# 1911 Encyclopædia Britannica/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*(*x* − *a*) = yz, *y* (*y* − *b*) = *zx*, *z* (*z* − *c*) = *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* (*x* − *a*) = *yz* we obtain

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 3*x*^{2} + *xy* + *y*^{2} = 15, 31*xy* − 3*x*^{2} − 5*y*^{2} = 45. Substituting
*y* = *mx* in both these equations, and then dividing, we obtain
31*m* − 3 − 5*m*^{2} = 3 (3 + *m* + *m*^{2}) or 8*m*^{2} − 28*m* + 12 = 0. The roots of this
quadratic are *m* = 12 or 3, and therefore 2*y* = *x*, or *y* = 3*x*.

Taking 2*y* = *x* and substituting in 3*x*^{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* = 3*x*, 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*.