..2x-80= 47x-2100 .. 36x-1440-47x-2100 .. 11x=660 .. x=60 .. y=2x—80-120-80-40. 57. Second method. Find an expression for either of the unknown quantities from one of the equations, as in the preceding method. Substitute this expression in the other equation, and the result will evidently be an equation containing only one unknown quantity. (1) Thus in Ex. (1) the expression for x, from the first 16-2y; and this, substituted for x in the second equation is x= 3 equation, converts that second equation into х 4(16—2y)—3——7; that is 64-8y-9y=—21.—17y= 3 (2) In like manner the value of x, deduced from the second equation, in Ex. (2), being put for a in the first equation, converts that first equation into 3(3y—15) 5 +2y=48; that is 9y 45+10y=240.. 19y=285 .. y=15; and since a= 3y-15_3y = -3.. x=9—3-6. 5 5 58. Third method. Multiply the given equations by such numbers the smaller the better-or such quantities, as will make the coefficients of one of the unknowns the same in both equations. If the coefficients thus made equal have like signs, subtract one equation from the other; but if they have unlike signs, add the two equations together: the terms with equal coefficients will thus disappear, and therefore one of the unknowns will be eliminated in the result, which will be an equation containing only the other unknown. The value of this unknown being found, the solution may be completed as in the former methods, or we may return to the original pair of equations and eliminate, by a similar process, the unknown quantity before retained. The coefficients of the unknown to be eliminated may always be equalised by multiplying each by the other, as is obvious, so G that the equations may always be prepared for the intended addition or subtraction, after we have fixed upon the unknown to be eliminated, by multiplying the first equation by the coefficient of that unknown in the second, and the second by the coefficient of the same unknown in the first: thus, taking the general example, a1x+b1y=c1, α2x+b2y=c2, and multiplying the first equation by a, and the second by a,, in order to eliminate x, we have the equations Again, multiplying the same equations by b, and b1, to eliminate y, we have the equations the same formulæ as those obtained in 56. In numerical examples, it will in general be easier to complete the solution by substituting the value of the unknown first determined, in one of the proposed equations, and thence to deduce the other unknown, as in Art. 57, than to determine both the unknowns by this method of elimination; but when the coefficients are letters, the above mode of proceeding is remarkably neat and easy. Since, by this third method, either of the unknown quantities may be eliminated without the introduction of fractions into the process, it will usually be found the most convenient of the three. In the above general example, the coefficients are equalised by multiplying each by the other; but when the coefficients are numbers, smaller multipliers than these will often answer the purpose of equalising the coefficients: the smallest are always those which give for the result the least common multiple of the original coefficients, and are of course the best that can be employed with a view to the saving of figures: we obtain these smallest multipliers at once, in any proposed case, by depriving the two coefficients we are dealing with of their greatest common divisor, and using only the resulting quotients, as in examples (3) and (4) which follow. (1) 5x+3y=42 3x-2y=10 (3) 9x-4y= 7 Multiplying the first equation Multiplying the first by 2 and by 2 and the second by 3, to equalise the coefficients of y, we the second by 3 have 18x8y 14 10x+6y= 84 18x-15y=-21 9x-6y= 30 Subt. 7y= 35..y=5 NOTE.-Each of the three methods now explained, instead of being applied immediately to the given equations, may be postponed till those equations are modified in any way that may seem to confer on them greater simplicity; for instance, we may add them together, or subtract one from the other, and then take the resulting equation in connexion with either of the original equations, instead of the originals themselves; or we may multiply either by any quantity, add or subtract the product, and take the result in connection with either of the proposed equations. For example: the equations (3) above, give, by subtraction, 3x+y=14; so that, instead of the proposed, we may if we please take the pair 3x+y=14, 6x—5y——7, which are rather simpler. In Ex. 21, 22, and 24 following, the equations will be prepared for solution by dividing the second by the first. SIMPLE EQUATIONS WITH THREE UNKNOWN QUANTITIES. 59. Either of the methods for the solution of two simultaneous equations with two unknown quantities, may be applied to three simultaneous equations with three unknown quantities, and indeed to any number of such equations. By eliminating one of the unknowns as already taught, three equations may be reduced to two containing only the other unknowns, four equations may be reduced to three, and so on. However numerous be the simple equations in the set proposed, the elimination of the same unknown from all the pairs that can be formed from them, will lead to a new set of equations, one fewer in number than the original set, and containing unknowns one fewer in number: 1 1 y * In these examples we may first find the reciprocals of x and y, namely, and which may be represented by and y'; unity divided by the reciprocal of any quantity is, of course, that quantity itself. num. + den. + To these examples the principle at page 113 may be applied, namely num.-den. num.+den. In Ex. 26, x, y may be replaced by z', y'. num.-den. |