The eldest son's share, by the first formula, is 16000+ 2 × 2000+ 1000 7000 crowns. 3 The other shares may be found by the other two formulas. Let the learner solve this question by making x cqual to the less part, and also by making it equal to the mean. Exam. 5th, Art. III. may be solved by this formula. Let the learner generalize the questions in Art. III. as far as to Exam. 16th. The examples in Art. I. may be generalized still farther. 8. A man bought corn at 4s. (a) per bushel, rye at 6s. (b) per bushel, and wheat at 8s. (c) per bushel: there was an equal quantity of each sort. The whole came to 90s. (d). How many bushels were there of each ? It will readily be perceived that it is impossible actually to perform the operations of addition, subtraction, &c. on letters; but it is easy to represent these operations. We however frequently speak of adding, subtracting, multiplying, and dividing algebraic quantities, by which we mean, representing these operations. We have seen that to express 3 times x or 3 times a we write 3, 3 a, that is, x or a multiplied by 3. In the same manner, if we wish to express a times x, that is, a multiplied by a, we write a x; and if we wish farther to express that a x (that is, a times x) is to be multiplied by b, we write ab x. x the number of bushels of each. *Let Here x is taken a times, and b times, and c times, that is, (a+b+c) times. This may be expressed thus, (a+b+c) x, * Let the learner perform this example first by the numbers. enclosing the three coefficients connected by their signs in a parenthesis. This will be plain if we put it in numbers. 4x+6x+8x is the same as (4+6+8) x, that is, 18 x. If we had we should divide by 18, (a+b+c) x=d d In the same manner divide by (a + b + c), Particular Ans. 5 bushels. This general formula is expressed in words as follows: Divide the price of the whole by the price of a bushel of each sort added together, and it will give the number of bushels of each sort. 9. A father dying left $25000 (or a) to be divided betweer. his wife, son, and daughter; his son was to have 3 (or b) times as much as the daughter, and the wife 2 (or c) times as much as the son. What was the share of each? In this example observe that a is taken 1 time, and b times, and bc times. When a letter is written without a coefficient, it is always understood to have 1 for its coefficient; thus x is the same as 1 x. Having found the share of the daughter, it is easy to find the shares of the other two. The son's share is 3 x = 7500, or b x = The wife's do. is 6 x = 15000, or b c x= ab 1+ b + b c The learner may now generalize some of the examples in Art. I. in this manner. 10. A gentleman, distributing some money among some beggars, found, that in order to give them 8 (or a) cents apiece, he should want 5 (or b) cents; he therefore gave them 7 (or c) cents, and he had 4 (or d) cents left. How many beggars were there? 11. There is a cistern which is supplied by two pipes; the first will fill it alone in 7 (or a) hours, the second will fill it alone in 5 (or 6) hours. In what time will it be filled if both run together? Let the number of hours in which both together will fill it. 1 The first will fillor of it in one hour, and the second will 1 α fill or of it in one hour; both together will fill ÷ + } or 1 b + of it in one hour. In x hours they will fill x times as Ꮖ a Ђ But a hours is the whole time, therefore, the cistern being 1, Suppose one pipe would fill the cistern in 8 hours, and the other in 43 hours, and find the answer by the general formula Ans. 3 hours 12. Suppose it were required to make a rule for Fellowship. First take a particular case. Three men, commencing trade together, furnished money in the following proportions; A $8 as often as B $5, and as often as C $3. They gained $800. What is each man's share of the gain? It is evident that they must receive in the proportion of the capital that they respectively furnished. Now, instead of 8, 5, and 3, suppose they furnished in the proportion of m, n, and p; and let the whole gain be a. 1 m Since a fraction is divided by dividing its numerator, the part of ma by m. ma will be found by dividing the numerator m + n + p a multiplied by m is m a, therefore, m a divided by m + n + p |