PROBLEMS WHICH PRODUCE stittrLn ECLUATIONS.
From certain quantities which are known, to investigate others Ythich have a given relation to them, is the business of Algebra.
When a question is proposed to be re solved, we must first consider fully its meaning and conditions. Then substi. tuting for such unknown quantities as ap pear most convenient, we must proceed as if they were already detertnined, and we wished to try whether they would an. swer all the proposed conditions or not, till as many independent equations arise as we have assumed unknown quantities, vvhich yvill always be the case if the ques tion be properly limited ; an'd by the so lution of these equations,- the quantitiet sought will be determined.
Prob. 1. To divide a line of 15 inches i nto two such parts, that one may be three fourths of the other.
Let 4 x = one part, then 3 x = the other.
7 x = 15, by the question, 15 60 4 4 x = 8-- one part, 7 7, 45 3 3 x 6 - the other, 7 7', Prob. 2. If 3 can perform a piece ot work in 8 days, and B in 10days, in what time will they- finish it together ? Let x be the time required.
1 In one day, A performs . r; part of the work ; therefore, in x days, he performs 8 parts of it ; and in the same time, B performs parts of it ; and calling the work 1, , , 10x +8x=80 18 x=80 8 8 4 days.
x-= =-4= 4 18 9 Prob. 3. .4 and B play at bowls, and A bets B three shillings to two upon every game; after a certain number of wallies, it appears that .4 has won three shillings; but had he ventured to bet five shillings to two, and lost one game more out of the same number, he would have lost thirty shillings : how many games did they play ? c be the number of games Let x A won, y the number B won, then 2x is what A won ofB, 2 x 3 y=3, by the ques tion ; x-1 .2, C.9 would win on Z the 2d supposition y+1 .5, B would win, 5 y ± 5 2 x + 2=30, by the question ; or 5 y 2 5 2=23, therefore, 5 y 2 and 2 x 3 y=3 by addition, 2 5 2 y 3 y=26 y= 6 y = 13 2x =3 +3 y=3+39=42 x=21 + y = 34, the number of games.
OS qVADRATIC VICATIONS.
When the terms of an equation involve the square of an unknown quantity, but the first power does not appear, the value of the square is obtained by the preced ing rules ; and by extracting the square root on both sides, the quantity itself is found.
Ex.1. Let 5 45=0; to find .r. By trans. 5 xi a 45 9 .lierefore, x := v9=±3.
The signs + and are both prefixed to the root, because 'the square root of a quantity may be either positive or nega tive. The sign of x may'also be nega tive; but still x be either equal to -1-- 3 or 3, Ex. 2. Let a x'=.--b c d; to find x.
bed xi = a k If both the first and second powers of the unknown quantity be found in an equation : Arrange the terms according to the dimensions of the unknown quanti ty, beginning with the highest, and trans posethe known quantities to the other side ; then, if the square of the unknown quantity be affected with a coefficient, divide all the terms by this co-efficient, and if its sign be negative, change the signs of all the terms, that the equation may be reduced to this form, x' ± p x= ± q. Then add to both sides the square of half the co-efficient of the first power of the unknown quantity, by which means the first side of the equation is made a complete square, and the other consists of known quantities; and by extracting the square root on both sides, a simple equation is obtained, from which the value of the unknown quantity may be found.
Ex. 1. Let + p x now, we know that x' + p x is the square 4 of x , add thereforei- to both sides, 2 4and we have 1.2 +p x ; 4 then by extracting the square root on both sides, x + (q ) and by trans.
2 x= ±1 In the same manner, if .r2 p 7, x is found to be (q Ex. 2. Let x2 12 x + 35=0; to find x. By transposition, x2 12 x = 35, and adding the square of 6 to both sides of the equation, .r2 12 a- + .16 = 36 35 = 1 ; then extracting the square root on both x 6 = * 1 x = 6 *1 7 or 5 ; either of which, substituted for x in die original equation, answers the condition, that is, makes the whole equal to nothing.