Solution of any Equation of the First Degree. quantities being represented by x, y, z, &c., the equation becomes Ax+By+Çz + &c. + M = 0. 105. Problem. To solve any equation of the first degree. Solution. Having reduced the equation to the form Ax+By+Cz + &c. + M = 0, find, as in art. 101, the value of either of the unknown quantities, as x for instance, which, is, by art. 90, and any quantities at pleasure may be substituted for y, z, &c. 106. Problem. To solve several equations with several unknown quantities. First Method of Solution called that of Elimination by Substitution. Find the value of either of the unknown quantities in one of the equations in which it occurs, and substitute its value thus found, which is generally in terms of the other unknown quantities, in all the other equations in which it occurs. The new equations thus formed, together with those in which this unknown quantity does not occur, are one less in number than the given equations, and contain one unknown quantity less, and may, by a succession of similar eliminations be still farther reduced in number and in the number of their unknown quantities, until only one equation is finally obtained; Solution of Equations. Elimination by Substitution. and the solution of all the given equations is thus reduced to that of one equation. 107. Corollary. When there are just as many equations as unknown quantities, the final equation of the preceding solution will, in general, contain but one unknown quantity, the value of which may be thence obtained; and this value, being substituted in the values of the other quantities, will lead to the determination of the values of all the unknown quantities. 108. Corollary. When the number of unknown quantities is more than that of the given equations, the final equation will contain several unknown quantities, and will therefore be indeterminate; so that a problem is indeterminate, which gives fewer equations than unknown quantities. 109. Corollary. When the number of unknown quantities is less than that of the given equations, only as many of the given equations are required to determine the values of the unknown quantities as there are unknown quantities; and the problem is therefore impossible, when the values of the unknown quantities determined from the requisite equations do not satisfy the remaining equations. 110. Problem. To solve two equations of the first degree with two unknown quantities. Solution. Suppose, as in art. 104, the given equations to be reduced to the forms Case in which the roots of two Equations are Zero. Ax+By+M = 0, A'x+By+M=0: in which x and y are the unknown quantities. The value of x, obtained from the first of these equations, is The value of y is found from this equation, by art. 90, to be which, substituted in the above value of x, gives 111. Corollary. The value of x, obtained by the preceding solution, would be zero, if we had BM'B' M. But, in this case, if the first of the given equations is multiplied by B', and the second by B, these products become, by transposition and substitution, whence BB'y - B'M, A'Bx-BB'y - BM-BB'y-B'M; that is, the given equations involve the condition that two Case in which the roots of two Equations are infinite and indeterminate. different multiples of x are equal. But this is impossible, unless x = 0. The value of y would, likewise, be zero, if we had A' M-AM', which leads to similar conclusions with regard to y as those just obtained with regard to x. 112. Corollary. The denominators of the values of both the unknown quantities would be zero, if we had A B'A' B. But, in this case, if the first of the given equations is multiplied by B' and the second by B, these products become, by transposition and substitution, that is, they involve the impossibility that the two unequal quanties B'M and BM' are equal. 113. Corollary. Both the terms of the fractional value of x would be zero, if we had BM' = B' M, and AB' = A' B. But, in this case, if the first of the given equations is multiplied by B' and the second by B, the products become, by substitution, AB'x+BB'y+B'M=0, A'Bx+BB'y+BM'=AB'x+BB'y+B'M=0; |