The equation xyz = 1500 can be thought of as a system of equations. This particular system has 3 unknowns and 1 equation. As a result of this, there is not enough information to determine the exact values of x, y and z.
Consider this (incorrect) argument. Let's suppose that x = 1, y = 1 and z = 1500. Then xyz = (1)(1)(1500) = 1500. This satisfies our given equation, so these must be the values of our variables.
But this is incorrect, because if x = 10, y = 10, and z = 15, then xyz = (10)(10)(15) = 1500. This still satisfies our equation, but none of the variables have the same value as previously concluded.
In general, if you have n equations and k unknowns with k > n, then there is not a unique solution to the system.
