I've got this problem, with relation to rates. [A] is the concentration of one substance, is the concentration of another, etc. [R] is a general reactant, [P] is a general product, is a general substance (which thus relates to both reactants and products). But you won't actually need chemistry for this, just thought it might help conceptualize. -d[A]/dt=k*[A]n1*n2*[C]n3 ... n1...3 are orders of reaction, and they are there with respect to every , a different n for each substance. This may take any real value including 0, fractional or negative values. These orders are constant, as is k. However, because of the rates changing, each substance will be dependent on time t, and by the definition of the rate as above, on the concentration of each other substance. Typically I have seen this rearranged to: -d[A]/([A]n1*n2*[C]n3 ...)=k*dt And then "solved". But I don't know what this solving means! Can anyone help? There is another system which helps along the solution, though again I don't know how. This is: if we are finding the rate of loss for a reactant, then for each and every reactant we can write [R]=[R]0-((vA/vR)*([A]0-[A])), where vA and vR are once again constants (stoichiometric coefficients, if you're interested - they will be real and positive), but there is a different vR for every reactant (as with the orders - e.g. there is vC, where [C]=[C]0-((vA/vC)*([A]0-[A])), etc.). And for each product, we can write similarly: [P]=[P]0+((vA/vP)*([A]0-[A])), the only difference being the plus I have placed in bold. So, to summarize: Variables: [A], , [C], [D], ..., t Constants: k, n1 (pertains to [A]), n2 (pertains to ), n3, ..., vA (pertains to [A]), vB (pertains to ), vC (pertains to [C]), ..., [A]0 (initial value of [A]), 0, [C]0, [D]0, ... Equations: -d[A]/([A]n1*n2*[C]n3 ...)=k*dt [R]=[R]0-((vA/vR)*([A]0-[A])) [P]=[P]0+((vA/vP)*([A]0-[A])) Our final answer would be a solution for [A] in terms of all of the constants listed above and t. Obviously I don't necessarily seek an exact, complete answer for the general case, just to learn the method.