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XXXVIII. Extraction of the roots of compound quantities of

any degree

XXXIX. Extraction of the roots of numerical quantities of

193

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XLIV. Binomial Theorem, continued from Art. XLI.

XLV. Continuation of the same subject

XLVI. Progression by difference, or Arithmetical progression

XLVII. Progression by quotient, or Geometrical progression
XLVIII. Logarithms

221

226

228

233

239

XLIX. Same subject continued

242

I. Logarithms of fractions

249

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PREFACE.

THE first object of the author of the following treatise has been to make the transition from arithmetic to algebra as gradual as possible. The book, therefore, commences with practical questions in simple equations, such as the learner might readily solve without the aid of algebra. This requires the explanation of only the signs plus and minus, the mode of expressing multiplication and division, and the sign of equality; together with the use of a letter to express the unknown quantity. These may be understood by any one who has a tolerable knowledge of arithmetic. All of them, except the use of the letter, have been explained in arithmetic. To reduce such an equation requires only the application of the ordinary rules of arithmetic ; and these are applied so simply, that scarcely any one can mistake them, if left entirely to himself. One or two questions are solved first with little explanation in order to give the learner an idea of what is wanted, and he is then left to solve several by himself.

The most simple combinations are given first, then those which are more difficult. The learner is expected to derive most of his knowledge by solving the examples himself; therefore care has been taken to make the explanations as few and as brief as is consistent with giving an idea of what is required.

In fact, explanations rather embarrass than aid the learner, because he is apt to trust too much to them, and neglect to employ his own powers; and because the explanation is frequently not made in the way, that would naturally suggest itself to him, if he were left to examine the subject by himself. The best mode, therefore, seems to be, to give examples so simple as to require little or no explanation, and let the learner reason for himself, taking care to make them more difficult as he proceeds. This method, besides giving the learner confidence, by making him rely on his own powers, is much more interesting to him, because he seems to himself to be constantly making new discoveries. Indeed, an apt scholar will frequently make original explanations much more simple than would have been given by the author.

14-6= 8, is read 6 subtracted from 14, or 14 less 6, or 14 minus 6 is equal to 8.

Observe that the signs and affect the numbers which they stand immediately before, and no others. Thus

14-6816; and 14 +86 = 16;

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and 86+14= 16; and, in fine, 6+8+14= 16. In all these cases the 6 only is to be subtracted, and it is the same, whether it be first subtracted from one of the numbers, and then the rest be added, or whether all the others be added and that be subtracted at last.

(X) (.) An inclined cross, or a point, is used to express multiplication; thus, 5 x 315, or 5.3

15.

(-) A horizontal line, with a point above and another below it, is used to express division. Thus 15-3 = 5, is read 15 divided by 3 is equal to 5.

But division is more frequently expressed in the form of a fraction (Arith. Art. XVI. Part II.), the divisor being made the denominator, and the dividend the numerator.

Thus 1

= 5,

is read 15 divided by 3 is equal to 5, or one third of 15, is 5, or 15 contains 3, 5 times.

Example.

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This is read, 9 times 6 and 15 less 3 are equal to 8 times 7 less 16 divided by 4, and 14.

To find the value of each side; 9 times 6 are 54 and 15 are 69, less 3 are 66. Then 8 times 7 are 56, less 16 divided by 4, or 4 are 52, and 14 more are 66.

In questions proposed for solution, it is always required to find one or more quantities which are unknown; these, when found, are the answer to the question. It will be found extremely useful to have signs to express these unknown quantities, because it will enable us to keep the object more steadily and distinctly in view. We shall also be able to represent certain operations upon them by the aid of signs, which will greatly assist us in arriving at the result.

Algebraic signs are in fact nothing else than an abridgment of common language, by which a long process of reasoning is presented at once in a single view.

The signs generally used to express the unknown quantities above mentioned are some of the last letters of the alphabet, as x, y, z, &c.

1)/2607 (237)

22

I. 1. Two men, A and B, trade in company, and gain 267 dollars, of which B has twice as much as A. What is the share

of each?

In this example the unknown quantities are the particular shares of A and B.

Let a represent the number of dollars in A's share, then 2x will represent the number of dollars in B's share. Now these added together must make the number of dollars in both their shares, that is, 267 dollars.

x+2x=267 Putting all the x's together, 3x=267

If 3 x are 267, 1 x is of 267 in the same manner as if 3 oxen were worth $267, 1 ox would be worth

of it.

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A's share.

B's share.

2. Four men, A, B, C, and D, found a purse of money containing $325, but not agreeing about the division of it, each took as much as he could get; A got a certain sum, B got 5 times as much; C, 7 times as much; and D, as much as B and C both. How many dollars did each get?

Let x represent the number of dollars that A got; then B got 5 x, C7x, and D (5x+7x) = 12 x. These, added together, must make $325, the whole number to be divided.

x+5x+7x+12x=325

Putting all the x's together,

25x=325

x= 13 A's share.
5x =
65= - B's

7x= 91 C's
=
12 x 156=D's

66

66

Note. All examples of this kind in algebra admit of proof. In this case the work is proved by adding together the several shares. If they are equal to the whole sum, 325, the work is right. As the answers are not given in this work, it will be well for the learner always to prove his results.

In the same manner perform the following examples.

3. Said A to B, my horse and saddle together are worth $130, but the horse is worth 9 times as much as the saddle. What is the value of each?

4. Three men, A, B, and C, trade in company, A puts in a certain sum, B puts in 3 times as much, and C puts in as much

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as A and B both; they gain $656. What is each man's s of the gain?

5. A gentleman, meeting 4 poor persons, distributed cents among them, giving the second twice, the third times, and the fourth four times as much as the first. many cents did he give to each?

6. A gentleman left 11000 crowns to be divided betw his widow, two sons, and three daughters. He intended the widow should receive twice the share of a son, and each son should receive twice the share of a daughter. quired the share of each.

Let a represent the share of a daughter, then 2x will re sent the share of a son, &c.

7. Four gentlemen entered into a speculation, for wh they subscribed $4755, of which B paid 3 times as much as and C paid as much as A and B, and D paid as much as Ba C. What did each pay?

8. A man bought some oxen, some cows, and some she for $1400; there were an equal number of each sort. the oxen he gave $42 apiece, for the cows $20, and for t sheep $8 apiece. How many were there of each sort?

In this example the unknown quantity is the number of cac sort, but the number of each sort being the same, one charac ter will express it.

Let x denote the number of each sort.

Then x oxen, at $42 apiece, will come to 42 x dolls., and cows, at $20 apiece, will come to $8 apiece, will come to 8x dolls. make the whole price.

20 x dolls., and a sheep, a These added together mus

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70 x 1400
x= 20

Ans. 20 of each sori

9. A man sold some calves and some sheep for $374, the calves at $5, and the sheep at $7 apiece; there were three times as many calves as sheep. How many were there of each?

Let x denote the number of sheep; then 3x will denote the number of calves.

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