1 3abc - (a^3 + b^3 + c^3)
-------- = ------------------------------;
[lambda] (a^2 - bc)(b^2 - ac)(c^2 - ab)
and therefore x, y and z are known from (3). The same artifice solves
the equations x^2 - yz = a, y^2 - xz = b, z^2 - xy = c.
2. _Equations which are homogeneous and of the same degree._--These
equations can be solved by substituting y = mx. We proceed to explain
the method by an example.
_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_.