Elements of Algebra

198 ELEMENTS SECT. IV. less than 40 by $40-5x$, and this quantity must be equal to $12-x$. We have, therefore, $40-5x=12-x$. Adding $5x$, we have $40=12+4x$; and subtracting 12, we obtain $28=4x$; lastly, dividing by 4, we have $x=7$, the number sought. 594. Question 11. To divide 25 into two such parts, that the greater may be equal to 49 times the less. Let the less part be $x$, then the greater will be $25-x$; and the latter divided by the former ought to give the quotient 49: we have therefore $\frac{25-x}{x}=49$. Multiplying by $x$, we have $25-x=49x$; adding $x$, we have $25=50x$; and dividing by 50, gives $x=\frac{1}{2}$. The less of the two numbers is $\frac{1}{2}$, and the greater is $24\frac{1}{2}$; dividing therefore the latter by $\frac{1}{2}$, or multiplying by 2, we obtain 49. 595. Question 12. To divide 48 into nine parts, so that every part may be always $\frac{1}{2}$ greater than the part which precedes it. Let the first, or least part be $x$, then the second will be $x+\frac{1}{2}$, the third $x+1, \ldots$. Now, these parts form an arithmetical progression, whose first term is $x$; therefore the ninth and last term will be $x+4$. Adding those two terms together, we have $2x+4$; multiplying this quantity by the number of terms, or by 9, we have $18x+36$; and dividing this product by 2, we obtain the sum of all the nine parts $=9x+18$; which ought to be equal to 48. We have, therefore, $9x+18=48$; subtracting 18, there remains $9x=30$; and dividing by 9, we have $x=3\frac{1}{3}$. The first part, therefore, is $3\frac{1}{3}$, and the nine parts will succeed in the following order: 1 2 3 4 5 6 7 8 9 $3\frac{1}{3} + 3\frac{5}{6} + 4\frac{1}{3} + 4\frac{5}{6} + 5\frac{1}{3} + 5\frac{5}{6} + 6\frac{1}{3} + 6\frac{5}{6} + 7\frac{1}{3}$. Which together make 48. 596. Question 13. To find an arithmetical progression, whose first term is 5, the last term 10, and the entire sum 60. Here we know neither the difference nor the number of terms; but we know that the first and the last term would enable us to express the sum of the progression, provided only the number of terms were given. We shall therefore suppose this number to be $x$, and express the sum of the