60. To reduce a fraction to its lowest terms, we divide the two terms of the fraction by their greatest common divisor.

61. Algebraic fractions being of the same nature as fractions in arithmetic, the rules for the fundamental operations are the same. We shall merely subjoin these rules, with some examples under each.

MULTIPLICATION OF ALGEBRAIC FRACTIONS.

RULE.

Multiply the numerators together for a new numera

tor, and the denominators for a new denominator.

RULE. Invert the divisor, and then proceed as in multipli

RULE.

ADDITION OF ALGEBRAIC FRACTIONS.

Reduce the fractions to a common denominator; then add the numerators together, and place their sum over the common

RULE.

Reduce the fractions to a common denominator; then

place the difference of their numerators over the denominator, and it will be the difference required.

62. The above rules are sufficient for the solution of all

equations of the first degree, however complicated.

Let it be proposed, for example, to find the value of x in the equation

Indicating the operations required in order to make the denominators disappear, we have,

(10x+bc) (13 a—b) b— (12 x — c2) a b = a (9x+3bc) (13 a—b)

performing the operations, transposing and reducing, we have

The following examples will serve as an exercise for the

身

63. Most of the questions, which we have hitherto considered, involve more than one unknown quantity. We have been able to solve them, however, by representing one of the unknown quantities only by a letter, since, by means of this it has been easy, from the conditions of the question, to express the other unknown quantity. In many questions the solution becomes more simple by representing more than one of the unknown quantities by a letter, and in complicated questions, it is frequently necessary to do this.

The question, art. 1, viz. To divide the number 56 into two such parts, that the greater may exceed the less by 12, presents itself naturally with two unknown quantities. Thus, denoting the less part by x and the greater by y, we have by the conditions of the question

Deducing the value of y from the second equation, we have y=x+12; substituting for y in the first equation its value x+12, we have x + x + 12 = 56, an equation, which contains only one unknown quantity, and from which we obtain x = 22.

A person has two horses and a saddle, which of itself is worth $10. If the saddle be put upon the first horse, his value will be twice the second; but if the saddle be put upon the second, his value will be three times the first. What is the value of each ?

Let the value of the first horse, and y that of the second, we have by the question

x+10=2y

y+10=3x

Deducing the value of y from the second of these equations, and substituting it for y in the first, we have

Substituting next for x its value 6 in the second equation, we have y+10=18

The process by which one of the unknown quantities in an equation is made to disappear, is called elimination. The method of eliminating one of the unknown quantities, pursued above, is called elimination by substitution.

64. Since the two members of an equation are equal quantities, it is evident, 1°. that we may add two equations member to member without destroying the equality. 2°. we may subtract the members of one equation from those of another without destroying the equality.