Elements of Algebra

CHAP. III. OF ALGEBRA. 201 Three nutmegs, therefore, cost $4\frac{1}{2}d$., and four cost $6d$.: wherefore each costs $1\frac{1}{2}d$. 601. Question 18. A certain person has two silver cups, and only one cover for both. The first cup weighs 12 ounces; and if the cover be put on it, it weighs twice as much as the other cup: but when the other cup has the cover, it weighs three times as much as the first. Required the weight of the second cup, and that of the cover. Suppose the weight of the cover to be $x$ ounces; then the first cup being covered, it will weigh $x+12$; this weight being double that of the second, the second cup must weigh $\frac{1}{2}x+6$; and, with the cover, it will weigh $x+\frac{1}{2}x+6$, or $\frac{3}{2}x+6$; which weight ought to be the triple of 12; that is, three times the weight of the first cup. We shall therefore have the equation $\frac{3}{2}x+6=36$, or $\frac{3}{2}x=30$; so that $\frac{1}{2}x=10$ and $x=20$. The cover, therefore, weighs 20 ounces, and the second cup weighs 16 ounces. 602. Question 19. A banker has two kinds of change: there must be $a$ pieces of the first to make a crown; and $b$ pieces of the second to make the same. Now, a person wishes to have $c$ pieces for a crown. How many pieces of each kind must the banker give him? Suppose the banker gives $x$ pieces of the first kind; it is evident that he will give $c-x$ pieces of the other kind; but the $x$ pieces of the first are worth $\frac{x}{a}$ crown, by the proportion $a:x::1:\frac{x}{a}$; and the $c-x$ pieces of the second kind are worth $\frac{c-x}{b}$ crown, because we have $b:c-x::1:\frac{c-x}{b}$. So that, $\frac{x}{a}+\frac{c-x}{b}=1$; or $\frac{bx}{a}+c-x=b$; or $bx+ac-ax=ab$; or, rather $bx-ax=ab-ac$; whence we have $x=\frac{ab-ac}{b-a}$, or $x=\frac{a(b-c)}{b-a}$; consequently, $c-x$, the pieces of the second kind, must be $=\frac{bc-ab}{b-a}=\frac{b(c-a)}{b-a}$.