he had drawn 18 gals. from each, he mixed the remainder together and added 33 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 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 the less part.

The equation will be

x+3x=500
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 + 3 x = a

3x=

The work is now preserved in the result, and it appears that one part will be 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.

α

3 a

Then = 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;

α

= 1883 3; and =5649.

4

3 a
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

Then

And

the number of oranges.
the number of pears.

6 x = the number of apples.

By the conditions, 12x+6x + 5 x = a

23 x = a

=

Suppose a 184 cents, then 2 of 1848 = the number of oranges; 2 × 8 = 16 the number of pears; and 6 × 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;

404896 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. Andif 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.

88, then

Any number that is divisible by 11, being put in the place of a, will give an answer in whole numbers. Let a of it is 48, agreeing with the answer in Art. II. In the course of the solution it appears, that a is of x; and the result shows, that x is equal to TT is, the value of x is found by multiplying a by the fraction inverted.

equal to of a.

That

4. In an orchard of fruit-trees, of them bear apples, of them cherries, and the remainder, which is a, bear peaches. How many trees are there in the orchard?

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

of a.

Proof.

=

5. The 8th example of Art. II. is solved as follows: Instead of 100 put a, and let the whole number of

Leta 135, and find the answer in the same way.
The answer will be 53.

Proof.

5353+261 +24=135.

The learner may now generalize the examples in Art. II. The preceding examples admit of being generalized still