CHAP. XVIII.] ELEMENTARY ILLUSTRATIONS. 289 



In like manner we have from the second equation, 

 (y + yxz) 1, + ty (xz + xz + xz) = ; 



and from the sum of these two equations we are to eliminate y 

 and z. 



If in that sum we make y = 1, z = 1, we get the result ~s + t. 



If in the same sum we make y = 1, z = 0, we get the result 



xs + s x + t. 

 If in the same sum we make y = 0, z = 1, we get 



xs + sx + xt -f tx. 



And if, lastly, in the same sum we make y = 0, z = 0, we find 

 xs + sx + tx + tx 9 or #? + sx + t. 



These four expressions are to be multiplied together. Now 

 the first and third may be multiplied in the following manner : 



(?+ ) (x's + sx + xi + tx) 



= xs + xt + (s + i) (sx + tx) by (IX. Prop, n.) 



j = x~s + xt + ~sxt + sxl. (2) 



Again, the second and fourth give by (IX. Prop, i.) 



(xs + SX + 1) (xJ+ sx + t) 



= x~s + sx. (3) 



Lastly, (2) and (3) multiplied together give 

 (xJ + sx) (x~s + sxJ + xt + tUTs) 

 = xs + sx (sxt + xt + tx~s) 



= x~s + sxt. 

 Whence the final equation is 



(l-s)x + s(l -#)(1-^) = 0, 

 which, solved with reference to x, gives 



X = 



(1-0 -(!-) 

 = ^st + * (1 - 1) + (1 - *) < + (1 - s) (1 - t), 



