Page:Carroll - Tangled Tale.djvu/129

Rh Let us then eliminate lemonade and sandwiches, and reduce everything to biscuits—a state of things even more depressing than "if all the world were apple-pie"—by subtracting the 1st equation from the 2nd, which eliminates lemonade, and gives $$\scriptstyle y\ +\ 3z\ =\ 3$$, or $$\scriptstyle y\ =\ 3\ -\ 3z$$; and then substituting this value of y in the 1st, which gives $$\scriptstyle x\ -\ 2z\ =\ 5$$, i.e. $$\scriptstyle x\ =\ 5\ +\ 2z$$. Now if we substitute these values of $$\scriptstyle x$$, $$\scriptstyle y$$, in the quantities whose values are required, the first becomes $$\scriptstyle (5\ +\ 2z)\ +\ (3\ -\ 3z)\ +\ z$$, i.e. 8: and the second becomes $$\scriptstyle 2\ (5\ +\ 2z)\ +\ 3\ (3\ -\ 3z) + 5z$$, i.e. 19. Hence the answers are (1) 8d., (2) 1s. 7d.

The above is a universal method: that is, it is absolutely certain either to produce the answer, or to prove that no answer is possible. The question may also be solved by combining the quantities whose values are given, so as to form those whose values are required. This is merely a matter of ingenuity and good luck: and as it may fail, even when the thing is possible, and is of no use in proving it impossible, I cannot rank this method as equal in value with the other. Even when it succeeds, it may prove a very tedious process. Suppose the 26 competitors, who have sent in what I may call accidental solutions, had had a question to deal with where every number contained 8 or 10 digits! I suspect it would have been a case of "silvered is the raven hair" (see