2\
C. A -B A B A JB
Straight Angle BAC. Right Angle BAC. Acute Angle BAC.
\
X\
A. B
Obtuse Angle BAC.
Perigon BA C.
Reflex Angle BAC.
Exercise. Classify each of the angles in the figures on p. 17.
14. Measurement of Angles. To state the size of an angle we adopt some standard angle as a unit, and say how many of these units are needed to fill the given angle. Two different units are in common use, one called the degree the other the radian.
Degree Measure of Angles. When a perigon is divided into 360 equal parts, each such part is called a degree.
II. § 14] MEASUREMENT OF ANGLES 19
So we have
360 degrees = a perigon. Then 180 degrees = a straight angle
and 90 degrees = a right angle.
The symbol for degrees is °, so that 10° means "ten degrees."
Angles are usually measured with a protractor (see inside
Protractor
of back cover) as shown in the above figure. To measure, a reflex angle, measure its excess over a straight angle, or measure what is lacking to make a perigon.
Exercise 1. Draw ten different angles, some acute, some obtuse, and some reflex. Measure each and write its value on your figure.
Exercise 2. Draw a triangle. Measure each angle. What is their simi? Repeat this with another triangle of different shape.
Exercise 3. Draw a triangle and tear apart as in the above figure. Place the three angles with their vertices
20 ALGEBRA — FIRST COURSE [ii, § 15
together, one angle next to the other, without overlapping. What is the sum of the angles of the triangle?
Exercise 4. Repeat Exercise 2, using a quadrilateral, that is, a figure bounded by four straight lines. Do not draw a square or a rectangle, but rather a figure whose sides and angles are quite unequal.]
Exercise 5. Repeat Exercise 3, using a quadrilateral.
Exercise 6. Repeat Exercises 2 and 3, using a pentagon, that is, a figure bounded by five straight lines.
Radian Measure of Angles. In this system the unit of measure is a radian, instead of a degree as in the system just considered. You can easily make a protractor gradu- ated in radians.
Exercise 1. On stiff paper, or, better, light cardboard, draw a circle with a radius of, say, two inches. Carefully cut it out and mark a point on the circumference or rim. On a good-sized sheet of paper draw a straight line and mark off on it parts, each equal to the radius of the circle. Roll the circle carefully along this line, starting with the marked point on the rim placed at the beginning of the first division on the line. Each time that a point on the rim of the rolling circle reaches a division point on the line, mark that point on the rim. Now draw lines from the center of the circle to the points marked on the rim. You then have a series of equal angles, each of which is one radian.
Exercise 2. Define a radian.
Exercise 3. By rolling the circle so that it makes just one complete turn, find approximately how many radians there are in a perigon. You will find a little more than six radians. Estimate the decimal part as well as you can.
15. The Number it. . The number of radian units in a perigon is not a whole number, as you found in the last exer- cise. Nor can this number be expressed either by a fraction or by a terminating decimal. It is a so-called incommen-
II. §15] MEASUREMENT OF ANGLES 21
surable number and by general agreement is always indi- cated by 2 TT, TT being a Greek letter called ''pi." We therefore have
2 TT radians = a perigon = 360 degrees. TT radians = a straight angle = 180 degrees.
180 Then 1 radian = — degrees. (About 57°.3.)
TT
The number for which w stands, to four decimal places, is 3.1416; less exactly it is ^j^-. It should be remembered that both of these values are only approximate.
Exercises.
1. From Exercise 3 above, find as nearly as you can the number of radii that must be taken to equal the length of the circumference of a circle. The number of diameters. (A diameter of a circle is a line through the center and terminated each way by the circumference.)
2. If c denotes the circumference of a circle, d the diam- eter, and r the radius, show that
c = 2 TTr,
and that
C = TTd,
State these equations in words and learn them.
3. Using the radian scale on your protractor, repeat Exer- cise 2 (p. 19).
4. Using the radian scale on your protractor, repeat Exercise 4 (p. 20).
5. Using -2y2- as the value of tt, find the number of degrees in one radian. Repeat, taking tt as 3.1416. In each case carry the result to three decimal places.
6. Draw an acute angle. Measure it with the degree scale on your protractor, then with the radian scale. Solve as in the example worked out on p. 16 to find tKe_ number of degrees in a radian.
22
ALGEBRA — FIRST COURSE
III. § 16
7. Draw an obtuse angle and do the work as instructed in Exercise 6.
8. Draw a reflex angle and repeat work of Exercise 6. How do the answers to Exercises 6, 7, 8 compare? How
do they compare with results of previous work?
9. How many radians are there in the sum of the angles of a pentagon ?
16. Measurement of Areas. If a unit of length is chosen, then a square whose side is that unit of length is our unit of area. To measure an area we must find how many such square units are required to cover it.
Exercise 1. On cross-section paper draw a rectangle of any convenient base and height. How many linear units are there in the base? In the height? How many square units in the area? To answer the last question count the squares. Do your numbers fit the formula
Area of rectangle = base times height?
In letters, this formula may be written a = bh.
Verify this formula by drawing several other rectangles such as: a = 3, 6 = 4; a = 5, 6 = 2; a = J, 6 = J. When the rectangle is a square whose side is h, its area is
a = hh. In place of hh we usually write b^, so that
a = h\ (a equals the square on 6.)
Exercise 2. Draw a parallelogram, that is, a quadrilateral whose opposite sides are parallel. Cut this out from your
II. § 16] MEASUREMENT OF AREAS 23
paper. Now cut off a triangle from one side and fit it on the other. Is the area of your parallelogram equal to the area of a rectangle of the same base and height?
Repeat Exercise 1, using a parallelogram. So verify the statement:
Area of parallelogram = base times height. Using letters: a = bh,
as for the rectangle.
Exercise 3. On cross-section paper draw a triangle, tak- ing one of its sides along one of the ruled lines. Call this side the base. Draw a line at right angles to the base and leading to the opposite comer of the triangle. Call this line the altitude. How many units are there in the length of the base? In the altitude? Find as well as you can the number of square units in the area of the triangle, add- ing parts of squares to make whole squares. Do your num- bers fit the formula
Area of triangle = J (base times height)?
Draw several other triangles and in each case verify the formula
a = ^hh.
Exercise 4. Draw a circle on your paper; also draw a triangle with its base equal in length to the circumference of the circle and its height equal to the radius of the circle. Count the number of squares in each and see how they compare, approximately.
How would you find the area of a circle? Letting a stand for the number of square units in the area, c for the number of linear units in the circumference, and r for the number of linear units in the radius, we have the statement
a = ^cr; but we have shown that c = 27rr; therefore a = J • 2 irr • r
= irr\
24
ALGEBRA — FIRST COURSE
[11, § 17
Exercise 5. On cross-section paper draw a circle with any convenient radius, such as five times the side of one square. Find as nearly as possible the number of square units in the area of the circle. Also find the area of a square whose side equals the radius of the circle. Do your num- bers fit the formula
Area of circle = tt times square on radius?
Draw several other circles of various sizes and in each case verify the formula
a = Trr^.
17. Measurement of Volume. To measure the cubical content of a solid body we choose as our unit of measure a cube whose edge is one linear unit. The number of such cubic units contained in the body to be measured is called its volume.
For example, a pint jar will contain about 27 cubic inches of water; that is, its volume is 27 cu. in.
The volumes of a few regular bodies are given below, each in terms of the dimensions of the body. The dimensions in each formula should be expressed in the same unit of length. State each formula in words.
Exercises.
1. Measure the dimensions of a rectangular parallelopiped. Find its volume by immersing it in water in a graduated beaker and noticing how much the water rises. Do your numbers fit the formula?
/
I
L.
....i
Bectanoular Parallelopiped Volume.'^ dhc.
Cylinder Volume ^Ti'T'^ b.
II. § 17]
MEASUREMENT OF VOLUME
25
Cone Volume^ ys"^^^^'
Sphere Volume ^ya'^'''^.
|
(6) Cylinder. |
|||
|
r = S,h = |
= 2; |
r |
= 1 |
|
(c) Sphere. |
|||
|
r = 5; |
r |
= 3§ |
2. Repeat Exercise 1 with a cylinder.
3. Repeat Exercise 1 with a cone.
4. Repeat Exercise 1 with a sphere.
5. Calculate the volume when the dimensions are as below. (Take tt as V" or 3|.)
(a) Cone.
r = 2, /i = 4; r = 1|, /i = 2J; r = 3i /i = f .
r = 2f.
6. Divide the volume of a cylinder by the volume of a cone having the same base and height. What is the quo- tient?
7. If a cylindrical glass is filled with water and then a solid cone of the same base and height is placed in the glass, what part of the water will be forced out?
8. A cone-shaped funnel is 8 inches across the top and 6 inches deep. How much water will it hold?
9. Measure the dimensions (height and radius) of a pint jar, using the inch as the unit of length. How many cubic inches of water will the jar hold? Is "a pint a pound"?
10. A sphere and a cylinder have the same radius. How high should the cylinder be to have the same volume as the sphere? First take the radius, say, 3 inches; then solve again taking the radius equal to r.
26 ALGEBRA — FIRST COURSE [Ii.§i7
Summary.
Measurement of Length
English Units * Metric Units
Inch (in.) Millimeter (mm.)
Foot (ft.) Centimeter (cm.)
Yard (yd.) Decimeter (dm.)
Rod (rd.) Meter (m.)
Mile (mi.) Kilometer (km.)
1 rod = 16i ft. 1 mile = 320 rods = 5280 ft. 1 meter = 10 decimeters = 100 cm. = 1000 mm. 1 km. = 1000 meters.
Angles: straight angle, right angle, acute angle, obtuse angle, reflex angle, perigon. Angle measure: 360 degrees make a perigon.
2 TT radians make a perigon.
The letter ir denotes the number of times the diameter of a circle is contained in the circumference; 2 tt is the number of times that the radius is contained in the circumference.
c = ird = 2Trr.
TT = 3.14159 + • • • = W approximately. One straight angle = 180 degrees = tt radians.
One right angle = 90 degrees = ^ radians.
The sum of the angles of a triangle is two right angles. The sum of the angles of a quadrilateral is four right angles.
Areas
|
Rectangle or Parallelogram |
Triangle |
Circle |
|
a = hh. |
a = ihh. |
a = wr^. |
|
' = r |
'-¥■ |
a = icr. |
|
"'i- |
-¥• |
II. §17] MEASUREMENT OF VOLUME 27
Volumes
Rectangiilar Parallelopiped Cylinder
Volume = abc. Volume = xr^/i.
Cone Sphere
Volume = J Tr%, Volume = | tt^.
CHAPTER III
100
212
MEASUREMENTS CONTINUED — TEMPERATURE, WEIGHT AND DENSITY, FORCE
18. Measurement of Temperature. A thermometer is an instrument for measuring temperature. It consists of a small bulb or reservoir filled with mercury and connecting with a very narrow vertical tube. When the mercury in the bulb is heated it expands and rises into the tube. So the height of the mercury in the tube indicates the degree of heat to which the bulb is exposed.
The Centigrade Scale of Temperature. Im- merse a thermometer in melting ice and mark the point on the tube where the mercury stands zero. This is called the freezing point. Next immerse in boiling water and mark the new point 100. This is called the boiling point. Divide the space between zero and 100 into 100 equal parts. Each such part is called one de- gree Centigrade. The graduations are extended below the zero point to indicate temperatures below zero.
The Fahrenheit Scale of Temperature. Pro- ceed as above, except that the freezing point is marked 32 and the boiling point 212. Divide the space between these into 180 equal parts. Each such part is called one degree Fahrenheit.
The adjacent figure shows two thermometers, one gradu- ated Centigrade, the other Fahrenheit. Both scales may be placed on the same thermometer.
28
so
III. 5 19] MEASUREMENT OF WEIGHT 29
Exercises.
1. Immerse two thermometers, one graduated C. and the other F., or one thermometer with both scales on it, in a cup of cold water and record the reading on each scale. Next immerse in a cup of hot water and record the readings on each. From these readings find how many degrees Fah- renheit equal one degree Centigrade. Write out each step in full as in the example worked out on p. 16.
2. How many degrees Fahrenheit are equal to 100 de- grees Centigrade? From this find how many degrees Fahren- heit equal one degree Centigrade.
3. When the Fahrenheit scale reads 70°, what is the Centigrade reading ?
4. When the Centigrade scale reads 35°, what is the Fahrenheit reading?
6. Let F denote the Fahrenheit reading, and C denote the corresponding Centigrade reading. Find C when F = 86°; 68°; 50°; 41°; 14°. Fmd F when C = 10°; 20°; 30°; 70°; 10° below zero.
6. Show that for all temperatures above freezing
JP' = 32 + I C; and that
C = i (jP - 32).
7. At what temperature Centigrade is the Fahrenheit reading equal to three times the Centigrade reading?
19. English Units of Weight. The weight of a body is measured by the pull of the earth upon that body. When we say that a body weighs five pounds, we mean that the earth pulls on it with five times the pull on a one-pound weight.
Now what is a one-poimd weight? It is the weight of a certain piece of platinum which is carefully preserved by the British government, and used to test other pound weights.
30 ALGEBRA — FIRST COURSE [111,5 20
When this weight is hung on a spring balance, the place to which the pointer moves is marked 1. Any other weight which pulls the pointer to the same place is then also a pound. By hanging on two such weights, then three, and so on, each time marking the place where the pointer stops, we get a spring balance gradu- ated in pounds. To weigh a body we need only hang it on the balance and notice where the pointer stops.
Based on the pound we have the English units of weight as follows:
1 pound (Avoirdupois) =16 ounces = 7000 grains. 100 pounds = 1 hundredweight. 2000 pounds = 1 ton. 2240 pounds = 1 long ton.
There is another pound, called the Troy pound, in less common use. It is divided into 12 ounces and 5760 grains.
We shall deal only with the pound avoirdupois. Exercises.
1. How many ounces in 2 lb.? In 3| lb.? In 4J lb.? In n lb.? In (a + h) lb.? How many grains in each of these?
2. How many pounds in 40 oz.? In 75 oz.? In m oz.? In Qi + h) oz.? How many grains in each of these?
3. How many pounds in 10,000 gr.? In 1400 gr.? In r gr.? In (c — d) gr.?
4. How many grains in a lb. + 6 oz.? How many ounces in r lb. + ^ gr.?
20. Metric (or French) Units of Weight. In France the unit of weight is the weight of a cubic centimeter of water at a temperature of 4 degrees Centigrade. It is called a gram. This is a rather small unit, for it takes nearly 500 grams to make a pound; therefore a larger unit, namely the kilogram or 1000 grams, is more commonly used. A
Ill, § 21]
MEASUREMENT OF WEIGHT
31
kilogram equals about 2.2046 pounds. A half kilo would then be a little more than one pound, and'*this is the official unit which takes the place of the English pound.
For deUcate weighings, such as are needed in physics and chemistry, the gram is divided into smaller units thus: one tenth of a gram = a decigram; one hundredth of a gram = a centigram; one thousandth of a gram = a milligram.
Exercises.
1. How many decigrams in 10 grams? In 1.5 grams? In g grams? In {m + n) grams? How many centigrams?
2. How many milligrams in 5 decigrams? In 3 centi- grams? In k decigrams? In k decigrams + I milligrams? How many grams in each of these?
21. The Beam Balance. For accurate weighing the spring balance is replaced by the beam balance. This is constructed on the following principle.
n
n
When a beam is supported at its middle on a knife edge and equal weights are placed at equal distances from the knife edge, the beam will balance.
For convenience a scalepan is usually fastened to the beam at each end; in the druggists' scales the pans rest on
32 ALGEBRA — FIRST COURSE tiii.§22
top of the beam; in the chemical balance the pans are hung from the beam by wires.
Exercises.
1. In one pan of a beam balance place a weight of one pomid; how many grams must be placed in the other pan to balance? How many grams in a pound?
2. Place a piece of iron or other substance in one scale- pan; balance it first by English units, then by metric units, noting the amount of each. From your notes calculate the number of grams in an ounce. Write out the solution in full as in the example worked out on p. 16.
Repeat this experiment several times with different amounts of material.
3. A certain weight is balanced by q lb. and r oz. How many grams does it weigh?
22. The Lever. A beam placed on a knife edge is usu- ally called a lever. The ordinary ''see-saw" is a rough sort of lever. The knife edge is called the fulcrum; the parts of the beam on each side of the fulcrum are called the arms of the lever.
If two boys, one considerably heavier than the other, were to play see-saw, would they sit at equal distances from the fulcrum? Why not? Which one would sit farther out?
This illustrates the following rule, called the 'principle of the lever:
|
n |
m' |
||||
|
r |
|||||
|
"^ d, |
d2 |
*1 |
Rule. In order that a weight Wi, placed at a distance di from the fulcrum, shall balance a weight W2 placed at a dis- tance dk from the fulcrum, we must have
Fi X c^i = TF2 X d^.
III. § 221 MEASUREMENT OF WEIGHT 33
Exercises.
1. State the last equation in words.
2. If a boy weighing 75 lb. sits 6 feet from the fulcrum, where should a boy weighing 100 lb. sit to balance the beam? Give neat solution, letting d stand for the number of feet in the required distance.
3. How far from the fulcrum must a 12-lb. weight be placed to balance a 30-lb. weight placed 5 feet from the fulcrum?
4. A weight of 200 grams is placed 25 centimeters from the fulcrum. How far from the fulcrum must a weight of half a kilogram be placed to balance?
When several weights are placed at various distances to one side of the fulcrum, and other weights at various dis- tances to the other side of the fulcrum, the beam will balance when the sum of the products formed by multiplying each weight on one side by its distance from the fulcrum is equal to the sum of such products from the other side.
Example. Weights of 5, 10 and 3 lb. respectively are placed at the distances shown in the figure. What weight W placed 8 feet from the fulcrum will balance the beam?
e,, 10 lb. Wlb.
B.
m.
L « ^
-io >|4-— -« J
Solution: According to the last rule we must have
5X10 + 10X6 = 3X2 + TFX8. That is, 110 = 6 + 8 W,
or 104 = SW.
Therefore, Tf = 13 lbs.
Check this answer.
Exercises. Find what is necessary for balance in each of the following cases.
34 ALGEBRA — FIRST COURSE [ill. §23
To left of fulcrum To right of fulcrum
1. TF = 3 lb. 7 lb.; 5 lb. ? lb.
d= 4 ft. 6 ft.; 8 ft. 2 ft.
2. W = 8oz. 5oz.; . ? oz. 10 oz.
d = 7 in. 4 in. ; 13 in:. 5 in.
3. IT = 100 gm. 75 gm.; 60 gm. 20 gm.
d = 5 cm. 8 cm.; ? cm. 10 cm.
4. Tf = 2 kgm. 5 kgm. 4 kgm.
d = 25 cm. 15 cm. ? cm.; 30 cm. b. W = 2ilb.; 3Hb. 1.4 lb. 21b.
d= 25 in.; 6 in. ? in. 9| in.
23. Density — Specific Gravity.
Exercise 1. Determine the weight of a solid rubber ball in grams. Determine the volume of the ball by dropping it into water in a graduated beaker and noticing the amount the water rises. Express this volume in cubic centimeters. How many grams does this amount of water weigh? Divide the weight of the ball by the weight of the same volume of water. The quotient is called the density of the ball.
Definition. The density of any substance, when water is taken as the standard, is the quotient obtained by dividing the weight of a given volume of that substance by the weight of an equal volume of water. The density of a liquid is also often called "specific gravity.''
Exercise 2. Determine the density of iron, wood, stone, and glass by repeating Exercise 1 with pieces of each of these substances. Make a table of your results. Other Exercises.
3. Find the density of a solution made by dissolving 10 gm. common salt in 20 gm. water. (Divide the weight of the solution by the weight of an equal volume of water.)
4. Find the density of a solution made by dissolving w grams of salt in n grams of water.
6. How many grams of salt should be dissolved in 25 cc. water to make a solution whose density is 1.4?
6. To 10 cc. of a liquid of sp. gr. 1.2 are added 20 cc. of a liquid of sp. gr. 0.8. What is the sp. gr. of the mixture?
in. § 2i]
SUMMARY
35
Table of Densities
Aluminum 2.6
Brass 8.4
Charcoal 1.6
Copper 8.8
Cork 14r-.24
Diamond. . . .
Glass
Gold
Gutta Percha ,
Iron
Ivory
Alcohol . Benzine Ether. .
3.53 4r-4.5 19.3
0.97
7.8 1.8
Lead 11.4
Marble 2.6
Mercury 13 . 6
Nickel 8.7
Paraffin 0.89
Platinum 21.5
Silver 10.5
Tin 7.3
Zinc 7.2
Water 1.0
Liquids
0.81 Glycerine 1.27
0.90 Turpentine 0.88
0.73 Water 1.00
24. Summary.
Temperature
Centigrade scale: freezing = 0 degrees;
boiling = 100 degrees. Fahrenheit scale: freezing = 32 degrees;
boiling = 212 degrees.
English
Units of Weight
Metrics
|
Grain (gr.) |
Milligram (mgm.) |
|
Ounce (oz.) |
Centigram (cgm.) |
|
Pound Gb.) |
Decigram (dgm.) |
|
Hundredweight (cwt.) |
Gram (gm.) |
|
Ton |
Kilogram (kgm.) |
The Lever. To balance a lever: Multiply each weight on one side of the fulcrum by its distance from the fulcrum
36 ALGEBRA — FIRST COURSE [iii.§25
and form the sum of these products; do the same with the weights on the other side and form the sum of these products; the two sums must be equal.
The density, or specific gravity, of a substance is the quotient obtained by dividing the weight of a given volume of that substance by the weight of an equal volume of water.
25. Questions and Problems for Review.
1. What are the units used to measure angles? Describe each.
2. What are the common English units used to measure lengths, areas, volumes, and weights?
3. What are the French or metric units?
4. Define density.
5. Define the meaning of the words "lever'' and "ful- crum.''
6. What is the sum of the angles of a triangle in degrees? In radians? What is the sum of the angles of a quadri- lateral?
7. On cross-section paper draw a rectangle of any con- venient width and height. Draw a triangle of the same width and height as the rectangle. , By counting squares verify the formula
Area of triangle = J (area of rectangle).
Express this formula in letters, letting t stand for the area of the triangle and r for the area of the rectangle.
8. Draw a circle; also draw a triangle whose base equals the perimeter of the circle and whose height equals the radius of the circle. By counting squares compare the areas of the two figures as well as possible. If a stands for the number of square units in the area of the circle, c for" the number of linear units in the circumference, and r for the number of linear units in the radius, do your numbers fit the equa- tion
III. § 25] PROBLEMS 37
We have also found experimentally that
c = 2 Trr. Usmg these two equations, show that a = Trr 2 (p. 23).
9. If one angle is 2 times the size of another angle, and the sum of the two angles is x radians, what is the number of radians hi each angle? Draw these angles, after you have found the size of each.
Solution:
Let a = the number of radians in the first angle.
Then 2 a = the number of radians in the second angle,
and 3 o = the number of radians in the sum of the two angles.
But X = the number of radians in both angles.
Therefore 3 a = tt, since 3 a and x stand for the same number of radians.
Then a = 5 » ^^® number of radians in the first angle,
and 2 a = -^ , the number of radians in the second angle,
o
Therefore the first angle contains 5 radians, and the second angle
o
contains -^ radians.
Check: The next step is to check the correctness of our answers. To do this we go back to the original statement of the problem and see whether every statement is fulfilled. To begin with, the one angle must be twice as large as the other.
-K- radians = 2X5 radians.
Also, the sum of the two angles must be ir radians.
5 radians + -^ radians = ir radians.
6 6
Therefore our answers are correct.
10. If one angle is 2 times the size of another angle, and the sum of the two angles is 180 degrees, what is the number of degrees in each angle? Draw.
38 ALGEBRA — FIRST COURSE lin.§25
11. Take your answers to exercise 9 and solve to find the number of degrees in each angle. Make use of letters in your solution. How do the answers of exercises 10 and 11 compare?
12. The sum of two angles is f tt radians, and one of them is 3 times the size of the other. What is the number of radians in each? Draw.
13. State exercise 12 expressing the sum in degrees. Solve and show agreement of answers.
14. The sum of three angles is | radians. The first is J of the size of the second, and the third is J of the size of the first. What is the number of radians in each angle? Draw these angles.
16. State exercise 14, expressing the sum in degrees and compare, as in exercise 13.
16. The length of a rectangle is 3 times its width. The length of the line bounding it (called the perimeter) is 372 mm. What is the length of the rectangle? The width? Draw the rectangle.
17. State exercise 16, using the inch unit to express the combined length of the three lines. Solve, draw, and com- pare.
18. Of five lines the first is 2J times the second, the third is I as long as the first, the fourth is as long as the sum of the first and third, and the fifth is | as long as the sum of the first and fourth. What is the length of each line if the combined length is 27 millimeters?
19. If one of the angles of a triangle is J as large as an- other, and the third is j of the sum of the first and second, what is the number of radians in each angle? Draw such a triangle.
20. If the first of three angles of a triangle is ? as large as the second and the third is f as large as the first, what is the number of radians in each angle? Draw such a triangle.
21. The first of the three angles of a triangle is half as large
III. 1 25] PROBLEMS 39
as the second, and the third is equal to the sum of the first and second. What is the number of degrees in each angle? Draw.
22. The first angle of a quadrilateral is f as large as the second, the third is equal to the first, and the fourth is equal to the second. What is the number of radians in each angle of the quadrilateral? Draw such a quadrilateral.
23. One of the four angles of a quadrilateral is ^ as large as another, a third is equal to the sum of the two, and the fourth is equal to the difference between the two. What is the number of degrees in each angle? Draw such a quadri- lateral.
24. In the following state the problem in good English, using the word instead of the letter which stands for that word. Solve by substituting the number instead of the letter in the formula on p. 26, and answer the question asked in your problem. Illustrate each problem.
For the rectangle or parallelogram:
(a) h = 7, h = 2, a =? (See formulas on p. 26.)
Illustration. If a rug is 7 feet long and 2 feet wide, what is the num- ber of square feet in the rug?
Solution: Formula, a = hh.
Since 6 = 2 and h = 7,
a = 2 . 7 = 14.
Therefore there are 14 square feet in the area of the rug.
(6) a = 100, 6=8, h =2
(c) a = 57, /i=3, b =?
id) a = TV, b=h h=?
(e) 6=53f, A = 7f, a =?
For the triangle:
(/) 6 = 7.5, /i=40, a=?
(g) a = 286, h =459, h =?
{h) a = 2h /i = lf, h =?
40 ALGEBRA — FIRST COURSE llll,§25
|
) ALGEBRA - |
-FIRST COURSE |
|
For the circle: |
|
|
(0 r = 110, |
c =? a=? |
|
0*) c = 3.5, |
r =? a=? |
|
(k) c? = .42, |
c =? a=? |
|
(I) a = 49 TT, |
r =? c =? |
25. If a certain sum of money is put at simple interest at a certain rate, for a certain time, you have learned from your arithmetic that
interest = principle X rate X time.
Using the first letter of each word as we have done in previous examples,
i = p 'T *t.
(a) What does p equal in terms of i, r, Vt What does t equal in terms of p, i, r? What does r equal in terms of p, i, Vt
(b) If a sum of 200 dollars is placed at simple interest at 6% for 3 years, what is the interest due? Solve by using formula and answering question asked.
(c) The interest on $1200 placed at the rate of 5% was $240. What was the time of the note?
(d) At what rate of interest must $1400 be placed in order that it bring $294 interest by the end of three years?
26. An ivory ball and an India-rubber ball of the same size together weigh 60 grams. Ivory is twice as heavy as India rubber. What is the weight of each ball?
27. A solution of alum and water weighs 456 grams. Alum weighs 1.7 times as much as water. How many grams of alum are there in the solution?
28. Marble is 1.5 times as heavy as ivory. 9 marble balls weigh 46.8 grams more than 7 ivory balls. What is the weight of each ball?
29. A brass ornament plated with gold weighs 50 grams. There are 5 times as many cubic millimeters of brass as there are of gold, and one cubic millimeter of gold weighs
Ill, §25] PROBLEMS 41
.0193 grams, and one cubic centimeter of brass weighs 8.4 grams. How many cubic millimeters of each metal are there in the ornament?
30. If we know the velocity of a moving body, and the length of time it moves, how may we find the distance it moves?
Let V stand for the velocity, t for the time, and d for the distance; state the answer to the last question in letters.
State the formula for t in terms of d and v. State the for- mula for V in terms of d and L Make problems for each of these cases.
CHAPTER IV GRAPHIC REPRESENTATION OF QUANTITY
26. The Amount of any Quantity Represented by the Length of a Line. The result of the measurement of any quantity may be expressed by a number which tells us how many units or parts of units of measure are contained in the quantity measured. We may also represent the amount of our quantity by the length of a line; this is especially use- ful when several quantities are to be compared, because the eye takes in at a glance the various lengths of lines.
Example 1. On your section paper let one division of the line rep- resent 1 pound; then four divisions will represent 4 pounds; six divisions will represent 6 pounds; and so on.
lib.
klh.
6 lb.
Example 2. Let two divisions represent a distance of 1 mile. Then 4 divisions will represent 2 miles; 7 divisions will represent 3| miles.
1 mi.
2 mi.
sVs mi
42
IV. §26] GRAPHIC REPRESENTATION OF QUANTITY 43
On the scale of ^ inch to the mile, how wide would a map of the United States be? (The distance east and west is about 3000 miles.)
How wide would a map of your state be? How high?
Example 3. If a line 1 inch long represents 10,000 people, then a line 2 1 inches long represents 25,000 people, a line | inch long repre- sents 2500 people.
10,000
^S,000
2500
Example 4. To compare unequal quantities of the same kind, it is very convenient to represent them by parallel hnes drawn under each other. This is illustrated below.
Lengths of rivers.
Mississippi 4300 miles
Amazon 3300 miles
Nile 3400 miles
Volga 2400 miles
M
i^ssi:
£L
NMe
Vdlga
Example 5. We may also draw the lines representing the values of quantities perpendicularly. Thus in the following diagram the heavy vertical lines represent the relative heights of four mountain peaks.
44
ALGEBRA — FIRST COURSE
[IV. § 26
Heights of mountains.
Pike's Peak 14,000 feet
Mt. Blanc 16,000 feet
Chimborazo 20,000 feet
Mt. Everest 29,000 feet
Exercises. In the following select an appropriate unit and represent by drawings.
1. Draw lines whose lengths shall represent 3 lbs. ; 2§ lbs. ; 2f lbs.; 20 oz.; 50 oz.
2. On the scale of one unit of division on your section paper to the foot, what lengths are repre- sented by the following: 3 divi- sions; 6 divisions; 12 divisions; 36 divisions; 90 divisions. Draw lines representing 5 feet; 3| feet; 1§ yards; J rod.
Make a diagram of the floor of a room whose dimensions are 10 feet by 15 feet, on a scale of one division to five feet.
3. Selecting an appropriate length to represent one day, tell what durations of time the following lengths would represent? 7; 20; jV; ^is-
How long a line would be required to represent 10 days? one year? Draw lines to represent 4 hours; 75 minutes; 1 hour and 40 minutes.
4. Express by drawings the yearly interest on $100 at 1%; at 2%; at 4%; at 7%; at 10%.
On the same scale represent the yearly interest on $200 at 1%; on $400 at 1%; on $300 at 2%.
5. Selecting a length to represent 100,000 people, represent by lengths of lines the population of the following cities:
|
so 000 |
ft. |
||||
|
25 000 |
|||||
|
20 000 |
|||||
|
15 000 |
|||||
|
<w |
|||||
|
10 000 |
o |
^ |
|||
|
•n |
g |
S s |
^ |
||
|
5000 |
? |
D |
|||
|
g |
|||||
|
0 |
IV. §26] GRAPHIC REPRESENTATION OF QUANTITY 45
Pittsburg 375,000
Cleveland 480.000
City: Milwaukee Buffalo Baltimore
Population: 320,000 400,000 550,000
6. By measurement on a map, using the scale given on the map, find the distances between various cities. Make a table of the results and then show the distances by lines.
7. From the following diagram read off the population of the United States for the various dates.
■-BvO
mo-
1890-
• -mo-
50 PopulUtion in million
100
8. From the following diagram read off the amount of national debt per capita in different years.
|
ion ft |
|||||||||
|
0 |
6 |
) dolU |
irs |
L |
National debt per Capita in CT.S.
46
ALGEBRA — FIRST COURSE
IIV. § 26
^10.
r~i
I I 5
-f — -f 1^ 1-
World's Production of Gold
9. From the preceding diagram read off the world's pro- duction of gold in the various years.
Make diagrams to show the data given in the tables below.
10. Distance from New York to Chicago, 910 miles,
to Boston, 160 miles,
to Cleveland, 580 miles, to Washington, 230 miles.
11. Population of countries, 1910:
United States, 93 milUon people, England, 45 million people,
France, 39 million people,
IV. § 26] GRAPHIC REPRESENTATION OF QUANTITY 47
Germany, 63 million people,
Italy, 34 million people,
Japan, 51 million people.
12. Number of people to the square mile, 1910:
United States, 26 persons,
England, 374 persons,
France, 191 persons,
Germany, 311 persons,
Italy, 313 persons,
Japan, 344 persons.
13. Distances from the sun to the four inner planets:
Sun to Mercury, 36 million miles,
Sun to Venus, 67 million miles,
Sun to Earth, 93 million miles.
Sun to Mars, 141 million miles.
14. Make diagrams showing various data. The following are suggested.
Population of your state for every ten years.
Amount of public debt of your state for several years.
Amount of crops — com, wheat, etc.
Value of crops.
Amount of various manufactures.
Population of leading cities.
The comparison of quantity is also expressed by the com- parative sizes of angles. The amount of turning or circular motion is quite as effective in its power to convey compari- sons as is the straight line. It is not used so extensively, because a straight line is easier to make. The following pictures illustrate its use.
48
ALGEBRA — FIRST COURSE
[IV, § 26
Population of Cities
Speed Records
London, 7 . 2 million, New York, 5 . 2 million, Paris, 2.8 million,
Chicago, 2.2 million, Tokio, 2.2 million.
Berlin,
2.0 million.
Automobile, 142 miles per hour, Locomotive, 120 miles per hour, Aeroplane, 106 miles per hour, Pidgeon, 86 miles per hour,
Horse, 43 miles per hour,
100 yd. dash, 21 miles per hour.
CHAPTER V POSITIVE AND NEGATIVE NUMBERS
27. Exercises.
1. A boy strikes a ball at exactly opposite points with two mallets at the same time and with the same force. What is the result as to the movement of the ball?
2. If he strikes first with the right-hand mallet, and then, after the ball has come to rest, with the left=hand mallet with an equal force, what is the result?
3. If he strikes with both mallets at the same time, strik- ing with one mallet with a force which would send the ball 5 feet and with the other mallet with a force which would send the ball 20 feet, what is the result?
4. If in Exercise 3 the boy strikes first with one mallet, then, after the ball has come to rest, with the other mallet, what is the result?
5. Does it make any difference whether two forces act at the same time or one after the other?
6. Two men are driving a stake; one strikes with a force that sends the stake down two inches, the other with a force that sends the stake down three inches. What is the combined effect?
7. Three boys are pulling a load on a sled, one with a force of 25 pounds, another with a force of 58 pounds, and the other with a force of 97 pounds. With what force is the load being pulled?
8. Two small boys are pulling a small wagon along; one pulls with a force of 25 pounds, and the other pulls with a force of 34 pounds. A boy comes up behind and pulls
49
50 ALGEBRA — FIRST COURSE [V. § 28
with a force of 57 pounds in the opposite direction. What is the result?
From the exercises given, and with a httle further experi- ment and thought of your own, you will be thoroughly con- vinced that forces are constantly acting against each other. An object cannot stop itself ^when it is once in motion, neither can it change the rate at which it is moving, nor the direc- tion in which it is moving, any more than it can start itself to moving. This truth is known as Newton's first law of motion.
28. Opposite Qualities. The exercises given have dealt with forces. Almost all quantities with which we deal may bethought of, and commonly are thought of, as having two opposite qualities which tend to counteract or neutralize each other.
East is opposite to west; if we go east any distance, then west the same distance, we come back to the starting point.
In the same way we have :
North opposite to south; Up opposite to down; Right opposite to left; Forward opposite to backward; Profit opposite to loss; Assets opposite to liabilities; Future time opposite to past time; Temperature above zero opposite to temperature below zero.
Example 1. A man starts from a certain town A and goes 10 miles due east; from the point where he now is he goes 15 miles due west. Where is he then with respect to the original starting point A?
The answer is: 5 miles west from A. We must state not only the distance from A, but also the direction.
Example 2. A man has $10,000 assets and $15,000 liabilities. What is his financial status?
The answer is: He owes $5000. It is not enough to say merely $6000. We must also say whether it is an asset or a liability.
V.§29] POSITIVE AND NEGATIVE NUMBERS 51
Example 3. If at noon the temperature is 20 degrees above zero, and if the temperature rises 10 degrees during the afternoon, then falls 40 degrees during the night, what is the temperature the next morning?
Answer: 10 degrees below zero. It is not enough to say 10 degrees. We must also say whether it is above or below zero.
As we have seen, forces, whether acting at the same time or at different times, may act in opposite directions to one another. In mathematics we express the idea that one force acts in the opposite direction to another force by saying that one is the negative of the other, negative meaning opposite.
Definition. Whenever a quantity has two opposite quali- ties, we call one of them positive and the other negative.
Notation. The symbol for positive is +, and the symbol for negative is — . When no sign is written, the + sign is understood.
Illustrations.
If + 100 dollars means gain, then — 100 dollars means loss.
If + 10 feet means 10 feet up, then — 10 feet means 10 feet down.
If + 10 days means 10 days later, then — 10 days means 10 days earlier.
If + m degrees means an angle measured counterclockwise, then — m degrees means an angle measured clockwise.
If + n means count n units forward, then — n means count n units backward.
Exercise.
(a) If earning money is positive, what is negative?
(6) If going west is positive, what is negative?
(c) If upstream is positive, what is negative?
{d) If the pull of gravity is positive, what is the pull of a balloon?
29. Positive and Negative Numbers.
Definition. Numbers which express the measurement of positive quantity are called positive numbers. Numbers which express the measurement of negative quantity are called negative numbers.
52 ALGEBRA — FIRST COURSE [v. § 30
Notation. A positive number is indicated by the sign + written before the number; a negative number is indicated by the sign — written before it. When no sign is written, the positive sign is understood.
Thus if + 10 means 10 units of quantity of a certain kind, then — 10 means 10 units of quantity of the opposite kind.
Let the student give a number of illustrations.
Absolute Value of a Number. When we wish to indicate merely the value of a number, without regard to its sign, we use the symbol | |.
Thus I 5 I =+5, and also | - 5 | =+ 5.
30. Graphic Representation of Positive and Negative Numbers. We have already shown how we use lengths of lines to represent numbers, and how the comparative lengths of lines give a clear illustration of comparative values of numbers. We have dealt, however, with numbers of the same quality or sign.
We now show how to represent numbers of opposite quality, so that our diagram will show both value of the number and its quality.
Let us mark off a series of equal distances on an indefinite straight line. Mark one of the points of division zero. This is the point from which we begin to count. Mark the points to the right of this 1, 2, 3 and so on. Mark the points to the left, — 1, — 2, — 3, and so on. Thus
-6-5-/^-5-2-1 0 12 S k 5 6 ^-1 I I I I 1 1 1 J 1 I I 1_
On this scale we can show numbers of opposite quality. Example 1. Represent $3500 profit and $2500 loss.
> I ^ T 7 , ! T 7 . u
Loss Profit
Scale, 1 division to 1000 dollars.
V.§3i] POSITIVE AND NEGATIVE NUMBERS 53
Example 2. Represent 4.5 miles east and 3 miles west.
-U -3 -2 -] 0 1 2 3 I, 5
-^ '. ' ' I ' ■ ' ' , '
West East
Scale, 1 division to the mile.
Example 3. Represent 22 pounds pull and 45 pounds push.
^30 -^0 -10 0 10 2U '30 kO 50
— 1 1 1 1 -t 1 1 \ 1-
Pull ' Push
Scale, 1 division to 10 pounds.
Selecting a suitable unit, make diagrams of the following:
4200 births, 2500 deaths.
15 degrees rise, 35 degrees fall.
In all of these diagrams the amount of the quantity is expressed by the number and is represented by a line of the proper length; opposite qualities are expressed by lines dra^vn in opposite directions. In each case one of the lines represents a positive number and the other line a negative number. That is, positive and negative numbers are repre- sented by oppositely directed lines.
31. Summary.
Many quantities, such as force, distance, time, and tem- perature, admit of the notion of opposite qualities.
In mathematics opposite qualities are usually distinguished by the words positive and negative.
Measurements of quantities are expressed by numbers; measurement of positive quantities by positive numbers; measurement of negative quantities by negative numbers.
The symbol + indicates positive.
The symbol — indicates negative.
The absolute value of a number is expressed by | |.
Geometrically, the measurements of quantities of opposite qualities are expressed by oppositely directed Hnes.
54 ALGEBRA — FIRST COURSE [V,§31
Exercises. Represent by diagrams each of the following pairs of quantities:
1. 10 miles north; 15 miles south.
2. 15 pounds push; 25 pounds pull.
3. 60 degrees above zero; 20 degrees below zero.
4. $2000 income; $1500 expenditure. •
5. If a man earns $150 a month and spends $100 a month, show his total earnings and expenditures in one year.
6. If a tank is filled by a pipe which flows 100 gallons a minute and emptied by a pipe which flows 60 gallons a minute, represent the total inflow and outflow in 5 minutes when both pipes are open.
CHAPTER VI ADDITION AND SUBTRACTION
32. Notation. The signs of addition, subtraction, multi- plication and division are the same as in arithmetic.
To indicate that two numbers a and b are to be added we write a + 6; this means that h is to be added to a. Now the numbers a and h may be both positive; one positive, the other negative; or both negative. Thus:
(+ 3) + (+ 4) means add positive 4 to positive 3; (+ 3) + (— 4) means add negative 4 to positive 3; (— 3) + (+ 4) means add positive 4 to negative 3; (— 3) + (— 4) means add negative 4 to negative 3.
When a number is positive we usually do not write its sign; thus 3 means + 3, and 3 + 4 means (+ 3) + (+ 4).
We shall now illustrate geometrically what is meant by adding two numbers together, whatever their signs may be. We shall represent each number by a directed line; we then add these directed line segments.
33. Addition of Directed Line Segments. Let us call the point from which a line segment starts the initial point of the segment, -and the point where it ends the final point of the segment; the final point will be marked by an arrow- head to show the direction of the line.
Two directed line segments are added by placing the ini- tial point of the line we are adding on at the final point of the line we are adding to, keeping each line in its original direction. The line drawn from the initial point of the line we are adding to, to the final point of the line we are adding on, is the sum of the two lines. The direction of the sum
55
56 ALGEBRA — FIRST COURSE IVI. § 33
is always from the initial point of the hne added to, to the final point of the line added on.
This is illustrated in the following figures; here lines ex- tending toward the right are called positive, toward the left negative.
Illustration I.
To add the line RS to the line MN.
J2 + 51 K 4^ ^
^
Starting at the initial point, draw the line MN. Now add the line RS by placing the point R on the point N, taking care that each line is kept in its original direction.
E ± iL
The distance MS is the sum. It is a positive line. Therefore MN + RS = MS. Illustration II.
Case 1. To add the line RS to the line MN, when MN is longer than RS.
S^ ^ R M ± ^
As before, starting at the initial point, draw the hne MN; add the line RS by placing the point R on the point N, taking care to keep the lines in their original directions. (See note on p. 57.)
M ± ^
S - i2
Therefore MN + RS = MS. MS is a positive line.
Case 2. To add line RS to line MN, when MN is shorter than RS.
As before, start at the initial point, and draw line MN; add the^line
VI. § 331 ADDITION 57
RS by placing point R on point N, care being taken to keep the lines in their original directions.
Therefore MN + RS = MS. MS is a negative line.
IlliLstration III.
Case 1. To add line RS to line MN, when MN is longer than RS,
^ N - M
Adding according to the instructions given before, we have N - M
Therefore MN + RS = MS. MS in this case is negative.
Case 2. To add Hne RS to hne MN, when MN is shorter than RS,
B 4- £ N ~ Jg
Adding as before,
N^ — M
R
Therefore MN + RS = MS. MS ia & positive line.
Illustration IV.
To add line RS to line MN.
^^ — R i^ — M
Adding as before,
N^ -- M
S^ R
Therefore MN + RS = MS. MS is a negative line.
Note. The student must keep in mind that in Illustrations 11 and m, the lines MN and RS coincide — he one on the other — and that in the picture one is drawn a httle below the other to show the directions.
58 ALGEBRA — FIRST COURSE ivi.§34
34. Geometric Addition of Numbers. We are now ready to illustrate what we mean by adding two numbers together, whatever may be their sign. Let the numbers be a and b.
First make a number scale. Starting from zero on this scale, lay off a line segment to represent the number a in amount and direction; using the point that you have now reached as an initial point, lay off a line segment to represent the number b in amount and direction; then the number represented by the line from the origin or zero point to the point last reached is the sum a -\- b.
As shown above line segments can be added when both are positive, both negative, one positive and one negative; so the numbers which they represent may be added, when both are positive, both negative, one positive and the other negative.
If a and b are both positive, the addition corresponds to the addition of line segments as shown in Illustration I.
Suppose a is 3 and 6 is 4; we have
-10 1 2 3 U 5 6 7 8 1 1 \ I I I I I L_
3 + 4 = 7.
Suppose a is 7 and 6 is — 3; this corresponds to Illustra- tion II, Case 1, in the addition of line segments.
-10 123U5G78
^
7 +(-3) =4.
Suppose a is — 7 and 6 is 3; this corresponds to Illustra- tion III, Case 1, in the addition of line segments.
—8 -7 -6 -5, -U -5 -^ -1 0 I ^ I t ^ I I I I I L
-7 + 3 =-4.
VI, § 34J
ADDITION
59
In general, if the two numbers are a and 6, a being positive and h negative we have the two figures below. The first is for the case where the number of units in h is less than the number of units in a, and the second is for the case where the number of imits in 6 is greater than the number of units in a.
-3
a-^h
>!^
•^a
L
-I
^
c-j-6
■»ia
In these figures, as in the case of preceding figures, the fact that h stands for a negative number is shown by the arrow. The sum of a and 6, a + 6, is positive in the first figure and negative in the second.
We may now state the following rule:
If a is positive and h is negative then
the sum a -\- b is
( positive if\h\ is less than a;
i negative if\b\ is greater than a. If both numbers are negative, the two directed lines will both extend in the negative direction; hence the sum of two negative numbers is a negative number whose absolute value equals the sum of the absolute values of the two numbers.
"6 -5 -4* -s -2
I . I ! I ^_L_
W
_K-
a+6
This corresponds to Illustration IV in the addition of line segments.
Estimate the values of a and b in the above figures.
60 ALGEBRA — FIRST COURSE ivi,§35
Addition of more than two numbers.
Example. Add 4 + (- 6) + 8.
To do this we first add two of the numbers; then to their sum we add the third.
This is shown in the diagram below; 4 + (- 6) leads to - 2; add- ing + 8 to - 2 leads to + 6.
Algebraically expressed this is
4 + (-6) +8 = 6.
35. Exercises and Problems.
Add the following geometrically, representing the num- bers by lengths of lines. The best results will be obtained from the use of squared paper. Care should be taken in each example, especially when fractions occur, to select the unit which may be most conveniently used in the example. Also have care that the arrowheads are always placed to indicate the direction of the line. Write the algebraic solu- tion in each case.
|
1. |
6 + 3. 6. 140+ (-300). |
|
2. |
-12 + 2. 6. 42 +(-8) + (-3). |
|
3. |
2 +(-3) + 14. 7. -52 + 40 +(-3). |
|
4. |
i+(-i). 8. -15 + 4+ (-10). |
|
9. -3 + (-10) + (-2) +32. |
|
|
10. .7 + (- .3) + (- .05) + 1.4. |
|
|
11. 3.7 + 5.7 + (- 4) + 7. |
|
|
12. i+(-l) + (-i). |
|
|
13. -^ + (-1) + ! + (-§). |
In each of the following problems give graphic represen- tation, algebraic expression and result.
14. A boy is pulling a small wagon along with a force of 5 lbs. ; another boy comes up behind and pulls back with a force of 20 lbs. In what direction and with what force is the wagon pulled?
VI, § 35]
PROBLEMS 61
Solviion: Let the line — represent a pound force. Then the following wiU represent the forces and their sum.
-15 -10 -5 0 5 10
< < I 1 I I I I I I 1 I I 1 I I I l—J I I I I I I I — I — L.
The algebraic expression is
51b. + (-20 lb.) =-15 lb. The negative sign shows that the resulting pull is backward.
16. A man earned $5 on Monday and spent $3; on Tuesday he earned $2 and spent $6; on Wednesday he earned $7 and spent none; on Thursday he earned $10 and spent $4; on Friday he earned $7 and spent $7. How much had he at the end of each day? What does a negative an- swer mean?
16. When the mercury in a Fahrenheit thermometer reg- istered 73 degrees, the bulb of the thermometer was in a bottle of ether. It was taken out quickly. As the ether on the bulb evaporated, the mercury fell 2 degrees. The ther- mometer was then placed in hot water and the mercury rose 56 degrees. What temperature did the thermometer then register?
17. A thermometer (at 25 degrees) was placed in finely crushed ice, and the mercury fell 24.6 degrees. The ther- mometer was then placed in a mixture of salt and ice, and the mercury fell 19 degrees more. When it was finally placed in hot water, the mercury rose 108.2 degrees. What was the temperature of the hot water?
18. On consecutive days one winter the following was noticed: On the morning of the first day the thermometer registered zero; by the next morning the mercury had risen 23 degrees; by the following morning it had fallen 9 degrees; by the following morning it had fallen 25 degrees more; and by the next morning it had risen 7 degrees. What did the thermometer then register?
62 ALGEBRA — FIRST COURSE ivi. § 35
19. A boy wished to have the water in his beaker a cer- tain temperature. He tested the hydrant water and found that it registered 34 degrees. He poured in hot water and raised it 23 degrees; he next poured in cold water and lowered it 12 degrees; then hot water and raised it 3 degrees; then cold water, and lowered it 7 degrees; then cold water, and again lowered it 2 degrees, when he found that it was the temperature that he wished it to be. What was the final temperature?
20. Make a few experiments using the thermometer, not- ing changes of temperature. Write problems from these observations.
The rule for adding angles is similar to that for adding lines. Draw the initial arm of the first angle. It is custom- ary to have this extend from left to right. Using your pro- tractor count a angular units, and draw the final arm of your first angle; using this as the initial arm of the second angle count h angular units and draw the final arm of the second angle. The sum is the angle with Its initial arm the same as the initial arm of the first angle and its final arm the last line drawn.
In making the drawing lay off a positive angle so that the turning from the initial arm toward the final arm is opposite to the turning of the hands of a clock, that is, counterclockwise; lay off negative angles in the clockwise direction.
Illustration. Add an angle of — 44 degrees to an angle of 25 degrees.
Reading from the figure we have:
Z CAD-\- ZDAB = ZCAB.
The algebraic expression is:
25*' + (-44°) 19^
VI. § 36] SUBTRACTION 63
Add the following; give drawing and algebraic equation. Use a protractor, so as to have the work as accurate as possible.
21. 25° +(-34°).
22. - 13° + 56° + (- 12°).
23. 43° + (- 156°) + 73°.
24. 43° + (- 94°) + 141° + 40°.
26. J radian + f radian + (— i) radian.
26. — I radian + i radian -\- {— i) radian.
27. I IT radian + (— f tt) radian.
28. IJ X radian + (— 27r) radian + (— Jtt) radian.
36. Subtraction. In the preceding exercises about forces we assumed that we knew the number of pounds and the direction of each force, and we had to find the result of their combined action; that is, the number of pounds and the direction of the force that results from all the forces com- bined in pulling and pushing the body in one direction or the opposite. It may be just as necessary, in practical work, to be able to tell what force must be combined with another to secure a desired result. For example, what force must be combined with a downward force of 2 pounds to get a resultant upward force of 5 pounds? Plainly this is the inverse of the work that we have been doing. So we determine the answer to this problem by guessing the amount and then adding to test whether we have guessed correctly or not. After a while we become able to guess and test the result very rapidly.
Definition of Subtraction. The process of determining the amount which must he added to one number to produce another is called subtraction.
What we have said about forces will apply to all the other subjects that we treated in addition.
Exercises. Give answers to the following by guessing the answer and then adding to test the correctness of the guess.
64 ALGEBRA — FIRST COURSE [Vi,§36
1. A mass of iron filings lying on the scales weighs 23 grams. What mass of filings must be added in order to make the mass weigh 37 grams?
2. The thermometer registers 69 degrees. How many degrees must be added in order to make it register 40 degrees?
3. The pressure of the atmosphere at one time sustained in a barometer a column of mercury 73.1 centimeters high, and at another time 74.7 centimeters high. How many centimeters have been added by the change of pressure?
4. At what rate can a man row in still water if the rate of a stream is 3 miles an hour, and if he can row at the rate of 3 miles an hour upstream?
5. A boatman rowing upstream finds that at one point he is not moving. If his rate of rowing is 5 miles an hour, what is the rate of the stream at that point?
Subtraction of Lines.
lUtistration. Subtract the line RS from the line PQ.
^ p s
The question is — What length of line must be added to the line RS to get the line PQ?
Examining your graphic representation of the addition of two directed lines, you will observe that the line you added to and the line which is the sum of the two lines both have the same initial point. The line added on extends from the final point of the line added to, to the final point of the sum. From this we see that, when we are given a line and wish to determine the length and direction of a line that must be added to it in order to obtain another line, which is in reaUty the sum of the given line and the one to be found, we place the two given Unes with their initial points together, taking care that each keeps its original direction. Draw a line from the final point of the line to be added to, to the final point of the Une which is the sum. This is the line we seek.
That is place Une RS and line PQ with the point R on the point P taking care that each keeps its original direction. Draw a line from point S to point Q, The line SQ is the line which we are seeking;
VI. § 36] SUBTRACTION 65
that is, it is the line which must be added to the line RS to obtain the line PQ. (See note on p. 57.)
Line PQ — line RS = line SQ. SQ is a negative line. That is, SQ must be added to RS to give PQ.
Exercises. Subtract the first of the following lines from the second, observing carefully the above remarks.
B< A D S
M f-N i?-< K
Q^ p S-* R
Subtraction of Numbers.
Illustration. If you have a line 3 units long, what length of line must you add to it in order to have a Une — 2 units long?
This does not differ from the preceding work excepting that we use the scale of measurement in order to do the work more quickly. We follow the same rule as above.
-3 -2 ~l 0 1 2 3
^J 1 \ ! 1 I
Thus we see that we must add a line negative 5 units long to a line 3 units long in order to have a line negative 2 units long. The algebraic expression for this is
(-2) -(+3) =-5.
This is read, "Negative 2 minus 3 equals negative 5," or "3 sub- tracted from negative 2 gives negative 5," and means that which must be added to 3 to get negative 2.
The check to this is: 3 + ( - 5) = -2.
Exercises. In the following, give graphic representation, algebraic expression, result, and check.
1. What length of line must be added to a line negative 3 units long in order to have a line positive 8 units long?
2. Subtract a line positive 28 cm. long from a line positive 23 cm. long. We see readily that this would have little
66 ALGEBRA — FIRST COURSE [VI. §36
meaning if we did not hold to our definition of the word subtract. The question in reality is — What must be added to the line 28 cm. in length (taking into consideration both distance and direction) to have a line 23 cm. long?
3. What must be added to a line 11 units long to have a line 2 units long?
4. What must be added to a line negative 17 units in length to have a line negative 2i units in length?
5. What must be added to a line 54 mm. long to have a line 0 mm. long?
6. What must be added to a line — 7f cm. long to have a line — 2f cm. long?
7. What must be added to a line 4^ cm. long to have a line 2f cm. long?
8. Repeat the idea of the questions asked in Exercises 3 to 7 using the word subtract.
9. If a boy is pulling a block with a force of 6 lb., with how much force and in what direction must another boy pull, if the net result is to be a pull of 5 lb. in the opposite direction? How else may this question be asked?
A graphic representation may be made of this exercise in the same way that we represented the exercises in the addi- tion of forces. That is, we let a certain length of line repre- sent the unit of force, and by taking multiple lengths of this line or fractional parts of it, we can make a line repre- sentation of the forces, which lines may be subtracted in the same manner as is given above.
Let one unit of length represent one pound force. Then the graphic representation of the subtraction is as follows.
-5 Q 6
' ' I 1 I
-11 The algebraic expression for this is
(-5)- (+6) =-11 or, -5 -6 =-11,
VI. § 36] SUBTRACTION 67
Such a graphic representation may be made of any meas- urable quantities.
Make graphic representation of the following exercises; give algebraic expression and result as in preceding exer- cises.
10. If a man earns ten dollars, how much must he spend to be in debt four dollars? Since spending is the opposite of earning, this may be asked — How much must a m^n add to ten dollars in order to have negative four dollars?
11. If wheat is $1.05 a bushel on one day, and $.93 the next, what has been added to the price the first day to get that of the second?
12. If certain shares of stock sell at 5 cents above par one day, and 2 cents below the next, what has been added to the price of the first day to get that of the next?
13. A strip of iron when warm extended past a certain notch 3 cm.; when cooler, the end touched the notch. If we call the effects of expansion positive, how much was added to the length of the strip by contraction? If when cold the strip lacked 2 cm. of reaching to the notch, how much was added to the original length by contraction?
14. An iron nail lying in a pan on the scales weighs 10 grams. When a magnet is held over it, the scales register 2 grams. If we regard an upward pull as positive, what is the pull of the magnet upon the nail?
Write problems leading to some of the following algebraic expressions; give graphic representation and answers for all.
|
15. |
9-2. |
22. |
Subtract 8 from 7. |
|
|
16. |
- 22 - 8. |
23. |
Subtract 3J from 1|. |
|
|
17. |
- 22 - ( - |
9). |
24. |
Subtract 3.6 from .7. |
|
18. |
0 - (- 3). |
26. |
Subtract 20.8 from - .9. |
|
|
19. |
0-3. |
26. |
Subtract - .08 from .25. |
|
|
20. |
0 - (- 15). |
27. |
Subtract — 1.7 from — .7. |
|
|
21. |
-7-0. |
28. |
Subtract — f from — f . |
68 ALGEBRA — FIRST COURSE IVI. § 36
29. From an angle of 30 degrees subtract an angle of 45 degrees.
Illustration: The question is — What must be added to an angle of 45 degrees to get an angle of ^30 degrees?
The symbolic expressions are
ZBAC - Z BAD = ZDAC. 30° - 45° = - 15°. Check: ZBAD+ ZDAC= ABAC.
45° + (- 15°) =30°.
For the exercises below make geometric picture; give alge- braic expression and result.
30. — I TT radians — | tt radians.
31. i TT radians — x radians.
32. 27° - 125°.
33. - 45° - (- 60°).
34. — J TT radians — ( — f tt) radians.
35. What angle must be added to 10 degrees to get a right angle?
36. What angle must be added to 37 degrees to get a right angle?
37. What angle must be added to ix radians to get a right angle?
Two angles whose sum is a right angle are called comple- mentary angles. Either angle is said to be the complement of the other. In the last three exercises you have found the complements of the angles given.
38. What is the complement of 80 degrees? of 53 de- grees? of — i X radians? of — tt radians? of 175 degrees? of 0 degrees? of 90 degrees?
39. What angle must be added to 10 degrees to get a straight angle?
40. What angle must be added to fx radians to get a straight angle?
VI. § 361 SUBTRACTION 69
Two angles whose sum is a straight angle are called sup- plementary angles. Either angle is said to be the supple- ment of the other. In the last two exercises you have found the supplements of the angles given.
What are the supplements of the angles given in Exercise 38?
Exercises for Quick Oral DriU. The teacher may read the following or similar exercises, and the student should be able to give the answer as soon as the exercise is read. Read the subtraction exercises in three different ways.
1. 2 - (- 3). 4. - 2 - (- 6).
2. 5 - (- 4). 5. 8 - 9. 3.-5-4. 6. 3 + r + (,- 4).
7. 5 + (-2) + (-3) 4-2 + 7.
8. 3 - (- 7). 13. 3 H- 6 + (- 15) + 9.
9. - 3 - (- 7). 14. 0 - 12.
10. 5-8. 16. 0 - (- 12).
11. - 15 - (- 3). 16. 3 - 21.
12. 4 - 12. 17. 21 - (- 16).
18. - 13 + (-2) + 10 + 3.
19. -5 + 7+ (-8) +7 + 8.
20. - 2 + (- 3) +5. 22. - 2 - 0.
21. 0 - (- 4). 23. 0-2.
24. -4 + 5+ (-5) + (-3). 25. I - 1 - J 26. f - 2f .
27. -§+| + (-f) + |.
28. tV + (-t\) + (-^)+tV
29. - i^j + (- ^%) + B + (- ^y.
30. U - (- H). 34. 120 + (- 125).
31. 120 - 125. 35. - 120 - 125.
32. il + (- tV). 36. - 120 + 125.
33. \l - tV- 37. - 120 - (- 125).
38. -15 + 17+(-l) + (-3).
39. 22 + (- 27) + 5 + 3. 42. 5 - (- 6).
40. -5 + (-6). 43. 5 + 6.
41. -5-6. 44. 37 +(-41) + 5.
70 ALGEBRA — FIRST COURSE IVI.§37
37. Summary.
Addition of directed lines. To add a given directed line to another, place the initial point of the line you are adding, on the final point of the line you are adding to, having care that both lines keep their original direction. The distance and direction of the line extending from the initial point of the line added to, to the final point of the line added on, is the length and direction of the sum.
Comparison of measurable quantities may be expressed by means of lines as well as by numerals and letters.
To express the addition of quantities by the addition of lines, assume a unit length of line to express a unit of quan- tity. By taking multiples or fractional parts of this unit, draw lines to express the quantities to be added. Add these lines by the above rule.
Subtraction of directed lines. Subtraction is the inverse of addition. It consists in finding the quantity which when added to one stated quantity will give another stated quantity.
To subtract one directed line from another directed line, place the two lines with their initial points together having care that the lines keep their original direction. The length and the direction of the line extending from the final point of the line to be added to, to the final point of the sum, is the line required.
To express subtraction of quantities by subtraction of lines, express their values by means of lines as instructed in addition, then subtract these lines by the rule just given.
The terms minuend and subtrahend are used in the same sense as in arithmetic.
CHAPTER VII MULTIPLICATION AND DIVISION
38. Meaning of Multiplication. The terms multiplicand and multiplier are used in the same sense as in arithmetic.
The subject of multiplication adds nothing new to the work we have been doing. So far, the numbers that we have been dealing with have told us what to do with a unit of measure if we wished to express the measurement of quantity. For example, when we say that a line is negative 5 cm. in length, we mean to take the unit, turn it over, and use it five times to get the length of line called for. Now instead of a single unit we shall use multiples of or fractional parts of the unit, using them in either the same direction as the unit or in the reverse direction; these numbers are multiplicands, and the multiplier tells us what to do with them just as the number in the example above told us what to do with the unit to get it. In other words.
To multiply a second number by a first, we are to do to the second number the same thing that we did to the unit to get the first.
39. Graphic Illustrations of Multiplication.
Illustration I. A line three units long lying in the positive direc- tion must be used five times to measure a line; what is the length of the Une?
Unit Measuring Line
0 S 6 9 IS 15
•^ 1 1 1 1 1 1 1 1 1 1 1 1 1 1 >4
0 I'S 2-3 3-3 lc$ 5-3
Our hne is 15 units long. The algebraic expression of this operation is 5 . 3 = 15. 71
72 ALGEBRA — FIRST COURSE [Vil.§39
It i3 read 5 times 3 equals 15.
Did we use the 3, in this case, in the same way that we would have used the unit had we been asked to lay off a line 5 units long? Does our definition of multiplication hold in this case?
Illustration II. A line 3 units long is lying in the positive direction. Reverse it and use it 5 times to measure a hne. What length of line is measured? What is its direction? What sign is used to indicate "reverse" direction?
The geometric representation of this is
Unit Measuring Line
'15 -Jf -9 -6 -3
I I I I 1 \ 1 1 t ,. I
-Ji-2 -S-2 -2-3 "1-3
•IT"
The algebraic expression of this is
- 5 . 3 = - 15.
It is read negative 5 times 3 equals negative 15.
In this case did you do the same thing to 3 that you would have had to do to the unit to get negative 5?
Illustration III. A line 3 units long is lying in the negative direc- tion (opposite direction to the unit). Keeping it in this direction lay it off 5 times along the line to be measured. What is the length of line measured? What is its direction? What is the sign used to indi- cate " keep same direction " ?
The geometric representation of this is
p5ti<
Measuring Line \< 1 ' ^
-Jl -9 -« -»
5(23) U-S) S(.-3) Si-S) 3 (-3) 0
The algebraic expression of this is
5. (-3) =-15.
It is read 5 times negative 3 equals negative 15.
Illustration IV. A hne 3 units long is lying in the negative direc- tion. Reverse it, and lay it off 5 times along the line to be measured. What is the length of Une measured? What is its direction? What sign is used to denote reverse direction?
VII, § 39] MULTIPLICATION 73
The geometric representation of thia is
Unit Measuring Line H *< ' ' '
0 3 6 9 IS 15
I I I I I I I I I I I I I I I ^1
0 -l{-3) -2 (-3) -3 (-2) -Ui-S)
-5tr3)
The algebraic expression of this is
^5. (-3) = 15. It is read, negative 5 times negative 3 equals 15.
Exercises.
1. Repeat Illustrations I, II, III, IV, using a line f of a unit in length as the measuring line. Give geometric rep- resentation, and algebraic expression.
2. Give geometric representation and result of the fol- lowing:
f.4; f.(-4); -f.4; - f • (- 4).
3. If a piece of iron weighs 25 5-gram weights, and a piece of wood weighs 7 5-gram weights, and a piece of stone weighs 10 5-gram weights, how many 5-gram weights do the pieces together weigh? How many grams do they weigh?
4. If three boys, capable of pulling an average of 75 pounds apiece, are pulling a sled, and seven boys, with the same average pulling capacity, come up and pull in the oppo- site direction, in what direction will the sled be pulled, and with how many times 75 pounds force? With how many pounds force?
6. If you go 3 times 3J units forward, then 6 times 3| units backward, how many 3J units are you from the start- ing point?
6. If you go 3 times — 2 units forward, then 6 times — 2 units backward, how many times — 2 units are you from the starting point? How many units?
74 ALGEBRA — FIRST COURSE ivii.§40
7. If you go 3 times — 3f units forward and 6 times — 3J units backward, how many times — 3^ units are you from your starting point?
8. If you go 3 times n imits forward and 6 times n units backward, how many times n units are you from your starting point?
9. A load is being pulled in a certain direction with a force of 3 100-pounds. What force must be added in order that it may be pulled in the opposite direction with a force of 17 100-pounds?
10. Two forces are acting in opposite directions with a resultant force of 6.4 5-pounds. If one of the forces is known to be 151.2 5-pounds, what is the amount and direc- tion of the other force?
11. If an engine has gone 17 m-meters from the station, how far and in what direction must it go to be — 3 m-meters from the station.
12. If the temperature rises 4 d degrees, then falls 7 d degrees, what is the total change in temperature?
40. Multiplication of General Numbers. In the preced- ing exercises we brought out the following facts:
If the sign of the multiplier is positive, the sign of the product is the same as the sign of the multiplicand.
If the sign of the multiplier is negative, the sign of the prod- uct is opposite to the sign of the multiplicand.
In other words:
A positive number times a positive number gives a posi- tive nmnber.
A positive number times a negative number gives a nega- tive number.
A negative number tunes a positive number gives a negative number.
A negative number times a negative number gives a positive number.
Would these facts have been brought out had we used
vri, § 40]
MULTIPLICATION 75
other numbers in our exercises? Yes, you will say, any numbers might have been used. Let us then use the letter m to stand for the multiplier, and the letter n to stand for the multiplicand. Such letters may then be called general numbers, since they may be used to stand for any number we choose.
Our last four rules, in algebraic form, will then be
m * n = fnn.
m • (— n) = — mn. — m* n = — mn.
— m ' {— n) = mn. Exercises.
1. The distance from floor to floor in an office building ia on the average h feet. The elevator boy, starting at the main floor, runs the cage up two stories, then T down three, then up seven, when he is at the top [_ floor. How many feet is it from the main floor to the top floor?
The geometric solution is shown in the adjacent diagram. The algebraic expression is:
2A + (-3/i) + 7A =6A.
6h- 574-
Sh
Ih
Furthermore, we could bring out the mathemati- cal thought in a still more general way by having the example read thus:
The distance from floor to floor in an office build- ing is on the average h feet. The elevator boy, starting from the main floor, runs the cage up a stories, then down h stories, then up c stories, when he is at the top floor. How many feet is it from the main floor to the top floor?
The algebraic expression for the second state- ment, which we might call the general statement, is
ah +{~bh)-\-ch=(a-b + c) h.
. ' '
■ih^-^
76 ALGEBRA — FIRST COURSE ivii.§40
When we express numbers by means of letters, we have no single symbol as we have in arithmetic to express their sum, so, when necessary to show that a combination of numbers is to be regarded as one quantity w^e inclose the numbers in parentheses as shown above (§8). However, either expression in the second statement may be regarded as the answer to our example.
2. A man made it a point to have his average daily ex- penses the same as the rent from a certain piece of property which he owned. After t days during the month of October the renter of the property moved and the rent stopped. Compute the net amount the man had for the month, if the average rent per day was r dollars.
The work of writing problems leading to given arithmetic or algebraic expressions is of great value. It is the best way of testing a student's insight into the meaning of algebraic symbols.
When a problem is given and the solution is called for, the student arrives at the answer to the question asked by first translating the English statements into algebraic sym- bol language and then doing the work called for by these symbols.
The writing of a problem is the reverse of this process. The algebraic expression is given and the student is asked to translate its meaning into English.
The correctness of the problem should always be tested by a translation back into the algebraic language; that is, by the solution of the problem.
Example 1. Write a problem leading to the expression 6 . 40 + 3 . 40.
Solution : Upon examining this expression we see that there are two terms, each of which is a multiple of 40. So we must write a problem in which 40 is repeated.
Six boys, exerting a force of 40 pounds each, are pulling on a load. Not being able to move it, they call to their assistance 3 other boys
VII. § 411 DIVISION 77
each capable of exerting a force of 40 pounds. They succeed in start- ing the load in motion. What was the amount of force necessary to move the load?
Check : Translating into algebraic symbols:
6 • 40 = the number of pounds force the first 6 boys exert; 3 • 40 = the number of pounds force the 3 boys exert;
so that
6 • 40 + 3 • 40 = the number of pounds necessary to move the load.
6 • 40 + 3 • 40 = 9 • 40, the number of pounds necessary to move the load.
Give a statement of a problem like this, using general numbers.
Example 2. Write a problem leading to the expression 6. (- 15) -6-25, or, - 6 . 15 - 6 . 25.
Solution : A train runs due north for 6 hours at the rate of 25 miles an hour. At what rate must it now go for 6 hours so as to arrive at the same place as if it had gone south from the original starting point for 6 hours at 15 miles an hour?
Here we regard north as positive. Check this statement. Give a similar problem involving general numbers.
The best way to learn to write these problems is to study carefully the expressions obtained from problems in the preceding lists. Such lists are given not only for the solu- tion of the problems but to show how such problems arise.
Write problems for some of the following. Give graphic representation and result for all.
3. 15.4 + 3.4. 9. -6a + 7a+(-3)a.
4. 9. (-6) -3. (-6). 10. -2(-c)+(-5)(-c). 6. 7 . (- 2/) - 10 . (- y). 11. m{-n) +r(- n).
6. 3.f + (-7).f. 12. -9(-p)-5(-p).
7. 7wJ+ (-3)w?+(-4)iy. 13. a(-6) - (-a)(-6).
8. -/+(-4)/. 14. 2/b-8/b.
41. Division.
Division 0} a first number by a second is the process of deter-
78 ALGEBRA — FIRST COURSE IVII.§41
mining the number by which you must multiply the second number to get the first.
It is the inverse operation of multiplication, and is a purely guessing process. You guess the answer and mul- tiply to see if your guess is correct. In this respect it does not differ from the work in arithmetic, for long division is nothing more than a series of guesses and multiplication to test the correctness of the guess.
No new rules need be given for division, as you have but to observe the rules for multiplication when it comes to testing.
Example. You are to divide — ahhy b; that is, you want to know by what you must multiply h to get — ab. You have learned that this is — a. So — ab divided by b equals — a.
The symbols for division are the same as in arithmetic. So sym- boUcally we write
— ab -7- b] or — r —
Of these two forms the latter is almost always used. We then have — ab
Notice that ordinary cancellation, as in arithmetic, can be used here to reduce the given fraction. The only difference is that in arith- metic you used only positive numbers.
Problems. In these problems give algebraic expression and result. Multiply to check your answer.
1. A chauffeur finds that he has gone 7 1 miles in t hours. What is his speed per hour? What is his speed per hour if he goes m miles in t hours?
2. A -boy agrees to work for $1.50 per day; how many days must he work to earn $6? How many days must he work to earn n dollars, if he earns d dollars a day?
3. A cup made of tin in the form of a cone holds v grams of water, and a cup in the form of a cylinder whose height and circumference around the bottom are the same as that
VII, §411 DIVISION 79
of the cone, holds 3 v grams. What is the ratio of the volume of the cylinder to that of the cone?
By ratio of one number to another we mean the number by which we must multiply the second to get the first.
4. One cup holds mw grams and another holds w grams. What is the ratio of the capacity of the two cups?
5. If a boatman rows at the rate of v miles an hour in a stream that flows at the rate of s miles an hour, what is the ratio of his rate of rowing to the rate of the stream?
6. If an automobile goes r miles an hour for t hours, and another automobile goes r miles an hour in the opposite direction for s hours, what is the ratio of the distance the first goes to the distance the second goes?
7. If a train is going east at the rate of r miles an hour, what is the ratio of its velocity to that of a train going west at the rate of r miles an hour?
8. What is the ratio of the length of a line — 2 cm. long to the length of a line 7 cm. long?
9. By what must you multiply a force of 51 lbs. in order to have a force of f lb.? Answer this, using the word ratio. Check.
10. If 12 men can do a piece of work in a day, how much can 3 men do in a day if the average amount of work done by each is the same?
11. If a men can do a piece of work in m days, what part of the work will 1 man do in 1 day, if the average amount of work done by each is the same?
12. By what must a line — 5 units long be multiplied in order to have a line — 12 units long? Answer this, using the word ratio.
13. Divide a line — 1 cm. long by a line r cm. long. Read this, using the word ratio.
14. The same number of boys were trying to keep a door shut as the number that were trying to pull it open. The amount of force exerted was ma — mb pounds. What
80 ALGEBRA — FIRST COURSE [Vii,§42
is the number of boys on each side, if the average number of pounds exerted by each boy on one side is a, and the average number of pounds exerted by each on the other side is 6?
SimpHfy each of the following expressions as much as possible. State problems leading to some of them.
15. ^J. 20. ^^. - 26. ^- °>^'.
8t Uab^ -am
"• 5a ^^- -b , ^^- xH-yy
1,. 32»_^ 22. _6_. ' 27. ^^^'(-'•^
8 71 — ab 2^ pV
18 4« „, (-h){~k) ia{-b)'
„ 2«6 (-x)(-j/)(-z) „. (jymH-ny
5 a —xy (f)2m% — n)
42. Summary.
Definition of multiplication. To multiply a second num- ber by a first, do to the second number what you must do to the unit to get the first.
Law of signs in multiplication. If the sign of the multi- plier is positive, the sign of the product is the same as that of the multiplicand.
If the sign of the multiplier is negative, the sign of the product is the opposite to that of the multiplicand.
Literal numbers are used in mathematics when we wish to bring out general truths, that is, truths which do not depend upon the exact amounts, but upon the relation of the quantities under consideration.
Division. This is the inverse of multiplication.
CHAPTER VIII ADDITION AND SUBTRACTION OF POLYNOMIALS
43. Definitions. An expression consisting of several parts connected by the signs + or — is called a polynomial.
For example,
a + 6, 2h — 3k + 1, m-{-4:n — pq + Srs
are polynomials.
Each part of a polynomial, including its sign, is called a term of the polynomial.
Thus 4 n is a term of the last polynomial written above. Another term is — pq.
An expression consisting of a single term is called a monomial.
A polynomial consisting of two terms is called a binomial.
A polynomial consisting of three terms is called a tri- nomial.
Two terms which differ only by a numerical factor, as 3 mn and — 5 mn, are called similar terms.
44. Addition and Subtraction of Polynomials.
Example 1. You learned in your arithmetic and reviewed in the chapter on measurement the fact that a quantity may be measured by a unit and subdivisions of that unit.
For example a distance is measured in yards and feet. Say that it measures 9 yards and 2 feet. This may be written 9 yards + 2 feet.
Another distance measures 3 yards and 7 feet. This may be written
3 yards + 7 feet.
Then the sum of these two distances is 12 yards and 9 feet, which may be written
12 yards + 9 feet. 81
82 ALGEBRA — FIRST COURSE [viii,§44
That is, we add the parts of the distances measured by one kind of units together, then add the parts measured by another kind of units together, and combine the results to get the measurement of the sum of the lines.
If we regard the inch as the unit of measure in this exer- cise, and the yard and the foot as multiples of this unit, our exercise becomes:
Add a line 3 • 36 inches + 7 • 12 inches long to a line 9 • 36 inches + 2 • 12 inches long. Adding as before the sum is 12 . 36 inches + 9-12 inches. So that
(9 . 36 + 2 . 12) + (3 • 36 + 7 • 12) = 12 . 36 + 9 . 12.
Expressing this more generally by letting y stand for the number of inches in a yard and / stand for the number of inches in a foot, we have
(92/ + 2/) + (3 2/ + 7/)=12 2/ + 9/.
Example 2. Suppose that a Une measures 9 yards and 1 foot and 4 inches. Let y stand for a length of one yard, / for a length of one foot, and i for a length of one inch, each measured in terms of another unit, as a centimeter. Then the length of the line in centimeters is
expressed by
92/+/ + 4i.
This is a trinomial.
Suppose that a second Une measures 3 yards and 2 feet and 6 inches. Its length is then expressed by
Now the sum of the lengths of these lines will be expressed by
(9y+/ + 4i) + (32/ + 2/ + 6i),
and we know that we can get this sum by adding yards to yards, feet to feet, and inches to inches. So we have
(92/ +/ + 4i) + (3y + 2/ +:6z) = 12 2/ + 3/ + 10 I.
In this way any two polynomials are added together, by adding like terms.
The difference of the lengths of our two lines is expressed by
(92/+/ + 4i)-^3y + 2/ + 6i).
VIII. § 441
ADDITION OF POLYNOMIALS
83
and again we know that this difference can be found by subtracting terms measured in like units. So we have
Let the student check this by adding.
In the same way we might have such expressions as the following; state each of these as a problem in measurement, and check each result.
(9y + 4i) + (62/ + 2/-8i) = 151/ + 2/- 4 i.
{9y + 4i) -(Qy + 2f + Qi) = Sy-2f-2i.
iSy-3f + 5i) -ilQy + 2f-Si) =-2y-5f + Si.
Make up for yourself several more exercises of this kind. Geometric Illustration.
Add: (2m-3/) + (-3m + 4/-2) + (-m + 6/). Solviion:
I
|
ill! |
IS 1 |
5 |
1 1 1 |
0 1 ! 1 |
1 |
1 1 1 1 |
10 1 1 1 1 |
|
|
-3 m 1 ) 1 r |
-5 m [ 1 ^ |
~? |
1 )a |
1 "ijii |
3 m |
|||
|
^^^ |
||||||||
|
I 1 1 |
\ 1 ; 1 |
j ' |
||||||
|
1 1 1 1 |
||||||||
|
-A/ |
\r'^ |
0 |
3f |
hf |
i" |
i 5/ ! |
||
|
^ |
||||||||
|
^ |
||||||||
|
K— |
Algebraic expression:
(2w-3/) + (-3m + 4/-2) + (-m + 6/) =-2m + 7/ - 2.
In the diagram the measuring line for / might have been laid off on the original scale line, but it is much less confusing to draw a new scale to count by. The student is again cautioned not to forget that all of the segments representing the numbers are supposed to lie on the original scale line. In order not to forget that the thing we are in- terested in is the distance we are from the point where we began to count, it is well to mark that distance on the scale line, every time we change the measuring line, as shown in the figure.
Check : Let m = 3, / = 1^ or f , as shown in the figure.
84 ALGEBRA — FIRST COURSE [VIII. §44
Substituting in the given expression,
(2.3-3-f) + (-3.3 + 4.|-2) + (-3 + 6.f) = (6 - 4i) + (- 9 + 6 - 2) + (- 3 + 9) = H + C-5)+6
Substituting in the answer,
-2.3 + 7-li-2=-6 + 10^-2
Since we get 2^ from each expression, the expressions are equal. This is also the result in the figure.
In the following exercises make diagrams for the addition of some of them; add the others and check all.
It is of great importance that fractional and negative numbers be used in at least part of the checks, since it im- presses the fact that the letters given may have either integral or fractional values and that these may be either positive or negative; — a may be either a positive or a negative number according to the value of a.
Exercises.
1. Add 2r-sto5r-2s.
2. Add Qx + 5y-7to2x-7y + 5.
3. Add 3a- 166 + 2 to 4a + 96 + 4c. Add:
4. (2r-3p+-6) + (4r + 8p-2) + (6r + 9).
6. (- 6a + 5c -46 -6) + (-8a + 66) + (16a + 9).
6. (3r-6s + 10 + 20 + (6r + 7^-24) + (-5s-12r).
7. (7s + 50 + (-4a + 60+(7^-4a + 95)+(-19i-4s).
In the subtraction of polynomials the same idea enters as in addition. We subtract the terms separately and add the results of the subtractions.
Example. Subtract 2 r — 3 s from r — 4 s + 2 <. Writing in algebraic form:
(r-4s + 20 - (2r-3s) =?
VIII. §441 SUBTRACTION OF POLYNOMIALS 85
The first question is — What must be added to 2 r to get r ? The answer is — r. The second question is — What must be added to — 3 s to get — 4s? The answer is — s. The third question is — What must be added to 0 to get 2 t ? The answer is 2 t. So we have (r-4s + 2 0-(2r-3s)=-r-s + 2«i
Check : The check is to add the result to the second number to see if you get the first.
(2r-3s) + (-r -s + 20 =r-4r + 2«.
It would be well to make a drawing of a few of the checks. The student must be careful to do his checking conscientiously; that is, he must actually do the adding or he will be led to make many blunders.
8. Subtract 7 r + 2 s from 2 r + 6 s.
9. Subtract -Zs-\-2t from 7 s - bt.
10. Subtract —2a — Zh from a + 6. In the following do the work indicated.
11. (2 c + 7 d - 6) - (5 c - 2 e + 7).
12. (-5^-36- 10) -(-2^-46- 19).
13. (76-4a-6c + 4) - (-96 -4a + 6c- 4).
14. (2 a - 3 c) + (5 a + 16 c) - (3 a - 8 c).
15. (6-6(^-4a + 9) + (7r + 86- 17d)
- (4r-7(^ + 3a- 10).
16. (-7/+8sf-10/i-p) + (5/-8/i + /i-10p)
-(-5/-9sr-6).
17. (7r-6^-9s- 15) + (-4r-12i+12s)
- (9r + 2s- 17).
18. Write the supplements of the following angles:
a degrees; 2 a — 6 degrees; a + 3 6 degrees; 2 a — 4 6 +3 degrees; 2 r radians; J tt — r radians; If x + 2 r radians.
19. Write the complements of the angles of Exercise 18.
20. Write the complement of angle a, then write the supplement of this complement. Draw this. Do you get the same answer in your drawing as in your algebra?
21. Write the complement of the angle a, then write the supplement of this complement, then write the complement of this supplement. Draw and see if you get the same result.
86 ALGEBRA — FIRST COURSE [Vlll.§46
45. Rules for Adding and Subtracting Polynomials.
(1) To add two polynomials, add like terms, each term to be taken with the sign before it.
(2) To subtract two polynomials, subtract like terms, each term to be taken with the sign before it.
Also, since subtracting a quantity is equivalent to adding its negative, we may use the following rule for subtracting polynomials.
(3) To subtract one polynomial from another, change the signs of all the terms of the polynomial to be subtracted and add the result to the other polynomial.
Try Rule 3 in some of the preceding exercises.
46. Removal of Parentheses. From what you have now learned about polynomials you will easily see that such equa- tions as the following are true.
-\-{a — b — c)=a — b — c. — {a — b — c) —— a-\rb -{- c.
Rule. When a polynomial is inclosed in parentheses pre- ceded by a positive sign, the parentheses may be omitted.
When a polynomial is inclosed in parentheses preceded by a negative sign, we may omit the parentheses only if we at the same time change the sign of each term of the polynomial.
We shall apply this to an example involving several signs of aggregation (§8).
Example. Remove all signs of aggregation from the expression - {a -[(6 + 3d) -(4e-/)] + (26-7d + 3e)j.
Here we may first remove the parentheses inside the brackets [ ]; we then have
- 5a-[6 + 3d-4e+/] + (26-7d + 3e)i.
Now remove brackets after changing the sign of each term inclosed; also omit the parentheses; we then have
-Sa-6-3d + 4e-/ + 2&-7d + 3ei.
Combining similar terms, we have
- |a + &-10d + 7e-/}.
VIII. 5 47] SUMMARY . 87
Now remove braces after changing the signs of all terms inclosed; we finally have
-a-h + 10d-7e+f.
In this work it should be noticed that brackets and braces are merely different forms of parentheses.
Exercises. Simplify the following, removing all signs of aggregation.
1. [a+{c-d)- (3a + 2d)].
2. [- (2 m - n) + (5 m + 6 n) - (8 m - 3 n) + 5 m].
3. -[2x+(4:y-7x)- (Sy + Sx)- (5x-2y)].
4. -[-iu + 2v-Sw)-\-{3u-2v + w)-u + Sv-^w], 6. -\[-(7x-2y)-{-{Sx-4.y)]-[2x-5y-(3x-y)]\.
N.B. Whenever the expression inclosed in parentheses can be simplified by a reduction or combination of its terms, this should be done first.
47. Summary.
The amount of a quantity may be measured by comparing it with different units. The units may be of the standard kind as used in arithmetic, or they may be general units.
When only one length is used this measurement is ex- pressed by means of one term and the expression is called a monomial.
When two different lengths are used in the same measure- ment, the result is expressed by two terms and the expression is called a binomial.
When three different lengths are used in the same meas- urement, the result is expressed by means of three terms and the expression is called a trinomial.
In general, when two or more lengths are used in the meas- urement of a quantity, there are two or more terms in the expression of the measurement and the expression is called a polynomial.
To add polynomials. Write the polynomials in parentheses with the plus sign between them. Add the like terms and
88 ALGEBRA — FIRST COURSE [Vlll,§47
express the answer with different terms, one following the other, connected by the sign obtained by the addition.
To subtract polynomials. Write the polynomials in paren- theses, the one to be subtracted following the one from which it is to be subtracted, with the minus sign between them. Subtract the like terms in the manner that you learned to subtract monomials, and write the result with the different terms, one following the other, connecting them by the sign obtained from the subtraction.
Removal of parentheses. If a parenthesis is preceded by a plus sign, the parenthesis may be removed without changing the signs of the terms in the parenthesis. If a parenthesis is preceded by a minus sign the parenthesis may be removed, provided the sign of each term in the parenthesis is changed.
Problems in Addition and Subtraction.
1. On a winter day the temperature at noon was twice the temperature at 6 a.m. and the temperature at 6 p.m. was two-thirds of the temperature at noon. The sum of the three was 65 degrees. Find the temperature at each time.
2. The income from a certain business during the second year was double the income during the first year; during the third year it again doubled. The total income was $4500. What was. the income each year ?
3. The volume of a cylindrical cup in cubic inches is 3 times the volume of a conical cup of the same height and base. Using these as measures it is found that 2 measures of the first and 5 of the second just fill a gallon can. What is the volume of each cup ?
4. If 2 a dollars and 5 h dimes are taken from 4 a dollars and 3 b dimes the remainder is $4.60. Also 3 b dollars equal 15 6 dimes. Find the numbers for which a and b stand.
5. Show by drawing that
(2 + 3) - (8 - 5 - 6) = (2 + 3) + (- 8 + 5 + 6).
VIII. 1 47] SUMMARY 89
6. Three children are digging dandelions. One digs at the rate of a dozen per hour; the second at the rate of 6 dozen per hour; the third at the rate of c dozen per hour. The first morning the first works 4 hours, the second 3 hours, and the third 5 hours. The second morning the first works 5 hours, the second 3 hours, and the third 5 hours. How many more dandelions were dug the second day than the first?
7. There are four sizes of shot. The largest size weighs s grams, the second weighs r grams, the third p grams. On one pan of the scales are 7 of the first kind, 15 of the second, 18 of the third, and 25 of the fourth. On the other pan are 13 of the first, 6 of the second, 30 of the third, and 4 of the fourth. How many grams weight are on the first pan ? How many are on the second ? How much must be added to the weight on the second pan to have the scales balance?
CHAPTER IX MULTIPLICATION AND DIVISION OF POLYNOMIALS 48. Multiplication of a Polynomial by a Monomial.
Example 1. The length of a rectangle is 5 inches, its width is 3 inches; the length of a second rectangle is 5 inches and its width is 4 inches. What is the sum of their areas?
^111 II 5
Area of the first rectangle is 5 • 3 square inches. Area of the second rectangle is 5 • 4 square inches. Then the sum of these areas is (5 • 3 + 5 • 4) =35 square inches. This is equivalent to the area of a rectangle whose dimensions are 6 by 7, as shown in the figures above. That is, we have
5. 3 + 5. 4 = 5 (3 + 4) = 5-7.
In exactly the same way, if the dimensions of the first rectangle are a by 6 and the dimensions of the second rectangle are a by c, the sum of their areas is
a'h -\-a' c,
but this must be equivalent to a single rectangle whose dimensions are a by (,6 + c\ That is, we always have
a6 + oc = a (6 + c).
Example 2. If '3 boys pull a sled forward with a force of 30 pounds and 3 other boys pull the sled forward with a force of 35 pounds each, what is the total pull on the sled?
90
IX. §48] MULTIPLICATION OF POLYNOMIALS 91
Solution: We can get the result in two different ways. First: Total pull of the first 3 boys is 3 • 30 pounds.
Total pull of the other 3 boys is 3 • 35 pounds. Therefore the whole is the sum of these,
(3 . 30 + 3 • 35) pounds = (90 + 105) pounds = 195 pounds.
Second: Taking one boy from each set of three, their combined pull is (30 + 35) pounds. Since there are three such pairs of boys, their total pull is
3 (30 + 35) pounds = 3-65 pounds = 195 pounds. Hence we have
3 (30 + 35) = 3 . 30 + 3 . 35. Why? This is another illustration of the rule
a (6 + c) = ah -\- ac.
State this rule In words.
It is easy to illustrate this rule when some of the numbers are nega- tive.
Example 3. If 3 boys pull a sled forward with a pull of 30 pounds each and if 3 boys pull the sled backward with a force of 35 pounds each, what is the effective pull on the sled?
Solution:
Use + for pull forward and — for pull backward.
First: 3 • 30 + 3 • (- 35) = 90 + (- 105)
= - 15.
That is, the net effect is a backward pull of 15 pounds. Second: Take the boys in pairs, one pulling forward and the other pulling backward. The net effect of the pulling of each pair is 30 + (-35) =-5. Since there are three such pairs, the total effect is
3 [30 + (-35)] = 3- (-5) = - 15.
Since the expressions in the first and second solutions stand for the same thing, namely total pull, we have
3 [30 + (- 35)] = 3 . 30 + 3 (- 35).
Using letters, suppose n boys pull the sled forward with a force of
92 ALGEBRA — FIRST COURSE UX. § 48
p pounds each, and n other boys pull it backward with a force of q pounds each, then in the first solution we would have
n-v + n(- g); In the second solution we would have
n[v\-{- q)l
Here again we must have
n[p + i-q)] =n'p+n{-q),
which is the same form, except for a difference in the letters, as
a (6 + c) = a6 + ac.
If, in the last example, both sets of boys pull backward, show that the total pull may be expressed in either of the forms
n K-p) + (-g)] or n{-p) +n(-g),
so that we have
n[{-p) -f- i-q)] =n{-p) +n(-g).
Another example. of this sort is the following.
Example. A man who can row at the rate of Vr miles an hour in still water enters a stream which has a velocity of Vs miles an hour. How far can he row against the stream in one hour? How far can he row in three hours?
Solution: In this example if we regard the direction in which the man is going as positive, then the direction in which the stream is flowing is negative.
Therefore the distance gone in an hour is
Vr + i-Va) =Vr- V,.
The distance gone in 3 hours is
3 {Vr — Va).
But the man would be just as far at the end of 3 hours if it were possible for him to stop the stream and he rowed for 3 hours in still water, then he stopped rowing and allowed the stream to flow at its accustomed rate and pull him along for 3 hours. In the first case the two velocities are acting simultaneously, while in the latter case they would be acting consecutively.
The algebraic expression of the first case is as given above. For the latter case it would be
3t;r-3t;,.
IX, §49] MULTIPLICATION OF POLYNOMIALS 93
Therefore,
3 {Vr — Va) = SVr — S Vg.
This is illustrated graphically below.
Vr
I — . — '■ ' 1 1 — .->^
; S(Vt-Vs) ^
-l-*l
J I t_
-V5
1 \ L
3V,
SV^SV,
>| -SVa
49. Distributive Law. The equation a{b -\-c) = ah -}- ac and the similar equations just considered express what is known as the distributive law. This law asserts that the product of a single number by the sum of two numbers is identical with the sum of the products of the first number by the other two numbers taken singly.
As to which of the two ways this shall be written will depend entirely upon the use we are going to make of it in our work.
Exercises.
1. A girl earned I cents an hour helping in the laboratory, and c cents an hour for correcting papers. She spent 4 hours a day in the laboratory, and 3 hours a day marking papers. How much did she earn during a month of 20 days? First write the algebraic expression for the entire amount earned in one day, then write it for the amount earned in 20 days. Also write the amount earned for lab- oratory work during the month, then the amount earned for correcting papers during the month, and add the two amounts together, as the total amount earned. These two expressions for the amount earned during the month are equal to each other. Write this equality.
2. A teacher prepared for her class b bottles of a solution, each of which contained w grams of water and 15 grams of
94 ALGEBRA — FIRST COURSE rix,§50
salt. How many grams of the solution were there? Write the expression for this in two different ways placing the equality sign between them.
3. If a student ^ends h dollars a month for board and r dollars a month for room rent for 6 months, how much does he spend? Write the equality of the two different algebraic expressions for the answer to this.
4. From each of a bottles of water, each containing g grams, an average of 3 centigrams evaporated. How many centigrams were left in the bottles?
6. If a boy earns w dollars a day and pays c cents a day for his board, how much has he by the end of the month?
6. A measuring stick is 3 c feet and 2 b inches long. When applied to a line it goes 3| times. How long is the line?
7. A measuring line lacks I feet of being 20 feet long. When applied to a line it goes / times, how long is the line?
8. A cubic centimeter of gold weighs w grams and a cubic centimeter of lead weighs I grams less. What is the weight of I cubic centimeters of lead?
Write problems leading to the following expressions; change the letters to any suitable to the problem, and write the two algebraic expressions of the answers equal to each other.
9. 3 (a + 4). 12. 5(8a + 7&).
10. 5 (- 2 a + 3). 13. a (a- 1).
U. 7 (6 a- 1). 14. a(a + b).
15. b(a + 2h - c).
50. The Commutative Law of Addition. This law as- serts that the value of the sum of two numbers does not depend upon the order in which the numbers are taken when they are added.
Algebraically expressed, a + b =b-\- a, no matter what the sign of a and 6.
IX, §51] MULTIPLICATION OF POLYNOMIALS 95
Geometrically expressed,
J 1 I I 1 I I ' t
Thus we see that if the two numbers are positive the sums are the same.
In like manner : a -{■ (— b) =— h -\- a.
0
t I I I I \ I I I I
|
-b |
. |
a. |
^1 |
|||
|
^i |
||||||
|
^-" |
-6 |
|||||
|
1 |
- a+(-6) [ |
|||||
|
1 |
||||||
|
1 |
a |
In like manner: — a -\-h = h -\- (— a).
0
^ — r-
' s ^^
■ a+b
k
In like manner : ^ a -{- (— h) = — h -\- (— a).
-a+(>&) U 1 —
-b-i-i-a) K"
-6
51. The Commutative Law of Multiplication. This law asserts that the value of the product of two numbers does
96 ALGEBRA — FIRST COURSE [ix.§52
not depend upon the order in which the numbers are taken when they are multiphed.
Algebraically expressed: db = ba, no matter what the sign of a and b.
Sylvester's illustration of the commutative law for multi- pHcation. Take a baskets containing b apples each. Now since there are b apples in one basket, in a baskets there are a times b apples or ab apples. Take an apple from each of the baskets and place in a new basket; that basket will have a apples in it. Again take an apple from each of the original baskets, and you will have another new basket with a apples in it. If you keep on this way, you will at last transfer all the apples in the original baskets to b new baskets, and there will be a apples in each basket. Since there are a apples in each basket, in b baskets there will be b times a apples or ba apples.
Therefore, ab = ba.
52. Exponents. The area of a rectangle is the product of its two dimensions. If these dimensions are a and b, the area is ab. If the rectangle is a square, whose dimen- sion is a, then its area is a • a. The volume of a cube whose edge is a, is a • a • a.
To shorten the writing of such products, we use the fol- lowing symbols.
In place of a • a write a^.
In place of a • a • a write a^.
In place of a • a • a • a write a^.
In place of a • a • a • a • a write a^, and so on.
These are read a square, a cube, a fourth power, a fifth power, and so on.
The little number written above and to the right of an- other number shows how many times the other number is to be used as a factor; the little number is called the ex- ponent, and the other number is called the base.
IX, §52] MULTIPLICATION OF POLYNOMIALS 97
We also write
a% = a* a ' b; a%^ = a* a- a 'h 'b; and so on.
Exercises. As drill to become familiar with this notation, write the following; make use of the exponent. Read what you have written.
1. 3 aaabb. 5. 4| hhhhk • 4| hhkkst.
2. 7 • 7 mrrmn. 6. 4 • 4 • 4 • a • 4 bbba.
3. 8 r • 8 sstttr. 7. 2 r • 2 r • 5 r.
4. i'iccdddfd. 8. - 7 a - (- 7 a) ab.
Write the following without making use of the exponent.
9. 9 • 3^ m • 3 r^m^n. 11. 32 r^s%^ • (- S^rh%^).
10. ^cd'UcdJ. 12. (-3)2(-3a)3.
13. Write the answers to Exercises 9, 10, 11, 12, making use of the exponent.
14. Write as products of prime factors, 20, 72, 36, 75, 98, 312.
15. Calculate 23 . 3^ . 52.
Meaning of word coefficient. In expressions of the kind with which we have just been dealing, we call the figures the numerical factors and the letters the literal factors. The numerical factors are usually written first, that is, pre- ceding the literal factors.
When we wish to refer to one particular factor, we call the other factors coefficients of this one. Thus, in the ex- pression 4 abCf if we are interested in c, then 4 ab is its co- efficient; 4 be is the coefficient of a; 4:ae is the coefficient of 6; 4 6 is the coefficient of ae; 4 is the coefficient of abc; abe is the coefficient of 4. Coefficient means co-factor. How- ever, the numerical factor is referred to unless otherwise stated. If you were called upon to give the coefficient in the term 3 abcj you would say that it is 3.
98 ALGEBRA — FIRST COURSE [IX, 5 53
63. Multiplication of One Polynomial by Another.
Example 1. A teacher in a laboratory prepared for a class w bottles, each containing w grams of water and 15 grams of salt. The class used 13 bottles. How many grams of the solution were there left?
Solution: We can compute this in two different ways, and give algebraic expressions for the different ways of computing. The easier way to compute it is to subtract the number of bottles used from the number of bottles prepared, and then compute the number of grams of solution in those left. The algebraic expression for this is
{w — 13) {w + 15) = the number of grams of the solution left;
or we can compute the number of grams of solution that we had at first and from this subtract the number used, and thus arrive at the number of grams left. The algebraic expression for this is
w {w -\- 15) — 13 (ly + 15) = the number of grams left;
therefore (w - 13) (w + 15) = w {w + 15) - 13 (w; + 15); (Why?) or, writing in the other form, = {w^ + 15 w) — {ISw + 195).' Subtracting as indicated, = w^ -\- 2 w — 195, the number of grams
left.
Check by letting w = 60. Substituting in the original problem: If a teacher in a laboratory prepared for a class 60 bottles each containing 60 grams of water and 15 grams of salt, there were in each bottle the sum of 60 grams and 15 grams which is 75 grams. There were at first 60 bottles, but the class used 13 bottles; there were then 47 bottles left. Since there were 75 grams in each bottle, in 47 bottles there would be 47 times 75 grams which is 3525 grams. This is the number of grams left after the class used 13 bottles.
Substituting w = QO in the answer, which is supposed to be the number of grams left, we have
wj2 + 2 wj - 195 = 602 + 2 • 60 - 195 = 3600 + 120 - 195 = 3525.
Since this is the number which we got by our arithmetic solution, w^ + 2w — 195 must be the correct number of grams left after the class used 13 bottles.
Example 2. A man who can row at the rate of Vr miles an hour in still water, rows up a stream which flows at the rate of Vg miles an hour. He rows m hours in the morning and a hours in the afternoon. How many miles does he row upstream during the day?
IX, §53] MULTIPLICATION OF POLYNOMIALS 99
Solution: As before, we can compute this in two different ways to arrive at the correct result. First add together the number of hours he rows in the morning and in the afternoon, finding the number of hours he rows during the day; then multiply the number of miles he progresses per hour by this. Second, find the number of miles he rows in the morning and the number of miles he rows in the afternoon ; add these, obtaining the number of miles he rows during the day. The algebraic expressions of these two computations are equal to each other. This gives the equation
(w + a) (Vr -Vs) =m {Vr -Vs) +a {Vr - Vs); multiplying on the right, = {mvr — mva) + {avr — avg) ;
adding as indicated, = mVr — mva + aVr — avg,
the number of miles the man rows up the stream during the day.
Check by letting m = S, a = 2, Vr = 5, Vg =2. Substituting in the original problem:
If a man who can row at the rate of 5 miles an hour in still water, rows up a st'-eam which flows at the rate of 2 miles an hour, he will go up the stream at the rate of 3 miles an hour. Since he rows 3 hours in the morning and 2 hours in the afternoon, he rows during the day the sum of 3 hours and 2 hours which is 5 hours. Since he advances 3 miles in 1 hour and rows for 5 hours, he will advance 5 times 3 miles, which is 15 miles.
Substituting in the answer:
mVr — mVa -\- avr — ava = 3«5 — 3«2 + 2«5 — 2»2 = 15-6 + 10-4 = 15.
Since this is the number we got from our arithmetic solution mAr — mva + avr — avs must be the correct number of miles he rows.
Exercises. Solve the following as illustrated above.
1. In an alloy of metal there are in each cubic centimeter g grams of one metal and 4 grams of another. In one brick of the metal there are g cubic centimeters, and in another there are 15 cubic centimeters. How many grams do the two bricks together weigh?
2. Two boys each agree to work for w dollars a day, out of which 25 cents is taken for dinner. If one works t days and the other 5 days, how much do they together receive?
100 ALGEBRA — FIRST COURSE [ix,§53
3. How much more does the boy who works t days receive than the boy who works 5 days?
4. Two men start out to walk. One walks t days a week for 2 weeks and the other walks m days. If they both walk d miles in the morning and t miles in the afternoon, how much farther does the one go than the other?
5. There are two pieces of land. One is square and the other is rectangular. The rectangular piece has a length 6 meters longer than the side of the square, and a width 11 meters shorter than the side of the square. How many more square meters are there in the area of the square than in the area of the rectangle? •
6. The length of one side of a square is s meters. A rec- tangle is 1 meter more in length and 1 meter less in width, what is the difference in the area of the two figures? Which is the larger?
7. The side of a square is s meters. A rectangle is 3 meters more in length and 3 meters less in width. How much more is the area of the square than that of the rectangle?
8. The side of a square is s meters. A rectangle is 5 meters more in length and 5 meters less in width. How much more is the area of the square than the area of the rectangle?
9. The side of a square is s meters. A rectangle is 7 meters more in length and 7 meters less in width. How much greater is the area of the square than the area of the rectangle?
10. The side of a square is s meters. A rectangle is 12 meters more in length and 12 meters less in width. How much greater is the area of the square than that of the rectangle?
11. In Exercises 8, 9, 10, 11 and 12, are the perimeters the same in each case? How do the areas of the different rec- tangles compare? Draw them, arranging them in the order
IX. §54] DIVISION OF POLYNOMULS 101
of their size (use a millimeter instead of meter as the unit). If rectangles have equal perimeters, which is the greatest?
Write problems leading to five of the following. Solve all, expressing the results in the different algebraic forms as in the preceding exercises:
.12. (m + 2) (m + 5). 16. i2k-7)&k- 15).
13. (^1 - « (si + S2). 17. (5r-/i)(3r + 2/i).
14. (r + 6) (r - 2). 18. (c - 1) (c^ + c + 1). 16. (p-8)(p-7). 19. (62 + 6 a6) (2 62 - 3a6).
20 (m + 3 rs) {m} — 3 mrs + 9 rh^).
21. (a + 3 6) (a + 3 6).
22. (m^n + 2 a%) {m^n + 2 a%).
23. (-4-6a26)(-4 + 6a26).
24. (-6 a;^^/ + 2 xy^) (-Qx^y-{-2 xy^).
25. i-d'-2cd^){d^-2cd^).
26. (a:2 + 5) (x2 + X - 30).
54. Division of Polynomials.
Example 1. A man, who can row at the rate of 5 miles an hour in still water, rows upstream against a 2-mile current. He rows 24 miles the first day and 15 miles the second. How many hours did he row?
Solution : He advances in one hour a distance of (5 — 2) miles. He rowed (24 + 15) miles.
Therefore the time he rowed is
f±^ = f (Wh.V) = 13 hours.
Here we can simplify our first expression by combining the numbers, and afterwards by dividing out.
Suppose the same problems to be stated in a literal form.
A man, who can row at the rate of r miles an hom* in still water, rows upstream against the current flowing c miles an hour. He rows n miles the first day and m miles the second. How long did he row?
Solution: He advances in one hour a distance (r — c) miles. He rows (n + m) miles. Therefore the time used in rowing is
^-t^ hours. (Why?)
102 ALGEBRA — FIRST COURSE [ix.§54
We cannot now simplify the result further so we leave it as it stands. It is the indicated quotient of one binomial, namely (n + m), by another binomial, namely (r — c). Such indicated quotients are called fractions.
Example 2. The cost of seeding a piece of ground was 4 dollars per acre, the cost of cultivating was 2 dollars per acre, and the cost of harvesting was 3 dollars per acre. The total cost was $270. How many acres were there?
Solution: The total cost per acre is (4 + 2 + 3) dollars.
Therefore the number of acres is
270 270 ^3„_
4+2+3 9
If the cost of seeding is s dollars, the cost of cultivating is c dollars, the cost of harvesting is h dollars, and if the total cost is a dollars, we shall have the number of acres
s + c + /i
We shall study such form farther on in the chapter on factoring.
Problems in Division of Polynomials.
1. If a man can row at the rate of r miles an hour in still water, and if he rows downstream in a current flowing at the rate of c miles an hour, how many hours will it take him to go d miles? If he rows upstream, how long does it take him to go d miles ?
2. If it costs a cents to set the type for a printed circular, and if it costs h cents a hundred for the printing, and c cents a hundred for the paper, how many circulars can be made for n dollars ?
3. To construct an office building costs for each story / dollars for the floor, w dollars for the walls, p dollars for the partitions. The total cost of the building is c dollars, which includes h dollars for the basement and r dollars for the roof. How many stories are there ?
4. A room is a feet long and h feet wide, and another room is c feet long and d feet wide. It costs k dollars to lay a
IX.§S5] SUMMARY 103
floor in these two rooms. What is the price of the floor per square foot ?
6. If a man loans p dollars to one person and q dollars to each of two others, and receives i dollars interest for the two years, what is the rate of interest?
6. If the temperature at a certain place was d, e, and / respectively for three consecutive days, what was the aver- age temperature for the time?
7. If three rectangles having the same altitude and bases which are 6, c, and d respectively, are added so that their bases are in a straight line, their area is a square inches. What is their altitude ?
8. A cube, a cylinder, a cone, and a sphere of iron weigh Vi Q) '"'i s grams respectively, and have a combined volume of V cubic centimeters. What is the density of iron?
9. The momentum of a boy riding his bicycle is m mile- pounds per hour. The weight of the boy was Wi pounds and the weight of the bicycle was W2 pounds. What was the rate per hour at which they were moving? (Momentum in mile-pounds per hour = weight in pounds X speed in miles per hour.)
55. Summary.
Commutative law of addition. This law asserts that the value of the sum of two numbers does not depend on the order of summation.
Commutative law of multiplication. This law asserts that the value of the product of two numbers does not depend on the order of multiplication.
A positive integral exponent of a number is a small number written to the right and a little above the number to show how many times the number is used as a factor. The num- ber to which the exponent is attached is called the base.
A coefficient of a number is a cofactor with that number.
To multiply a polynomial hy a monomial. Write the num-
104 ALGEBRA — FIRST COURSE [ix.§55
bers as factors — the monomial followed by the polynomial inclosed in parentheses. Make this equal to the product of each term of the poljrnomial multiplied by the monomial, the partial products being connected by the proper sign to form the complete product.
To multiply a polynomial hy a polynomial. Write the two polynomials in parentheses as factors. Multiply the mul- tiplicand by each term of the multiplier, writing the partial products in parentheses connected by the proper sign. Add or subtract as indicated by the sign.
CHAPTER X PROBLEMS LEADING TO SIMPLE EQUATIONS
56. Problems.
Example 1. Apparatus needed — a pair of balances and some shot (any substance of which you can have a number of the same size and weight will do). Place a quantity of shot in one pan of the balances, and a different quantity in the other together with weights enough to balance. Find the weight of each shot. See that at all times the scales are balanced.
Now the process which you go through in the actual use of the scales in order to find the weight of each shot can be expressed in algebraic symbols.
Suppose you had placed 23 shot in one pan, and 12 in the other, and then found it necessary to place 22 gram weights on the second pan in order to balance the scales. Since you are interested in the weight of one shot only, the thing that you would do in the actual operation of weighing would be to take out of both pans the same number of shot, until you had shot only in one pan. Then you would count the num- ber of shot left in the pan and the number of gram weights left in the other pan, and from this compute the weight of one shot. We shall now represent this in algebraic language.
Let w = the number of grams in the weight of each shot;
then 23 w = the number of grams in the weight of 23 shot,
and 12 w = the number of grams in the weight of 12 shot.
Therefore 23 w = the number of grams in one pan, while 12 w +22 = the number of grams in the other pan.
Since these two amounts balance each other, they are equal to each other. Therefore we write
2Sw = 12w + 22.
Now, just as we did in the actual weighing, we can take out 12 w from each pan and have
llw? = 22. 105
106 ALGEBRA — FIRST COURSE ix,§56
Since w is multiplied by 11 to be equal to 22, and 2 is multiplied by 11 to be equal to 22, therefore,
w =2.
Therefore, since w stood for the weight of each shot, each shot must weigh 2 grams.
Check: We can test the correctness of this result by computing the weight in each pan.
If 1 shot weighs 2 grams, 23 shot will weigh 23 • 2 or 46 grams, which is the weight in one pan.
12 shot will weigh 12 • 2 or 24 grams. To this add the 22 grams, and we have 46 grams as the weight on the other pan.
The two amounts are equal and hence balance each other, therefore our answer must be correct.
1. Place different quantities of shot in each pan and then balance the scales by placing gram and fractional gram weights on each pan. Proceed to find the weight of each shot. See that at each moment the scales are kept balanced.
State this experiment in words; represent in algebraic symbols; solve and check.
2. A boy found that if he placed 2 iron balls of an equal size on one pan of the scales, he must place 50 gram weights on the other, in order to balance them. What is the weight of each ball?
3. In weighing a stone block in the laboratory, a student found that if he placed a 20-gram weight on the side of the balances with the stone, it would balance a 100-gram weight on the other side. What is the weight of the stone? Ex- plain this without using symbols, then solve by means of symbols.
4. If 5 marbles of equal size and 5 gram weights balance 2 marbles and 40 gram weights, what is the weight of each marble?
5. If your book and 20 grams balance 625 grams, what is the weight of your book?
X,§56] SIMPLE EQUATIONS 107
Example 2. If 21 cc. of marble weigh 56.7 grams, what is the density of marble? That is, what is the weight of one cc. of marble?
Solution:
Let d = the density of marble.
Then 21 d = the number of grams in the weight of 21 cc.
But 56.7 = the number of grams in the weight of 21 cc,
therefore 21d = 5Q.7, and d = 2.7, the density of marble.
Check : If one cubic centimeter of marble weighs 2.7 grams, 21 cc. will weigh 21 times 2.7, which is 56.7. This is the amount that the example states that it should weigh. Therefore the result is correct.
6. How many cubic centimeters are there in a brass ball which weighs 109.2 grams? The density of brass is 8.4.
7. If you place on one pan of the scales a piece of wood and another f as large, they together balance 16 lbs. in weights placed on the other pan. What is the weight of each piece of wood?
8. The density of platinum is 2.17 more than the density of gold. Two bars, one containing 20 cc. of platinum and the other containing 20 cc. of gold, together weigh 826 grams. What is the density of platinum ? of gold ?
9. A cubic centimeter of ebony weighs 1 gram more than a cubic centimeter of cork. A piece of cork containing 16 cc. and a piece of ebony containing 7 cc. placed together on one pan of the scales balance 11.14 gram weights placed on the other pan. What is the density of cork? of ebony? What is the weight of each piece ?
10. Iron is 4 times as heavy as ivory. 14 ivory balls and 6 iron balls of the same dimensions are placed on the scales and are found to balance 456 gram weights. What is the weight of each ball? What is the density of ivory if each ball contains 6f cc? What is the density of iron?
11. If 20 shot and a 5-gram weight together balance 4 shot and 29 grams, what is the weight of each shot ?
12. If 2 iron blocks of the same size, together with 4 oz. weights, just balance a block f as large as one of them
108 ALGEBRA — FIRST COURSE [x. § 56
together with 12 oz. weights, what is the weight of each block?
13. A boy placed in one pan of the scales 110 shot and in the other 50 shot and 25 grams in weights. He then re- moved from the first pan enough to make the two pans balance. He found afterward by calculation that he had removed shot enough to weigh 15 grams. What was the weight of each shot?
14. The weight of one cubjc centimeter of copper is 1.1 grams more than that of the same amount of iron. 5 iron balls weigh 278.3 grams more than 3 copper balls of the same size. What is the density of iron, if each ball contains 23 cc. ? What is the density of copper?
15. The density of lead lacks 4.3 of being 2 times that of iron. 6 iron balls weigh 1.6 grams more than 4 lead balls of the same size. What is the weight of each iron ball if each ball contains 1 cc. ? What is the weight of each lead ball ?
16. An iron pail covered with tin weighs 1.1366 kilograms. There are 125 cc. of iron and 22 cc. of tin. If iron has a density of .5 more than tin, what is the density of iron and of tin?
17. If 3 cc. of a substance weigh 9 grams, what does 1 cc. weigh. If 16 cc. of a substance weigh 38 grams, what is the density of the substance? If 84 cc. of a substance weigh 21 grams, what is the density of the substance? If 6 cc. weigh 1 gram, what is the density of the substance ? If V cc. weigh m grams, what is the density of the substance?
The answer to this last question is usually expressed in
the form
, m d = —•
V
This can be used as a formula by which to solve other exercises. Using this as a formula solve the following:
18. A block of lead in the shape of a cube whose edge is 10 cm, weighs 11,300 grams. What is the density of lead?
X.§57] SIMPLE EQUATIONS 109
19. A block of marble 6 dm. wide, 6 dm. thick and 1.5 m. long weighs 1,468,800 grams. What is the density of marble ?
20. A block of silver 2 cm. long, c cm. wide and c cm. thick weighs 21 grams. What is the density of silver?
21. The mass of a substance is 7 grams and its volmne is 10 cc. What is its density?
22. The mass of a substance is 7ab grams and its volume is 14 b cc. What is its density?
23. Solve — not using formula:
Alcohol is .8 as heavy as water. A bottle containing 354.3 cc. of alcohol and 118.1 cc. of water weighs 401.54 grams. If the weight of the glass is not included in this weight, what is the density of alcohol ? of water ?
24. From the Table of Densities on p. 35, write three problems leading to equations and solve.
57. Formation of Equations. In all of these exercises you will have noticed that certain amounts are known and certain amounts are unknown. You will also have noticed that we get the values of those that are unknown through their relations to those that are known. Those quantities that are given by experiment and measurement are called known quantities. Those whose values are to be found through their relation to the known quantities are called unknown quantities.
Whenever we have enough data on the relations of one unknown quantity to known quantities to be able to form two sets of symbols standing for the same amount, we can place the sets equal to one another, and solve for the unknown quantity as we have done in the preceding exercises.
The expression formed by placing these two sets of syra- bols equal to one another is called an equation.
The truth which we employ in forming the equation is called an axiom. It is usually worded:
110 ALGEBRA — FIRST COURSE [X. § 58
Things which are equal to the same thing are equal to each other.
An axiom is a statement whose truth we assume without proof.
58. Solution of Equations. Having formed our equa- tion, we further assume that we may add equal amounts^to both sides of the equation without destroying the equaUty, or that we may subtract equal amounts from both sides of the equation without destroying the equality. Thus we get an equation with the unknown terms on one side of the equation and the known terms on the other. Likewise we assume that both sides of an equation may be multiplied or divided by equal quantities without destroying the equality. This leads to the solution of the equation.
So we have the following five axioms by means of which we solve our equation:
I. Things equal to the same thing are equal to each
other. II. If equals be added to equals, the results are equal. ni. If equals be subtracted from equals, the results are
equal. IV. If equals be multiplied by equals, the results are
equal. V. If equals be divided by equals, the results are equal.
Example 1. A boy riding his wheel starts for his home 20 miles away. He rode 2 miles farther the second hour than he did the first, and 3 miles farther the third hour than he did the first, and reached his destination. How far did he ride each hour?
Solution: Let d = the number of miles the boy rode the first hour. Then d -\- 2 = the number of miles he rode the second hour,
and d + 3 = the number of miles he rode the third hour.
Then 3 cZ + 5 = the number of miles he rode;
but 20 = the number of miles he rode.
Therefore 3 d + 5 = 20. (Things which are equal to the same thing are equal to each other.)
X.§58] SIMPLE EQUATIONS 111
Now subtract 5 from each side of the equation; we have
Sd = 15. (If equals are subtracted from equals, the results are equal.) Then d = 5, (If equals are divided by equals, the results are equal.) and d + 2 = 7, (If equals are added to equals, the results are equal.) and d + 3 = 8. (For the same reason.)
^o the boy rode 5 miles the first hour, 7 miles the second hour and 8 miles the third hour.
Check: If a boy rode his wheel at the rate of 5 miles an hour the first hour and at the rate of 7 miles an hour the second hour, he rode 2 miles farther the second hour than the first. If he rode 8 miles the third hour, he rode 3 miles farther the third hour than the first. If he rode 5 miles the first hour, 7 miles the second hour and 8 miles the third hour, he rode altogether (5 + 7 + 8) miles, which equals 20 miles, the entire distance. Therefore since these answers check each state- ment of the original example, they must be correct.
Problems.
1. Two boys returning from a fishing trip wished to divide the fish caught equally between them. Upon count- ing them the one who had caught the most said to the other, *' If I had 6 more, I should have 2 times as many as you have. I will give you 11, then we shall have an equal number." How many fish did each have ?
2. A class committee appointed to look up the price of a piece of statuary reported that the money in the treasury was but 50 cents more than half as much as was needed for the purchase, and that it would be necessary for the class to raise 62 dollars more in order to buy the piece selected. How much did the piece of statuary cost ?
3. Two boys pulled a small wagon in opposite directions, the forward pull being three times the backward pull. The result was a forward pull of 74 lbs. What was the amount and direction of each boy's pull ?
4. Three men on a building are trying to raise a beam with ropes. One of the men exerts a force of f as many pounds as the strongest of the three, and the other a force of ^ as many pounds as the strongest. They succeed in lifting
112 ALGEBRA — FIRST COURSE [x.§58
the beam which weighs 470 pounds. What is the number of pounds force of each man's pull ?
5. A young man wishing to earn money to pay his board while attending school decided to build fires in furnaces at private residences in the neighborhood. He found that the number of engagements that he could make for this kind of work was just } of the number of cents that he asked per day at each residence, and that if he would lower his price 5 cents per day at each residence he could get 2 more en- gagements, and thus earn 15 cents more per day. How much did he ask per residence?
6. A lady arranging a flower bed found that she had plants enough to arrange exactly in the form of a square, but if she attempted to arrange them in the form of a bed with 4 more in the length and 2 less in the width, she would lack 16 of having enough plants. How many plants has she?
7. Of three lines one is 6 cm. longer than f of the length of another, and the third is 7f cm. less than the sum of the other two, while their combined length is 9 cm. What is the length of each line?
8. There are two angles whose sum is 97 degrees. If 13 degrees be subtracted from the larger and added to the smaller, the two angles will be equal. What is the size of each angle?
Example 2. A solution of alcohol and water is 95% strong. How much water must be added to have it 75% strong?
Remark. By 95% strong we mean that .95 of any selected amount or part is alcohol, and that 5% is water.
Then in starting an exercise of this kind, unless some definite amount is stated, we can speak of one part, a part being any convenient amount.
SoltUion:
Let w = the number of parts of water that must be added to
one part of the given solution. Then 1 -{■ w = the number of parts in the new solution.
.75 (1 -\- w) = the number of parts of alcohol.
X,§58] SIMPLE EQUATIONS 113
But .95 = the number of parts of alcohol, smce the amount of
alcohol is not changed. .-. .75 (1 + w) = .95. Why?
w = ^5, the number of parts of water that must be added to reduce a 95% solution to a 75% solution. Check as in previous problems.
9. Upon examining a bottle of rose-water and glycerine which she had just bought, a lady said to the druggist, "There seems to be too much glycerine in this." He an- swered "I thought you said that you wished equal parts." The lady answered "No, I said that I wished 35% glycerine." How much rose-water must be added if the lady is willing that it be added to the four ounces that are in the bottle?
10. A solution of sulphuric acid and water is 33 J % sul- phuric acid. How much water must be added to have it 10% sulphuric acid?
11. In a solution of iodine and alcohol, there is 6.25% of iodine. How much alcohol must be added in order to have 1% iodine?
12. A 2-ounce bottle of peroxide is 95% strong. This is too strong for general use. How much water must be added to make a 10% solution?
13. A pint of ammonia is purchased for cleaning purposes. It is a 90% solution. This is too strong. How much water must be used if a 50% solution is desired ?
14. If sea-water is 12% salt, how much water must be evaporated in order to have it 90% salt?
16. Write three problems relating to solutions with which you are acquainted. Metals in alloy may be used. Solve your problems.
16. If one angle of a triangle is 15 degrees more than another, and a third is 2 times the sum of these two, what is the number of degrees in each angle of the triangle ?
17. If one of the angles of a triangle is 15 degrees more than 3 times another, and J of the difference between these
114 ALGEBRA — FIRST COURSE ix.§58
two is equal to J of the third angle, what is the number of degrees in each angle of the triangle ?
18. If you subtract 5 degrees from the number of degrees in the first angle of a triangle and multiply the remainder by 3, you will get the number of degrees in the second angle of the triangle. The third angle is 4 times the second. What is the number of degrees in each angle of the tri- angle ?
19. The second side of a triangle is 5 times the first, and the third is 5 cm. longer than the first. If the third is sub- tracted from the second, the difference will lack 2 cm. of being equal to the first. What is the length of each side ?
20. One side of a triangular lot is 200 feet longer than an- other, and the third side is 2 times this one. It requires 1100 feet of fence to enclose the lot. What is the length of each side?
21. A triangle has two of its sides equal. If 1 were added to the number of centimeters in the length of one of them, the sum would be 3 times the number of centimeters in the length of the third side. If 3 cm. were subtracted from each of the equal sides and added to the third side, the triangle would be equilateral. What is the length of each side of the triangle ?
22. If one angle of a triangle is ^ tt radians more than f of another, and the third is equal to 4 times the difference between these two, what is the number of radians in each angle of the triangle ?
Write problems concerning a triangle leading to the fol- lowing equations:
23. s-f(s-M) + (2s-2)=27.
24. 5a - 30 = 180. 25. 2ia + Jtt = t.
Example 3. If an angle is 2f times its complement, what is the number of degrees in the angle? in the complement? Solviion: Let o = the number of degrees in the angle.
X.§58] SIMPLE EQUATIONS 115
Then 90 — o = the number of degrees in the complement. Why? 2f (90 — a) = the number of degrees in the angle. Why? 2f (90 -a) =a. Why? .*. 240 — 21 a — a, doing the multiplication indicated.
240 = 3| a, adding 2| a to both members. Why?
a = -52-. Why?
= Vr, multiplying numerator and denominator by 3, = 65xr, number of degrees in the angle. 90 — a = 24/r, number of degrees in the complement of the angle.
Check: Since the angles are complementary their sum should equal
90 degrees. (65r\ + 24r\) degrees is 90 degrees. The angle is as
/^f'"5_ 7 on
many times its complement as ^^t^j which is ^^=7;, which is 2f . Since
24j f z70
the answers obtained satisfy all the conditions of the problem, they
must be correct.
26. If an angle is 200 degrees more than 2 times its com- plement, what is the size of the angle and of its complement ?
27. If — 15 degrees be added to | of the supplement of an angle, the sum will be equal to the angle. What is the number of degrees in the angle? in the supplement of the angle ?
28. How many radians are there in an angle if its comple- ment is 3 times its supplement ?
29. Two angles are complementary. If x\ tt radians be added to one and subtracted from the other, the two angles will be equal. What is the number of degrees in the angles ?
30. Write three problems concerning complementary an- gles and supplementary angles. Solve to see that they are correct.
31. One acute angle of a right-angled triangle is IJ times the other. What is the number of radians in each of the angles ?
32. Of two adjacent sides of a parallelogram one is | as long as the other. The perimeter is 49 cm. What is the length of each side ?
116 ALGEBRA — FIRST COURSE IX,§58
33. Of the six angles of a hexagon, A, B, C, D, E, F, angle 5 is 5 degrees more than 2 times angle A ; angle C is 14 degrees more than ^ of angle B; angle D is 2 times as large as angle C; angle E lacks 10 degrees of equalling J angle A ; angle F equals the difference between angle A and angle E. The sum of the angles of a hexagon is 8 right angles. Find the number of degrees in each angle of the hexagon.
34. The sum of two angles is 150 degrees; f of the smaller equals J of the larger. What is the number of degrees in each angle?
35. In a triangle whose sides are a, h, c, if the side c con- tained 1 unit more it would be just J of side a; if it were 1 unit less it would be J the length of side h. Furthermore, if one unit b^ subtracted from side a and added to side h, sides a and h would be equal. Find the length of each side of the triangle.
36. Three men, A, B, C, on a building, are trying to raise a beam with ropes. A pulls with a force of f as many pounds as C, and B pulls with a force of f as many pounds as C. The beam weighs 471 pounds. With what force does each man pull if we regard a downward pull as positive ?
37. The density of wrought silver is 7.945 more than that of agate. A silver ornament set with agate contains 8 cc. of each, and weighs 110.6 gm. Find the density of silver and of agate.
38. A glass bottle containing alcohol is closed with a cork. There are 23 cc. of glass in the bottle and 50 cc. of alcohol; the cork measures 5 cc. The whole weighs 100.12 grams. If a cubic centimeter of alcohol and one of cork together weigh 1.08 grams, and the density of glass is .7 less than 4 times that of alcohol, what is the density of each substance ?
39. Two men start from a point 200 miles apart and travel toward each other, one at the rate of 5 miles an hour, and
X.§58] SIMPLE EQUATIONS 117
the other at the rate of 10 miles an hour. After how many hours will they meet ?
40. ^, traveling at the rate of 20 miles a day, has 4 days start of B, traveling at the rate of 26 miles a day, in the same direction. After how many days will B overtake A ?
41. The circumferences of the front wheel and of the hind wheel of a wagon are 2 and 3 yards respectively. What distance has the wagon moved when the front wheel has made 10 revolutions more than the hind wheel?
Hint. — Let r = number of revolutions the front wheel will make.
42. The sum of two digits of a number is 12. If the digits be interchanged, the resulting number will be equal to the original. What is the number ?
43. The sum of the two digits of a number is 12. If the digits be interchanged, the resulting number exceeds the original number by | of the original number. What is the number ?
44. The sum of two digits of a number is 11. If the digits are interchanged, the resulting number is 45 less than the number. What is the number ?
45. Write three problems on numbers and their digits and solve.
46. One boy says to another: "Think of a number; add 7; double result; take away 8; tell me your answer and I will tell you the number thought of.'' How can he do it?
47. Make up a problem like number 46.
48. In sending packages