Elements of Algebra

CHAP. VI. OF ALGEBRA. 229 consequently the equation, $\frac{15x}{100-x} = \frac{2000-20x}{3x}$, which becomes $25x^2 = 200000 - 4000x$; and, lastly, $x^2 = -160x + 8000$; whence we obtain $x = -80 + \sqrt{(6400 + 8000)} = -80 + 120 = 40$. So that the first girl had 40 eggs, the second had 60, and each received 10 pence. 655. Question 10. Two merchants sell each a certain quantity of silk; the second sells 3 ells more than the first, and they received together 35 crowns. Now, the first says to the second, I should have got 24 crowns for your silk: the other answers, And I should have got for yours 12 crowns and a half. How many ells had each? Suppose the first had $x$ ells; then the second must have had $x+3$ ells; also, since the first would have sold $x+3$ ells for 24 crowns, he must have received $\frac{24x}{x+3}$ crowns for his $x$ ells. And, with regard to the second, since he would have sold $x$ ells for $12\frac{1}{2}$ crowns, he must have sold his $x+3$ ells for $\frac{25x+75}{2x}$; so that the whole sum they received was $\frac{24x}{x+3} + \frac{25x+75}{2x} = 35$ crowns. This equation becomes $x^2 = 20x - 75$; whence we have $x = 10 \pm \sqrt{(100-75)} = 10 \pm 5$. So that the question admits of two solutions: according to the first, the first merchant had 15 ells, and the second had 18; and since the former would have sold 18 ells for 24 crowns, he must have sold his 15 ells for 20 crowns. The second, who would have sold 15 ells for 12 crowns and a half, must have sold his 18 ells for 15 crowns; so that they actually received 35 crowns for their commodity. According to the second solution, the first merchant had 5 ells, and the other 8 ells; and since the first would have sold 8 ells for 24 crowns, he must have received 15 crowns for his 5 ells; also, since the second would have sold 5 ells for 12 crowns and a half, his 8 ells must have produced him 20 crowns; the sum being, as before, 35 crowns.