By this substitution is obtained the equation distinguished by the mark C (Art. 236); if it is required to make the second term disappear, it is necessary to determine a by the equation ma+P=0, or a=——

Р

m

Observing that, in this expression, P is the coefficient of the second term of equation A, and m is the exponent of the highest power of x, it follows that the second term of an equation is made to vanish by substituting, for the unknown quantity of that equation, a new unknown quantity, to which is added the coefficient of the second term of the equation, taken with a contrary sign, and divided by the degree of the equation.

The third term of equation A is made to disappear by assuming

m(m−1)a2+(m−1)Pa+Q=0.

From this equation two values of a can be obtained; consequently there are two different quantities by the substitution of which an equation can be transformed into another which shall be deprived of its third term.

To make the fourth term vanish, it is necessary to resolve an equation of the third degree, and so on to the last term, which (to make it disappear) requires the determination of a from an equation like that proposed; namely,

a+Pam-1+Qam-2

....

+Ta+U=0.

The reason is this: to equal to zero the last term of the equation in y is to suppose that one of the values of y is zero; for when this equation has no last term, it is evidently verified by making y=0; consequently the relation xy+a becomes x=a; therefore for a may be taken any of the values of x, therefore a ought to be determined by the same equation as x.

When one term of an equation is made to vanish it may happen that other terms may disappear at the same time; but that this may happen it is necessary that there be certain relations between the coefficients of the proposed equation.

Let it, for example, be required to find the relation which must subsist among the coefficients of an equation in order that the second and third terms may vanish together?

It is evidently necessary (Eq. C) that

ma+P=0 and 1m(m−1)a2+(m−1Pa)+Q=0.

P

The first equation gives a=; It is necessary that this value satisfy

the second equation, or that

Q

(m−1)P2

m

=0, which is the relation required. 2m

If the value of Q, deduced from this expression, is substituted in equation A, the general form of equation, whose second and third terms ought to disappear together, will be obtained.

It is not possible, by successive transformations, to make first one term vanish, then a second, then a third, &c., for each operation would reproduce the term destroyed by the preceding.

Taking the equation +Q-2+, &c.=0, which wants the second term, the substitution of y+a for x in this equation gives

(y+a)+Q(y+a)2+, &c. =0.

Developing this equation, the result is

y”+mayTM−1+[1m(m−1)a2+Q]yTM−2+, &c. =0.

Now it is evident that the value of a, which in this equation makes {m(m—1)a2+Q=0, must be different from 0, consequently the term in y must be reproduced by the process employed for the removal of the third term.

452

1st Example. It is required to transform the equation,

x2+px=q, or x2+px−q=0,

into another which shall want the second term, assuming x=y—

y

px=p (y-2)

2

p2

-q=

Whence the transformed equation is y2

Resolving this equation, the values of y are,

2

+

a

2d Example. It is required to take away the second term of the equation,

x3-9x+26x-34=0,

assuming x-3=y, or y+3=x?

x=(y+3)=y3+9y2+27y+27

-9x2=-9(y+3)=-9y2-54y-81

26x=26(y+3)=

-34=

+26y+78

-34
y-10=0

Whence y3-y-10=0 is the transformed equation required.

3d Example. It is required to take away the second term of the equation, x3+6x2-3x+4=0?

Ans. The transformed equation is y3-15y+26=0.

4th Example. It is required to take away the second term of the equation, x-12x+17x2-9x+7=0?

Ans. The transformed equation is ya-37y2-133y-110=0.

238. The success of certain transformations can be made evident by means of the composition of equations.

Admitting that the equation X=0 decomposed into factors is

if x is replaced by -x, and the signs of all the factors are afterwards changed, the result is

(x+a)(x+b)(x+c)... =0;

an equation, the roots of which are evidently equal to those of the proposed equation, but of contrary signs.

This is the transformation of Article 232.

If in the equation (x-a)(x-b)(x-c)..

multiplying the factors respectively by &c., and changing the sign of each factor, the resu

y
n'

If again r is made equal to and each factor is multiplied by n (which amounts to the multiplication of every term by n") the result is

(y-na)(y-nb) (y—nc)

an equation of which the roots are na, nb, nc,

This transformation is that of Article 234..

If x=x+a the equation (-a)(x—b)(x−c)

=0,

=0 becomes

(x'+a'—a)(x'+a'—b)(x'+a'—c) ... =0, which has evidently for roots a-a', b-a', c-a',. . . . that is to say, the roots of the proposed equation, each diminished by a'.

When, to make the term Pa" of the equation X=0 to vanish, x is made the roots of the transformed equation are

equal to y

P

m'

therefore the coefficient of the second term of the transformed equation is equal to zero, or the equation in y has no second term.

In these instances a very simple relation exists between the original unknown quantity x and the new unknown quantity y or '; but whatever the nature of this relation the process to be followed is always the same; the relation is first to be expressed by an equation; then the value of x is to be deduced in terms of y; then, lastly, this value is to be substituted in the proposed equation.

The equation of relation between x and y may be too complicated to be resolved with regard to x. In this case it becomes necessary to eliminate x between the equation of relation and the proposed equation X=0.

SECTION XIII.

OF ELIMINATION AND ITS APPLICATIONS.

239. Two equations of the first degree, containing two unknown quantities, being given, the elimination of one of the unknown quantities is easily accomplished; and the final equation (that is, the equation which gives the other unknown quantity) is also of the first degree. But such is not the case when the equations are of a degree superior to the first.

Let there be between the unknown quantities x, y two complete equations, one of the degree m, the other of the degree n.

These two equations can always be reduced to the form

....

+Tx+V=0

x"+P′x"¬+Q'x”—2. +Tx+V'=0

In which P, P' are respectively of the form a+by, a'+b'y;

Q, Q'
R, R'
&c.

c+dy+ey, c'+d'y +e'y2;
f+gy+hy2+ky3, f'+g'y+h'y2+k'y3 ;

&c.;

and the terms V, V' contain y to the powers m, n respectively.

To resolve equations A, B is to assign all the pairs of numbers which, being substituted for x, y, reduce A, B at the same time to the form 0=0. These numbers are termed roots or solutions of the proposed equation.

The principle of the resolution of these equations may be presented under two different points of view.

1st. Let a, ẞ be values of x, y, such that being substituted simultaneously in the equation A, B, these equations are reduced to 0=0; then it is evident, that if ẞ is first substituted for y, and in the results (which contain only x), x is replaced by a, the same result 0=0 is obtained.

But since the polynomials which form the first members of the equations A, B ought, after the substitution of ß for y, to become zero at the same time for x=a, they must have the form,

Equation A of Q(x—a),

Equation B of Q'(x—a).

Therefore the substitution of ẞ for y introduces into the equations A, B a common factor, x-a, the equalling of which to 0 gives the solution x=a corresponding to the solution y=ß.

This is a property of every root y, and every value of y which does not possess it ought to be rejected as incapable of forming part of the solutions.

2d. Let it be supposed that the equations A, B are resolved separately, with regard to r (y being taken, for the time, as a known quantity); let the roots of equation A be represented by a, a', a", &c. ; the roots of equation B by a, a', a", . . . .; and let the equations be put under the form

A of (x—a)(x—a')(x—a'')
B of (x-a)(x-a′)(x—a′′)

=0,

=0.

is made equal to any one of

If now any one of the roots a, a', a" the roots a, a', a'", . . . (as, for example, a to a,) the result is an equation in y which will give for y one or more values ß, B', B", &c., such that, being substituted in A, B, these equations will acquire a common divisor, x-a, or x-a; and the value y=3 deduced from a=a, with the corresponding value x=a=a, (which becomes a number by the substitution of 3 for y) will simultaneously reduce the equations A, B to the form 0=0.

Then, if 3, one of the values of y, is substituted in the two polynomials A, B, the coefficients P, Q, R . . . P', Q', R'. . . will be changed into numbers, which may be expressed by (P), (Q), (R) . . . (P'), (Q), (Ř') ... and the resulting equations,

x2+(P)x+(Q)x−2
x2+(P' ́)x”¬'+(Q'′)x2−2

'+(T)x+(V)=0
+(T)x+(V′)=0.

C,

D.

containing only x, will be satisfied by x=a, or x=a; that is to say, they will acquire a common divisor of the first degree x-a. Seeking, therefore, the common measure of the polynomials which form the first members of equations C, D, and making it equal to zero, the result obtained is x=a=a, a being a number.

In the same manner is found another value of x, corresponding to the second values' of y; and so on for other values.

240. When, therefore, two proposed polynomials are operated upon as if to find their common measure -a (a being a quantity which depends on y, or a function of y) and a remainder is found which involves y only, it is this remainder which, for all the values ß, B′, B′′ . . . of y, becomes zero. Making the remainder equal to zero, the last divisor becomes the greatest common measure of the proposed polynomials. Let R=0 be the equation formed by making the remainder in y equal to zero; the values y=ß, ß′, "... are obtained by resolving this equation, and the corresponding values of by substituting successively the values B, B', 6"... of y in the last divisor (or remainder which is of the first degree in x, and which may be represented by Ax-B, A, B being functions of y), and making, after each substitution, Åx-B=0 a condition from which the proposed polynomials become zero. By this means numbers a, a', a"... are obtained as values of

x, which, taken with B, B', B',... the values of y, satisfy the proposed equations.

It is to be observed that for the two proposed equations between x and y two others are substituted, the one in y only, the other in x and y.

Example 1. Let r3 + 3yx + 3yr-98=0

x2+4yx-2y-10=0

x2+4yx-2y-10) x3+3yx2+3y2x-98 (x-y

1,

2.

9y2x+10x-2y3—10y—98 or (9y2+10)x−2y3-10y-98

(9y2+10)x—2y3—10y—98) (92+10)2+(361/3+40) x 18y1-1102–100 (x

(9y2+10)x2-(2y3+10y+98)x

(38y3+50y+98)z-18y4-110y2-100 or (19y3+25y+49)x— 9y1— 55y2- 50

972+ 10

(9y2+10)x−2y3—10y-98 (92+10) (19y3+25y+49)x−81y6-585y+-100072-500 (1973+25y+49 (9y2+10) (19y3+25y+49)x−38y6-240y1-1960y3-250y2-2940y-4802

—43y6—345y1+1960y3—750y2+2940y+4302

Making this remainder equal to zero, and rendering the first term positive, the final equation is

43y+345y+-1960y3+750y2-2940y-4302=0,

and the common measure of the first degree, or the equation which gives the values of x, is

(9y2+10)x-2y3-10y-98=0.

241. The same final equation and the same common divisor may belong to an indefinite number of systems of two equations of the same degree or of different degrees. If it is proposed to form two equations which shall have x-y for common divisor of the first degree, and y-1=0 for final equation. To the product of x-y by any factor which is a function of x, or of x and y (and, for example, of the first degree in x), let y2-1 be added, the result is a polynomial of the second degree; to the product of this polynomial by a function of x or of x and y let the polynomial divisor of the first degree be added, and so on. Any two whatever of these consecutive polynomials are of different degrees; and, under the same degree, their composition may be varied at pleasure. But any two of them, being made equal to zero, will admit the same solutions, and in the same number, as a necessary consequence of the law of their formation.

242. The accuracy of the preceding method may be established by this demonstration:

Let A, B be the proposed polynomials,
R, R ... the successive remainders,
the successive quotients.

Q, Q

Then A=BQ+R
B=RQ'+R'
R=R'Q"+R"

It follows from the first equality that each of the systems of values of x, y which gives A=0, B=0, must also give R=0, and, reciprocally, that the solutions in x, y of B=0, R=0, are also solutions of A=0, B=0. There fore the question, to find the roots common to A=0, B=0, is reduced to this other; to find the roots which satisfy the system of the two equations B=0, R=0, in one of which a ory is raised to a degree less by one than in the polynomials A, B.