x-y=2, Ex. 3. Given x z=3, to find x, y, and z. y-z=1, 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, 0 0 then we shall find x=- y 0 and z= 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 equations. We shall illustrate a few of those artifices, by the following examples. By adding the three equations, we shall have 2 2 Z 1 1 1 121' -+-+ + + x Y z 8. 9 10 360 Or, dividing by 2, OC Y 2 720 From this subtracting each of the three first equations, and we shall have By adding x to each member of the first equation, to the second, and z to the third, we shall get x+y+z+u=3x=4y=5%; 3x 5 3x y= ; 4 and from thence, z=- and " which values being substituted in the first equation, we have but, by the fourth equation, u=x-14; ... x. -14. or 20x-280-13x; 13x 20 Ex. 6. Given 4x-4y-4z=24, 6y-2x-2z=24, and 7z-y-x=24, to find the values of x, y, and z. By putting x+y+z=S, the proposed equations become 8x-4S-24, 8y-2S-24, 3z-S=24; ... x=3+1S, y=3+1S, z=3+1S. By adding these three equations, we have x+y+z=9+7S; whence S=72. Substituting this value for S, in x, y, and z, we shall find to find the va lues of x, y, and z. Ex. 7. Given x+y+z=90, 2x+40=3y+20, and 2x-4x+40=10, Ans. x=35, y=30, and z=25. to find the va lues of x, y, and Ex. 8. Given +ay+z, y+a=2x+2z, and z+a=3x+3y, z. 11 Ex. 9. It is required to find the values of x, y, and 2, in the following equations : x+y=13, x+2=14, and y+z=15. Ans. x=6, y=7, and z=8. Ex. 10. In the following it is required to find the values of x, y, and z. 3 y -= + 124, +=+ x=48, 94, Ans. y=120, z=240. Ex. 12. Given x+y+ z= 9, x+2y+32=16, and xy-22= 3, 2. y=8, and z=6. to find the va lues of x, y, and z. Ans. x=4, y=3, and 2=2. Ex. 13. Given x+y+ z=12, to find the +2y+32=20, values of x, and 1x+y+2= 6, y, and z. 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. x=8, y=9, and z=91. Ex. 15. Given x+y=100, y+12=100, and z+x=100; to find the values of x, y, and z. Ans. x=64, y=72, and z=84. 4x+3y+z 2y+2z-x+1 Ex. 16. Given 10 C -2- .5 15 =5+ 5 9x+5y-22 2x+y-3z_7h+z+3, 1 + 6' to find the values of x, y, and z. Ans. x=9, y=7, and 2=3. Ex. 17. Given x+y=357, y=3z=476, z+1u 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 alge braic 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 re |