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he had drawn 18 gals. from each, he mixed the remainder together and added 37 gals. of water, and the mixture was worth 8s. per gal. How many gallons of each sort were there at first?

Equations, Generalization.

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 remained in the result. But algebra has a more importanti 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 x = the less part.
The equation will be x + 3x = 500

4 x = 500

x = 125

3*

= 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

4x = a

a

4

3a 3 x =

a

a

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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.

3a Then = 125; and

=375. 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 = 5649.

4

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

23 = No. of oranges.

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6 a

6x =

“ of apples.

23 Suppose a = 184 cents, then as of 184 = 8= the number of oranges; 2 8 = 16 = the number of pears; and 6 x 8 = 48 = the number of apples. This is easily proved. 8 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

ways come out a whole number when it is wished; for if the numbers were taken at random, there would frequently be fractions in the result. The method is to solve it first with a letter, as has been done in the two preceding examples. If any number, which is divisible by 4, be put in the place of a, in the first example, the answer will be in whole numbers. And if any number, which is divisible by 23, be put in the place of a, in the second example, the answer will be in whole numbers.

Let the learner now generalize the examples in Art. I., by substituting a letter instead of the number; and after the result is obtained, put in the numbers again, and see if the answers agree. Let him also try other numbers.

The examples in Art. II. may be generalized in the same

manner.

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

his age.

Instead of 88 put a, and let x = the number required.

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11

Any number that is divisible by 11, being put in the place of a, will give an answer in whole numbers. Let a=88, then i of it is 48, agreeing with the answer in Art. II.

In the course of the solution it appears, that a is equal to of x ; and the result shows, that x is equal to i of a. That is, the value of x is found by multiplying a by the fraction 4 inverted.

4. In an orchard of fruit-trees, f of them bear apples, & of them cherries, and the remainder, which is a, bear peaches. How

many trees are there in the orchard ? ,

5*

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43

33 12

Then

+

ta

4 12 x

+

ta
12 12
5%

12
5x 12 a

12 a

5 Any number that is divisible by 5, may be put in the place of a. If a = 15, the answer is 36.

36 36 Proof.

+ + 15 = 36.

3 4 5. The 8th example of Art. II. is solved as follows :

Instead of 100 put a, and let x = the whole number of geese. Then

+21= a

2 Multiplying by 2, 5* +5= 2 a By transposition,

5x = 20-5

2a-5

5
2 a 5
X =

1

5 5 Let

a = 100.

2 X 100 - 5 195 Then

= 39; 5

5 2 x 100

1=40

-1= 39.

5 Let a= 135, and find the answer in the same way.

The answer will be 53. Proof.

53 +53 +261 +21= 135. The learner may now generalize the examples in Art. II. The preceding examples admit of being generalized still

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more, but the process would be too difficult for the learner at present. The following question admits it more easily.

6. (Art. III. Exam. 1.) Two men, A and B, hired a pasture for $55, and A was to pay $13 more than B. How much did each pay?

This question is, to divide the number 55 into two such parts, that one may exceed the other by 13.

* Let us represent 55 by a, and 13 by b. The question now is to divide the number a, into two such parts, that one may exceed the other by the number b: a and b being any two numbers, of which a is the larger. Let

x = the less part. Then x +b = the greater part. And

x + x +b=a

2 +b=a By transposition,

2 x = a

b
6

6 Dividing by 2,

2

2 When a number, consisting of two or more parts, as a--b, is to be divided, it is evident that all the terms must be divided, 6

b But the fractions in and having a common de2 2

2 nominator, one numerator may be subtracted from the other. 6

b Hence is the same as

This is easily seen in 2 2

2 numbers. See below, where 55 and 13 are substituted for a and b.

Hence it appears, that the less part is found by subtracting half of the excess of the greater above the less from half

the number to be divided; or by taking half the difference between the number to be divided and the excess. The greater part is equal to : +6; hence if 6 be added to b

it will give the greater part : 2 2

a

a as

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a

a

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* Whenever the learner finds any difficulty in comprehending the operations in the general solutions, let him first solve the questions with the numbers.

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