GEOMETRICAL CONTINUITY) shows us that in such cases some of the roots are imaginary. To represent equations involving three unknowns x, y, z, a third axis is introduced, the z-axis, perpendicular to the plane xy and passing through the intersection of the lines x, y. In this notation a linear equation represents a plane, and two linear simultaneous equations represent a line, i.e. the intersection of two planes; a quadratic equation represents a surface of the second degree. In order to graphically consider equations containing only one unknown, it is convenient to equate the terms to y; i.e. if the equation be [f](x) = 0, we take y = [f](x) and construct this curve on rectangular Cartesian co-ordinates by determining the values of y which correspond to chosen values of x, and describing a curve through the points so obtained. The intersections of the curve with the axis of x gives the real roots of the equation; imaginary roots are obviously not represented. In this article we shall treat of: (1) Simultaneous equations, (2) indeterminate equations, (3) cubic equations, (4) biquadratic equations, (5) theory of equations. Simple, linear simultaneous and quadratic equations are treated in the article ALGEBRA; for differential equations see DIFFERENTIAL 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 a b c -- + -- + -- = 0 ... (1). z x y Multiply by z, x and y, and divide the sum by xyz; then a b c -- + -- + -- = 0 ... (2). y z x From (1) and (2) by cross multiplication we obtain 1 1 1 1 ----------- = ----------- = ----------- = -------- (suppose)(3). y(b^2 - ac) z(c^2 - ab) x(a^2 - bc) [lambda] Substituting for x, y and z in x(x - a) = yz we obtain 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.