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10. Two men, driving their sheep to market, A says to B, give me one of your sheep and I shall have as many as you ; B says to A, give me one of your sheep and I shall have twice as many as yoy. How many had each?
Let x = the number A had,
= the number B had. If B gives À one, their numbers will be
x + 1 and y-1. If A gives B one, their numbers will be
- 1 and y + 1, ec. 11. If A gives B $5 of his money, B will have twice as inuch as A has left; but if B gives A $5 of his money, A will have three times as much as B has left. How much has each?
12. A man bought a quantity of rye and wheat for £6, the rye at 4s. and the wheat at 5s. per bushel. He afterwards sold í of his rye and i of his wheat at the same rate for £2. 17. How many bushels were there of each?
13. A man bought a cask of wine, and another of gin for $210 ; the wine at $1.50 a gallon, and the gin at $0.50 a gallon. He afterwards sold of his wine, and of his gin for $ 150, which was $15 more than it cost him. How many gallons were there in each cask ?
14. A countryman, driving a flock of geese and turkeys to market, in order to distinguish his own from any he might meet with on the road, pulled three feathers out of the tail of each turkey, and one out of the tail of each goose, and found that the number of turkeys' feathers exceeded twice those of the geese by 15. Having bought 10 geese and sold 15 turkeys by the
way, he was surprised to find that the number of geese exceeded the number of turkeys in the proportion of 7 to 3. Re quired the number of each at first. Let
x = the number of turkeys, and
3x = 2y + 15 2.
- 105 .y + 10 =
3 3. Freeing the 2d from fractions, 3y + 30= 7x— 105
Instead of the method employed above for eliminating one of the unknown quantities, we may find the value of one of them in one equation, as if the other were known; and then
this value may be substituted in the other, and an equation will be obtained, containing only one unknown quantity, which may be solved the usual way. 4. Divide the first by 3,
2y + 15
3 5. Multiply the 4th by 7,
14 y + 105
3 7. Multiply by 3, 9y+90 = 14 y + 105 - 315 8. Transposing & uniting, 300 = 5 y
y = 60. The value of x may be found by substituting 60 for y in
7 x =
120 + 15 9.
= 45. 3
Ans. 45 turkeys, and 60 geese. Let the learner go back and solve, in this manner, the preceding examples in this Art. Sometimes one method is preferable and sometimes the other.
15. A person expends $1 in apples and pears, buying his apples at 3 for a cent, and his pears at 2 cents apiece; afterwards he accommodates his neighbour with 1 of his apples and of his pears for 30 cents. How
of each did he buy?
Let x = the number of apples.
16. A market-woman bought eggs, some at the rate of 2 for a cent, and some at the rate of 3 for two cents, to the amount of 65 cents ; she afterwards sold them all for 120 cents, and thereby gained one half cent on each egg. How many of each kind did she buy?
17. It is required to find two numbers such, that if 1 of the first be added to the second, the sum will be 30, and if f of the second be added to the first, the sum will be 30.
18. It is required to find two numbers such, that of the first and of the second added together will make 12, and if the first be divided by 2 and the second be multiplied by 3, 을 of their sum will be 26.
19. Two persons, A and , talking of their ages, says A to B, 8 years ago I was three times as old as you were, and 4 years hence I shall be only twice as old as you. Required their present ages.
20. There is a certain fishing rod, consisting of two parts, the
upper of which is to the lower as 5 to 7; and 9 times the . upper part, together with 13 times the lower part, is equal to 11 times the whole rod and 8 feet over. Required the length of the two parts.
21. A vintner has two kinds of wine, one at 5s. a gallon, and the other at 12s. of which he wishes to make a mixture of 20 gallons, that shall be worth 8s. a gallon. How many gallons of each sort must he use ?
22. A vintner has 2 casks of wine, from each of which he draws 8 gallons; and finds that the number of gallons remaining in the less, is to that in the greater as 2 to 5. He then puts 1 gallon of water into the less, and 5 gallons into the greater, and then the quantities are in the proportion of 5 to 13. What quantity did each contain at first ?
23. A farmer, after selling 13 sheep and 5 cows, found that the number of sheep he had remaining, was to that of his cows in the proportion of 4 to 3. After three years he found that he had 57 more sheep, and 10 more cows than he had at first ; and that the proportions were then as 3 to 1. What number of each had he at first ?
24. When wheat was 8 shillings a bushel, and rye 5 shillings, a man wished to fill his sack with a mixture of wheat and rye, for the money he had in his purse. If he bought 15 bushels of wheat, and laid out the rest of his money in rye, he would want 3 bushels to fill his sack; but if he bought 15 bushels of rye, and then filled his sack with wheat, he would have 15 shillings left. How much of each must he purchase in order to lay out his money and fill his sacks ?
25. A grocer had 2 casks of wine, the smaller at Ts. per gallon, the larger at 10s. The whole was worth $112. When he had drawn 18 gals. from each, he mixed the remainder together and added 3; gals. of water, and the mixture was worth 8s. per gal. How many gallons of each sort were there at first ?
IX. In the examples hitherto proposed a numerical result has always been obtained. The solution with numbers has been performed at the same time with the reasoning ; and when the work was finished, no traces of the operations re mained in the result. But algebra has a more important purpose. Pure algebra never gives a numerical result, but is used to trace general principles and to form rules. In order to preserve the work so that the operations may appear in the result, it will be necessary to introduce a few more signs.
1. It is required to divide $500 between two men, so that one of them may have three times as much as the other.
Let r = the less part.
4 x = 500
X = 125
3x = 375
Ans. One part is $125, and the other $375. This question is to divide 500 into two such parts, that one part may
be three times as much as the other. It is evident that the process will be the same for any other number, as for 500.
Let the number to be divided be represented by the letter a. This will stand for any number.
Then the question will be, to divide any number, a, into two such parts, that one part may be three times as much as the other. The equation will be x + 3x = a
4 x = a
3a 3x =
The work is now preserved in the result, and it appears that one part will be 1 of the number to be divided; and the other,
of it. This is a rule that will apply to any number. Suppose a = 500 as in the example.
= 125; and - 375. 4
4 Ans. One part is $125, and the other $375; the same as above.
Suppose it is required to divide $7532 in the same proportions. Then a= 7532; = 1883; and
4 Ans. One part is $1883, and the other is $5649. 2. A man sold some apples, some pears, and some oranges for a number a of cents, the apples at two cents apiece, the pears at three cents apiece, and the oranges at five cents apiece. There were twice as many pears as oranges, and three times as many apples as pears. How many were there of each ? Let x = the number of
oranges, Then 2x = the number of
pears. And 6 x = the number of apples. By the conditions, 12 x + 6x + 5x = a
23 x = a
= No. of oranges. 23
2 a 2 x =
a X =
66 of pears.
6 x =
6 of apples. Suppose a = 184 cents, then 'g of 184 = 8= the number of oranges; 2 x 8 = 16 = the number of pears; and 6 X 8 = 48 = the number of apples. This is easily proved. oranges, at 5 cents apiece, come to 40 cents ; 16 pears, at 3 cents apiece, come to 48 cents ; and 48 apples, at 2 cents apiece, come to 96 cents;
40 + 48 + 96 = 184. The learner may be curious to know, how it is possible to make the examples in such a manner, that the answer may al