not imply whether the time is before or after the present date. Suppose that it is before; then the equation is a + x = 2 (b + x) = 2 b + 2 x, x = a 2b. If 2b is greater than a, the first supposition is correct and leads to an arithmetical value of x, since, by equation (1) the value of 2ba is; but the second supposition is incorrect and leads to a negative value for x, since by equation (2) the value of a 2 b is If, however, 2 b is less than a, the second supposition is correct and leads to an arithmetical value for x, and the first supposition is incorrect and leads to a negative value for x. It happens sometimes that a negative result indicates that the wrong choice has been made out of two possible suppositions which the problem allowed. When such a wrong choice has been made, it is not necessary to go through the whole investigation again, for the result obtained from the wrong supposition can be used. It is necessary only to take the absolute value of the negative result and place the time before the present date if it was supposed to be after, or after the present date if it was supposed to be before. 198. The equation a + x = 2 (b+x) may be regarded as representing symbolically what is enunciated in the following: Let a and b be two quantities. What quantity must be added to each in order that the first sum may be twice the second? Here the words sum, quantity, and added may be used in the algebraic senses, so that x, a, b, may be + or One of the admissible senses of this algebraic statement is found in the arithmetical question concerning the ages of A and B. More may be included in the algebraic statement than in the statement of the problem. It appears then, that when a problem is translated into an equation, the same equation may be the symbolical expression of a more comprehensive problem than that from which it was obtained. ZERO SOLUTIONS 199. A zero result may, in some cases, be the answer to a question. In other cases it proves the impossibility of the equation. PROBLEM I.—A man is 50 years old and his son is 10 years old. After how many years will the father be 5 times as old as the son? Let the required number of years. This result is the correct answer to the problem. time the father is five times as old as his son. At the present PROBLEM II.-The denominator of a fraction is four times its numerator; if 9 is added to the numerator and 15 to the denominator, the fraction is . What is the fraction? Let Then 4x or the numerator of the fraction. the denominator of the fraction. From the conditions of the problem the equation is and the fraction is 4x = which is indeterminate; that is, no determinate fraction will fulfill the required conditions. PROBLEM III.-One kind of flour can be bought at $9.50 per barrel, and another at $6.25 per barrel. How many barrels of each kind of flour must be purchased in order to make a mixture of 100 barrels worth $625? Let the number of barrels of the first kind of flour. Then 100 x= the number of barrels of the second kind; and 9x+6 (100-x) = the number of dollars the mixture is worth. Hence, no mixture which contains the first kind of flour can be made to satisfy the conditions. INDETERMINATE SOLUTIONS 200. PROBLEM I.-A man is 50 years old and his son 15 years old. After how many years will the father be 35 years older than his son? Let x = the number of years required. Then, by the conditions of the problem, 50+ x = 15+ x + 35 or 50+ x = 50+x. The two members are identical, and the equation is satisfied for any finite value of x whatever (155), i. e., the problem is indeterminate. From solving the equation in the usual way, the equation follows: or The symbol means therefore that the condition of the problem is satisfied when x = any finite number. It is evident from the problem that the father will be at any time 35 years older than his 1. If a-c c is equal to 0. c is 0, but db is not zero, formula (2) gives which has no meaning. possible, because if a which can not be true, It is easy to show that the equation is imc = 0, then a = c an l equation (1) becomes ax + b = ax+d, since b is not equal to d. 2. Let a c and b = d; then formula (2) becomes x = 0 b b = [273, 1] which has no meaning. It is easy to see that in this case, equation (1) is satisfied for any finite value of x, because it becomes ax + b = ax + b, the members of which are identical. INFINITE SOLUTIONS 201. PROBLEM I.-What number must be added to the numerator and the denominator of the fraction to make the fraction equal to 1? This equation is impossible in an arithmetical sense, as there is no number which added to 5, is equal to 8 plus the same number. The greater the value of x, the nearer the value of the fraction approaches 1, or the more nearly is equation (1) satisfied. 5+ 8+ x The impossibility of satisfying equation (1) by a finite value of x means that the problem from whose conditions this equation was derived was impossible. PROBLEM II.-One pump can fill a reservoir in 15 days; another can fill it in 25 days; and a third can empty it in 9 days. If all the pumps are set working at the same time, how long will it take to fill the reservoir? Let x = the number of days required to fill the reservoir when all the pumps are working. = the part filled in one day when all the pumps are working; = the part filled in 1 day by the first pump, = the part filled in one day by the second pump, = the part emptied in 1 day by the third pump. 1 1 Hence, + 15 25 = the part filled in 1 day when all the pumps Simplify the first member of equation (1); it then becomes zero. 1 x There is no finite quantity such that 1 divided by x is equal to But x can be taken so large that the value of differs from 0 by less than any assignable quantity. Such a value of x is called numerical infinity and is represented by the symbol co. This result shows that the reservoir will never be filled, since the part filled in one day is just the same as the part emptied in one day; i. e., 202. In accordance with what is given in 203, when the formula for the solution of an equation in one unknown quantity gives for the value of this unknown quantity an expression of the form, we say that the equation is impossible; but it does not follow that the problem is impossible-we can affirm, and only affirm, that the quantity taken for the unknown ceases to exist. 203. When the denominator of a fraction decreases, the fraction increases and will increase indefinitely if the denominator diminishes indefinitely. However, it is customary to say that, in case the denominator becomes zero, the fraction becomes infinite. This relation is expressed, x = ∞. This is an incorrect conclusion, because the fraction whose denominator is zero does not represent anything. If the given quantities of a problem vary in such a way that the denominator of the unknown quantity approaches zero, the unknown quantity itself increases indefinitely; but, when the denominator is actually zero, the solution does not exist and the equation is impossible. 204, PROBLEM III.-Two particles, with given velocities, have uniform motions along an indefinite straight line, LL'; at the same time that one of them passes A, the other passes a second point A'. Find the point on the straight line where the two particles meet. There are several cases to consider in this problem according to the position of the points A and A', the magnitudes of the velocities of the two particles, and the direction of these velocities. I' A A' R L Suppose, first, that the two points A and A' are situated on the same side of 0, and respectively at the distances A0 = a and A'0 a'. Suppose that both of the particles move in the same direction, from left to right, the first with the velocity v, the second with the velocity, v', which is less than v. Evidently, the two particles will meet at some point R to the right of the point A'. Let this distance OR be x. |