Page:Encyclopædia Britannica, Ninth Edition, v. 1.djvu/579

Rh SIMPLE EQUATIONS.] ALGEBRA 541 pressed by equations, or translated from the common language into that of algebra, we must consider whether the problem be properly limited; for in some cases the conditions may be such as to admit of innumerable solu tions, and in others they may involve an absurdity, and thus render the problem altogether impossible. Now, by considering the examples of last section, it will appear, that to determine any number of unknown quantities, there must be given as many equations as there are unknown quantities. These, however, must be such as cannot be derived from each other, and they must not involve any contradiction ; for in the one case the problem would admit of an unlimited number of answers, and in the other case it would be impossible. For example, if it were required to determine x and y from these two equa tions, 2x - Zy 13, 4a; - 6y = 26; as the latter equation is a consequence of the former (for each term of the one is the half of the corresponding term of the other), it is evident that innumerable values of x and y might be found to satisfy both equations. Again, if x and y were to be determined from these equations, x + 2y = 8, 3x + Qy 26, it is easy to see that it is impossible to find such values of x and y as will satisfy both ; for, from the first, we find 3x = 24 6y ; and from the second, 3x = 26 6y; and therefore 24- Qy = 26 - 6y, or 24 = 26, which is absurd; and so also must have been the conditions from which this conclusion is drawn. 69. But there is yet another case in which a problem may be impossible ; and that is, when there are more equations than unknown quantities ; for it appears, that in this case, by the rules of last section, we should at last find two equations, each involving the same unknown quantity. Now, unless these happened to agree, the pro blem would admit of no solution. On the whole, therefore, it appears that a problem is limited when the conditions furnish just as many independent equations as there are unknown quantities to be determined : if there be fewer, the problem is indeterminate ; but if there be more, the problem in general admits of no solution whatever. 70. We shall now apply the preceding observations to some examples, which are so chosen as to admit of being resolved by simple equations. Ex. 1. What is that number, to which if there be added its half, its third, and its fourth parts, the sum will be 50] Let x denote the number sought ; then its half will be ^, its third - , and its fourth - ; !0 t) 4 Hence Or 12x + 8x + Qx = 1200, 50^=1200; Thus it appears that the number sought is 24, which upon trial will be found to answer the conditions of the question. Ex. 2. A post is J of its length in the mud, ^ in the water, and 10 feet above the water; what is its whole length 1 Let its length be x feet, then the part in the mud is OJ /y* -, and that in the water -; therefore, from the nature of the question, xx - + - From this equation we find 1x + 120 = I2x, and x = 24. Ex. 3. A market-woman bought a certain number of eggs at 2 a penny, and as many at 3 a penny, and sold them all out again at 5 for 2d. ; but ; instead of getting her own money for them, as sne expected, she lost 4d. : what num ber of eggs did she buy ? Let x be the number of eggs of each sortj Then will ^ be the price of the first sort; 2t CC &quot;* And - = the price of the second sort. o Now, the whole number being 2x, we have 5 : 2x : : 2 : = price of both sorts at 5 for 2d, - + - - = 4, by the question. it O Q Hence I5x + I0x- 24= 120, And x = 120, the number of each sort. Ex. 4. A person at play lost of his money, and then won 3s. ; after which he lost of what he then had, and then won 2s. ; lastly he lost of what he then had, and, this done, found he had only 12s. left : what had he at first] Suppose he began to play with x shillings. He lost of his money, or -, and had left x-- =. 4 44 He won 3s. and had then + 3 = 4 4 He lost 305 + 12 x + 4 . 3o;+12 of- 444 He won 2s. and had then 23+16 and had left + 2 = He lost of 4 12x+96 or 28 and had left 4 28 28 And because he had now 12s. left, we have this equation, 12cc+96 28 = 12. Hence I2x = 240, and x = 20. Jx. 5. To divide the number 90 into 4 such parts, that if the first be increased by 2, the second diminished by 2, the third multiplied by 2, and the fourth divided by 2, the sum, difference, product and quotient shall be all equal to each other. In this question there are four quantities to be deter mined; but instead of introducing several letters, having put x to denote the first of them, we may find an expres sion for each of the remaining ones, as follows Because x + 2 = second quantity - 2, x + 4 = the second quantity; And because x + 2 = third x 2 ; cc + 2 - = the third quantity. a And in like manner 2(# + 2) = the fourth quantity. Now by the question, the sum of all the four = 90; z + * + 4 + ^ + 2(^ + 2) = 90. ft Hence 9# = 162, and x = 18; Therefore the numbers required are 18, 22, 10, and 40. Ex. 6. A and B together can perform a piece of work in 12 hours, A and C in 20, and B and C in 15 hours; in what time will each be able to perform it when working separately ? That we may have a general solution, let us suppose A and B can perform the work in a hours, A and C in b hours, and B and C in c hours. Let x, y, and z, denote the times in which A, B, and C, could perform it respec tively, if each worked alone; and let the whole work be represented by 1. The question gives at once
 * 1) = 24.