Subtracting from this, a, for the expenses of the second year, we have, The profits of the second year is the nth part of this sum, or Adding this amount to the stock at the beginning of the year, we have, for the amount at the close of the second year. Subtracting a from this, we have, , Adding to this remainder its nth part, we have for the stock at the end of the third year, n2x + 2nx + x - Зига - Зпа + n2 a which becomes, by bringing the fractions to a common denominator, and reducing similar terms, n2x+3n2x + 3nx + x-3n3а - 6n2а-4na-a But, by the conditions of the problem, the whole stock is exhausted at the end of the third year; therefore, n3x + 3n2x + 3nx + x -3n3а - 6n2а - 4na - a 0. n3 Clearing of fractions and transposing, n3x + 3n2x + 3nx + x = 3n3a+6n2a +4na + a This result may be simplified, by observing that both the numerator and denominator of the fraction may be divided by n+1, and we have, x= a(3n2 + 3n + 1) (n + 1) = a(3n2 + 3n+1) This same result might have been obtained by a shorter method; for, if at the end of the third year nothing remained, the whole stock must have been exhausted in deducting from the stock at the beginning the sum of a dollars, for the expenses of this year; for had any thing remained after this sum was deducted, what remained, plus the nth part of it, must have remained at the close of the year, and the stock would not have been exhausted; whence we have, n2x + 2nx + x - 3n2а — 3па - а n2 = 0, 14. There are two numbers whose sum is the sixth part of their product, and the greater is to the less as 3 to 2. Required those numbers. Ans. 15 and 10. 15. Find a number, such, that whether it be divided into two or three equal parts, the continued product of the parts shall be equal to the same quantity. Ans. 61. 16. There are two numbers whose sum is 37, and if three times the less be subtracted from four times the greater, and this difference divided by 6, the quotient will be 6. What are the numbers? Ans. 21 and 16. 17. What number is that to which, if 20 be added, and from of the sum 12 be subtracted, the remainder will be 10? Ans. 13. 18. Divide the number 72 into three parts, so that one-half the first shall be equal to the second, and three-fifths of the second equal to the third. Ans. 40, 20, and 12. 19. A person after spending one-fifth of his income, plus $10, had then remaining one-half of it, plus $35. Required his income. Ans. $150. 20. A and B engaged in trade, A with $240, and B with $96; A lost twice as much as B; and upon settling their accounts it appeared, that A had three times as much remaining as B. How much did each lose? Ans. A lost $96, В $48. 21. A person being asked the hour, answered that it was between five and six, and the hour and minute hand were together. What was the time? 22. It is required to divide the number 34 into two such parts, that the difference between the greater and 18, shall be to the difference between the less and 18, as 2 to 3. Ans. 22 and 12. 23. A cistern into which water was let by two cocks, A and B, will be filled by both, running together, in 12 hours, and by the cock A alone, in 20 hours. In what time will it be filled by the cock B alone? Ans. 30 hours. 24. Two persons, A and B, have the same annual income, A saves one-fifth of his; but B, by spending $80 per annum more than A, at the end of 4 years finds himself $220 in debt. What was their annual income? Ans. $125. 35. A shepherd, in time of war, was plundered by a party of soldiers, who took one-fourth of his flock and one-fourth of a sheep more; another took one-third of what he had left and one-third of a sheep; then a third party took one-half of what then remained and one-half a sheep. After which he had but twenty-five sheep left. How many had he at first? Ans. 103. 26. A person passed th of his age in childhood, Toth in youth, th + 5 in matrimony; he then had a son whom he survived four years, and who reached only one-half the age of his father. At what age did this person die ? Ans. 84. 27. A trader maintained himself for three years, at the expense of £50 a year, and in each of those years augmented that part of his stock which was not so expended by onethird thereof. At the end of the third year his original stock was doubled. What was that stock? Ans. £740. 28. Two men, at the distance of 150 miles, set out to meet each other; one goes three miles in the time that the t other goes seven. travel? What part of the distance does each 29. Out of a certain sum a man paid his creditors $96, half of the remainder he lent a friend; he then spent onefifth of what now remained, and after all these deductions had one-tenth of his money left. How much had he at first? Ans. $128. 30. The expense of paving a square court at 50 cents per square yard, is the same as that of surrounding it with an iron fence at $1 75 per foot. How many square feet does it contain? Ans. 15,876. SOLUTION OF SIMPLE EQUATIONS WHICH INVOLVE MORE THAN ONE UNKNOWN QUANTITY. (35.) When, by the conditions of a problem, two or more unknown quantities are to be determined, it is necessary that there should be as many independent equations as there are unknown quantities, otherwise the problem would be indeterminate. To illustrate this take the following example:A laborer having worked for a person twelve days, and having with him, during the first seven days, his son, received 100 shillings; he afterwards worked eight days more, five of which he had his son with him, and received 68 shillings. How much did he earn per day himself, and how much did his son earn? Let x = the daily wages of the man, and y = those of his son; then 12x is the whole of the father's wages, and 7y the son's; and the sum of these is, by the conditions of the problem, 100 shillings; we have, therefore, the equation 12x+7y = 100. But this equation cannot be determined, since the value of x depends on that of y. If, for instance, y = 4, then 12x+7x4 = 100, 12x + 28 = 100, x = 6; but if y = 6, then we shall have 12x + 42 = 100, and 58 5 12-46. But if we have another condition and another equation, we can determine the value of x by combining the two equations together, and there will be but one value of either of the unknown quantities that will answer the conditions of the problem. This other equation can be derived from the problem; for the laborer worked eight days afterwards at the same price, and had with him his son five of these days, and received 68 shillings. This condition gives the equation 8x + 5y = 68. By the first condition we had 12x+7y = 100. Now, in order to find the value of x and y, let it be observed, that an equation is not destroyed by multiplying both terms by the same quantity. Multiplying, therefore, the first of these equations by 7, and the second by 5, they become, 7×(8x + 5y = 68) = 56x + 35y = 476 5×(12x+7.y = 100) = 60x + 35y = 500. It is evident, moreover, that if one equation be subtracted from another, term by term, the result will be an equation, since it is only subtracting equals from equals, and the remainders must therefore be equal. If, now, we subtract the first of these last equations from the second, we shall have, an equation, which has but one unknown quantity, and which Substituting this value of x, in the equation 12x +7y=100, we have, 12x6 + 7y = 100, or 72+7y = 100, These values of x and y, introduced into the original equations, are the only numbers that will satisfy the conditions of the problem. By substituting 6 in the place of x, and 4 in the place of y, we have, 12 x 6+7 x 4 = 100, or 72+ 28 = 100, and, 8×6+5×4 = 68, or 48 + 20 = 68. Now, if in the first equation, we should make x = 5, and y = 5, we should have, 12×5+7×54 = 100, and the |