94 



Algebra. 



I, -R =o (i) ) 



L -R =o (2) \ ^ ^ 



Rewrite (a) and (b) so that they shall read 



Iv,-R. = o 



and L,-R, = o (^) ) ... 



S(V-Rr)+T(L.-RJ=o (^) \ ^^^ 



It is evident that any set of values which will satisfy (c) will 

 satisfy (d), for whatever makes ly,— R^ and L„— R. each zero will 

 satisfy (d). 



It is seen from (t^) that any set of values that satisfies (d) must 

 make L^— R, zero. Equation (^) will then become 



T(L-R,)=o. (5) 



Now, since T is a known quantity not zero, this cannot be sat- 

 isfied unless L^— R3 is zero. Hence any set of values that satisfies 

 (d) must make L,— R, and L^— R, each zero ; that is, must be a 

 solution of (c). 



Now, since any solution of (c) is a solution of (d) and any 

 solution of (d) is a solution of (c), the two systems are equivalent. 



7, Theorem. // from a system containhig two miknowji 

 quantities 



we derive the system. 



L=R (2) \ (") 



L=R. (2)1 



L.L=R.R, (^)\ '*^ 



the derivation is questionable if L,^ and R^ both involve imknown 

 quantities, but legitimate if either is a kyioum quantity not zero. 

 First, suppose that L^ and R^ each involve unknown quantities. 

 Any value of the unknown quantities which will satisfy the 

 equation 



L=o 

 must satisfy equation (/^), since the relation L,= Ri must hold if 

 system (b) is to be satisfied. 



Also any value of the unknown quantities which will satisfy 

 the equation 



R=o 

 must satisfy (\), since the relation Li=R, must hold if the 

 system is to be satisfied. 



