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B gained a sum equal to his stock and £153 over. Now the amount of both their gains was equal to 5 times the stock of either. What was the stock ?

Let x denote the stock. Then A's gain was 2 x + 27, and B's was x + 153. These added together must make 5 times the stock, that is, 5 x.

5x = 2 x + 27 + x + 153 Uniting the r's in 2d member, and the numbers

5x = 3 û + 180 Subtracting 3 x from both sides,

2x = 180

90 11. A young man being asked his age, answered that if the age

of his father, which was 44 years, were added to twice his own,

the sum would be four times his own age. What was his age ?

12. A man meeting some beggars, gave each of them 4 pence, and had 16 pence left ; if he had given them 6 pence apiece, he would have wanted 12 pence more for that purpose. How many beggars were there, and how much money had he?

Let x represent the number of beggars.

13. A man has six sons, each of whom is 4 years older than his next younger brother; and the eldest is three times as old as the youngest. Required their ages.

14. Three persons, A, B, and C, make a joint contribution, which in the whole amounts to £76, of which A contributes a certain sum, B contributes as much as A and £10 more, and C as much as A and B both. Required their several contributions.

15. A boy, being sent to market to buy a certain quantity of meat, found that if he bought beef, which was 4 pence per pound, he would lay out all the money he was entrusted with ; but if he bought mutton, which was 34 pence per pound, he would have 2 shillings left. How much meat was he sent for ?

16. A man lying at the point of death left all his estate to his three sons, to be divided as follows : to A he gave one half of the whole wanting $500; to B one third ; and to C the rest,

which was $100 less than the share of B. What was the whole
estate, and what was each son's share ?
Let x represent the whole estate.
A's share will be 500

2

X

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C's share

100

3 These together will be equal to the whole estate, which was represented by a.

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Uniting x's and numbers in the first membe

60
- 600
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6
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3600 The whole estate is $3600; the shares are $1300, $1200, and $1100, respectively.

17. A father intends by his will, that his three sons shall share his property in the following manner; the eldest is to receive 1000 crowns less than half the whole fortune ; the second is to receive 800 crowns less than $ of the whole; and the third is to receive 600 crowns less than 1 of the whole. Required the amount of the whole fortune, and the share of each.

18. A father leaves four sons, who share his property in the following manner; the first takes 3000 livres less than one half the fortune ; the second, 1000 livres less than one third of the whole ; the third, exactly one fourth ; and the fourth takes 600 livres more than one fifth of the whole. What was the whole fortune, and what did each receive ?

19. In a mixture of copper, tin, and lead ; 16 lb. less than one half of the whole was copper ; 12 lb. less than one third of the whole was tin, and 4 lb. more than one fourth of the whole was lead. What quantity of each was there in the mixture ?

20. A general having lost a battle, found that he had only 3600 men more than one half of his army left, fit for action; 600 more than one eighth of them being wounded, and the rest, which amounted to one fifth of the whole army, either slain or taken prisoners. Of how many men did his army consist before the battle ?

21. Seven eighths of a certain number exceeds four fifths of it by 6. What is that number?

22. A and B talking of their ages, A says to B, one third of my age exceeds its fourth by 5 years. What was his age ?

23. A sum of money is to be divided between two persons, A and B, so that as often as A takes £9, B takes £4. Now it happens that A receives £15 more than B. What is the share of each ?

24. In a mixture of wine and cider, 25 gallons more than half the whole was wine, and 5 gallons less than one third of the whole was cider. How many gallons were there of each ?

IV.

1. A man having some calves and some sheep, and being asked how many he had of each sort, answered, that he had 20 more sheep than calves, and that three times the number of sheep was equal to seven times the number of calves. How Inany were there of each ?

Let x denote the number of calves.
Then : +20 will denote the number of sheep.

7 times the number of calves is 7 x; 3 times the number of sheep is 3x + 60; for it is evident that to take 3 times x + 20, it is necessary to multiply both terms by 3. By the conditions these must be equal,

3x + 60. Subtracting 3 x from both members,

4 x = 60

x = 15 = number of calves. x + 20 = 35 = number of sheep.

Ans. 15 calves, and 35 sheep.

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2. Two men talking of their ages, the first says, your age is 18 years more than mine, and twice your age is equal to three times mine. Required the age of each.

3. Three men, A, B, and C, make a joint contribution, which in the whole amounts to £276. A contributes a certain sum, B twice as much as A and £12 more, and C three times as much as B and £12 more. Required their several contributions. 4. A man bought 7 oxen and 11 cows for $591. For the

gave $15 apiece more than for the cows. How much did he give apiece for each?

Let x denote the price of a cow.
Then the price of an ox will be x + 15.
11 cows at x dollars apiece will come to 11 x dollars.

If one ox cost x + 15 dollars, 7 oxen will cost 7 times x + 15, which is 70 + 105.

The price of the oxen and of the cows added together will make $591, the whole price.

11 x + 7-8 + 105 = 591 Uniting r's,

18 x + 105 = 591 Subtracting 105 from both members,

18 x = 486 Dividing by 18,

x = 27 = price of cows.

x + 15 = 42 = price of oxen. 5. A man bought 20 pears and 7 oranges for 95 cents. For the oranges he

gave 2 cents apiece more than for the pears. What did he give apiece for each ?

6. A man bought 20 oranges and 25 lemons for $1.95. For the oranges he gave 3 cents apiece more than for the lemons. What did he give apiece for each?

7. Two persons engage at play, A has 76 guincas, and B 52, before they begin.

After a certain number of games lost and won between them, A rises with three times as many guineas as B. How many guineas did A win of B?

Let x denote the number of guineas that A won of B.
Then A, having gained x guineas, will have 76 + x
B, having lost x guineas, will have only 52

A has now three times as many as B, that is, 3 times 52— X, which is 156 — 3x. It is evident that both 52 and x must be multiplied by 3, because 52 is a number too large by x, therefore 3 times 52 will be too large by 3r.

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76 + x= 156-3x
x = 156

76
x + 3x = 156 — 76

4 x = 156 76
4 x = 80
20

Ans. 20 guineas Proof. If A won 20 guineas of B, A will have 96 and B 32 3 times 32 are 96.

This equation is rather more difficult to solve than any of the preceding. In the first place I subtract 76 from both members, so as to remove it from the first member. Then to get 3x out of the second member, which is there subtracted, I add 3 x to both members; then the x's are all in the first member, and the known numbers in the other.

N. B. Any term which has the sign +, either expressed or understood, may be removed from one member to the other by giving it the sign —; for this is the same as subtracting it from both sides. Thus x+3=10; x is not so much as 10 by 3, we therefore say x = 10 — 3. Again, 5 x = 18 + 3x. Now 5 x is more than 18 by 3x, therefore we may say 5 x — 3x = 19.

Any term which has the sign — before it may be removed from one member to the other by giving it the sign +. This is equivalent to adding the number to both sides. Thus 5x

3= 17. In this it appears that 5 x is more than 17 by 3; therefore we say 5x=17+ 3. Again, 5 x = 32 — 3 x. Here it appears that 5 x is not so much as 32 by 3x; therefore we say 5x + 3x= 32. This is called transposition.

Hence it appears that any term may be transposed from one member to the other, eare being taken to change the sign.

In the last example, 76 was transposed from the first member to the second, and the sign changed from + to —; and 3 r was transposed from the second member to the first, and the sign changed from — to +. This has been done in many of the preceding examples.

When a number, consisting of two or more terms, is to be multiplied, all the terms must be multiplied, and their signs preserved. In the last example, 52 - I, multiplied by 3, gave a product 156

8. A person bought two casks of wine, one of which held exactly three times as much as the other. From each he drew

3 x.

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