and substituting this value and its square, in the place of y and ya in the first equation, we obtain a result involving only x.

186. If both of the proposed equations involved the second power of each of the two unknown quantities, the above method could be applied in resolving only one of the equations, either with respect to x or y.

Let there be, for example, the equations

Substituting this value of y, and its square in the first, we obtain

x2) = m2.

a x2 + bx Vn2 x2 + c (n2 Our purpose appears to be answered, since we have arrived at a result, which does not involve the unknown quantity y, but we are unable to resolve the equation containing x, without reducing it to a rational form, by making the radical sign, under which the unknown quantity is found, to disappear.

It will be readily seen, that if this radical expression stood alone in one member, we might make the radical sign to disappear by raising this member to a square. Collecting together all the rational terms then in one member, by transposing the terms bx √n2 x2 and m2, we have

a x2 + c (n2 — x2) — m2 = b x √ n = x2; taking the square of each member, we form the equation a2x2+c2 (n2x2)2+m2 +2acx2(n2 —x2)—2am3 x2-2cm2 (n2 —x2

which contains no radical expression.

=b2x2 (n2 —x3),

The method, we have just employed for making the radical sign to disappear, deserves attention, on account of the frequent occasion we have to apply it; it consists in insulating the quantity found under the radical sign, and then raising the two members of the proposed equation to the power denoted by the degree of this sign.

187. The complicated nature of this process, which increases in proportion to the number of radical expressions, added to the difficulty of resolving one of the proposed equations with reference to one of the unknown quantities, a difficulty, which is often insurmountable in the present state of algebra, has led those, who have cultivated this science, to seek a method of effecting the

elimination without this; so that the resolution of the equations shall be the last of the operations required for the solution of the problem.

In order to render the operation more simple, we reduce equa. tions with two unknown quantities to the form of equations with only one, by presenting only that, which we wish to eliminate. If we have, for example,

x2 + ax y + bx = cy2 + dy + e,

we bring all the terms into one member, and arrange them with reference to ; the equation then becomes

x2 + (ay + b) x

abridging this, by making

x2 + P x + Q = 0.

The general equation of the degree m with two unknown quantities must contain all the powers of x and y, which do not exceed this degree, as well as those products, in which the sum of the exponents of x and y does not exceed m; this equation then may be represented thus;

xm+(a+by)xm−1+(c+dy+ey2).xm−2+(ƒ+gy+hy2+ky3)xm−3 +uym-1)x+p+q'y+r'y3...

+(p+qy+ry2

...

No coefficient is assigned to am in this equation, because we may always, by division, free any term of an equation we please, from the number, by which it is multiplied. Now if we make a+by=P, c+dy+ey3 = Q, ƒ+gy+hy2+ky3 =R,

=

p+qy....+u ym-1 T, p' + q' y .... + v' ym = U, the above equation takes the following form,

x2 + Pxm-1 + Q xm−2 + R xm3.....+ Tx + U = 0. 188. It should be observed, that we may immediately eliminate x in the two equations of the second degree,

x2 + P x + Q = 0, x2 + P' x + Q' = 0,

by subtracting the second from the first. This operation gives (P — P') x + Q — Q′ = 0,

whence

-

substituting this value in one of the two proposed equations, the first for example, we find

(Q-Q')2
(PP)2

P(Q-Q)
P-P

+ Q = 0;

making the denominators to disappear, we have

(Q-Q)3P (P — P') (Q — Q) + Q (P — P')2 = 0, then developing the two last terms, and making the reduction (Q — Q')2 + (P — P') (P Q' — Q P') = 0.

We have then only to substitute for P, Q, P', and Q', the particular values, which answer to the case under consideration.

189. Before proceeding further, I shall show, how we may determine, whether the value of any one of the unknown quantities satisfies at the same time the two equations proposed. In order to make this more clear, I shall take a particular example; the reasoning employed will, however, be of a general nature. Let there be the equations

which we shall suppose furnished by a question, that gives y = 3. In order to verify this supposition, we must substitute 3 in the place of y, in the proposed equation; we have then

equations, which must present the same value of x, if that, which has been assigned to y, be correct. If the value of x be represented by a, the equation (a) and the equation (b) will, according to what has been proved in art. 179, both of them be divisible by x a; they must, therefore, have a common divisor, of which x a forms a part; and in fact, we find for this common divisor x 2 (48); we have therefore a = 2. Thus the value y= 3 fulfils the conditions of the question, and corresponds to

x = 2.

If there remained any doubt, whether or not the common divisor of the equations (a) and (b) must give the value of x, we might remove it, by observing, that these equations reduce themselves to

(x2+11x+49) (x — 2) = 0,

(x+14) (x-2) = 0,

from which it is evident, that they are verified by putting 2 in the place of x.

190. The method I have just explained, for finding the value of x, when that of y is known, may be employed immediately in the elimination of x.

Indeed, if we take the equations (1) and (2), and go through the process necessary for determining whether they have a common divisor involving x, instead of finding one, we arrive at a remainder, which contains only the unknown quantity y and numbers, that are given; and it is evident, that if we put in the place of y its value 3, this remainder will vanish, since by the same substitution, the equations (1) and (2), become the equations (a) and (b), which have a common divisor. Forming an equation, therefore, by taking this remainder and zero for the two members, we express the condition, which the values of y must fulfil, in order that the two given equations may admit, at the same time, of the same value for x.

The adjoining table presents the several steps of the operation relative to the equations,

on which we have been employed in the preceding article. We find for the last divisor,

and the remainder, being taken equal to zero, gives 43y+345 y 1960 y3 +750 y2

2940 y

4302 = 0,

an equation, which admits, besides the value y = 3 given above, of all the other values of y, of which the question proposed is susceptible.

The remainder above mentioned being destroyed, that preceding the last becomes the common divisor of the equations proposed; and being put into an equation, gives the value of x when that of y is introduced. Knowing, for example, that y = 3, we substitute this value in the quantity

(9 y2 + 10) x -2y3 10 y 98; then taking the result for one member, and zero for the other, we have the equation of the first degree

191. The operation to which the above equations have been subjected, furnishes occasion for several important remarks. First, it may happen that the value of y reduces the remainder preceding the last to nothing; in this case, the next higher remainder, or that which involves the second power of x, becomes the common divisor of the two proposed equations. Introducing then into this the value of y, and putting it equal to zero,

(38y+50y+98)(9y+10)x-162y-1170y-2000y3-1000

-(38y3+50y+98) (9y2+10)x+ 76y+ 480y+3920y3+ 500y2+5880y+9604

86y690y+3920y3-1500y2+5880y+8604

Putting this remainder equal to zero, then dividing all its terms by 2, and changing the signs in order to make the first term positive, we have

43y+345y*—1960y3+750y3—2940y-43020,