compared, will furnish two equations each involving only y and z; from which the values of y and z may be deduced by any of the rules in the preceding section, and hence, the value of x can be readily ascertained. The same observation applies to this method of solution, as did to the last. In some particular equations, two unknown quantities may be eliminated at once. Ex. 2. Given x+y+z=31 x+y―z=25 to find the values of x, y, & z. Adding the first and third equations, 2x=40; Subtracting the second from the first, 2z=6; and subtracting the third from the second, x-y=2, Ex. 3. Given x=20. :3 ; 2y=16; .. y=8. x-z=3, to find, x, y, and z. Here, subtracting the first equation from the second, we have y-z=1; which is identically the third. Therefore, the third equation furnishes no new condition; but what is already contained in the other two; and, consequently, the proposed equations are indeterminate; or, what is the same, we may obtain an infinite number of values which will satisfy the conditions proposed. This can be easily verified, by comparing the proposed equations with those of (Art. 207), and substituting in the formulæ of roots, (Art. 215); for, then we shall find a 0 0 ==== 0 0 254. It is proper to remark, that in particular cases, Analysts make use of various other methods, besides those pointed out in the practical rules; in the resolution of equations, which greatly facilitate the calculation, and by means of which, some equations of a degree superior to the first, may be easily resolved, after the same manner as simple equàtions. We shall illustrate a few of those artifices, by the following examples. By adding the three equations, we shall have 2 2 2 1 1 1 121 x Or, dividing by 2, +=+- = y 2 8 9 10 360 1 1 1 121 X y 2 720 From this subtracting each of the three first equations, and we shall have Ex. 5. Given 2x=y+z+u, 3y=x+z+u, ( to find the values of x, y, 4z=x+y+u, ( z, and u. and u-x-14, By adding x to each member of the first equation, y to the second, and z to the third, we shall get which values being substituted in the first equation, we have 13x + -+u ; ··· u=. ; 4 5 but, by the fourth equation, u=x-14; 13x 20 ..x-14=- or 20x-280=13x ; 20 3x whence x=40: consequently y 30, z=24, and u=x 4 14=26. Ex. 6. Given 4x-4y - 4z=24, to find the values of x, 6y-2x-2x=24, and 72-y-x=24, y, and z. By putting x+y+2=S, the proposed equations become 8x-4S-24, 8y-2S-24, 8z-S-24; •*. *=3+3S, y=3+1, z=3+1S. By adding these three equations, we have x+y+2=9+7S; whence S=72. Substituting this value for S, in x, y, and z, we shall find x=39, y=21, and z=12. to find the values of x, y, and z. Ans. x=35, y=30, and z=25. and z+a=3x+3y,) to find the values of x, y, and z. Ex. 9. It is required to find the values of x, y, and z, in the following equations; 1 x+y=13, x+2=14, and y+2=15. Ans. x=6, y=7, and 2=8. Ex. 10. In the following it is required to find the values of x, y, and z. =4, y=3, and z =2, to find the values of x, y, and z. Ans. x=6, y=4, and z=2. Ex. 14. Given x+y-z-8, x+z-y=9, and y+z-x= 10; to find the values of x, y, and z. Ans. x8, y=9, and z=9! Ex. 15. Given x+y=100, y+z=100, and z+4x=100; to find the values of x, y, and z. Ex. 16. Given 9x+5y-22 12 4x+3y+z 10 Ans. x=64, y=72, and z= =84. 11 + and 12 15 5 2x+y-3z7y+z+3 1 5y+32 4 = 2x+3y-z +2z=y−1+ 12 x, y, and z. ; to find the values of 6 Ans. x=9, y=7, and z=3. ̧ Ex. 17. Given x+3y=357, y+}z=476, z+÷u=595, and u+3x=714; to find the values of x, y, z, and u. Ans. x 190, y=334, z=426, and u=676. CHAPTER V. ON THE SOLUTION OF PROBLEMS, PRODUCING SIMPLE EQUATIONS. 255. 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 condition 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. 256. 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. 257. In the solution of a problem, if the conditions be properly limited, there will be as many independent equations as unknown quantities, (Art. 287), in which case, the problem is said to be determinate; but if the conditions of the problem are not properly limited, that is, are not sufficient in number, or not sufficiently independent of each other, the resulting equations will either exceed in number the unknown quantities, and will therefore some of them be identical or inconsistent, or will be fewer in number than the unknown, and (Art. 240), consequently will admit of an indefinite number of solutions in this last case, the problem is unlimited, or it is called an indeterminate problem; and if the conditions are incongruous, or, what is the same thing, if the equations are contradictory to each other, the problem is (Art. 239), not only unlimited, but also impossible. Having hitherto laid down such rules as are necessary for the investigation and solution of problems producing equations of the first degree, as well as for discovering when they are truly limited, the different methods of solution shall be fully illustrated in the two following sections, by à great variety of practical examples. § I. SOLUTION OF PROBLEMS PRODUCING SIMPLE EQUATIONS, Involving only one unknown Quantity. 258. 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 ; 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. |