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Problem.—Given that one glass of lemonade, 3 sandwiches, and 7 biscuits, cost 1s. 2d.; and that one glass of lemonade, 4 sandwiches, and 10 biscuits, cost 1s. 5d.: 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) 8d.; (2) 1s. 7d.

Solution.—This is best treated algebraically. Let $$\scriptstyle x$$ = the cost (in pence) of a glass of lemonade, $$\scriptstyle y$$ of a sandwich, and $$\scriptstyle z$$ of a biscuit. Then we have $$\scriptstyle x\ +\ 3y\ +\ 7z\ =\ 14$$, and $$\scriptstyle x\ +\ 4y\ +\ 10z\ =\ 17$$. And we require the values of $$\scriptstyle x\ +\ y\ +\ z$$, and of $$\scriptstyle 2x\ +\ 3y\ +\ 5z$$. 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.