Systems of Equations. 95 



Moreover, aii}^ value of the unknown quantities which will 



satisfy 



milst satisfy (4.) since the relation L, = R-, must hold. 



Therefore, from these considerations, it is evident that the 

 system (d) is not equivalent to system (a), but to the three 

 systems 



L=R. / . 



L=o s (''■^ 



R=o S ('''^ 



L=R. \ ^*3^ 



Second, suppose that either L, or R^ is a known quantity not 

 zero. 



One of them, say R^, is the known quantity. Therefore L^ 

 cannot be zero, since the relation Li=R, must hold. Hence the 

 introduced system (bj and (bj are absurdities, since they require 

 that ly^ and R^ be zero. Consequently the derivation is legitimate 

 since the introduced S3'Stems are incompatible. 



8. Examples. As an illustration of the theorem, consider the 

 sj^stem 



- ' X— 4=6— r / 



2-r+r=i3 \ 

 This is satisfied by .1 = 3 and j'=7. Now form the system 



-V— 4=6— J^' ) 



(x-^)(2x+v)=i2,(6-y) \ 

 which is satisfied by either of the sets of values, .1 = 3, )'=7 or 

 .r=4, j/=6. The additional solution may be obtained from either 

 of the systems 



.1 — 4=6—1' ) 

 .1—4=0 I 



.V— 4=6— r) 

 6— r=o \ 



As another example consider the system 

 :v— 2J=3) 

 .V-f-2J'=7 ) 



