29. x+y+= l +m+n, lx + my + nz = mn + nl + lm, (m − n) x + (n − 1) y + (1 − m) z = 0. 30. lx+ny+mz = nx + my + lz= mx + ly + nz 31. l*x + m2y + n2 = lmx + mny + nlz = nlx + lmy + mnx 147. We now proceed to consider simultaneous equations, one at least of which is of the second or of higher degree. We first take the case of two equations containing two unknown quantities, one of the equations being of the first degree and the other of the second. Substitute this value of x in the second equation; we then have From the above example it will be seen that to solve two equations of which one is of the first degree, and the other of the second degree, we proceed as follows: From the equation of the first degree find the value of one of the unknown quantities in terms of the other unknown quantity and the known quantities, and substitute this value in the equation of the second degree; one of the unknown quantities is thus eliminated, and a quadratic equation is obtained the roots of which are the values of the unknown quantity which is retained. The most general forms of two equations such as we are now considering are lx + my + n = 0, ax2 + bxy + cy2 + dx+ey +ƒ = 0. From the first equation we have Hence on substitution in the second equation we have to determine y from the quadratic equation a (my + n)2 - lby (my + n) + cl3y2 − dl (my + n) + el3y + fl2 = 0. Having found the two values of y, the corresponding values of x are found by substitution in the first equation. 148. It should be remarked that we cannot solve any two equations which are both of the second degree; for the elimination of one of the unknown quantities will in general lead to an equation of the fourth degree, from which the remaining unknown quantity would have to be found; and we cannot solve an equation of higher degree than the second, except in very special cases. For example, to solve the equations ax2+ bx+c=y, x2 + y2 = d. Substitute ax + bx + c for y in the second equation, and we have x2+(ax + bx + c)2=d, which is an equation of the fourth degree which cannot be solved by any methods given in the previous chapter. 149. There is one important class of equations with two unknown quantities which can always be solved, namely equations in which all the terms which contain the unknown quantities are of the second degree. The most general forms of two such equations are and ax2 + bxy + cy2 = d a'x2 + b'xy + c'y2 = d'. Multiply the first equation by d', and the second by d and subtract; we then have (ad' — a'd) x2 + (bd′ – b'd) xy + (cd′ — c'd) y2 = 0. The factors of the above equation can be found either by inspection, or as in Art. 81; we therefore have two equations of the form la+my=0 either of which combined with the first of the given equations will give, as in Art. 147, two pairs of values of x and y. 150. The following examples will shew how to deal with some other cases of simultaneous equations with two unknown quantities; but no general rules can be given. Square the members of the first equation, and add four times the second; then (x+y)2=64. |