CHAPTER XII. INTERPRETATION OF THE SOLUTIONS OF PROBLEMS. 1. In solving equations we do not concern ourselves with the meaning of the results. When, however, an equation has arisen in connection with a problem, the interpretation of the result becomes important. In this chapter we shall interpret the solutions of some linear equations in connection with the problems from which they arise. Positive Solutions. 2. Pr. A company of 20 people, men and women, proposed to arrange a fair for the benefit of a poor family. Each man contributed $3, and each woman $1. If $55 were contributed, how many men and how many women were in the company? Let x stand for the number of men; then the number of women was 20-x. The amount contributed by the men was 3 x dollars, that by the women 20 x dollars. By the condition of the problem, we have The result, 17, satisfies the equation, but not the problem. For the number of men must be an integer. This implied condition could not be introduced into the equation. The conditions stated in the problem are impossible, since they are inconsistent with the implied condition. If the problem be generalized, its solution will show how the given data can be modified so that all the conditions, expressed and implied, shall be consistent. The generalized problem may be stated thus: A company of m people, men and women, proposed to arrange a fair for the benefit of a poor family. Each man contributed a dollars, and each woman b dollars. If n dollars were contributed, how many men and how many women were in the company? The solution of the equation of this problem is In order that x may be an integer, n - - bm must be exactly divisible by a - b. Thus, if, in the given problem, the number of people were 21 instead of 20, the other data being the same, we should have If all the conditions of a problem, expressed and implied, be consistent, a positive solution will satisfy these conditions and therefore give the solution of the problem. Negative Solutions. 3. Pr. A father is 40 years old, and his son 10 years old. After how many years will the father be seven times as old as his son ? Let x stand for the required number of years. Then after x years the father will be 40+ years old, and the son 10+ years old. By the condition of the problem, we have 40+x=7(10+ x), whence x = - 5. (1) This result satisfies the equation, but not the condition of the problem. For since the question of the problem is “after how many years?" the result, if added to the number of years in the ages of father and son, should increase them, and therefore be positive. Consequently, at no time in the future will the father be seven times as old as his son. But since to add 5 is equivalent to subtracting 5, we conclude that the question of the problem should have been, "How many years ago?" The equation of the problem, with this modified question, is: 40x=7(10x); whence x = 5. (2) Notice that equation (2) could have been obtained from equation (1) by changing x into -X. 4. The interpretation of a negative result in a given problem is often facilitated by the following principle: Ifx be substituted for x in an equation which has a negative root, the resulting equation will have a positive root of the same absolute value; and vice versa. 5. Pr. Two pocket-books contain together $100. If one-half of the contents of one pocket-book, and one-third of the contents of the other be removed, the amount of money left in both will be $70. How many dollars does each pocket-book contain? Let x stand for the number of dollars contained in the first pocketbook; then the number of dollars contained in the second is 100 - x. When one-half of the contents of the first, and one-third of the contents of the second are removed, the number of dollars remaining in the first is 2, and in the second (100-x). By the conditions of the problem, we have x + (100 - x)= 70, whence x = 20. Substituting x for x in the given equation, we obtain − {x + }(100 + x)=70, or (100 + x) − } x = 70. This equation corresponds to the following conditions: If x stand for the number of dollars in one pocket-book, then 100 + x stands for the number of dollars in the other; that is, one pocket-book contains $100 more than the other. The second condition of the problem, obtained from the equation, is: two-thirds of the contents of one pocket-book exceeds one-half of the contents of the other by $70. Therefore the modified problem reads as follows: Two pocket-books contain a certain amount of money, and one contains $100 more than the other. If one-third of the contents be removed from the first pocket-book, and one-half of the contents from the second, the first will then contain $70 more than the second. How much money is contained in each pocket-book? 6. These problems show that the required modification of an assumption, question, or condition of a problem which has led to a negative result, consists in making the assumption, question, or condition the opposite of what it originally was. Thus, if a positive result signify a distance toward the right from a certain point, a negative result will signify a distance toward the left from the same point; and vice versa; etc. Zero Solutions. 7. A zero result gives in some cases the answer to the question; in other cases it proves its impossibility. Pr. A merchant has two kinds of wine, one worth $7.25 a gallon, and the other $5.50 a gallon. How many gallons of each kind must be taken to make a mixture of 16 gallons worth $88? Let stand for the number of gallons of the first kind; then 16. will stand for the number of gallons of the second kind. Therefore, by the condition of the problem, we have 7.25x+5.5 (16x) = 88; whence x = 0. That is, no mixture which contains the first kind of wine can be made to satisfy the condition. In fact, 16 gallons of the second kind are worth $88. Indeterminate Solutions. 8. Pr. A merchant buys 4 pieces of goods. In the second piece there are 3 yards less than in the first, in the third 7 yards less than in the first, and in the fourth 10 yards less than in the first. The number of yards in the first and fourth is equal to the number of yards in the second and third. How many yards are there in the first piece? 7; in the Let stand for the number of yards in the first piece; then the number of yards in the second piece is x3; in the third piece, x fourth piece, x-10. Therefore, by the condition of the problem, we have x + (x − 10) = (x − 3) + (x − 7), or 2 x 10 = 2 x 10. This equation is an identity, and is therefore satisfied by any finite value of x. If it be solved in the usual way, we obtain 10 - 10 0 (22)x = 10 −10, or x= 2 2 0 - That is, the conditions of the problem will be satisfied by any number of yards in the first piece. Infinite Solutions. 9. Pr. A cistern has three pipes. Through the first it can be filled in 24 minutes; through the second in 36 minutes; through the third it can be emptied in 143 minutes. In what time will the cistern be filled if all the pipes be opened at the same time? x Let x stand for the number of minutes after which the cistern will be filled. In one minute of its capacity enters through the first pipe, and x hence in x minutes of its capacity enters. For a similar reason, - of 24 36 its capacity enters through the second pipe in x minutes; and in the same Б х 72 time of its capacity is discharged through the third pipe. Therefore, after x minutes there is in the cistern of its capacity. But by the condition of the problem, that the cistern is then filled, we have This result means that the cistern will never be filled. This is also evident from the data of the problem, since the third pipe in a given time discharges from the cistern as much as enters it through the other pipes. The Problem of the Couriers. 10. Pr. Two couriers are traveling along a road in the direction from M to N; one courier at the rate of m1 miles an hour, the other at the rate of me miles an hour. The former is seen at the station A at noon, and the other is seen h hours later at the station B, which is d miles from A in the direction in which the couriers are traveling. Where do the couriers meet? Assume that they meet to the right of B at a point C1, and let ≈ stand for the number of miles from B to the place of meeting C1 (Fig. 3). The first courier, moving at the rate of my miles an hour, travels d + z d + x miles, from A to C1, in hours; the second courier, moving at the m1 rate of m2 miles an hour, travels x miles, from B to C1, i in hours. By х M2 the condition of the problem it is evident that, if the place of meeting be to the right of B, the number of hours it takes the first courier to travel from A to C1 exceeds by h the number of hours it takes the second courier to travel from B to C1. We therefore have (i.) A Positive Result. The result will be positive either when hmid and m2 > m1, or when hm1<d and m2 <m1. A positive result means that the problem is possible with the assumption made; i.e., that the couriers meet at a point to the right of B. (ii.) A Negative Result. The result will be negative either when hmid and m2 <m1, or when hm1<d and m>mi. Such a result shows that the assumption that the couriers meet to the right of B is untenable, since, as we have seen, in that case the result is positive. That under the assumed conditions the couriers can meet only at some point to the left of B can also be inferred from the following considerations, which are independent of the negative result: If hm1 >d, the first courier has passed B when the second courier is seen at that station; that |