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

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?

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. quired the share of each.

Re

Let x represent the share of a daughter, then 2x 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?

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 cach 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 a sheep, at These added together must

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

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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 a denote the number of sheep; then 3x 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.

7x+15x=374

Putting the x's together, 22 x = 374
Dividing by 22,

x = 17 = number of sheep. 3x 51= 66 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 equa

tion.

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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. (7x+ 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 x 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 r. 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 so1ing 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 7 x + 15 x are 22 r. 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 is

found by dividing the known quantity by the coefficient of the unknown quantity.

10. A man bought some oranges, some lemons, and some pears, for 156 cents; the oranges at 6 cents each, the lemons at 4 cents, and the pears at 3 cents; there was an equal nun ber of each sort. Required the number of each.

11. In fencing the side of a field, the length of which was 450 yards, two workmen were employed; one fenced 9 yards, and the other 6 yards per day. How many days did they work?

12. Three men built 780 rods of fence; the first built 9 rods per day, the second 7, and the third 5; the second worked three times as many days as the first, and the third, twice as many days as the second. How many days did each work?

13. A man bought some oxen, some cows, and some calves for $348; the oxen at $38 each, the cows at $18, and the calves at $4. There were three times as many cows as oxen, and twice as many calves as cows. How many were there of each sort?

14. A merchant bought a quantity of flour for $132; for one nalf of it he gave $5 per barrel, and for the other half $7 How many barrels were there in the whole?

Let x denote one half the number of barrels.

15. From two towns, which are 187 miles apart, two travellers set out at the same time with an intention of meeting; one of them travels at the rate of 8, the other of 9 miles each day In how many days will they meet?

II. 1. A cask of wine was sold for $45, which was only of what it cost. Required the cost.

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is the same as of 3x. Now if of 3 x

4

is 45, 3x itself must be 4 times 45, or 180; 3 x being 180, x must be of 180, which is 60.

2. A man, being asked his age, answered, that if its half and its third were added to it the sum would be 88. What was his age?

Let a denote his age then,

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of 11 x being 88, 11 x will be 6 times 88, 11 x = 528

Dividing by 11,

x=48
Ans. 48 years.

3. If & of a hogshead of wine cost $65; what will a hogshead cost at that rate?

4. There is a pole and under water, and 5 feet out of water; what is the length of the pole?

Let a denote the whole length. Then ++5 must be

equal to the whole length. Hence,

Reducing to a common denominator,

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3

2

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Since the two members are equal, if5 be subtracted from

both, they will still be equal; hence,

20

6

6

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Proof. One half of 30 is 15, and one third of thirty is 10. Now 30 15 + 10 + 5.

=

There is another mode of reducing the above equation which in most cases is to be preferred. It is the same in principle. If both members of an equation be multiplied by the same number they evidently will still be equal.

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First multiply both members by 2, the denominator of one the fractions, and it becomes,

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Next multiply both members by 3, the denominator of the other fraction, and it becomes,

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5. In an orchard of fruit trees of them bear apples, of them pears, of them plums, 7 bear peaches, and 3 bear cherries; these are all the trees in the orchard. How many are there?

6. A farmer, being asked how many sheep he had, answered, he had them in four pastures; in the first he had of them. in the second, in the third, and in the fourth he had 24 sheep. How many had he in the whole?

7. A person having spent and of his money, had $26% left. How much money had he at first?

8. A man driving his geese to market, was met by another, who said good morrow, master, with your hundred geese; said be, I have not a hundred, but if I had as many more, and half

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