Page:Encyclopædia Britannica, Ninth Edition, v. 1.djvu/578

Rh 540 ALGEBRA [SIMPLE EQUATIONS. To eliminate x, let the first equation be multiplied by 5, and the second by 2 ; thus we have IQx- 4y = 20. Here the term involving x is the same in both equations ; and it is obvious, that by subtracting the one from the other, the resulting equation will contain only y, and known numbers; for by such subtraction we find 19y = 95, and therefore y = 5. Having got the value of y, it is easy to see how x may be found from either of the given equations ; but it may also be found in the same manner as we found y. For let the first of the given equations be multiplied by 2, and the second by 3, we have = 46, By adding these equations, we find 19#=76, and # = 4. 05. The following examples will serve further to illus trate these different methods of eliminating the unknown quantities from equations. Ex. 1. Given *, r  to determine x and y. To eliminate y, let the first equation be multiplied by /, and the second by b, and we have bdx + bfy = bg. Taking now the difference between these equations, afx bdx = cf bg , Or (of - ld}x = cf - bg, And therefore And therefore - . af-bd In the same manner may y be determined, by multiplying the first of the given equations by d, and the second by a ; for we find adx + bdy = cd , adx + afi/ = ay . And taking the difference as before, we get My - afy = cd-ag, cd-ag bd-af 66. This example may be considered as a general solu tion of the following problem. Two equations expressing the relation between the first powers of two unknown quantities being given, to determine those quantities ; for whatever be the number of terms in each equation, it will readily appear, as in Art. 55, that by proper reduction they may be brought to the same form as those given in the above example. 67. Let us next consider such equations as involve three unknown quantities. y-i Ex. 2. Given S+- 47 to find x, y, and z. Here the given equations, when cleared from fractions, become I2x+ 8y + 6^-1488, 2* = 2820, 4560. To eliminate z by the third method, let the first equa tion be multiplied by 10, the second bv 5, and the third by 3, the results will be these : z=14880, 75?/ + 60^=14100, r = 13680. Let the second equation be now subtracted from the first, and the third from the second, and we have Next, to eliminate y, let the first of these equations bo multiplied by 3, and the second by 5 ; hence, Subtracting now the latter equation from the formti:, And i _420-10a;_ 1448- 12s -8T/ 6 J^jc 3 Given 5?&quot; 7/ ? tx* ?/&quot;* uC == u~ * y * j &quot;&quot; ,&amp;gt;rv/ c to find a;, y, and 0. By subtraction, we have - x) (x a-b* (1), c 2 - a 2 (2) , Squaring, adding, and dividing by 2, W3 get (x + y z + 2 2 - xy ~ xz - yz} ( But x 2 + y + z 2 xy - xz - yz is the sum of the three given expressions, and. . equal to a 2 + 5 2 + c 2. Hence (a 2 + b* + c&quot;) (x + y + zf = a 4 + + c* - a 2 6 2 - a 2 c 2 - 6 2 c 2, which gives x + y + z . Equations (1) and (2) are now two simple equations, which, combined with the value of x + y + z as determined, give x, y, and z. Ex. 4. Given Jx = Jyz ( - - H - y a b Multiply the first by Jx, the second by Jy, and the third by *Jz, and add two and two. There results 2c 26

i/ , - 2a = *Jxyz 5C = 1^1 xyz Zc 25 yz = Jxyz (b + c-a) (a + c b) Multiplying any two of these we get one of the unknown quantities : x = (a + e - V) (a + b - c), &c. SECT. VIIL QUESTIONS PRODUCING SIMPLE EQUATIONS. 68. When the conditions of a problem Lave been ex-