117 CHAPTER IV. ON THE SOLUTION OF PROBLEMS, PRODUCING SIMPLE EQUATIONS. 175. The solution of a problem is the method of discovering by analysis, quantities which will answer its several conditions; for this purpose, there are four things to be distinguished: I. The given, that is to say, the known quantities, enunciated in the problem, and the quantities that are to be found. II. The translation of the problem into algebraic language, which is composed of the translation of every distinct condi tion that it contains into an algebraic equation. III. The resolution of the equations, that is, the series of transformations which the immediate translation must undergo, in order to arrive at an equation containing in the first member one unknown quantity alone in its simple state, and in the other a formula of operations to be performed upon the representations of given numbers. IV. Finally, the numerical valuation, or the geometrical construction of this formula. 176. Algebraic problems and their solutions may be considered as of two kinds, that is, numerical and literal, or particular and general. In the numerical, or particular method of solution, unknown quantities are represented by letters, and the known ones by numbers, as in arithmetic. In the literal, or general solution, all quantities, known and unknown, are represented by letters, and the answers given in general terms. A problem solved in this way, furnishes a theorem, which may be applied to the solution of all questions of the same kind. § I. SOLUTION OF PROBLEMS PRODUCING SIMPLE EQUATIONS, Involving only one unknown Quantity. 177. If from certain quantities which are known, another quantity be required which has a given relation to them, let the unknown quantity be represented by r; then, the condition enunciated in the problem being clearly understood, it can be easily translated into an algebraic equation, by means of the signs pointed out in the Introduction. Having now brought the question into an algebraic form, the value of the unknown quantity can be readily found by the application of the rules delivered Chap. III. Or, if there be more than one unknown quantity required, and that they bear given relations to one another, instead of assuming a symbol to represent each of them, it is more convenient to assume one only, and from the conditions of the problem to deduce expressions for the others in terms of that one and known quantities. And as the number of conditions ought to be one more than the number of quantities thus expressed, there will remain one to be translated into an equation; from which the value of the unknown quantity may be determined as above; and this being substituted in the other expressions, their values also may be discovered. PROBLEM I. What number is that, to which 17 being added, the sum will be 48 ? Let the required number be represented by z: by transposition, x=48-17: ..z=31. PROB. 2. What number is that, from which a being subtracted, the remainder is b? Let x represent the number required. Then by the problem r-a-b; by transposition, x=a+b. Here, if a=16, and b=14; then x=16+14=30; that is, 30 is a number, from which 16 being subtracted, the remainder is 14. PROB. 3. To find a number which, being subtracted from a, leaves b for a remainder. Designating the unknown number by x, we shall have this translation, a-x=b, .. x-a-b. 178. If we suppose a=10, b=4, we shall have x=6; then the subtraction is arithmetically performed. But if we had a=10, b=14, we must subtract 14 from 10, which cannot be 'done except in part, or that with respect to the portion of 14 equal to 10. The excess, in as much as it exists subtractively, will indi cate that the number a of which it is the representation must enter negatively in the enunciation where it is already subtracted from the number a, so that the enunciation of the problem is corrected and brought to these terms: to find a number which being added to 10, the sum will be 14; a problem whose translation is, designating the unknown quantity by x, 10+x=14; .;. x=14—10=4; whereas, the translation in the former case would be 10—x=14; ‚°. x—10—14, or x= - 4. The negative root 4, satisfies the equation of the problem, besides it announces a rectification in the enunciation; this is what appears evident, since the subtraction of a negative quantity is equivalent to the addition of a positive, (Art. 63). In fact, as has been already observed, (Art. 174), it makes known that the enunciation ought to be taken in an opposite sense to that which we first proposed in the problem. PROB. 4. A person lends at interest for one year a certain capital at 5 per cent; at the end of the year, according to agreement, he is to receive a sum b, besides the principal and interest, and the whole sum he receives must be equal to the capital. I demand what is the capital? Let the capital be designated by : Since 100 dollars becomes at the end of the year 105 dollars, we shall have the capital at the same time by this proportion, +b, by the problem, must be equal to x, we have therefore the equation 105x 100 +b=x; .. 105x+100b-100x; by transposition, 5x=-100b; by division, x=- -206. This answer does 179. Thus the capital shall be -206. not agree with the problem, and still if this value —206, be substituted for x in the equation found, we obtain 105 × 206 100 +b=20b, and, performing the operations indicated in the first member, it becomes - 206206, which is true. This value of x, although it is negative, satisfies the equation of the problem, as has been already observed (Art. 174), since its two members become identically equal by making the proper substitution. If we return again to the enunciation, we discover that it is impossible that a capital augmented by the interest would remain equal to itself, and that much more this impossibility takes place, if, besides the interest, we add to it a sum b; it is necessary therefore that one of these two parts, namely, the interest at 5 per cent, and b, be subtracted. In fact, if we carry into the first equation this circumstance -x, which is but x=- a number, we find 105.x -b=x, 100bx, a translation of the enunciation, by supposing the interest additive to the capital, in which case, the sum b ought to be subtracted. This equation, treated as the preceding, shall give x=20b,. If the interest at 5 per cent be subtracted from 100, in which case 100 reduces itself to 95, we have the capital x at the end of the year, by the proportion 95x 100 consequently, +b= x; multiplying by 100, and transposing, we shall have 100b=5x, ... x=20b. X, The negative isolated result, that is, the negative value of would announce a rectification or a correction in the terms of the enunciation, and the problem proposed could be re-established in two ways. PROB. 5. What number is that, the double of which exceeds its half by 6 ? .. multiplying by 2, 4x—x=12, or 3x=12, .. by division, x=4. PROB. 6. From two towns which are 187 miles distant, two travellers set out at the same time, with an intention of meeting. One of them goes 8 miles, and the other 9 miles a day. In how many days will they meet? Let x= the number of days required; then 8x= the number of miles one travelled, and 9x the number the other travelled; |