341 
1896 - 97 .] Dr Muir on Quaternary Qnaclrics . 
The solution thus suggested is to establish the pair of equations 
2B 
E 
6 . 
E 
2C 
K 
= 0, 
<9 
Iv 
2L 
D 
2B 
6 
F 
E 
K 
= o , 
G 
<9 
2L 
8. 
This 
is readily 
first of the pair we have only got to substitute in it the values of 
E and Iv got from the 2nd and 3rd of the triad. To verify the 
second of the pair we do the same, when it is found that the 
result is obtainable from the first of the triad by dividing by /?y8, 
o g 
and then multiplying by L -g- - B — • The first of the pair is thus 
seen to be derivable from the last two of the triad, and the second 
from all three. 
As a step towards a direct mode of elimination, it may be noted 
that the equations are transformable into 
FL/3 2 - FB8 2 + 2GByS + (DK - GE)S 8 - 2DL/3y = 0 ^ 
C/3 2 + By 2 - E/3y = 0 j- 
Ly 2 +CS' 2 - ICyS =0 j , 
that is to say, into a set of ternary quadrics. 
