21. A person bought 612 gallons of beer, which exactly filled three casks, the second held 44, and the third 24 times as much as the first. What did each hold?

7. The difference between two numbers is 25 and the greater is 4 times the less; required the numbers.

Let x represent the less, then +25 will represent the greater, but since by the question the greater is 4 times the less, 4 x will also represent the greater, these two expressions for the same thing will therefore be equal to each other, and we have

x+25= 4 x

An expression for the equality of two things is called an equation. The two equal quantities, of which an equation is composed, are called members of the equation; the one on the left of the sign of equality is called the first member, and the other the second.

If a member consist of parts separated by the signs and these parts are called terms.

Thus in the equation x+25= 4 x, the expression x + 25 is the first member, and 4x the second.

The quantities r and 25 are the terms of the first member. A 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 4 x, 7 x, 4 and 7 are the coefficients of x.

Equations are distinguished into different degrees. An equation, in which the unknown quantity is neither multiplied by itself, nor by any other unknown quantity, is called an equation of the first degree.

8. In the solution of a question by the aid of algebraic signs there are, it is evident from what has been done, two distinct parts. In the first, we form an equation by means of the re

lations established by the nature of the question between the known and unknown quantities. This is called putting the question into an equation.

In the second part, from the equation, thus formed, we deduce a series of other equations, the last of which gives the value of the unknown quantity. This is called resolving or reducing the equation.

9. No general and exact rule can be given for putting a question into an equation. When however the equation of a question is formed, there are regular steps for its reduction, which we shall now explain.

In order to this we remark, that when equal operations are performed upon equal quantities the results will be equal. This is self evident. It follows, therefore, since the two members of an equation are equal quantities that, 1°. the same quantity may be added to both sides of an equation without destroying the equality; 2°. the same quantity may be subtracted from both sides of an equation without destroying the equality; 3°. both sides of an equation may be multiplied, or 4o. both sides may be divided by the same quantity without destroying the equality.

10. Let it be proposed to resolve the equation

To resolve this equation, it will be necessary to transfer the terms 25 and 4 x from the members, in which they now stand, to the opposite. In order to this, let us first subtract 25 from both members, we then have

Comparing the last equation with the proposed, the following rule, for transposing a term from one member of an equation to the other, will be readily inferred, viz. Efface the term in the member in which it stands, and write it in the other with the contrary sign.

11. Let us take next the equation +=20

To resolve this equation, we must first free it from denominators; in order to this multiplying first by 3, we have

To free an equation therefore from denominators, we multiply the equation by the denominators successively.

Ex. 1. To free from denominators the equation

Since in this equation the denominator 12 is a multiple of 3, multiplying by 12, we have

Thus by multiplying first by 12, the number of multiplications necessary to free the equation from denominators is diminished, and the equation itself, when freed from denominators, will be left in a more simple state.

Ex. 3. To free from denominators the equation

The least number divisible by each one of the denominators of the proposed, it is easy to see, is 20. Multiplying by 20, we have

thus the proposed is freed at once from denominators, and the equation which results, it is evident, is the most simple to which it can be reduced free from denominators.

From what has been done we have the following rule, to free an equation from its denominators, viz. Find the least common multiple of the denominators, multiply each term by this common multiple, observing to divide, as we proceed, the numerator of each fractional term by its denominator.

12. Let it be proposed next to resolve the equation

Freeing from denominators, we have

10 x - 32 x 31221 — 52 x

transposing and reducing, we have

30 x = 333

whence dividing both sides by 30 we obtain

x = 1110

The unknown quantity in equations of the first degree can be combined with those which are known in four different ways only, viz. by addition, subtraction, multiplication and division. From what has been done, we have therefore the following rule for the resolution of equations of the first degree with one unknown quantity, viz. 1°. Free the proposed equation from its denominators; 2°. bring all the terms, which contain the unknown quantity into the first member and all the known quantities into the other; 3o. unite in one term the terms which contain the unknown quantity, and the known quantities in another; 4o. divide both sides by the coefficient of the unknown quantity. 13. Applying the above rule to the equation

x

6

+10=3+11, we obtain x 12.

In order to verify this result we substitute 12 for x in the proposed, it then becomes

The value x = 12 satisfies therefore the proposed equation. In general, to verify the value of the unknown quantity deduced from an equation, we substitute this value for the un

known quantity in the equation. If this renders the two members identically the same, the answer is correct.

14. The following examples will serve as an exercise for the learner in the reduction of equations.

The equations above have been taken at random. It should be observed, however, that an equation may always be considered as derived from the enunciation of some question. Thus the first equation in the preceding article may be considered as derived from the following enunciation, viz. to find a number such that one half, one third and one fifth of this number may together be equal to 31.

15. Though no general and exact rule can be given for putting a problem into an equation, yet the following precept will be found very useful for this purpose, viz. Indicate by the aid of algebraic signs upon the unknown and known quantities the same reasonings and the same operations, that it would be necessary to perform, in order to verify the answer, if it were known. Let us illustrate this precept by some examples.

1. A gentleman distributing money wanted 10 shillings to be able to give 5 shillings to each person; he therefore gave