1911 Encyclopædia Britannica/Equation/Simultaneous Equations

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11731251911 Encyclopædia Britannica, Volume 9 — - Equation Simultaneous Equations

I. Simultaneous Equations.

Simultaneous equations which involve the second and higher powers of the unknown may be impossible of solution. No general rules can be given, and the solution of any particular problem will largely depend upon the student’s ingenuity. Here we shall only give a few typical examples.

1. Equations which may be reduced to linear equations.—Ex. To solve x(xa) = yz, y (yb) = zx, z (zc) = xy. Multiply the equations by y, z and x respectively, and divide the sum by xyz; then

(1).

Multiply by z, x and y, and divide the sum by xyz; then

(2).


From (1) and (2) by cross multiplication we obtain

(suppose) (3).

Substituting for x, y and z in x (xa) = yz we obtain

and therefore x, y and z are known from (3). The same artifice solves the equations x2yz = a, y2xz = b, z2xy = 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 3x2 + xy + y2 = 15, 31xy − 3x2 − 5y2 = 45. Substituting y = mx in both these equations, and then dividing, we obtain 31m − 3 − 5m2 = 3 (3 + m + m2) or 8m2 − 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 3x2 + xy + y2 = 0, we obtain y2 (12 + 2 + 1) = 15; ∴ y2 = 1, which gives y = ±1, x = ±2. Taking the second value, y = 3x, and substituting for y, we obtain x2 (3 + 3 + 9) = 15; ∴ x2 = 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.