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ALGEBRA.

INTRODUCTION

THE operations explained in Arithmetic are sufficient for the solution of all questions in numbers, that ever occur; but it is to be observed, that in every question there are two distinct things to be attended to; first, to discover, by a course of reasoning, what operations are necessary; and, secondly, to perform those operations. The first of these, to a certain extent, is more easily learned than the second; but, after the method of performing the operations is understood, all the difficulty in solving abstruse and complicated questions consists in discovering how the operations are to be applied.

It is often difficult, and sometimes absolutely impossible to discover, by the ordinary modes of reasoning, how the fundamental operations are to be applied to the solution of questions. It is our purpose, in this treatise, to show how this difficulty may be obviated.

It has been shown in Arithmetic, that ordinary calculations are very much facilitated by a set of arbitrary signs, called figures; it will now be shown that the reasoning, previous to calculation, may receive as great assistance from another set of arbitrary signs.

Some of the signs have already been explained in Arithmetic ; they will here be briefly recapitulated.

(=) Two horizontal lines are used to express the words "are equal to," or any other similar expression.

(+) A cross, one line being horizontal and the other perpendicular, signifies" added to." It may be read and, more, plus, or any similar expression; thus, 7 +5 12, is read 7 and 5 are 12, or 5 added to 7 is equal to 12, or 7 plus 5 is equal to Plus is a Latin word signifying more.

12.

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(—) A horizontal line, signifies subtracted from. It is sometimes read less or minus. Minus is Latin, signifying less. Thus

1468, is read 6 subtracted from 14, or 4 less 6, or 14 minus 6 is equal to 8.

14

-

6 +8: = 16; and 14 +8

=

Observe that the signs and affect the numbers which they stand immediately before, and no others. Thus 16; 6 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.

15

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 = 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. 6 x 9 + 15 - 37.8+ 14. 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.

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 2 x 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.

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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 a represent the number of 5 x, C 7 x, and D (5 x + 7 x) ther, must make $325, the whole x + 5 x + 7 x + Putting all the x's together,

dollars that A got; then B got = 12 x. These, added togenumber to be divided. 12 x =

25 x =

x=

325

325

13 - A's share. 65= B's 66

5x = 7x=

91= : C's

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=D's

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. the value of each?

What is

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

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

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5. A gentleman, meeting 4 poor persons, distributed 60 cents among them, giving the second twice, the third three times, and the fourth four times as much as the first. How many cents did he give to each?

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

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

7. Four gentlemen entered into a speculation, for which they subscribed $4755, of which B paid 3 times as much as A, and C paid as much as A and B, and D paid as much as B and C. What did each pay?,1

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

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

Let x denote the number of each sort.

20 x dolls., and ≈ sheep, at These added together must

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 8 x dolls. make the whole price.

42 x 20 x +8x= 1400

Putting the x's together,
Dividing by 70,

70 x= = 1400

x= 20

Ans. 20 of each sort.

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?

x

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

Then a sheep, at $7 apiece, will come to 7 x dolls., and 3 r calves, at $5 apiece, will come to 5 times 3 x dolls., that is, 15 x dolls.

These added together must make the whole price.

7 x + 15 x = 374

Putting the x's together, 22 x

Dividing by 22,

374

x= 17 3x=

51

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number of sheep. calves.

The learner must have remarked by this time, that when a question is proposed, the first thing to be done, is to find, by means of the unknown quantity, an expression which shall be equal to a given quantity, and then from that, by arithmetical operations, to deduce the value of the unknown quantity.

This expression of equality between two quantities, is called an equation. In the last example, 7 x 15 x 374 is an equation.

The quantity or quantities on the left of the sign are called the first member, those on the right, the second member of the equation. (7 x 15 x) is the first member of the above equation, and 374 is the second member.

--

Quantities connected by the signs and are called terms. 7 and 15 x are terms in the above equation.

The figure written before a letter showing how many times the letter is to be taken, is called the coefficient of that letter. In the quantities 7x, 15 x, 22 x ; 7, 15, 22, are coefficients of x. The process of forming an equation by the conditions of a question, is called putting the question into an equation.

The process by which the value of the unknown quantity is found, after the question is put into an equation, is called solving or reducing the equation..

No rules can be given for putting questions into equations; this must be learned by practice; but rules may be found for solving most of the equations that ever occur.

After the preceding questions were put into equation, the first thing was to reduce all the terms containing the unknown quantity to one term, which was done by adding the coefficients. As 7x+15x are 22 x. Then, since 22 x = 374, 1 x must be equal to of 374. That is,

When the unknown quantity in one member is reduced to one term, and stands equal to a known quantity in the other, its value

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