first equation would be verified; but these same values will not verify the other equation, since they would give, 8×5+5×5= 684. By an extension of this same reasoning, it might be shown, that where there are three unknown quantities, there must be three equations; and in general, that the number of unknown quantities and the number of equations must be equal. The process of casting out an unknown quantity as in the preceding example, is called elimination. Rules for the resolution of equations containing two or more unknown quantities. RULE I. (36.) If there be two equations and two unknown quantities, multiply or divide the given equations by such quantities as will make the coefficients of the same unknown quantity equal in both equations; then, if the signs of this quantity be alike, subtract one equation from the other; and if the signs be unlike, add the equations together; the resulting equation will contain but one unknown quantity, the value of which may be found by the preceding rules. Substitute this value in one of the original equations, which will then contain but one unknown quantity; and the resolution of which will give the value of that unknown quantity. If there be more than two equations, and more than two unknown quantities, eliminate one of the unknown quantities by comparing one of the equations with all the rest, according to the preceding rule: there will result a number of equations one less, containing one unknown quantity less than the given equations. Proceed in the same manner with this new set of equations, and continue the operation till an equation is obtained which contains but one unknown quantity. Having found the value of this, substitute it in one of the equations which contains but two unknown quantities, and the value of a second unknown quantity can be determined; then substitute these two values in an equation containing three unknown quantities; proceed in this way till the values of all the unknown quantities are determined. RULE II. Observe which of the unknown quantities is least involved, and let the value of that quantity be found in each equation, by considering the rest as known. Let the values thus found be put equal to each other, and new equations will be obtained, out of which that quantity will be excluded. Repeat the operation with these new equations till an equation is obtained, which contains but one unknown quantity. RULE III. Find the value of one of the unknown quantities in one equation in terms of the others, and substitute this value in each of the other equations; and with the new equations thus arising, repeat the operation till you arrive at an equation with only one unknown quantity. EXAMPLES. 1. Given 5x + 2y = 19 to find x and y. { If we multiply the first of these equations by 3, the co efficients of y will be equal, and the equations become, To eliminate y, we add these equations (the signs of the terms containing y being unlike), and there results the equa tion, from which we find, 22x = 66, x=3. Substituting this value of x, in the equation 5x + 2y = 19, it becomes, whence, 15+ 2y = 19; 15 From the equation 5x + 2y = 19, we find y = = 4, and from the equation 7x - 6y = 9, we have, Putting these values of y equal to each other we have the equation Clearing of fractions, 14x - 18 = 114 - 30x. Transposing and reducing, 44x = 132, From the equation 5x + 2y = 18, we have, and substituting this value of y, in the equation 7x - 6y 9, we have, 7-6(1955) = 9; and performing the operations indicated, reducing, and 2 7x-57 +15x = 9; 22x = 66, x=3. Putting this value of x in the equation 7x-6y = 9, we { } 2. Given ax + by = n cx + dy = m to find z and y. To eliminate x, multiply the first equation by c, and the second by a; they will then become, асх + bcy = сп, acx + ady = am. Subtracting the second of these equations from the first, there remains the equation To find the value of x, multiply the first of the two given equations by d, and the second by b; they become, adx + bdy = dn, bcx + bdy = bm. Subtracting the first of these equations from the second, there remains, The value of x may also be obtained by substituting the value of y already found in one of the original equations; thus, taking the equation ax + by = n, and putting for y its value, we have, Reducing the second member of this equation to a fraction, we have, |