177. If the numerator be of lower dimensions than the denominator, we may still perform the division, and express the result in a form which is partly integral and partly fractional.

Here the division may be carried on to any number of terms in the quotient, and we can stop at any term we please by taking for our remainder the fraction whose numerator is the remainder last found, and whose denominator is the divisor.

Thus, if we carried on the quotient to four terms, we should have 162x9 1+ 3x2*

2x
1+3x2

=

2x-6x3+18x5 - 54x7 +

The terms in the quotient may be fractional; thus if x2 is divided by a3-a3, the first four terms of the quotient are

a3

ая

x10

24x

and the remainder is

a12 x10°

178. The following exercise contains miscellaneous examples which illustrate most of the processes connected with fractions.

21. {(2

5x28x-21 2x2-3x-9 35x2+24x-35 x2+7x2 - 8x7x2+51x-40

r+p-q

p+q-r

+

q+r-p
(p-q)(p-r) (q− r)(q− p) (r−p)(r−q)'

}}

-y (x + y _ x − y) } _ x3+x2y+xy2+y3 X- -Y x + y \x-y x+y/

+

a2 - (b-c)2 b2 − ( c − a)2 c2 - ( a − b)2
(c+a)2 - b2 (a+b)2 - c2 ' (b+c)2 - a2°

y2 1

=

2x2y+2xy2

1

x2+xy xy — y2 x+y

a2-4a+3. S a2-9

a2+a-6 a2 - 4a+4

1

За

25. (2a+2a-2) (3a ̄3α-2) ̃ ̄ 6a+(x−2a)(x – 3a)'

a2-a

\a1 + a2 + 1

a+1

29.

1

1

+

m3 +m

-

- m3

30.

(12-)(-)(--)

m

+

α

с

1

1

CHAPTER XXIII.

HARDER EQUATIONS.

179. SOME of the equations in this chapter will serve as a useful exercise for revision of the methods already explained; but we also add others presenting more difficulty, the solution of which will often be facilitated by some special artifice.

The following examples worked in full will sufficiently illustrate the most useful methods.

Note. By a simple reduction many equations can be brought to the form in which the above equation is given. When this is the case, the necessary simplification is readily completed by multiplying across or "multiplying up," as it is sometimes called.

180. When two or more fractions have the same denominator, they should be taken together and simplified.

This equation might be solved by at once clearing of fractions, but the work would be laborious. The solution will be much simplified by proceeding as follows.

The equation may be written in the form

(x-10)+2, (x − 6) + 2 _ (x − 7) +2, (x−9)+2 ̧

+

+

Hence, since the numerators are equal, the denominators must be

equal;

that is,