circumstance of the case, we can resolve the problem without a minus sign in the result. Thus : Let x= B's debt, then 3x= A's debt Per question, 400-3x-300-2x. Or x=100. 3. What number is that whose fourth part exceeds its third part by 12? Ans. 144. But there is no such abstract number as -144, and we cannot interpret this as debt. It points out error or impossibility, and by returning to the question we perceive that a fourth part of any number whatever cannot exceed its third part; it must be, its third part exceeds its fourth part by 12, and this enunciation gives the positive number, 144. Thus do equations rectify subordinate errors, and point out special conditions. 4. A man when he was married was 30 years old, and his wife 15. How many years must clapse before his age will be three times the age of his wife? Ans. The question is incorrectly enunciated; 7 years before the marriage, not after, their ages bore the specified relation. 5. A man worked 7 days, and had his son with him 3 days; and received for wages 22 shillings. He afterwards worked 5 days, and had his son with him one day, and received for wages 18 shillings. What were his daily wages, and the daily wages. of his son? Ans. The father received 4 shillings per day, and paid 2 shillings for his son's board. 6. A man worked for a person ten days, having his wife with him 8 days, and his son 6 days, and he received $10.30 as compensation for all three; at another time he wrought 12 days, his wife 10 days, and son 4 days, and he received $13.20; at another time he wrought 15 days, his wife 10 days, and his son 12 days, at the same rates as before, and he received $13.85. What were the daily wages of each? Ans. The husband 75 cts., wife 50 cts. The son 20 cts. expense per day. 7. A man wrought 10 days for his neighbor, his wife 4 days, and son 3 days, and received $11.50; at another time he served 9 days, his wife 8 days, and his son 6 days, at the same rates as before, and received $12.00; a third time he served 7 days, his wife 6 days, and his son 4 days, at the same rates as before, and he received $9.00. What were the daily wages of each? Ans. Husband's wages,$1.00; wife 0; son 50 cts. S. What fraction is that which becomes when one is added to its numerator, and becomes when 1 is added to its denoininator? Ans. In an arithmetical sense, there is no such fraction. The algebraic expression,, will give the required results. (Art. 58.) By the aid of algebraical equations, we are enabled not only to resolve problems and point out defects or errors in their enunciation, as in the last article, but we are also enabled to demonstrate theorems, and elucidate many philosophical truths. The following are examples: Theorem 1. It is required to demonstrate, that the half sum plus half the difference of two quantities give the greater of the wo, and the half sum minus the half difference give the less. Subtract (B) from (A) and divide by 2, and we have y = 1 s - 1 d These last two equations, which are manifestly true, demonstrate the theorem. Theorem 2. Four times the product of any two numbers, is equal to the square of their sum, diminished by the square of their difference. Let the greater number, and y= the less, as in the last theorem. 2x=s+d Many other theorems are demonstrable by algebra, but we de fer them for the present, as some of them involve quadratic equations, which have not yet been investigated; and we close the subject of simple equations by the following quite general problem in relation to space, time and motion. To present it at first, in the most simple and practical manner, let us suppose Two couriers, A and B, 100 miles asunder on the same road set out to meet each other, A going 6 miles per hour and B 4. How many hours must elapse before they meet, and how far will each travel? = Let x A's distance, y= B's, and t the time. As the miles per hour multiplied by the hours must give the distance each traveled, therefore, From equation (3,) we learn that the time elapsed before the couriers met was the whole distance divided by their joint motion per hour, a result in perfect accordance with reason. From equations (4,) we perceive that the distance each must travel is the whole distance asunder multiplied by their respective motions and divided by the sum of their hourly motions. Now let us suppose the couriers start as before, but travel in the same direction, the one in pursuit of the other. B having 100 miles the start, traveling four miles per hour, pursued by A, traveling 6 miles per hour. How many hours must elapse before they come together, and what distance must each travel? Take the same notation as before. Then x-y=100 (1.) As A must travel 100 miles more than B. But equations (2,) that is, x=6t and y=4t, are true under all circumstances. The result in this case is as obvious as an axiom. A has 100 miles to gain, and he gains 2 miles per hour, it will therefore require 50 hours. But it is the precise form that we wish to observe. It is the fact that the given distance divided by the difference of their motions gives the time, and their respective distances must be this time multiplied by their respective rates of motion. Now the smaller the difference between their motions, the longer the time before one overtakes the other; when the difference is very small, the time will be very great; when the difference is nothing, the time will be infinitely great; and this is in perfect accordance with reason; for when they travel equally fast one cannot gain on the other, and they can never come together. If the foremost courier travels faster than the other, they must all the while become more and more asunder; and if they have ever been together it was preceding their departure from the points designated, and in an opposite direction from the one they are traveling, and would be pointed out by a negative result. (Art. 59.) Let us now make the problem general. Two couriers, A and B, d miles asunder on the same road, set out to meet each other; A going a miles per hour, B going b miles per hour. How many hours must elapse before they meet, and how far will each travel? Taking the same notation as in the particular case, Let x= A's distance, y= B's, and t the time. If a=b, then x=d and y=d. A result perfectly obvious, the rates being equal. Each courier must pass over one half the distance before meeting. If a=0 x= 0x d 0+b bd =0 and y==d. That is, one will be at rest, and the other will pass over the whole distance. (Art. 60.) Now let us consider the other case, in which one courier pursues the other, starting at the same time from different points. Let the line CD represent the space the couriers are asunder when the pursuit commences, and the point E where they come together. C D E The direction from C towards D we call plus, the other direc tion will therefore be minus. Now as in the 2d example, (Art. 58.) |