268. Solve 2x2+3xy+y2=70, 6x2+xy-y2 = 50. In the ́equation a2(2+3v+v2)=70 put for v; thus x=3; and since y=vx we have y±4. The value v=- - 2 we shall find to be inapplicable; for it leads to the inadmissible result 2 × 0=70. In fact the equations from which the value of v was obtained may be written thus, x2(2+v)(1+v)=70, x2(2+v)(3-v)=50; and hence we see that the value of v found from 2+v=0 is inapplicable, and that we can only have 269. Equations may be proposed which do not fall under either of the two cases which we have discussed, but which may be solved by artifices which can only be suggested by trial and experience. We will give some examples. then from this equation combined with x+y=5_we can find x and y by the first case. Or we may complete the solution thus, x+y=5, We have now to find x and y from the simple equations x-y=±3; these lead to x=1 or 4, y=4 or 1. xy=20. 271. Solve 2+y2==41, These equations can be solved by the second case; or they may be solved in the manner just exemplified. For we can deduce from them x2+ y2+2xy=41+40=81, x2+ y2-2xy=41-40=1; then by extracting the square roots, x+y==9, x-y=±1. And thus finally we shall obtain Then proceeding as in Art. 271, we shall find x= ±3 or +2, y=2 or 3. therefore x2+y1+xy (x2 + y2)+x2y2 = 121 x-y=2; x2-2xy+y=4; x2+ y2=2xy +4 +2x2y2+y1 = 4x2y2+16x+16; x2 + y2 = 2x2y2+16xy+16 ......... (3). Substitute from (2) and (3) in (1); thus 2x2y+16xy+16+xy (2xy +4)+a22=121; that is, therefore 5x2y2+20xy = 105; x2y2+4xy=21. From this quadratic equation we shall obtain xy=3 or -7. Take xy=3, and from this combined with x-y=2, we shall obtain x=3 or -1, y=1 or -3. If we take xy=-7, we shall find that the values of x and y are impossible; see Art. 236. 8. (x-6)2+(y-5)2+2xy=60, 5y-4x=1. |