_Ex._ To solve 3x^2 + xy + y^2 = 15, 31xy - 3x^2 - 5y^2 = 45. Substituting y = mx in both these equations, and then dividing, we obtain 31m - 3 - 5m^2 = 3(3 + m + m^2) or 8m^2 - 28m + 12 = 0. The roots of this quadratic are m = 1/2 or 3, and therefore 2y = x, or y = 3x. Taking 2y = x and substituting in 3x^2 + xy + y^2 = 0, we obtain y^2(12 + 2 + 1) = 15; :. y^2 = 1, which gives y = [+-]1, x = [+-]2. Taking the second value, y = 3x, and substituting for y, we obtain x^2(3 + 3 + 9) = 15; :. x^2 = 1, which gives x = [+-]1, y = [+-]3. Therefore the solutions are x = [+-]2, y = [+-]1 and x = [+-]1, y = [+-]3. Other artifices have to be adopted to solve other forms of simultaneous equations, for which the reader is referred to J.J. Milne, _Companion to Weekly Problem Papers_. II. _Indeterminate Equations._ 1. When the number of unknown quantities exceeds the number of equations, the equations will admit of innumerable solutions, and are therefore said to be _indeterminate_. Thus if it be required to find two numbers such that their sum be 10, we have two unknown quantities x and y, and only one equation, viz. x + y = 10, which may evidently be satisfied by innumerable different values of x and y, if fractional solutions be admitted. It is, however, usual, in such questions as this, to restrict values of the numbers sought to positive integers, and therefore, in this case, we can have only these nine solutions, x = 1, 2, 3, 4, 5, 6, 7, 8, 9; y = 9, 8, 7, 6, 5, 4, 3, 2, 1; which indeed may be reduced to five; for the first four become the same as the last four, by simply changing x into y, and the contrary. This branch of analysis was extensively studied by Diophantus, and is sometimes termed the Diophantine Analysis. 2. Indeterminate problems are of different orders, according to the dimensions of the equation which is obtained after all the unknown quantities but two have been eliminated by means of the given equations. Those of the first order lead always to equations of the form ax [+-] by = [+-]c, where a, b, c denote given whole numbers, and x, y two numbers to be found, so that both may be integers. That this condition may be fulfilled, it is necessary that the coefficients a, b have no common divisor which is not also a divisor of c; for if a = md and b = me, then ax + by = mdx + mey = c, and dx + ey = c/m; but d, e, x, y are supposed to be whole numbers, therefore c/m is a whole number; hence m must be a divisor of c. Of the four forms expressed by the equation ax [+-] by = [+-]c, it is obvious that ax + by = -c can have no positive integral solutions. Also ax - by = -c is equivalent to by - ax = c, and so we have only to consider the forms ax [+-] by = c. Before proceeding to the general solution of these equations we will give a numerical example.