Elements of Algebra

CHAP. II. OF ALGEBRA. 315 will contain 180 ounces of fine silver; and thence results the second equation, $7x + 5\frac{1}{2}y + 4\frac{1}{2}z = 180$, or $14x + 11y + 9z = 360$. If we now subtract from this equation nine times the first, or $9x + 9y + 9z = 270$, there remains $5x + 2y = 90$, an equation which must give the values of $x$ and $y$ in integer numbers; and with regard to the value of $z$, we may derive it from the first equation $z = 30 - x - y$. Now, the former equation gives $2y = 90 - 5x$, and $y = 45 - \frac{5x}{2}$; therefore, if $x = 2u$, we shall have $y = 45 - 5u$, and $z = 3u - 15$; which shews that $u$ must be greater than 4, and yet less than 10. Consequently, the question admits of the following solutions: If $u = 5, 6, 7, 8, 9$, Then $\left\{ \begin{array}{l} x = 10, 12, 14, 16, 18, \\ y = 20, 15, 10, 5, 0, \\ z = 0, 3, 6, 9, 12. \end{array} \right.$ 29. Questions sometimes occur, containing more than three unknown quantities; but they are also resolved in the same manner, as the following example will shew. Question 4. A person buys 100 head of cattle for 100 pounds; viz. oxen at 10 pounds each, cows at 5 pounds, calves at 2 pounds, and sheep at 10 shillings each. How many oxen, cows, calves, and sheep, did he buy? Let the number of oxen be $p$, that of the cows $q$, of calves $r$, and of sheep $s$. Then we have the following equations: 1. $p + q + r + s = 100$; 2. $10p + 5q + 2r + \frac{1}{2}s = 100$; or, removing the fractions, $20p + 10q + 4r + s = 200$: then subtracting the first equation from this, there remains $19p + 9q + 3r = 100$; whence $3r = 100 - 19p - 9q$, and $r = 33 + \frac{1}{3} - 6p - \frac{1}{3}p - 3q$; or $r = 33 - 6p - 3q + \frac{1-p}{3}$; whence $1-p$, or $p-1$, must be divisible by 3; therefore if we make $p - 1 = 3t$, we have $p = 3t + 1$ $q = q$ $r = 27 - 19t - 3q$ $s = 72 + 2q + 16t$;