EQUATIONS OF OJTE TTNKNOWN QUANTITY.] MATHEMATICS. ALGEBRA. 



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CHAPTER III. 

 ALGEBRA. EQUATIONS. 



AT page 45C, a rule was given for the solution of a simple 

 equation, and several examples were proposed to show 

 its practical application. These examples, however, 

 were selected with a view to your acquaintance with the 

 operations of algebra at that stage of your progress. 

 You may now attempt equations of a higher order of 

 difficulty, and apply the following more general rule 

 for that purpose. 



GENEUAL llCLE/or the tolulion of a simple equation 

 with one unknown quantify. 



1. Clear the equation of fractions, if there be any. 

 This may be done by multiplying each numerator by all 

 the denominators except its own ; or by takiiig a common 

 multiple of the denominators and then multiplying each 

 numerator by this multiple, after suppressing that factor 

 in it which is equal to the denominator. 



2. Clear the equation of radical signs, if there be 

 any. This is done by causing the quantity under the 

 radical we wish to remove, to stand alone on one side of the 

 equation, and then performing the operation the reverse 

 of that indicated by the radical, on both sides of the 

 equation. Thus, if the radical be V we must T^ "* 

 both sides ; if it be V we must cute both sitle8 ; and so 

 on. This reverse operation evidently disengages the 

 quantity from the radical ; if two radicals enter, the 

 operation must be repeated in reference to the second 

 radical. Transposing, collecting, <kc., are sufficiently 

 explained in the former rule. It need only be added 

 here, that the example itself must suggest the order in 

 which the precepts given in either rule should follow one 

 another. 



L -6 ' 



Here, if each numerator 



= x ""--1-17 

 4 T C 1 17 ' 

 bo multiplied by the product of the denominators of all 



the other fractions (regarding 17 as y), we shall have 



24x + 302 = GO* 40* 4- 2040, .'. 24z 4- 30.c -f 40x 



2040 

 COx = 2040 ; that is, 34x = 2040, .% x = -^r = GO. 



But if we take the least common multiple of 5, 4, 2, 

 3, which is easily seen to be GO, we shall have to multi- 

 ply the first numerator by -g- or 12, the second by -j- or 



ir, t the third by y or 30, the fourth by y or 20, and 



GO 



the List (17) by y or GO. The equation cleared of frac- 

 tions iii this way will be 



12x -f 15x = 30x 20x + 1020, .-. 12x -f 15x 4- 20.C 

 30x = 1020. 



Tliat is, 17x = 1020, .'. x = -j y- = UO. 



It is plain, that, whichever method bo used, the equation 

 free from fractions always arises from multiplying both 

 sides of the original equation by the same quantity ; 

 thus, in the first way, each side is multiplied by 120, the 

 product of the denominators ; in the second way, each 

 side is multiplied by GO, the least common multiple of 

 the denominators. 



2. x + Jfa + y? = 3. Transposing the x, agreeably 



to precept 2, of the rule, ^/Gx + x* = 3 x; and squar- 

 ing both sides, 6-c + x 2 = 9 Cx + x 2 



9 3 

 .-.CJ5 = 9 Ox, .'. 12x = 9, .'.x = j^ = -j" 



3. iJx-\-7 Jx=l. Transposing, V j; 4'7= tJx-^-1. 



Squaring both sides, x + 7=x 4- 2 /x -\- 1. Transpos- 

 ing again, to get the radical alone on one side, 

 x x4-7 1=G=2 Jx, .'. 3= *Jx, .-. (squaring) 9 =x. 



4 -, o = 4- , ... Here it is evident 



that (u x) (a + x), or a 2 x 2 is the least common 

 multiple of the denominators. 



.-. 1 a(a 2 x' f ) = ax(a+x)-t-a(a x), 

 that is, 1 o 3 + ox 2 = a?x + ax 2 -fa 2 ax. Transpos- 

 ing the unknown terms to the left, . '. ax o 2 .c = a 1 

 fa 2 !, that is, (1 a) ax = a 3 + a' 1. 



