 We notice that the key to Descartes’ analysis is to make no distinction between the known and unknown quantities in the problem: Both kinds of quantities are granted a variable (generally, \(a, b, c\ldots\) for known quantities and \(x, y, z\ldots\) for unknown quantities), and thus, we treat the unknowns as if their values were already found. Or, as Descartes puts it, we “suppose the solution already effected.” The task then is to reduce the problem to an equation (in contemporary terms, to a polynomial equation in two unknowns) that expresses the unknown quantity, or quantities, in terms of the known quantities. For instance, take the following  problem:[5]