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+axy + bx cy2 + dy + e, = we bring all the terms into one member, and arrange them with reference to x; the equation then becomes x2 + (a y + b) x — cy2 — dy abridging this, by making we have ―dy—e=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; x2+(a+by)xm−1+(c+dy+ey2)xm−2+(f+gy+hy2+ky3)xm−3 +(p+qy+ry3 ... +uym-1)x+p'+q'y+r'y2. ... +v'ym=0. 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+ey2=Q, ƒ+gy+hy2 + ky3 = R, -1 .... +vym = U, p + qy.... + u ym1 = T, p'+q'y the above equation takes the following form, xm+P xm−1 + Q xm−2 + R xm−3..... + Tx+U= 0. 188. It should be observed, that we may immediately eliminate x in the two equations of the second degree, x2 + Px + Q = 0, x2 + P' x + Q' = 0, by subtracting the second from the first. This operation gives substituting this value in one of the two proposed equations, the first for example, we find (Q—Q')2 P(Q-Q) + Q = 0; making the denominators to disappear, we have = 0. (Q — Q')2 — P (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') We have then only to substitute for P, Q, P', and Q', the particular values, which answer to the case under consideration. 139. 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 reprcsented 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 divisor x for this common 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 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 climination 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 wil 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, (9 y2 + 10) x — 2 y3 — 10 y 98; and the remainder, being taken equal to zero, gives 43 y345y4— 1960 y3750 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 then taking the result for one member, and zero for the other, we have the equation of the first degree 91 x 1820, or x = 2. 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, x3+3x3y+3yx-98 |x2+4xy-2y2-10 or rather 2d rem. 1st rem.....+(9y2+10)x 2y3-10y-98 x2+ 4xy- 2y3-10|(9y+10)x-2y3-10y-98 or rather (9y2+10)x2+36xy3—18y+-110y3-100|(9y3+10)x-2y 3—10y—98 +40xу (38y3+50y+98)(9y+10)x-162y-1170y-2000y3-1000 (38y3+50y+98)(9y2+10)x+ 76y+ 480y+3920y3+ 500y2+5880y+9604 86y690y+3920y3-1500y+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+750y2-2940y-4302=0, we have an equation of the second degree, involving only x, the two values of which will correspond to the known value of y. If this value still reduce to nothing the remainder of the second degree, we must go back to the preceding, or that into which the third power of x enters, because this, in the case under consideration, becomes the common divisor of the two proposed equations; and the value of y will correspond to the three values of x. In general, we must go back until we arrive at a remainder, which is not destroyed by substituting the value of y. It may sometimes happen, that there is no remainder, or that the remainder contains only known quantities. In the first case, the two equations have a common divisor independently of any determination of y; they assume then the following form, PX D= 0, Q X D = 0, D being the common divisor. It is evident, that we satisfy both the equations at the same time, by making in the first place D= 0; and this equation will enable us to determine one of the unknown quantities by means of the other, when the factor D contains both; but if it contains only given quantities and x, this unknown quantity will be determinate, and the other will remain wholly indeterminate. With respect to the factors, which do not contain x, they are found by what is laid down in art. 50. Next, if we make at the same time P=0, Q = 0, we have still two equations, which will furnish solutions of the question proposed. Let there be, for example, (ax + by c) (mx + ny — d) = 0, (a' x + b'y- · c') (m x + ny — d) = 0; by supposing, first, the second factor, common to the two equations, to be nothing, we have with respect to the unknown quantities x and y only the equation and in this view the question will be indeterminate; but if we suppress this factor, we are furnished with the equations and in this case the question will be determinate, since we have as many equations as unknown quantities. |
