XXIII. SIMULTANEOUS EQUATIONS INVOLVING QUADRATICS. 346. We will now give some examples of simultaneous equations where one or more of the equations may be of a degree higher than the first; various artifices are employed, the proper application of which must be learned by experience. From the second equation y=20-x; substitute in the first, thus therefore therefore x2 - 2 (20-x)2=71; - x2+80x800 = 71, x2-80x-871. From this quadratic we shall obtain x=13 or 67; then from the equation y = 20−x we obtain the corresponding values of y, namely, y = 7 or -47. We have now four cases to consider; namely, (3) Suppose 2y-4xy+3x=17, y2x2=16. Let y=vx, and substitute in both equations; thus y = =vx=±5. Again, taking the second value of v we have The artifice here used may be adopted conveniently when the terms involving the unknown quantities in each equation constitute an expression which is homogeneous and of the second degree; see Art. 24. α therefore x + y + 2x2y2 = (a2 — 2xy)3 = a* — 4a3xy + 4x3y3 ; therefore x+y=a* - 4a3xy + 2x3y3. By substituting the values of x*+y* and x2 + y2 we obtain We may obtain this result also in another way. shewn that From this quadratic we can find two values of xy; let c denote one of these values, then we have thus that is, therefore xy = c; x + y = α, (x − y)2 = a3 — 4c ; x − y = ± √(a2 - 4c). Thus since x + y and x-y are known, we can find immediately the values of x and y. Or we may proceed thus. Assume x-y=z, then since x+y=a, we obtain Substitute in the second of the given equations; thus therefore (a + z)3 + (a − 2)3 = 3265, 5az+10a32 = 16b5 — a3. From this quadratic we may find 2, and hence z, that is, x-y; and hence finally x and y. More examples will be found in Chapter LIV. |