Elements of Algebra

314 ELEMENTS PART II. Now, if we suppose that among the numbers $f, g$, and $h$, the first, $f$, is the greatest, and $h$ the least, since we have $fx + fy + fz$, or $(x + y + z)f = fa$, (because $x + y + z = a$) it is evident, that $fx + fy + fz$ is greater than $fx + gy + hz$; consequently, $fa$ must be greater than $b$, or $b$ must be less than $fa$. Farther, since $hx + hy + hz$, or $(x + y + z)h = ha$, and $hx + hy + hz$ is undoubtedly less than $fx + gy + hz$, $ha$ must be less than $b$, or $b$ must be greater than $ha$. Hence it follows, that if $b$ be not less than $fa$, and also greater than $ha$, the question will be impossible: which condition is also expressed, by saying that $b$ must be contained between the limits $fa$ and $ha$; and care must also be taken that it may not approach either limit too nearly, as that would render it impossible to determine the other letters. In the preceding example, in which $a = 100$, $f = 3\frac{1}{2}$, and $h = \frac{1}{2}$, the limits were 350 and 50. Now, if we suppose $b = 51$, instead of 100, the equations will become $x + y + z = 100$, and $3\frac{1}{2}x + 1\frac{1}{3}y + \frac{1}{2}z = 51$; or, removing the fractions, $21x + 8y + 3z = 306$; and if the first be multiplied by 3, we have $3x + 3y + 3z = 300$. Now, subtracting this equation from the other, there remains $18x + 5y = 6$; which is evidently impossible, because $x$ and $y$ must be integer and positive numbers *. 28. Goldsmiths and coiners make great use of this rule, when they propose to make, from three or more kinds of metal, a mixture of a given value, as the following example will shew. Question 3. A coiner has three kinds of silver, the first of 7 ounces, the second of $5\frac{1}{2}$ ounces, the third of $4\frac{1}{2}$ ounces, fine per marc†; and he wishes to form a mixture of the weight of 30 marcs, at 6 ounces: how many marcs of each sort must he take? If he take $x$ marcs of the first kind, $y$ marcs of the second, and $z$ marcs of the third, he will have $x + y + z = 30$, which is the first equation. Then, since a marc of the first sort contains 7 ounces of fine silver, the $x$ marcs of this sort will contain $7x$ ounces of such silver. Also, the $y$ marcs of the second sort will contain $5\frac{1}{2}y$ ounces, and the $z$ marcs of the third sort will contain $4\frac{1}{2}z$ ounces, of fine silver; so that the whole mass will contain $7x + 5\frac{1}{2}y + 4\frac{1}{2}z$ ounces of fine silver. As this mixture is to weigh 30 marcs, and each of these marcs must contain 6 ounces of fine silver, it follows that the whole mass * Vide Article 22. † A marc is eight ounces.