Ans. 3 ad-5dx+2a2-ax-2 ac+cx=ad+

Ans. 24 x2-18x2-36 x 20 x 40.

4. Free the equation

=

Ans. x2+2xy+y2—x2+2xy—y2=x+y−x+

To free a Fraction from negative Exponents.

115. Corollary. If the given equation contains negative exponents, it can be freed from them by arts. 80 and 82.

from fractions and negative exponents.

Ans. x4 a a2x a2x a4x x2 a
=

-

117. Theorem. A term may be transposed from one member of an equation to the other member, by merely reversing its sign; that is, it may be suppressed in one member and annexed to the other member with its sign reversed from to, or from to +.

Proof. For suppressing it in the member in which it at first occurs is the same as subtracting it from that member; and annexing it to the other member with its sign reversed is, by art. 26, subtracting it from the other member; and, therefore, by art. 109, the equality is preserved.

118. Corollary. All the terms of an equation may be transposed to either member, leaving zero in the other member; and the polynomial thus formed may be reduced to its simplest form, by arts. 20 and 110.

to its simplest form in a series of monomials.

Solution. This equation, freed from fractions by arts. 112 and 113, is

7x+1+7x=6x+2—5xn+1—xn—3x2-6x+2,

which becomes, by the transposition of its terms and by the reduction of art. 20,

12x+1+11 x = 0,

and, by striking out the factor x”,

=

x2 - 1 (x+1)2 X- - 1

to its simplest form in a series of monomials.

Equations of the First Degree.

SECTION III.

Solution of Equations of the First Degree, with one unknown quantity.

120. Theorem.

Every equation of the first de

gree, with one unknown quantity, can be reduced to the form

Ꮖ

+B=0;

in which A and B denote any known quantities, whether positive or negative, and x is the unknown quantity.

Proof. When an equation of the first degree with one unknown quantity is reduced, as in art. 118, its first member is composed of two classes of terms, one of which contains the unknown quantity, and the other does not. If the unknown quantity, which we may suppose to be x, is taken out as a factor from the terms in which it is contained, and its multiplier represented by A, the aggregate of the first class of terms is represented by Ax; and the aggregate of the terms of the second class may be represented by B; whence the equation is represented by

121. Problem.

A x + B = 0.

To solve an equation of the first

degree with one unknown quantity.

Solution. Having reduced the given equation to the form

Ax+B=0,

transpose B to the second member by art. 117, and we have

Dividing both members of this equation by A, gives

B

Cases in Equations in the First Degree.

Hence, to solve an equation of the first degree, reduce it, as in art. 120, and transpose its known terms to the second member, and all its unknown terms to the first member; and the value of the unknown quantity is equal to the quotient arising from the division of the second member by the multiplier of the unknown quantity in the first member.

122. Corollary. When A and B are both positive or both negative, the value of x is, by art. 35, negative; but when A and B are unlike in their signs, one positive and the other negative, x is positive.

123. Corollary. When we have

But the smaller a divisor is, the oftener must it be contained in the dividend, that is, the larger must the quotient be; and when the divisor is zero, it must be contained an infinite number of times in the dividend, or the quotient must be infinite. Infinity is represented by the sign ∞. We have, then, in this case,

The given equation is, however, in this case,

0x x+B=0,