Elements of Algebra

312 ELEMENTS PART II. or $3r = 6s + 2$: consequently, $r = \frac{6s+2}{3} = 2s + \frac{2}{3}$. Now, it is evident, that this can never become an integer number, because $s$ is necessarily an integer; which shews the impossibility of such questions *. *** CHAP. II. *Of the Rule which is called Regula Cæci, for determining by means of two Equations, three or more Unknown Quantities.* 24. In the preceding chapter, we have seen how, by means of a single equation, two unknown quantities may be determined, so far as to express them in integer and positive numbers. If, therefore, we had two equations, in order that the question may be indeterminate, those equations must contain more than two unknown quantities. Questions of this kind occur in the common books of arithmetic, and are resolved by the rule called *Regula Cæci*, *Position*, or *The Rule of False*; the foundation of which we shall now explain, beginning with the following example: 25. Question 1. Thirty persons, men, women, and children, spend 50 crowns in a tavern; the share of a man is 3 crowns, that of a woman 2 crowns, and that of a child is 1 crown; how many persons were there of each class? If the number of men be $p$, of women $q$, and of children $r$, we shall have the two following equations; 1. $p + q + r = 30$, and 2. $3p + 2q + r = 50$, from which it is required to find the value of the three letters $p, q$, and $r$, in integer and positive numbers. The first equation gives $r = 30 - p - q$; whence we immediately conclude that $p + q$ must be less than 30; and, substituting this value of $r$ in the second equation, we have $2p + q + 30 = 50$; so that $q = 20 - 2p$, and $p + q =$ \* See the Appendix to this chapter, at Art. 3. of the Additions by De la Grange.