+7 
t, = (6a+ 20's, + a's, 
2, — 5t, — (10b +2!) s, + (5a+-4b') s, + 2a's, 
Bt, — 4t, —4es, + (8b+3c') s, + (4a4 66’) s, + 3a's, 
At, — 3t, = Bos, + (6b+4c’) s, + (8¢+80') s, + 4a's, 
Bt, — 2t, == Zes, + (40+5c')s, + (2a4100') s, 
— t,=cs, + (26+6c)s, 
which may always be satisfied, and the coefficients of the second are 
related to those of the first by the following system 
u, = (7a+20') t, + at, 
2u, — 6u, —= (1264 2c) t, + (6a+ 4b') t, + 2a't, 
3u, — bu, = det, + (106+ 8c) t, + (5a+ 60’) t, + 3a't, 
Au, — 4u, — 4et, + (8b+4c’) t, + (4a4 80’) t, 4+ da't, 
5u, — 3u, = Set, 4- (6b+ 5c’) t, + (8a+100') t, + 5a't, 
6u, — 2u, — Zet, + (464+ 6c) t, + (2a+120/)¢, 
— u,=et, + (26+7c) t,. 
This system is impossible unless 
Bu, + (8u,—5u,) + (5u,—3u,) + 5 (—u,) = 0 
or 
(35a4-106' 45e) t, + (5a'4-10b4-8c') t, + (5a4 60'+ 3c) t, + 
+. (8a' + 66-+-5¢) ¢, + (8a+100'+- 5c) t, + (5a! +1084 85c')¢, =O 
which may be written 
At,+B (2t,—5t,)+C (3¢,—4t,) + D (4, — 3) +E (5t,—2t,)+ F (—t,)=0 
if 
be 
A=— (Va +14b-41%c) , B= — (Ta+28'+9), 
1 1 
C age neet 1E) e= Sang IE BOOM Se). 
E=a +264 7¢ , Fs = (Ta 140470). 
Thus, choosing as before s,=—= 0, the sought condition takes this form 
s,[ dA 4+ (Bat 4b') B + (864 3c') C + 3e D] 
+ 5, [2a'B + (4a4 6b') C + (6b44c) D + 2c E] 
+ s, [8a'C + (Bat 86) D+ (4b 45) EH e F] 
Hs, [4a'D + (204100!) E + (264 6c') F] — 0. 
Writing this equation 
Bi Ss, fs 8, ue 83 +f, 84 ll 
and eliminating a’ 6’ c’ we obtain 
20 i Ai 
A= —-( 8a+8+8e) , B= —2(5a43e) , C= — (10a —B+10e) 
