Elements of Algebra

CHAP. III. OF ALGEBRA. 199 progression by $\frac{15x}{2}$. We know also, that this sum is 60; so that $\frac{15x}{2} = 60$; or $\frac{1}{2}x = 4$, and $x = 8$. Now, since the number of terms is 8, if we suppose the difference to be $z$, we have only to seek for the eighth term upon this supposition, and to make it equal to 10. The second term is $5 + z$, the third is $5 + 2z$, and the eighth is $5 + 7z$; so that $5 + 7z = 10$ $7z = 5$ and $z = \frac{5}{7}$. The difference of the progression, therefore, is $\frac{5}{7}$, and the number of terms is 8; consequently, the progression is $5 + 5\frac{5}{7} + 6\frac{3}{7} + 7\frac{1}{7} + 7\frac{6}{7} + 8\frac{4}{7} + 9\frac{2}{7} + 10$, the sum of which is 60. 597. Question 14. To find such a number, that if 1 be subtracted from its double, and the remainder be doubled, from which if 2 be subtracted, and the remainder divided by 4, the number resulting from these operations shall be 1 less than the number sought. Suppose this number to be $x$; the double is $2x$; subtracting 1, there remains $2x - 1$; doubling this, we have $4x - 2$; subtracting 2, there remains $4x - 4$; dividing by 4, we have $x - 1$; and this must be 1 less than $x$; so that $x - 1 = x - 1$. But this is what is called an identical equation; and shews that $x$ is indeterminate; or that any number whatever may be substituted for it. 598. Question 15. I bought some ells of cloth at the rate of 7 crowns for 5 ells, which I sold again at the rate of 11 crowns for 7 ells, and I gained 100 crowns by the transaction. How much cloth was there? Supposing the number of ells to be $x$, we must first see how much the cloth cost; which is found by the following proportion: As $5 : x :: 7 : \frac{7x}{5}$ the price of the ells. This being the expenditure; let us now see the receipt: in order to which, we must make the following proportion: