Performing the operations indicated, we have 12+7x+x2-60+6x=7x+x2 Omitting the quantities on each side which are equal, we have 12-60+6x=0, from which x=8. x2+a2+b2+c2 a+b-c+x Ans. x= Ans. x 20. Ans. x= Ans. x-c-26. _a(c2—b2) a2+b2 c2-ab Ans. x a+b' 11. The difference between two numbers is 2, and their product is 8 greater than the square of the less; what are the numbers? Ans. 4 and 6. 12. It is required to divide the number a into two such parts, that the difference of their squares may be c. Ans. a2 -C 2a and a2+c 2a 13. If a certain book contained 5 more pages, with 10 more lines on a page, the number of lines would be increased 450; but if it contained 10 pages less, with 5 lines less on a page, the whole number of lines would be diminished 450. Required the number of pages, and the number of lines on a page. Ans. 20 pages, and 40 lines on a page. NEGATIVE SOLUTIONS. ART. 172.-It has been stated already (Art. 23), that when a quantity has no sign prefixed, the sign plus is understood; and also (Art. 64), that all numbers or quantities are regarded as positive, unless they are otherwise designated. Hence, in all problems, it is understood, that the results are required in positive numbers. It sometimes happens, however, that the value of the unknown quantity in the solution of a problem, is found to be minus. Such a result is termed a negative solution. We shall now examine a question of this kind. 1. What number must be added to the number 5, that the sum shall be equal to 3? Let x the number. And x-3-5=-2. Now, -2 added to 5, according to the rule for Algebraic Addition, gives a sum equal to 3; thus, 5+(-2)=3. The result, -2, is said to satisfy the question in an algebraic sense; but the problem is evidently impossible in an arithmetical sense, since any posi. tive number added to 5, must increase, instead of diminishing it; and this impossibility is shown, by the result being negative, instead of positive. Since adding -2, is the same as subtracting +2 (Art. 61), the result is the answer to the following question: What number must be subtracted from 5, that the remainder may be equal to 3? Let the question now be made general, thus: What number must be added to the number a, that the sum shall be equal to b? Let x the number. And x-b-a. Now, since a+(b—a)=b, this value of x will always satisfy the question in an algebraic sense. While b is greater than a, the value of x will be positive, and, whatever values are given to b and a, the question will be consistent, and can be answered in an arithmetical sense. Thus, if b-10, and a 8, then x=2. But if b becomes less than a, the value of x will be negative; and whatever values are given to b and a, the result obtained, will satisfy the question in its algebraic, but not in its arithmetical sense. Thus, if b=5, and a=8, then x—— -3. Now 8+(-3)=5; that is, if we subtract 3 from 8, the remainder is 5. We thus see, that when a becomes greater than b, the question, to be consistent, should read, What number must be subtracted from the number a, that the remainder shall be equal to b? From this we see, 1st. That a negative solution indicates some inconsistency or absurdity, in the question from which the equation was derived. 2d. When a negative solution is obtained, the question, to which it is the answer, may be so modified as to be consistent. Let the pupil now read, carefully, the "OBSERVATIONS ON ADDITION AND SUBTRACTION," page 43, and then modify the following questions, so that they shall be consistent, and the results true in an arithmetical sense. 2. What number must be subtracted from 20, that the remainder shall be 25? (x=—5.) REVIEW.-172. What is a negative solution? When is a result said to satisfy a question in an algebraic sense? In an arithmetical sense? What does a negative solution indicate? 3. What number must be added to 11, that the sum being multiplied by 5, the product shall be 40? (x=-3.) 4. What number is that, of which the exceeds the by 3? (x=-36.) 5. A father, whose age is 45 years, has a son, aged 15; in how many years will the son be as old as his father? (x=-5.) DISCUSSION OF PROBLEMS. ART. 173.-When a question has been solved in à general manner, that is, by representing the known quantities by letters, we may inquire what values the results will have, when particular suppositions are made with regard to the known quantities. The determination of these values, and the examination of the various results which we obtain, constitute what is termed the discussion of the problem. The various forms which the value of the unknown quantity may assume, are shown in the discussion of the following question. 1. After subtracting b from a, what number, multiplied by the remainder, will give a product equal to c? Now, this result may have five different forms, depending on the values of a, b, and c. NOTE. In the following forms, A denotes merely some quantity. 1st. When b is less than ɑ. form +A. This gives positive values, of the 2d. When b is greater than a. This gives negative values, of the form -A. 3d. When b is equal to a. This gives values of the form f. gives values of the form . This 5th. When b is equal to a, and c is equal to 0. This gives values of the form 8. We shall examine each of these in succession. I. When b is less than a. In this case, a-b is positive, and the value of x is positive. To illustrate this form, let a=8, b=3, and c=20, then x=4, REVIEW.-172. When a negative solution is obtained, how may the question, to which it is the answer, be modified? 173. What do you understand by the discussion of a problem? The expression c divided by a―b, may have how many forms? Name these different forms. II. When b is greater than a. In this case, a-b is a negative quantity, and the value of x will be negative. This evidently should be so, since minus multiplied by minus produces plus; that is, if a-b is minus, x must be minus, in order that their product shall be equal to c, a positive quantity. To illustrate this case by numbers, let a=2, b=5 and c-12; then, a-b=-3, x=-4, and −3X-4=12. III. When b is equal to a. C In this case x becomes equal to. We must now inquire, what is the value of a fraction when the denominator is zero. с 2d. Suppose the denominator, then =10c. с 3d. Suppose the denominator, then 100c. .01 с 4th. Suppose the denominator 7000, then =1000c. .001 While the numerator remains the same, we see, that as the de nominator decreases, the value of the fraction increases. Hence, if the denominator be less than any assignable quantity, that is 0, the value of the fraction will be greater than any assignable quantity, that is, infinitely great. This is designated by the sign ∞, that is с co. This is interpreted by saying, that no finite value of x will satisfy the equation; that is, there is no number, which being multiplied by 0, will give a product equal to c.. IV. When c is 0, and b is either greater or less than a. 0 d If we put a-b equal to d, then x=- =0, since dX0=0; that is, when the product is zero, one of the factors must be zero. V. When b=a, and c=0. с 0 a-b0' In this case, we have x= or 2X0=0. Since any quantity multiplied by 0, gives a product equal to 0, any finite value of x whatever, will satisfy this equation; hence, x is indeterminate. On this account, we say that & is the symbol of indetermination; that is, the quantity which it represents, has no particular value. REVIEW.-173. When is x of the form +A? When is x of the form -A? When is x of the form &, or co? Show how the value of a fraction increases, as its denominator decreases. What is the value of a fraction whose denominator is zero? Of x when cis 0, and b greater or less than a? The form sometimes arises from a particular supposition, when the terms of a fraction contain a common factor. Thus, if cancel the common factor, a-b, and then make b=a, we have x=2a. This shows, that before deciding the value of the unknown quantity to be indeterminate, we must see that this apparent indetermination has not arisen from the existence of a factor, which, by a particular supposition, becomes equal to zero. The discussion of the following problem, which was originally proposed by Clairaut, will serve to illustrate further the preceding principles, and show, that the results of every correct solution, correspond to the circumstances of the problem. Two couriers depart at the same time, from two places, A and B, distant a miles from each other; the former travels m miles an hour, and the latter, n miles; where will they meet? There are two cases of this question. I. When the couriers travel toward each other. Let P be the point where they meet, A and a AB, the distance between the two places. Let x=AP, the distance which the first travels. Then a-x-BP, the distance which the second travels. B Then, the distance each travels, divided by the number of miles traveled in an hour, will give the number of hours he was traveling. Therefore, And n the number of hours the second travels. But they both travel the same number of hours, therefore the couriers travel at the same rate, each travels precisely half the distance. |