ANSWERS TO KNOT VII. _Problem._--Given that one glass of lemonade, 3 sandwiches, and 7 biscuits, cost 1_s._ 2_d._; and that one glass of lemonade, 4 sandwiches, and 10 biscuits, cost 1_s._ 5_d._: find the cost of (1) a glass of lemonade, a sandwich, and a biscuit; and (2) 2 glasses of lemonade, 3 sandwiches, and 5 biscuits. _Answer._--(1) 8_d._; (2) 1_s._ 7_d._ _Solution._--This is best treated algebraically. Let _x_ = the cost (in pence) of a glass of lemonade, _y_ of a sandwich, and _z_ of a biscuit. Then we have _x_ + 3_y_ + 7_z_ = 14, and _x_ + 4_y_ + 10_z_ = 17. And we require the values of _x_ + _y_ + _z_, and of 2_x_ + 3_y_ + 5_z_. Now, from _two_ equations only, we cannot find, _separately_, the values of _three_ unknowns: certain _combinations_ of them may, however, be found. Also we know that we can, by the help of the given equations, eliminate 2 of the 3 unknowns from the quantity whose value is required, which will then contain one only. If, then, the required value is ascertainable at all, it can only be by the 3rd unknown vanishing of itself: otherwise the problem is impossible. 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 _y_ + 3_z_ = 3, or _y_ = 3-3_z_; and then substituting this value of _y_ in the 1st, which gives _x_-2_z_ = 5, _i.e._ _x_ = 5 + 2_z_. Now if we substitute these values of _x_, _y_, in the quantities whose values are required, the first becomes (5 + 2_z_) + (3-3_z_) + _z_, _i.e._ 8: and the second becomes 2(5 + 2_z_) + 3(3-3_z_) + 5_z_, _i.e._ 19. Hence the answers are (1) 8_d._, (2) 1_s._ 7_d._ * * * * * 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 _im_possible, 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 "Patience") before any solution would have been hit on by the most ingenious of them.