degree with respect to r, and setting this equal to zero; each value of y gives, by means of this divisor, the corresponding value of x, which, substituted with it in the given equations, will satisfy them. Should this divisor reduce to zero by the substitution of the value of y, we must go back to the previous one of the second degree, which, put equal to zero, will furnish two values of x for each of y; if this reduce to 0, we must go to that of the 3° degree, and so on. 316. This conclusion may be arrived at in another manner. Denoting by A=0, for simplicity, the first of the two given equations F(x, y)=0, and by B=0 the second f(x, y)=0, by Q the quotient of A by B, and by R the remainder, we have

It follows from this equality that all the values of the unknown quantities r and y, which give A=0 and B=0, must also give R=0, since the quotient Q can not become infinite for finite values of x and y, the given equations being supposed to be entire functions, or capable of being rendered such with (See Art. 275, Cor. 2.)

respect to x and y.

For the same reason, all the values which will give B=0 and R=0, will also give A=0. The system of equations A=0, B=0 may, therefore, be replaced by the more simple system B=0, R=0.

If now B be divided by R, and a new remainder, R', be reached, it may be shown in a similar manner that the system B=0, R=0 can be replaced by the system R=0, R'=0, R' being of a lower degree with respect to r than R, and so on, till we arrive at a remainder independent of x. Let R" be this remainder. Then the original equations are replaced by the system R'=0, R"=0, in which R"=0 is the final equation in y only, and R' generally of the 1° degree with respect to x.

been sup

317. The same conclusion could not have been arrived at had y posed to enter into any of the denominators in the above process. Suppose, for instance, that Q in equation (2) contained denominators functions of y, then Q might possibly become infinite by the values of y reducing these denominators to zero, and BQ thus might be finite (see Art. 156, 3°), though B

were zero.

318. If, in order to prevent the occurrence of y in the denominator of the quotient when affecting the division of A by B, it had been necessary to multiply the polynomial A by some function of y, foreign roots might thus be introduced, not belonging to the proposed equation. For, call c this function, and represent by Q still the quotient obtained after this preparation, and by R the remainder, we shall have

CA=BQ+R.

This equality proves that the solutions of the equations B=0, R=0 are the same as those of the equations cA=0, B=0. But this last system divides itself into two others, A=0, B=0, and c=0, B=0, consequently the equations B=0, R=0 will admit all the solutions of the proposed equations; but they will admit, also, all those of the equations c=0, B=0, which can not belong to the equation A=0. The same may be shown for any foreign factor necessary to be introduced to effect any subsequent division.

On the other hand, factors are sometimes suppressed for convenience in the process for finding the common divisor. If these factors were such as would reduce to zero on attributing to y its proper values, the process ought to ter

minate, since the whole remainder becomes zero with one of its factors, and the preceding divisor would be a common measure of the two polynomials; and yet these values of y which produce this common measure would not have been presented by the final equation arrived at had the factor in question been suppressed without notice.

From the foregoing considerations we see that, to obtain the values of y which belong to the proposed equations, we must equate to zero the remainder which is independent of x, as also each of the factors in y which have been suppressed in the course of the operation, and resolve each equation separately; secondly, that among the values thus obtained there may be some which, on trial in the proposed equations, prove extraneous, and which must, therefore, be rejected.

319. Simplifications may sometimes be employed, the nature of which is explained conveniently by the aid of symbols, as follows: Let the polynomials A and B, the first members of the given equations, be put under the form

A=dd'd'uu'u", B=dd'd"vv'v",

in which drepresents a common divisor of A and B, containing r only; d' another, containing y only; and d" a third, containing both x and y. The other factors, u, u', u'', v, v', v'', have a similar meaning, except that they are not common to the two polynomials A and B. The proposed equations may be satisfied by placing d=0; this equation contains only r, and, when resolved, furnishes a limited number of values of this unknown quantity, to which may be joined any value whatever of y, and the given equations A=0 and B=0 will be satisfied. Again, d'=0 will satisfy them, which gives similarly limited values for y, unlimited for x. Finally, suppose d'=0; as d" contains both x and y, an arbitrary value may be given to one of the unknown quantities, and this equation will make known a corresponding one for the other.

The other modes of satisfying the given equations consist in equating to zero simultaneously one of the factors u, u', u' of the first, and one v, v', or v', of the other. But v and u can not be simultaneously equal to zero, since they each contain only x, and are supposed to have no common divisor, d having been understood to comprise all the common factors depending on x alone. For a similar reason, u' and ' functions of y alone can not at the same time be equal to zero. But u" and v", being put equal to zero, are to be proceeded with by the method of the common divisor, as already explained, and will furnish a limited number of values for y, and corresponding values limited also for x.

320. Should the remainder, in seeking for a common divisor, not contain y, but only known quantities, it could not be put equal to zero. In this case the given equations would be incompatible.

(1) Let the equations be

EXAMPLES.

(-2x+2)y3+(x1—2x3—2x2+2x+1)y2+(xo—2x2+x)y=0,
(-x+1)+(x2+x)y1+ (x3 —x2)y3 + (x^—x3)y2=0.

There are numerous simplifications of these, for they can be decomposed into factors like the following:

y(x-1)(x+y)× (x+1)(x2-2y-1)=0,
y(x-1)(x+y) y(x2—y2)=0.

Equating to zero first the common factors, each in its turn, we obtain

next equating to zero the other factors, we have four systems of equations, viz.,

In the first three systems, all the solutions, except x=-1, y=-1, have already been found; in the fourth, those in which we have x=-y are also already known; hence, in reality, we have only determined three new solutions, viz.,

(2) To resolve the two equations

x3 —3yx2 + (3y3—3y+1)x−y3+y2—2y=0,

These equations can not be decomposed into factors; hence we pass immediately to successive divisions. This remark will apply also to equations 3 and 4. First Division.

x3-3x2+(3y—y+1)x-y3+ y2-2y|x2-2yx+ y2 —y

Hence, the final equations are r—2y=0, y2—y=0. We deduce from these

and as we have neither introduced nor suppressed any factor, these two solutions are those of the proposed equations themselves.

(3) To resolve the two equations,

(y-1)x+2x-5y+3=0,
yx+9x-10y=0.

First Division.

(y-1) x2+2x −5y +3 |yx2+9x-10y
(y-1)yx2+2yx-5y+3yy-1
+(y-1)y.x2 — (—9y+9)x−10y2+10y

(−7y+9)x+ 5y2 — 7y.

As we have multiplied by y, it is necessary to resolve the equations y=0, yx2+9x-10y=0, which give x=0, y=0, and to examine whether these values make the dividend equal to zero. As this is not the case, it follows that they form a foreign solution, which it will be necessary to suppress.

Second Division.

yx2+9x-10y|(—7y+9)x+5y2 —7y

(−7y+9)yx2+(−63y+81)x+70y2—90y|yx+(—5y3+7y3—63y+81)

(-7y+9)yx-(— 5y3+7y)3x

(—5y3+7y2—63y+81)x+70y? —90y

(-5y+7y2-63y+81)(-7y+9)x-490y+1260-810y

(—5y3+7y2—63y+81)(−7y+9)x — 25y3 +70y1—364y3+846y2—567y

25y-70y-126y+414y-243y.

The equations which it is necessary to resolve are

(−7y+9)x+5y3—7y=0,

25y5-70-126y+414y2-243y=0.

The second gives the results, which may be readily verified,

By substituting these values in the first of the given equations, we obtain for r the corresponding values x=0, x=1, x=2, x=—5

√10.

In the second division we have been compelled to multiply by -7y+9, but no foreign solution has been introduced.

We have, then, only to suppress, in the five solutions above, that which has been introduced by the first division. There remain, then, for the given equations the four following solutions:

x2+(8y—13)x+ y2—7y+12\x2—(4y+1)x+y2+5y

+x2−(4y+ 1)x+y3+5y

(12y-12)x-12y+12

This remainder can be decomposed into the factors 12(y-1)(x—1); the calculations will be simplified, and we shall have these two systems of equa

x2-(4y+1)x+y2+5y=0, x2 - (4y+1)x+y2+5y=0.

Each of these can be at once resolved, and we find

Sy=1 Sy=1 Sy=0 Sy=-1
=3, | x=2, | x=1, \ x=1.

(5) x+2yx+2y(y-2)x+y2-4=0.

Sy=3 Ans.4,x=-5.

(6) 13-3yx+3x+3y2x-6yx-x-y3+3y+y-3=0, x+3yx3-3x2+3y2x-6yx-x+y3-3y2-y+3=0.

(7) x3+yx2-(y2+1)x+y—y3=0,

x3-yx2-(y2+by+9)x+y3+6y2+9y=0.

The first division gives the remainder

yx2+(3y+4)x-(y+3y2+4y).

To be able to perform the second division, we multiply the dividend by y; in the same way we prepare the first remainder to be divided. We thus arrive at a remainder of the first degree in r, which can be put under the form 32(y+3y+2)(x-y).

Dividing, then, the remainder of the second degree by x-y, we obtain the quotient

and there is no remainder.

yx+y2+3y+4=0,

From these calculations we conclude that the first members of the proposed equations are divisible by x-y, so that they can be verified by all the solutions of the indeterminate equation x-y=0. The other solutions are furnished by the system of two equations,

y2+3y+2=0, yx+y2+3y+4=0;

hence we obtain the solutions

y=-1, x=2; y=−2,x=+1.

METHOD OF LABATIE.

321. Having thus stated the principles on which the ordinary method of elimination depends, we shall now proceed to show how this method has lately been perfected by Labatie and Sarrus. By the aid of the theory which they have introduced, we shall be able to perform the required eliminations without introducing any foreign solutions.

Suppose that A and B represent the quotients which we obtain by dividing the first members of the given equations by all of their factors which depend only on y.

Let c be the factor by which it is necessary to multiply A, in order that we may be able to divide it by B; represent by q the quotient that we obtain in this division, and by Rr the remainder, r designating those factors of this remainder that are not dependent on x. Let c be the factor by which we must multiply B to render it divisible by R; represent by q, the quotient, and by Rr the quotient that we obtain in this second division, r, designating the product of those factors of this remainder which do not depend on r, and so on. Finally, suppose, for the sake of simplicity, that at the fourth division we obtain a remainder independent of x, and designate this remainder by r. We have the equalities