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$xy=0$

$ax +by +cxy +d=0$

$ax +by +cz +dxy +eyz +gxyz=0$

I made myself the examples, sometimes I face these equations and I do not know how to resolve them, all equations whose unknowns have exponent equals one but they can be multiplied together as I have put in the example $xy$, $yxzt$.... I want to know the name so I can find info and understand them because I google equations and many different come, mostly linear but I do not see these ones. I did not see the tag "equations" so I tagged differential-equations but I do not think It is that.

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    $\begingroup$ I suppose the term for such equations would be "systems of multilinear equations". $\endgroup$
    – Raskolnikov
    Jan 17, 2012 at 14:03

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In terms of nomenclature, equations of the form $$ c_1 x_1 + c_2 x_2 + \ldots $$

where $c$ is a constant are known as linear equations, equations: $$ c_1 x_1 x_2 + c_2 x_2 x_3 + \ldots $$

are bilinear equations. Though not as common, equations of the form: $$ c_1 x_1 x_2 x_3 + c_2 x_2 x_3 x_4 + \ldots $$ are referred to as trilinear equations. For higher orders @Raskolnikov suggested "systems of multilinear equations". Also, see the comment by @Lieven below who calls them "multilinear equations in homogeneous space".

Solving systems of linear equations is one of the goals of Linear Algebra. Less is known for bilinear equations, but such systems can often be "solved" in the sense of finding a near optimal solution (ie. as a constraint problem).

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  • $\begingroup$ The examples he posted aren't multilinear, because in that case, all terms should be a product of the same number of variables. All examples the OP posted are multivariate polynomials, although multivariate polynomials can of course also contain higher exponents of the same variable in the same term. They could be considered multilinear equations in homogeneous space though. $\endgroup$
    – Lieven
    Jan 17, 2012 at 14:42
  • $\begingroup$ Thank you, is there any way to solve them with exact solutions? $\endgroup$
    – user23090
    Jan 21, 2012 at 3:45
  • $\begingroup$ @user23090 I suspect each case has a different answer, but in general the answer is no. If you were looking to solve your first two equations (the bilinear ones), I would ask that as a separate question. $\endgroup$
    – Hooked
    Jan 22, 2012 at 5:46

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